• Sun. Oct 13th, 2024

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KCSE MATHEMATICS REVISION SERIES (QUESTIONS & ANSWERS)

MATHEMATICS I

PART I

SECTION I (50 MARKS)

  1. Evaluate without mathematical tables leaving your answer in standard form

0.01712 X 3

855 X 0.531ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

  1. Six men take 14 days working 8 hours a day to pack 2240 parcels. How many more men working

5 hours a day will be required to pack 2500 parcels in 2 daysย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3 Mks)

 

 

 

 

 

  1. Mย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  In quadrilateral OABC, OA = 4i – 3j. OC = 2i + 7j

AB = 3OC. cm: mB = 2:3. Find in terms ofย  i and j

Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  vector Omย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

 

 

 

 

 

Oย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A

 

  1. By matrix method, solve the equations

5x + 5y = 1

4y + 3x = 5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

 

 

  1. In the given circle centre O, รABC = 1260.

Calculate รOAC ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

 

Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  C

 

 

 

B

 

  1. Solve the equation

2(3x – 1)2 9 (3x – 1) + 7 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

  1. Maina, Kamau and Omondi share Shs.180 such that for every one shilling Maina gets, Kamau gets 50

Cts and for every two shillings Kamau gets, Omondi gets three shillings. By how much does Maina’s

share exceed Omondi’sย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

  1. Expand (2 + 1/2x)6 to the third term. Use your expression to evaluate 2.46 correct to 3 s.f (3 Mks)
  2. The probability of failing an examination is 0.35 at any attempt. Find the probability that

(i)ย ย  You will fail in two attemptsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1 Mk)

(ii)ย ย  In three attempts, you will at least fail once ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

  1. Line y = mx + c makes an angle of 1350 with the x axis and cuts the y axis at y = 5. Calculate the

equation of the lineย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

  1. During a rainfall of 25mm, how many litres collect on 2 hectares? (3 Mks)
  2. Solve the equation a 3a – 7 = a – 2 (3 Mks)

3ย ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย  6

  1. The sum of the first 13 terms of an arithmetic progression is 13 and the sum of the first 5 terms is

โ€“25. Find the sum of the first 21 termsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (5 Mks)

  1. The curved surface of a core is made from the shaded sector on the circle. Calculate the height of

the cone.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

 

 

 

 

 

O

20cmย ย ย ย ย  1250ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  20 cm

 

 

 

 

 

 

  1. Simplify (wx – xy – wz + yz) (w + z) (3 Mks)

z2 – w2

  1. The bearing of Q from P is North and they are 4 km apart. R is on a bearing of 030 from P and on

a bearing of 055 from Q. Calculate the distance between P and R.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

ย 

SECTION II (50 MARKS)

  1. In the given circle centre O, รQTP = 460, รRQT = 740 and รURT = 390

 

 

Uย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Tย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  P

 

 

Q

Sย ย ย  ย ย ย ย ย  390

ย ย ย ย ย  Calculateย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย R

(a)ย  รRSTย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1 Mk)

(b)ย  รSUTย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

(c)ย  Obtuse angle ROTย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(d)ย  รPSTย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

  1. The exchange rate on March 17th 2000, was as follows: –

1 US$ = Kshs.74.75

1 French Franc (Fr) = Kshs.11.04

ย ย ย ย ย  A Kenyan tourist had Kshs.350,000 and decided to proceed to America

(a)ย  How much in dollars did he receive from his Kshs.350,000 in 4 s.f?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b) The tourist spendย  ยผย  of the amount in America and proceeded to France where he spend Fr

16,200. Calculate his balance in French Francs to 4 s.fย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

(c) When he flies back to Kenya, the exchange rate for 1 Fr = Kshs.12.80. How much more in

Kshs. does he receive for his balance than he would have got the day he left?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

  1. On the provided grid, draw the graph of y = 5 + 2x – 3x2 in the domain -2 ยฃ x ยฃ 3ย ย  ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

(a) Draw a line through points (0,2) and (1,0) and extend it to intersect with curve y = 5 + 2x โ€“ 3 x 2

read the values of x where the curve intersects with the lineย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b)ย  Find the equation whose solution is the values of x in (a) aboveย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

  1. (a) Using a ruler and compass only, construct triangle PQR in which PQ = 3.5 cm, QR = 7 cm

and angle PQR = 300ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b)ย  Construct a circle passing through points P, Q and Rย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(c)ย  Calculate the difference between area of the circle formed and triangle PQRย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

  1. The given Region below (unshaded R) is defined by a set of inequalities. Determine the inequalities (8 Mks)

Y

 

4

 

 

 

2ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย  ย ย Rย ย ย ย ย ย ย  ย ย ย ย ย  (3,3)

ย ย 

 

X

-3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  ย ย ย ย 5

 

 

 

 

 

 

 

  1. The table below shows the mass of 60 women working in hotels

 

Mass (Kg) 60 โ€“ 64 65 โ€“ 69 70 โ€“ 74 75 – 79 80 โ€“ 84 85 – 89
No. of women 8 14 18 15 3 2

 

(a)ย ย  State (i)ย ย  The modal classย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1 Mk)

(ii)ย  The median classย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1 Mk)

(b)ย ย  Estimate the mean markย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

(c)ย ย  Draw a histogram for the dataย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

  1. XY, YZ and XZ are tangents to the circle centre O

at points A, B, C respectively. XY = 10 cm,

YZ = 8 cm and XZ = 12 cm.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 MKS)

Z

 

 

C

 

 

 

 

..ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B

X

 

Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Y

 

 

(a)ย  Calculate, length XAย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b)ย  The shaded areaย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (6 Mks)

  1. Maina bought a car at Kshs.650,000. The value depreciated annually at 15%

(a)ย  After how long to the nearest 1 decimal place will the value of the car be Kshs.130,000ย  ย ย ย ย ย  (4 Mks)

(b)ย  Calculate the rate of depreciation to the nearest one decimal place which would make the value of

theย  car be half of its original value in 5 yearsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

ย 

MATHEMATICS I

PART II

SECTION 1 (50 MARKS)

 

 

  1. Simplify 32a10ย ย  -2/5 รทย  9b4ย ย ย  ย ย 11/2

b15ย ย ย ย ย ย  ย ย ย ย ย  4a6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

 

  1. Use logarithm tables to evaluate

ร–0.375 cos 75

tan 85.6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

  1. The marked price of a shirt is Shs.600. If the shopkeeper gives a discount of 20% off the marked price, he makes a loss of 4%. What was the cost of the shirt? (3 Mks)
  2. The surface area (A) of a closed cylinder is given by A = 2pr2 + 2prh where r is radius and h is height of the cylinder. Make r the subject. (4 mks)
  3. In the circle centre O, chords AB and CD intersect at X. XD = 5 cm

ย ย ย ย ย  XC = 1/4 r where r is radius. AX:XO = 1:2 Calculate radius of the circle.ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย  (3 mks)

 

Aย ย ย ย ย ย ย ย ย ย ย ย  5cm ย ย ย ย ย ย D

 

 

Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  O

 

B

 

 

  1. Simplify ย ย ย ย 2ย ย ย ย  ย ย –ย ย ย  ย ย ย ย 1ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

5 – 2ร–3ย ย ย ย  5 + 2ร–3

 

 

  1. P is partly constant and partly varies as q2. When q = 2, P = 6 and when q = 3, P = 16. Find q when P = 64 ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 mks)
  2. The figure on the side is a tent of uniform cross-section Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  F

ABC. AC = 8m, BC = 8m, BD = 10mย ย  and (ACB = 1200.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  8m

If a scout needs 2.5 m3 of air, how many scouts can fitย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  120o Cย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย E

in the tent.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  8mย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (4 mks)

Bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  D

10m

  1. The length of a rectangle is given as 8 cm and its width given as 5 cm. Calculate its maximum % error in its perimeter ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)
  2. ABCD is a rectangle with AB = 6 cm, BC = 4 cm AE = DH = 4 cm BF = CG = 12 cm. Draw a

labelled net of the figure and show the dimensions of the net

  1. Expand (1 + 2x)6 to the 3rd term hence evaluate (1.04)6 (4 mks)
  2. The eye of a scout is 1.5m above a horizontal ground. He observes the top of a flag post at an

angle of elevation of 200. After walking 10m towards the bottom of the flag post, the top is observed at angle of elevation of 400. Calculate the height of the flag post ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (4 mks)

  1. A bottle of juice contains 405ml while a similar one contains 960ml. If the base area of the

larger Container is 120 cm2. Calculate base area of the smaller container.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. It takes a 900m long train 2 minutes to completely overtake an 1100m long train travelling at

30km per hour. Calculate the speed of the overtaking trainย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. Okoth traveled 22 km in 23/4 hours. Part of the journey was at 16 km/h and the rest at 5 km/h.

Determine the distance at the faster speedย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. P and Q are points on AB such that AP:PB = 2:7 and AQ:QB = 5:4 If AB = 12 cm, find PQ

(2 Mks)

SECTION B (50 MARKS)

 

  1. The income tax in 1995 was collected as follows:

ย ย ย ย ย  Income in Kshs. p.a ย ย ย ย  ย ย ย ย ย ย  ย ย ย rate of tax %

1 – 39,600ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  ย ย ย ย  ย 10

39,601 – 79,200ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  15

79,201 – 118,800ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  25

118,801 – 158,400ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย  ย 35

158,401 – 198,000ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย  ย 45

ย ย ย ย ย  Mutua earns a salary of Kshs.8,000. He is housed by the employer and therefore 15% is added to his salary to arrive at its taxable income. He gets a tax relief of Shs.400 and pay Shs.130 service charge. Calculate his net incomeย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (8 Mks)

  1. The probability Kioko solves correctly the first sum in a quiz is 2/5 Solving the second correct

is 3/5 if the first is correct and it is 4/5 if the first was wrong. The chance of the third correct is

2/5 if the second was correct and it is 1/5 if the second was wrong. Find the probability that

(a)ย  All the three are correctย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b)ย  Two out of three are correctย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

(c)ย  At least two are correctย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 Mks)

  1. A businessman bought pens at Shs.440. The following day he bought 3 pens at Shs.54. This

purchase reduced his average cost per pen by Sh.1.50. Calculate the number of pens bought earlier and the difference in cost of the total purchase at the two pricesย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (8 mks)

 

 

 

 

  1. In D OAB, OA = a, OB = b

OPAQ is a parallelogram.

ย ย ย ย ย  ON:NB = 5:-2, AP:PB = 1:3

Determine in terms of a and b vectors

(a)ย  OPย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b)ย  PQย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(c)ย  QNย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(d)ย  PNย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

 

  1. A cylindrical tank connected to a cylindrical pipe of diameter 3.5cm has water flowing at 150

cm per second. If the water flows for 10 hours a day

(a)ย  Calculate the volume in M3 added in 2 daysย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 ms)

(b) If the tank has a height of 8 m and it takes 15 days to fill the tank, calculate the base radius

of the tankย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 mks)

  1. A joint harambee was held for two schools that share a sponsor. School A needed Shs.15 million while

School B needed 24 million to complete their projects. The sponsor raised Shs.16.9 million while other

guest raised Shs.13.5 million.

(a) If it was decided that the sponsor’s money be shared according to the needs of the school

with the rest equally, how much does each school getย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (5 mks)

(b) If the sponsor’s money was shared according to the schools needs while the rest was in the ย ratio of

students, how much does each school get if school A has 780 students and school B 220

studentsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. Voltage V and resistance E of an electric current are said to be related by a law of the form

V = KEn where k and n are constants. The table below shows values of V and E

ย ย ย ย ย  V

0.35 0.49 0.72 0.98 1.11
E 0.45 0.61 0.89 1.17 1.35

ย ย ย ย ย  By drawing a suitable linear graph, determine values of k and n hence V when E = 0.75(8mks)

  1. The vertices of triangle P,Q,R are P(-3,1), Q (-1,-2), R (-2,-4)

(a)ย  Draw triangle PQR and its image PIQIRI of PQR under translation T =ย ย ย  3ย ย ย  on the provided grid ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย 4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

(b)ย  Under transformation matrix m =ย ย ย  4ย  3ย  , PIQIRI is mapped on to PIIQIIRII. Find the

co-ordinates of PIIQIIRII and plot itย ย  1ย  2ย ย ย  on the given gridย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 Mks)

(c)ย  If area of D PIQIRI is 3.5 cm2, find area of the images PIIQIIRII ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 Mks)

 

MATHEMATICS I

PART 1

MARKING SCHEME

 

  1. 171 X 171 X 3 X 10-5 M1

ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย 855 X 531

= 2 X 10-6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  2

 

  1. No. of men = 6 X 14 X 8 X 2500 M1

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  ย ย ย 2 X 5 X 2240

= 75ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

Extra menย ย ย ย ย ย ย  = 75 – 6 = 69ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

ย 3

  1. OM = 2i + 7j + 2/5 (4i – 3j + 6i + 21j – 2i – 7j) M1

= 2i + 7j + 2/5 (8i + 11j)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 26 i + 57 j

5ย ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  3

ย 

ย 

ย 

ย 

 

  1. 2 5ย ย ย ย ย ย  xย ย ย ย ย ย ย ย  =ย ย ย ย ย  1

3ย  4ย ย ย ย ย ย  yย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

 

xย ย ย ย ย ย ย ย ย  -1/7 ย ย 5/7ย ย ย ย ย ย  1

yย ย ย  =ย ย ย ย  3/7 ย ย -2/7ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

 

xย ย ย  =ย  3

yย ย ย ย ย ย  -1

 

x, 3, y = -1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

 

  1. Reflex รAOC = 126 x 2 = 2520 B1

Obtuse รAOC = 360 – 252 = 1080ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

= 1/2 (180 – 108)0

= 360ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

ย 3

  1. 18x2 – 39x + 18 = 0

6x2 – 13x + 6 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  B1ร– equation

6x2 – 9x – 4x + 6 = 0

3x(2x – 3) (3x – 2) = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

x = 2/3 ย orย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

x =1 ยฝ ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

4

ย 

  1. M :ย  Kย  :ย  Oย  =ย  4 : 2 : 3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– ratio

ย ย ย ย ย  Maina’sย  = 4/9 X 180

= 80/-ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– Omondi’s

ย ย ย ย ย  Omondi’s = 60/-ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  and Maina’s

ย ย ย ย ย  Difference = Shs.20/-ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 difference

3

  1. (2 + 1/2x)6 = 26 + 6(25) (1/2x + 15 (24) (1/2 x)2 M1

= 64 + 96x + 60x2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

2.46ย ย ย ย ย  = (2 + 1/2 (0.8))6

= 64 + 96 (0.8) + 60 (0.64)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 179.2

@179 to 3 s.fย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 4

  1. P (FF) = 7/20 X 7/20

= 49/100ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

P (at least one fail) = 1 โ€“ P (FI FI FI)

= 1- 13/20ย ย  3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 1 – 2197ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

8000

= 5803

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  8000 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 4

 

  1. grad = term 135

= -1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

yย  = mx + c

yย  = -x + 5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

ย 2

 

  1. Volume = 2 x 10,000 x 10,000 x 25 M1ร– x section area

1000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  10ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– conv. to litres

= 500,000 Ltsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

 

  1. 10a – 6(3a – 7) = 5(a -2) M1

10a – 18a + 42 = 5a – 10

– 13aย ย ย  = -52ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

aย ย ย ย ย ย ย  = 4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

  1. 2a + 12d = 2

2a + 4d = -10ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

8dย ย  = 12

dย ย  = 11/2 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

aย ย  = -8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

S21ย  = 21/2 (-16 + 20 X 3/2)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 147ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 5

 

  1. 2 p r = 120 x p x 40 M1

360

r = 6.667 cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

h =ย  ร– 400 – 44.44ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 18.86 cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 4

  1. = (w (x – z) โ€“ y (x – z)) (w + z) M1ร– factor

(z – w) (z + w)

= (w – y) (x – z) (w + z)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– grouping

(z – w) (z + w)

= (w – y) (x – z)

z – wย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

 

R

250ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– sketch

  1. 550

Qย  125ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย  PR = 4 sin 125ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย M1

Sin 25

A1

30

Pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

  1. (a) <RST = 1800 – 740 ย = 1060ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

(b) < RTQ = 900– 740ย ย ย ย ย ย ย ย ย ย  = 160ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

< PTR = 460 + 160ย ย ย ย ย ย ย ย  = 620ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

< SUT = 620 – 390 ย ย ย ย ย ย ย  = 230ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

(c)ย  Reflex รRQT = 180 – 2 x 16

= 180 – 32 = 1480ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

Obtuse ROT = 360 – 148 = 2120ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

(d)ย  < PTS = 46 + 180 – 129 = 970ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

< PST = 180 – (97 + 39) = 440ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

8

(a)ย  Kshs.350,000 = $ 350,000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

74.75

= $ 4682ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(b) Balanceย ย ย ย ย ย  ย ย ย ย ย  = 3/4 x 4682

= $ 3511.5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

$3511.5 ย ย ย ย  = Fr 3511.5 x 74.75ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

11.04

= Fr 23780ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

Expenditureย ย ย ย ย  = Fr 16 200

Balanceย ย ย ย ย ย ย ย ย ย ย  = Fr 7580

(c) Value on arrival = Kshs.7580 X 12.80

= Kshs.97,024

Value on departureย ย ย ย ย ย ย  = Kshs.7580 X 11.04ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 bothร–

= Kshs.83 683.2

Differenceย ย  ย ย  ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย = Kshs.97,024 – 83683.2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= Kshs.13,340.80ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 8

X -2 -1 0 1 2 3
Y -11 0 5 4 -3 -16

B1ร–values

 

y

S1ร– scale

8 —ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  P1ร– plotting

6 —ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  C1 ร– curve

4 —

2

 

-2 —ย ย ย  1ย ย  ย ย ย ย ย ย  ย ย 2ย ย ย ย ย ย ย  ย  3ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  x

-4 —

-6 —

-8 —ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  y=2x=2

-10 —

-12 —

-14 —ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  xย ย  =-0.53 + 0.1ย  BI

-16 —ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Nx = 1.87+ 0.1

 

y = 5+2x-3x2 =2-2xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  MI for equation

3x2-4x-4x-3=0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  AI equation

8

xย ย ย ย  = -0.53 ยฑ 0.1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

mxย ย  = 1.87 ยฑ 0.1

 

 

y = 5 + 2x โ€“ 3x2 = 2 โ€“ 2xย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– for equation

\ 3×2 โ€“ 4x โ€“ 3 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  MA1 ร– equation

ย 8

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

 

20.

 

 

 

 

B1 ร– 300

 

Rย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย  B1 ร– 2 ^ PQ, QR

B1 ร– 2 ^ bisectors

B1 ร– circle

 

 

9ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Q

 

 

Radius = 4.2 ยฑ 0.1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– radius

Area of circle = 22/7 x 4.22

= 55.44 ยฑ 3 cm2

Area of D PQR = 1/2 x 3.5 x 7.5 sin 30ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– D and circle

= 6.5625 cm2

Differenceย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 55.44 – 6.5625ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– sub

= 48.88 cm2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 8

  1. Line (i) y/2 + x/5 = 1

5y + 2x = 10ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร–equation

5y + 2x = 10ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– inequality

ย ย ย ย ย  Line (ii)ย ย ย ย ย  y/4 + x/-3 = 1

3y = 4x + 12ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– equation

3y < 4x + 12 or 3y – 4x < 12ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– inequality

ย ย ย ย ย  Line (iii)ย ย ย ย  grad = -1/3 y inter = 4

3y + x = 12 or 3y = -x + 12ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– equation

3y + x < 12ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– inequality

ย ย ย ย ย  Line (iv)ย ย ย ย ย  y – 3 = -3

x – 3ย ย ย ย ย  2

2y + 3x = 15ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– equation

\ย ย ย ย ย ย ย ย  2y + 3x ยฃ 15ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– equation

ย  8

CLASS

F x Fx Cf
60 โ€“ 64

65 โ€“ 69

70 โ€“ 79

75 โ€“ 79

80 โ€“ 84

85 โ€“ 89

8

14

18

15

3

2

62

67

72

77

82

87

ย 496

938

1296

1155

246

174

8

28

40

55

58

60

  Sf = 60   ย ย ย  Sfx 3809  

 

B1ร– x column

B1ร– f column

 

 

 

 

(a)ย  (i)ย  Modal classย ย  = 70 – 74ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  B1ร– model class

(ii) Median class = 70 – 74ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– median

 

(b)ย ย ย ย ย ย ย ย ย ย ย ย ย  Mean =ย  3809

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  60 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 63.48ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

 

S1ร– scale

B1 ร– blocks

59.5 – 64.5

64.5 – 69.5 e.t.c.

ย 8

(c)

ย 

Histogram

 

 

 

20ย  —

 

 

15ย  —

 

 

10 – –

 

 

5ย  —

 

 

 

 

55ย ย ย  60ย ย ย ย ย ย ย  65ย ย ย ย ย ย ย  70ย ย ย ย ย ย ย  75ย ย ย ย ย ย ย  80ย ย ย ย ย ย ย  85ย ย ย ย ย ย ย  90

 

  1. (a) XA = a, YA = 10 – a, YB = 10 – a, CZ = 10 – a = ZB

YZ = 10 – a + 12 – a = 8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

2a = 14

a = 7 cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

Cos X = 100 + 144 – 64

240 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– any angle of the D

= 0.75

X = 41.410

ย ย ย ย  1/2 X = 20.700ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1ร– 1/2 of the angle

 

r = OA = 7tan 20.7ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 ร– radius

= 2.645 cm

Shaded area = 1/2 X 10 X 12 sin 41.41 – 22/7 X 2.6452ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– D & circle

= 39.69 – 21.99

= 17.7 cm2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1ร–

ย 8

 

 

 

 

 

 

 

  1. (a) 650,000 (0.85)n = 130,000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– formula

1.15nย  ย  = 0.2

nย ย ย  = log 0.2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร–

log 0.85

=ย  1.3010

1.9294

= – 0.6990ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

– 0.0706

= 9.9 yearsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(b)ย  650,000 (1 – r/100) 5 = 325,000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

(1 – r/100) 5 = 0.5

1 – r/100ย  ย ย ย = 0.5 1/5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 0.8706

r/100 = 0.1294ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

rย ย ย  = 12.9 %ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

ย 8

MATHEMATICS I

PART II

MARKING SCHEME

ย 

SECTION I (50 MARKS)

 

 

  1. = b15ย ย ย  ย ย 2/5ย ย ย  Xย ย ย  4a6ย ย  3/2

32a10ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  9b4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– reciprocal

 

 

=ย ย ย ย ย ย ย ย ย  2a5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

27 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

 

ย ย ย ย ย  No.ย ย ย ย ย ย  ย ย ย ย ย  Log.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

0.375ย ย ย ย ย ย ย ย ย  1.5740 +

cos 75ย ย ย ย ย ย ย ย  1.4130

2.9870 _

tan 85.6ย ย ย ย ย  1.1138

3.8732 = ย 4 + 1.8732

2 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

2.9366

0.0864

 

  1. S. Price = ย 80ย  ย X 600

100

= Shs.480ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

Cost Price = x

96xย  ย ย ย ย  = 480ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

100

xย  =ย ย  Shs.500ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

  1. r2 + hr = A/2p ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

r2 + hr + (h/2)2 = A/2A + h/4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

(r + h/2)2 =ย  ร– 2A + h2

4pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

rย ย ย  = -h/2 ยฑย ย  ร–2A + h2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

4pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4

 

  1. (12/3r) (1/3 r) = (1/4 r) (5) M1

4r2 – qr = 0

r(4r – q) = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

r = 0

orย ย  rย  = 2.25ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

 

  1. = 2 (5 + 2ร–3) – 1 (5 – 2ร–3) M1

(5 – 2ร–3) (5 + 2ร–3)

= 10 + 4ร–3 – 5 +2ร–3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

13

= 5 + 6ร–3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

13ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

  1. P = Kq2 + c

6 = 4k + c

16 = 9k + cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– subtraction

5k = 10

k = 2

c = -2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1 ร– k and c

ย ย ย ย ย  P = 64ย ย ย ย  2q2 = 66

qย  = ร–33

= ยฑ 5.745ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 4

  1. Volume = 1/2 X 8 X 8 sin 120 X 10 M1 ร– area of x-section

ย ย ย ย ย  No. of scouts = 32 sin 60 X 10ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– volume

2.5 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 110.8

= 110ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 3

 

  1. Max. error = 2(8.5 + 5.5) – 2(7.5 + 4.5)

2

= 2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

% error = 2/26 X 100ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 7.692%ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

Gย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

 

 

  1. B1 ร– net

 

Hย ย ย ย ย ย ย ย ย ย ย ย  Dย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย Gย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Hย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 ร– dimen. FE must be 10cm

 

4cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย  4cm

 

B1 ร– labelling

E 4cm ย Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  12cmย ย ย ย ย  Fย ย ย ย  10cm ย ย ย Eย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย  ย 3

4cmย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย 12cm

E

F

  1. (1 + 2x)6 = 1 + 6(2x) + 15 (2x)2 M1

= 1 + 12x + 60x2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(1.04)6 = (1 + 2(0.02))6

= 1 + 12 (0.02) + 60(0.02)2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 1.264ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย 4

 

 

 

 

  1. BT = 10 cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

CT = 10 sin 40ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 6.428 mย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

A1 10cmย ย ย  Bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  h = 6.428 + 1.5

1-5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย  = 7.928ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

ย  4

 

 

  1. A.S.F = 405 2/3ย  =ย  27ย  2/3ย ย  =ย  ย 9ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1

960ย ย ย ย ย ย ย ย ย ย  64ย ย ย ย ย ย ย ย ย ย ย  10

smaller area = 29 ย X 120ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

164

= 67.5 cm2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  3

 

  1. Relative speed = (x – 30)km/h B1

2 kmย ย ย ย  =ย ย ย ย ย ย ย ย ย  2 hrs

(x – 30)km/hย ย ย ย ย  60ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

2x – 60 = 120

x = 90 km/hย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  3

  1. 16 Km/h 5 Km/hr

x Kmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (22 – x) Km

x + 22 – xย ย  = 11

16ย ย ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

5x + 352 – 16x = 220ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– x-multiplication

11xย  = 132

xย  = 12 kmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  3

 

  1. AP = 2/9 x 12 = 22/3 cm B1 ร– both AP & AQ

ย ย ย ย ย  AQ = 5/9 x 12 = 62/3 cm

\ PQ = 62/3 – 22/3 = 4 cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 ร– C.A.O

ย  2

 

  1. Taxable income = 115/100 x 8000 M1

= Shs.9200 p. m

= Shs.110,400 p.aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

Tax dues = 10/100 x 39600 + 15/100 x 39600 + 25/100 x 31200ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– first 2 slabs

= 3960 + 5940 + 7800ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– last slab

= Shs.17,700 p.a

= 1475 p.mย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

net tax = 1475 – 400

= Shs.1075ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 ร– net tax

Total deductions = 1075 + 130

= Shs.1205

net income = 8000 – 1205ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= Shs.6795ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

 

 

 

 

 

(a)ย  P (all correct) = 2/3 x 3/5 x 2/5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 12/125ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(b)ย  P (2 correct) = 2/5 x 3/5 x 3/5 + 2/5 x 2/5 X 1/5 + 3/5 x 4/5 x 2/5

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 18/125 + 4/125 + 24/125ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 46/125ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(c) P (at least 2 correct)

= P(2 correct or 3 correct)

= 46/125 + 12/125ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 46 + 12ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

125

= ย 58

ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  125ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

  1. Old price/pen = 440

x

New price/pen = 494ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร–both expressions

x + 3

440494ย ย  = 1.50

xย ย ย ย ย  x + 3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– expression

440(x + 3) – 494x = 1.5x2 + 4.5xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร– x-multiplication

x2 + 39x – 880 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1 ร– solvable quad. Eqn

x2 + 55x – 16x – 880 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– factors or equivalent

(x – 16) (x + 55) = 0

x = -55

or x = 16ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1 ร– both values

\ x = 16

difference in purchase = 19 X 1.50ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= Shs.28.50ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

  1. (a) OP = a + 1/4 (b – a) M1

= 3/4 a + 1/4 bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(b)ย  PQ = PO + OQ

= –3/4 a – 1/4 b + 1/4 (a – b)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= –1/2 a – 1/2 bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(c)ย  QN = QO + ON

= 1/4 (b – a) + 5/3 bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 23/12 b – 1/4 aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

(d)ย  PN = PB + BN

= 3/4 (b – a) + 2/3 bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 17/12 b – 3/4 aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

  1. (a) Volume in 2 days = 22 x 3.5 x 3.5 x 150 x 20 x 3600 M1 ร– area of x-section

7ย ย ย ย ย ย  2ย ย ย ย ย ย ย  2ย ย ย ย ย ย ย ย ย ย  1,000,000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– volume in cm3

= 103.95 m3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร– volume in m3

(b)ย  22 X r2 x 8 = 103.95 x 15 ย ย x 7ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

7ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

 

r2 = 103.95 x 15 x 7ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2 x 2 2x 8

= 31.01ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

r = 5.568 mย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

  1. (a) Ration of needs for A:B = 5:8

A’s share = 5/13 x 16.9 + 1/2 x 13.5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 13.25 Millionย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

B’s share = (13.5 + 16.9) – 13.25ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 13.25ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

  • Aโ€™s share 5/13 x 16.9 + 39/50 x 13.5

6.5 + 10.53

= 17.03 mย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

B’s share = 30.4 – 17.03ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1

= 13.37 Millionย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

  1. Log V = n Log E = log k
Log V -0.46 -0.13 -0.14 -0.01 0.05
Log E -0.35 -0.21 -0.05 0.07 0.13

B1ร– log V all points

B1ร– log E all points

S1 ร– scale

P1ร– plotting

Log V = n log E + log Kย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  L1 ร– line

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Log K = 0.08

K = 1.2 ยฑ 0.01ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 ร– K

N = 0.06/0.06ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1 ร– n

= 1 ยฑ 0.1

\ v = 1.2Eย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร– v

when E = 0.75, V = 0.9 ยฑ 0.1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  8

  1. (a) T 3 PQR ยฎ PIQIRI

4ย ย ย  PI (0,5), QI (2,2) RI (1,0)

PI QI RIย ย ย ย ย ย  PIIย  QIIย  RII

(b)ย  4ย  3ย ย ย  0ย ย  2ย ย  1ย ย  =ย ย  15ย ย ย  14ย ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1 ร–

1ย  2ย ย ย ย  5ย ย  2ย ย  0ย ย ย ย ย ย ย  10ย ย ย ย  6ย ย ย  1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1 ร–

 

PII (15,10), QII (14,6), RII (4,1)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ร–

(c)ย  Area s.f = det M

= 5

 

area of PII QII RII = 5 (area PIQIRI)

= 5 X 3.5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  M1ร–

= 16.5 cm2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1

ย  8

 

 

ย 

MATHEMATICS 2

PART I

ย 

SECTION A:ย 

 

  1. Use logarithm tables to evaluate ย  ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย (4 mks)

 

0.0368 x 43.92

361.8

 

  1. Solve for x by completing the square ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย (3mks)

2x2ย  – 5x + 1 = 0

 

  1. Shs. 6000 is deposited at compound interest rate of 13%. The same amount is deposited at 15% simple interest. Find which amount is more and by how much after 2 years in the bank ย ย ย ย ย  (3mks)

 

  1. The cost of 3 plates and 4 cups is Shs. 380. 4 plates and 5 cups cost Shs. 110 more than this. Find the cost of each item.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

 

  1. A glass of juice of 200 ml content is such that the ratio of undiluted juice to water is 1: 7 Find how many diluted glasses can be made from a container with 3 litres undiluted juice ย ย ย ย ย  (3mks)

 

  1. Find the value of ฮธ within ฮธย  < ฮธ < 360Oย  ifย  Cos (2 ฮธ + 120) =ย  ฮณ3ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (3mks)

2

 

  1. A quantity P varies inversely as Q2 Given that P = 4 When Q = 2.ย  , write the equation joining P andย  Q

hence find P when Q = 4ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  aย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. A rectangle measures 3.6 cm by 2.8 cm. Find the percentage error in calculating its perimeter. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

 

  1. Evaluate: ย ย ย ย ย ย ย ย  11/6ย ย  xย  ยพย  –ย  11/12ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

ยฝย  of 5/6

 

  1. A metal rod, cylindrical in shape has a radius of 4 cm and length of 14 cm. It is melted down and recast into small cubes of 2 cm length. Find how many such cubes are obtained ย ย ย ย ย ย ย ย  ( 3mks)

 

  1. A regular octagon has sides of 8 cm. Calculate its area to 3 s.f. ย ย ย ย ย ย ย ย ย ย ย  (4mks)

 

  1. Find the values of x andย  y ifย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ( 2 mks)

3ย ย ย ย ย ย ย ย ย  xย ย ย ย ย ย ย ย ย  1ย ย  =ย ย ย ย  2

2ย ย ย ย ย ย ย ย ย  1ย ย ย ย ย ย ย ย ย  -1ย ย ย ย ย ย ย ย  y

 

  1. An equation of a circle is given by x2 + y2 โ€“ 6x + 8y โ€“ 11 = 0 ย ย ย ย ย  (3mks)

Find its centre and radius

 

 

 

 

 

 

  1. In the figure given AB is parallel to DE. Find the value of x and y

 

 

 

 

 

 

 

 

  1. A line pass through A (4,3) and B(8,13). Findย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (6 mks)

(i)ย  Gradient of the line

(ii)ย  The magnitude of AB

(iii) The equation of the perpendicular bisector of AB.

 

  1. A train is moving towards a town with a velocity of 10 m/s. It gains speed and the velocity becomes 34 m/s after 10 minutes . Find its acceleration (2mks)

 

 

SECTION B:

 

  1. Construct without using a protractor the triangle ABC so that BC=10cm, angle ABC = 600 and

BCA = 450

  1. On the diagram , measure length of AC
  2. Draw the circumference of triangle ABC
  3. Construct the locus of a set of points which are equidistant from A and B.
  4. Hence mark a point P such that APB = 450 and AP = PB
  5. Mark a point Q such that angle AQB = 450 and AB = AQ

 

  1. (a) A quadrilateral ABCD has vertices A(0,2) , B(4,0) , C(6,4) and D(2,3). This is given a

transformation by the matrixย ย  -2ย  0ย  to obtain its image AI B I CI DI. under a second transformation

0 – 2

which has a rotation centre (0,0) through โ€“900 , the image AIIย  BII ย CIIย  DII ย of AIย  BIย  CIย  DI ย is

obtained.ย ย ย  Plot the three figures on a cartesian planeย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (6mks)

(b)ย  Findย  theย  matrix ofย  transformationย  thatย  mapsย  theย  triangleย  ABCย  where A (2,2)ย ย  B (3,4)ย ย  C (5,2)

ontoย  A B Cย ย  whereย  A( 6,10)ย  Bย  (10,19 )ย  C ( 12, 13).ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ( 2mks)

 

 

 

 

19.

 

 

 

 

 

 

In the triangle OAB, OA = 3a , OB = 4b and OC = 5/3 OA.ย  M divides OB in the ratio 5:3

  1. Express AB and MC in terms of a and b
  2. By writing MN in two ways, find the ratio in which N divides
  3. AB
  4. MC

 

 

 

 

 

 

  1. In the figure below, SP = 13.2 cm, PQ = 12 cm, angle PSR = 80O and angle PQR = 900. S and Q are the centres ย ย ย ย ย  (8mks)

 

Calculate:

The area of the intersection of the two circles

The area of the quadrilateralย  S P Q R

The area of the shaded region

 

 

 

 

 

 

 

 

 

 

 

  1. In an experiment the two quantities x and y were observed and results tabled as below
X 0 4 8 12 16 20
Y 1.0 0.64 0.5 0.42 0.34 0.28

 

  1. Byย  plottingย  1/yย  against x, confirm that y is related to x by an equation of the form

 

Y =ย ย ย ย ย  q

ย 

 

P + x

where p and q are constants.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

(b)ย  Use your graph to determine p and qย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

(c )ย  Estimate the value ofย ย  (i) y when x = 14

(ii) x when y = 0.46ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

 

  1. A racing cyclist completes the uphill section of a mountain course of 75 km at an average speed of v km/hr. He then returns downhill along the same route at an average speed of (v + 20) km/hr. Given that the difference between the times is one hour, form and solve an equation in v.

Hence

  1. Find the times taken to complete the uphill and downhill sections of the course.
  2. Calculate the cyclists average speed over the 150km.

 

  1. In the diagram below, X is the point of intersection of the chords AC and BD of a circle. AX = 8 cm, XC = 4cm and XD = 6 cm
  2. Find the length of XB as a fraction
  3. Show that XAD is similar to XBC
  4. Given that the area of AXD = 6cm2, find the area of BXC
  5. Find the value of the ratio

Area ofย ย ย ย ย ย  AXB

Area ofย ย ย ย ย ย ย  DXC

 

 

 

 

 

 

 

  1. A town B is 55 km on a bearing of 0500. A third town C lies 75km due south of B. Given that D lies on a bearing of 2550 from C and 1700 from A, make an accurate scale drawing to show the positions of the four towns.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

(scale 1cm rep 10 km)

From this find,

(a) The distance of AD and DC in kmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

(b) The distance and bearing of B from Dย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

(c)ย  The bearing ofย  C from Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)

 

MATHEMATICS 2

PART 1

MARKING SCHEME ย ย ย ย ย ย ย ย ย ย ย  (100MKS)

ย 

 

  1. No. Log

=ย ย  3.6502

0.3681ย ย ย ย ย ย ย ย ย ย ย ย ย  2.5660

0.3682ย ย ย ย ย ย ย ย ย ย ย ย ย  1.6427 +ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย  -4ย  =ย  1.6502ย ย ย ย ย  = 2.8251

0.2087ย ย ย ย ย ย ย ย ย ย ย ย ย  Logs ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย 2

361.8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2.5585ย ย ย ย ย ย ย ย ย ย ย ย ย  + – vย ย  ansย  (4)ย ย  ย ย ย ย ย  6.6850 x 10 -2

3.6502ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 0.06685

 

  1. 2 x2 โ€“ 5x + 1 = 0

x2 โ€“ 5 x + ยฝ = 0

2

x2 โ€“ 5 xย ย  = ยฝ

2

x โ€“ 5xย  +ย ย ย ย  5 2ย  ย ย =ย  ยฝย ย  +ย ย ย ย  5ย  2ย  ย  (m)

2ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4

 

= x –ย  5ย ย ย  = ยฝ +ย ย ย ย ย  25 ย ย ย =ย  17ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

4ย  ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย 16ย ย ย ย ย ย ย  16

 

= x โ€“ 5/4ย  =ย  17/16ย ย  =ย ย ย  1.0625

x โ€“ 5/4ย ย ย  ยฑย  1.031

X1 = -1.031 = 1.25 = 0.2192

X2 = 1.031ย  + 1.25ย  = 1.281

 

  1. A1 = P(1 + R/100)2 = 6000ย  xย  113/100 x 113/100 = Sh. 7661.40

 

A2 = P + PRT/100ย ย ย ย ย ย ย ย  =ย ย  6000 + 15 X 2 = 6000 + 1800

100

=ย ย  Shs. 7800

 

Amount by simple interest is more by Shs.ย  (7800 โ€“ 7661. 40)

Shs. 138.60

  1. Let a plate be p and a cup c.

3p + 4c = 380ย  x 5ย ย ย ย ย ย ย ย  ย ย ย  15p + 20cย  = 1900

4p + 5cย  = 490ย  x 4ย ย ย ย ย ย  16p + 20cย  = 1960ย 

-p ย  ย ย ย -60ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (m)

 

 

 

 

 

p = Shs 60

 

3(60) + 4 c = 380

4c = 380 โ€“180 = 2000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

c=ย ย  Shs. 50

Plate = Shs. 60 , ย ย ย ย ย ย ย ย ย ย  Cup = Shs. 50ย ย ย ย ย ย ย ย ย ย ย  (A both)

 

  1. Ratio of juice to water = 1ย ย ย ย ย ย ย ย ย  : ย ย ย ย ย ย ย ย ย  7

In 1 glass = 1/8 x 200 = Sh 25

3 litres = 300 ml (undiluted concentrate)ย ย ย ย ย ย ย ย ย ย  (3)

No. of glasses =vย ย ย  3000ย  =ย  120 glasses

25

 

  1. Cos (2 ฮธ + 120) = 3/2 = 0.866

Cos 30 , 330, 390, 690, 750 โ€ฆ.

ย ย ย ย ย ย ย ย ย ย ย  2 ฮธ + 120ย ย  ย ย ย ย ย ย  ย ย ย ย ย ย = 330

2 ฮธ = 210ย ย ย ย ย ย ย ย ย  ,ย ย ย ย  = 1050ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

2 ฮธ = 390 โ€“ 120ย ย  = 2700 ,ย  ย ย ย ย ย ย ย ย  ย ฮธ2 1350

2 ฮธ =ย  690 โ€“ 120ย  = 5700ย  ,ย ย ย ย ย ย  ฮธ3 2850ย ย  ย  ย ย (for 4 ans)

ฮธ4= 315oย  ย  ( for >2)

2 ฮธ =ย  750 โ€“ 120ย ย  = 6300 ,

ย 

  1. P =ย ย ย ย ย ย ย ย ย  kย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4ย  =ย  K/4ย ย ย ย ย ย ย ย ย ย  (substitution)

Q2ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 9

K = 4 X 4 ย ย ย ย ย ย ย  =ย ย ย ย ย ย ย ย ย  ย  16

9ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  9

P =ย  16ย ย  vย ย ย ย ย ย ย ย  when Q = 4

9Q2

ย 

P =ย ย ย ย ย ย ย ย  16ย ย ย ย ย  ย  =ย ย  1/9 ย ย ย ย ย ย ย ย ย ย  ย ย (A)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

9x4x4

 

  1. The perimeter = (3.6 + 2.8 ) x 2 = 12.8 cm

Max perimeter = (3.65 + 2.85) x 2 = 23 cmย  ย  Expressions

% error =ย ย  13 โ€“12.8ย ย ย ย  xย  100ย ย ย  mย ย ย ย ย ย ย ย  =ย ย ย ย  0.2ย ย ย ย ย ย ย  xย ย ย ย  100ย  (3)

12.8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  12.8

= 1.5620%ย ย ย ย ย ย ย  (A)

 

  1. ย ย ย ย ย 1 1/6 x ยพย  – 11/12ย ย  = (7/6 x ยพ )ย  -11/12ย ย ย ย ย ย ย ย  =ย  7/8 โ€“ 11/12ย ย  =ย ย  21-22ย ย 

ยฝย  of 5/6 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ยฝ of 5/6 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  5/12ย ย ย ย ย  ย ย ย ย ย ย ย  5/12

= -1/24ย ย ย  = -1ย  x 12ย ย ย  =ย  -1

5/12ย ย ย ย ย ย ย  24ย ย  5ย ย ย ย ย ย ย ย ย  10ย ย ย ย ย ย  (3)

 

  1. Volume of rod = ะŸ r2h = 22/7 x 4ย  x 14 = 704cm3ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  (m)

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Volume of each cube = 2x2x2 = 8 cm3 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย A

ย 

No. of cubes = 704 /8ย  = 88 cm3 ย  A

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

  1. < AOB = 360 = 450

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  8

Tan 67.5 =ย  h

4

h = 4 x 2.414ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A

=ย  9.650cm

Area of 1 triangle = ยฝ x 8 x 9.656 x 8 cm = 38.628 x 8ย ย  vm

Octagon areaย  =ย  38.628 x 8ย ย ย ย ย  m

=ย  309.0 cm2ย ย ย ย ย ย ย  (A)

 

  1. 3 ย  2ย ย ย ย ย ย ย  -1ย ย ย ย ย ย ย  ย ย ย ย  2

=

2ย ย ย ย ย ย ย ย ย ย ย  ย  1ย ย ย ย ย ย ย  ย  -1ย ย ย ย ย ย  ย ย ย  y

 

3 โ€“ x = 2ย ย ย ย ย ย  (1)ย ย  ย ย ย  x = 1 ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (2)

2 –ย  1 = yย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  y = 1ย  (A)

 

  1. x 2 + y2 โ€“ 6x + 8y โ€“ 11 = 0

x2 – 6x + (-3)2 + y2 + 8y + (4)2 = 11 + (-3)2 + (4)2 ย ย ย ย ย ย ย ย (completing the square)

(x โ€“ 3)2 + (y+4)2 = 11 + 9 + 16 = 36

(x โ€“ 3)2 + (y + 4)2 = 62 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

Centre isย  (3, -4)

Radius ย ย ย ย ย  = 6 units ย ย ย ย ย ย ย ย ย  Asย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

 

 

14.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figs A C Bย  and D C E are similar

ABย ย ย ย ย ย  =ย ย ย ย ย ย ย ย ย  ACย ย ย ย ย ย  =ย  andย ย  ABย ย ย ย ย ย  =ย ย ย ย ย ย ย ย ย  BC

ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย DEย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  DCย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  DEย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  CE

 

10ย ย ย  =ย  6 + x

3ย ย ย ย ย ย ย ย ย  6

= 10ย ย  =ย  15 + y,ย ย ย ย  m

3ย ย ย ย ย  ย ย ย ย ย  yย ย ย  ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  60 = 18 + 3x

10yย  = 15 + 3yย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย ย 3x = 42

7y = 15 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย x = 14

 

y = 15/7ย ย ย  ย ย ย ย ย ย ย  ย ย (A)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

A (4 , 3)ย ย ย ย ย ย ย ย ย ย  B(8,13)

 

  1. (i) gdtย ย ย ย ย ย ย ย ย  = change in yย ย ย  = 13-3 = 10ย ย ย ย  =ย  5

change in xย ย  ย ย ย  8-4ย ย  ย ย ย  4ย ย ย ย ย  ย ย ย  2

 

(ii)ย ย ย ย ย  Magย  AB ย =ย  8ย ย ย ย  -4ย ย ย ย ย ย ย ย  ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย  =

13 -3ย ย ย ย ย ย ย ย  10

Length =ย ย  ร–42 + 102ย  ย ย ย = ร–116 = 10.77 units

(iii)ย ย  Mid pointย  = 4 +8ย  ,ย ย ย  3 + 3

2ย ย ย ย ย ย ย ย ย ย ย ย  2

=ย  (6, 8)ย ย ย  (mid point)ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (5 mks)

gdt of perpendicular to AB = -ve rec. of 5/2

-2/5

Eqn isย  y = -2/5 x + c

8 = -2/5ย  x 6 + cย ย ย  =ย  40ย  = -12ย  +ย  5c

= c = 52/5

 

y = -2/5 x + 52/5ย ย ย ย ย ย ย  (A)

 

 

  1. Acceleration = Change in velocity

Time

= (34 โ€“ 10) m/sย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย  = 24 m/s

60 x 10ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  600

 

= 0.04m/s2-ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2)

 

 

ย 

 

 

 

 

 

 

 

 

 

 

 

 

17.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Triangleย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8)

AC = 9cm

Circumference Centre

Circle

Perpendicular bisector of AB

P

Q

 

 

 

  1. (b) a bย ย  2ย ย ย ย  ย  3ย ย ย ย ย ย ย  ย  5ย ย ย ย ย ย ย  6ย ย ย ย ย ย ย ย ย  10ย ย ย ย ย ย ย  12

cย ย ย ย ย ย ย ย ย  dย ย  2ย ย ย ย  ย  4ย ย ย ย ย ย ย  ย  2ย ย ย ย ย ย ย  10ย ย ย ย ย ย ย  19ย ย ย ย ย ย ย  13

 

2a +2b = 6ย  x 2ย ย ย ย ย ย  = 49 + 4b = 12

3a + 4b = 10ย ย ย ย ย ย ย ย ย ย ย ย  3a + 4b = 10

aย ย ย ย  = 2ย ย ย ย ย ย ย ย ย ย ย ย ย  4 + 2b = b

 

2c + 2d = 10×2 = 4c + 4d = 20ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2 b = 2ย  b = 1

3c + 4dย  = 19ย ย ย ย ย ย ย  3c + 4dย  = 19

cย ย ย ย ย ย ย ย ย ย  = 1

2 (1)ย  + 2d = 10

2d = 8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Matrix is ย ย ย ย ย ย ย ย ย  2ย ย ย ย ย ย ย ย ย  1ย ย ย ย ย  (A)

d = 4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1ย ย ย ย ย ย ย ย ย  4

 

 

 

 

 

OC = 5/3 (31) = 5A

 

19.

 

 

(a)ย  = AO + OBย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  MC = MO + OC

= -3aย  = 4b ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย = -5/8 (4b) + 5

= 5A โ€“ 5/2 b

 

(b) MN = 5 Mcย ย ย ย  = 3(5a โ€“ 5/2 b)

= 5 s a โ€“ 5/2 s b

 

MN = BN + BN

=ย  3/8 (4 b) + (1 โ€“ t) (-BA)

=ย  3/8 (4 b) + (1 โ€“ t)(3a โ€“ 4 b)

=ย  3/2ย  b + 3 ta โ€“4b + 4tb

= (3-3t) a (4t โ€“ 5/2)b

 

MN = MN

= 5 s a โ€“ 5/2ย  sb = (3-3t)a +ย ย  (4t โ€“ 5/2 )b

=ย  5 a =ย  3 โ€“ 3tย  ย ย ย ย  = 5s + 3t =3

= -5/2 s = 4t โ€“5/2ย  vย ย ย  ย 5s + 8t = 5ย 

-5t = -2ย ย ย ย ย ย ย ย ย ย ย  t = 2/5

5 sย ย  = 3 โ€“ 3(2/5)

= 3 โ€“ 6/5 = 9/5

= 3 โ€“ 6/5 = 9/5

s = 9/25

 

(i)ย ย ย  AN :ย ย ย ย  NB = 2 : 3

 

(ii)ย ย  MN :ย ย ย  9ย ย  :ย  16

 

 

 

 

 

 

 

 

20.

 

 

ฮธ x pr2

360

 

  1. Area of sector SPR =ย  80/360 x 13.2 x 13.2 x 3.142

=ย  121.6

Area of triangle SPR ยฝ x 13.2ย  x 13.2 x sin 80

= 85.8 cm2

(m of area of ) A (at least one)

(m of area)ย  A(at least one)

Area of segment = 121.6 โ€“ 85.8

= 35.8 cm2

Area of sector QPR = 90/360 x 3.142 x 12 x12

 

Area ofย  PQR = ยฝ x 12 x 12 = 722

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Area of segment = 113.1 โ€“ 72

= 41.1cm2

Area of intersection = (35.8 + 41.1) = 76.9 cm2

 

b).ย  Area of quadrilateralย  = Area ofย ย  PQR + SPR

=ย  85.8 + 72 = 157.8cm2

Area of shaded regionย  =ย  Area of Quadrilateral โ€“ Area of sector SPR

=ย  157.8 โ€“ 121.6

=ย  36.2 cm2

 

 

  1. y = qย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  p + x = qย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1ย  =ย  x + p

p + xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  yย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  yย ย ย ย ย  qย ย ย  q

 

Gradientย  = 1/qย ย  at (0, 0.95)ย  (8,2.0)ย  (8,2.0)ย  gradientย ย  =ย  2.0 โ€“ 0.95ย  =ย  1.05

8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  8

1ย ย ย ย ย ย ย ย ย  =ย  0.1312

q

qย  =ย  1ย ย ย ย ย  =ย  7.619

0.1312

q =ย  7.62.

 

y(1/y)ย  Interceptย ย  pย ย  ย =ย  0.95ย ย  ย ย Pย ย  =ย  0.95

qย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  7.62

 

p = 7.62 x 095ย  =ย  7.27

at x =ย  14,ย  y = 2.7

atย  y = 0.46,ย  1/yย  =ย  2.174

xย  =ย  9.6.

 

 

 

 

 

 

  1. a) Distanceย  =ย  75kmย ย  uphill speedย  =ย  vkm/h

uphill Timeย  =ย  75/v hrs

Downhill speedย  = ( + 20)ย  km/h

Downhill Timeย ย ย  =ย ย ย ย ย ย ย  75 ย ย ย ย ย ย ย ย hrs.

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย  v + 20

Takes larger uphill

75 ย –ย  75ย ย ย ย ย ย ย ย ย ย ย  ย =ย  1

vย ย ย ย ย ย ย ย  v+20

75 (v+20) โ€“ 75vย ย  ย ย ย ย ย ย ย ย  = 1

v(v + 20)ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1

75v + 1500 โ€“ 75vย  =ย  v(v + 20)ย  =ย  v2 + 20v.

v2 + 20vย  – 1500ย  =ย  0

vย  =ย  – 20 +ย  202 โ€“ 4(1)ย  (-1500)

2(1)

vย  =ย  –20 +ย  400 + 6000ย  = –20 + v6400

2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2.

V1ย ย ย ย  =ย  –20ย  +ย  80ย ย ย ย ย  =ย  30km/hr

2

V2ย ย ย  =ย ย  – 20 โ€“ 80ย ย ย ย ย  Xย ย  impossible

2

speed uphillย ย ย ย ย  =ย  30 km /hr,ย  T = 75ย  time =ย  2 ยฝ hrs

30

speed downhill =ย  50 km /hrย  Time = 75ย ย ย ย ย  Time =ย  2 ยฝ hr

50

Average speedย ย  =ย  Totalย  distanceย ย ย ย ย ย ย ย  =ย  150kmย ย ย ย ย ย ย ย ย  =ย  37.5 km/ hr

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย  Total timeย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  4hrs

 

X 0 4 8 12 16 20
Y 1.0 0.64 0.5 0.42 0.34 0.28
1/y 1.0 1.56 2.0 2.38 2.94 3.57

 

 

  1. A ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย B

 

 

 

 

Dย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  C

 

A x X x Cย  =ย  BX .ย  XD

8 x 4ย ย ย ย ย ย ย ย ย ย  =ย  6BX

BXย ย ย ย ย ย  =ย  8 x 142ย ย ย ย ย ย ย ย ย  =ย ย  16ย ย 

6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

X ADย ย  =ย  XBC

XA ย ย ย ย ย  =ย  8ย ย ย  =ย  24ย ย ย ย ย  =ย  3

XBย ย ย ย ย ย ย  16ย ย ย ย ย ย ย  16ย ย ย ย ย ย ย ย ย  2

XDย ย ย ย ย  =ย ย ย  6ย ย ย ย ย  =ย ย ย  3

XCย ย ย ย ย ย ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย  2

 

<ย ย  AXDย ย  =ย ย  BXCย ย ย ย ย ย ย ย ย ย ย  (vertically oppositeย  <s))

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  SAS holdsย  :ย  they are similar.

LSFย  =ย ย  3/2ย ย ย  ASFย  =ย  (3/2)2ย  =ย  9/4

Areaย  A x Aย  =ย  6cm2ย ย ย  Areaย  B x Cย  =ย  6 x 9 ย ย ย ย ย ย =ย  27ย ย  =ย  13.5cm2

4

 

24.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. a) AD =ย ย  50km

DCย ย  =ย ย  35km

BDย  = 90km

Bearing is 0200ย 

Bearing is 134oย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)

 

 

MATHEMATICSย  I

PART II

ย 

SECTION (52 MARKS)

 

  1. Without using tables, simplify

1.43 x 0.091 x 5.04

2.86 x 2.8 x 11.7ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. Make x the subject of the formula if

y = a/xย  +ย  bxย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. Give the combined solution for the range of x values satisfying the inequality

2x + 1<ย  10 โ€“ xย  <ย ย  6x โ€“ 1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. A man is employed at a KShs. 4000 salary and a 10% annual increment. Find the total amount of money received in the first five yearsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)

 

  1. A town A is 56 km from B on a bearing 0620.ย  A third town C is 64 km from B on the bearing of 140o.ย  Find

(i) The distance of A to Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

(ii) The bearing of A from Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

 

  1. Expand (x + y)6 hence evaluate (1.02) to 3d.p.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. Rationalise the denominator in ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

 

ร– 3

1 โ€“ v3

 

 

 

  1. The table below shows daily sales of sodas in a canteen for 10 days.

 

ย Day 1 2 3 4 5 6 7 8 9 10
No. of 52 41 43 48 40 38 36 40 44 45

 

Calculate the 4 day moving averages for the dataย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (3mks)

 

  1. Find the image of the line y = 3x = 4 under the transformation whose matrix is.

3mks

2ย ย ย ย ย ย ย ย ย ย  1

-1ย ย ย ย ย ย ย ย  2

 

  1. Three points are such that A (4 , 8), B(8,7), C (16, 5). Show that the three points are collinearย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)
  2. Write down the inverse of the matrix 2 โ€“ 3 hence solve for x and y if

4ย ย ย ย  3

2xย  – 3y = 7

4x + 3y +5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. Use the table reciprocals to evaluate to 3 s.f. 3mks

1/7ย  +ย  3/12ย  +ย  7/0.103

 

 

 

 

 

 

 

Given that O is the centre of the circle and OA is parallel to CB, and that angle

ABC =ย ย  1070,ย  find

(i) Angles AOC,ย  ย ย ย ย ย ย ย ย  ย ย ย ย ย (ii) OCBย ย ย ย ย ย ย ย ย ย ย ย ย ย  (iii) OABย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (3mks)

  1. Two points A and B are 1000m apart on level ground, a fixed distance from the foot of a hill. If the angles of elevation of the hill top from A and B are 60o and 30o respectively, find the height of the hill ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 mks)
  2. Two matatus on a dual carriageway are moving towards a bus stop and are on level 5 km from the stop. One is travelling 20 km/hr faster than the other, and arrives 30 seconds earlier. Calculate their speeds. ย ย ย ย ย  (5mks)
  3. If log x = a and log y = b, express in terms of a and b

Log ย x 3ย 

VYย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (2mks)

 

SECTION B:

ย 

  1. The table below gives the performance of students in a test in percentage score.
Marks 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79
No. of Students  

2

 

4

 

7

 

19

 

26

 

15

 

12

 

5

 

Using an assumed mean of 44.5, calculate

  1. The mean
  2. The standard deviation
  3. Find the median mark

 

 

 

  1. Draw the graph of y = 2x2 โ€“ x โ€“ 4 for the range of x -3ย  = xย  =ย  3.ย  From yourย  graph

State the minimum co-ordinates

  1. Solve the equations
  2. 2x2 โ€“ x โ€“ 4 = 0
  3. 2x2 โ€“ 3x โ€“ 4 = 0

 

 

 

 

 

 

  1. Two concentric circles are such that the larger one has a radius of 6cm and the smaller one radius of 4 cm. Find the probability that an item dropped lands on the shaded regionย ย ย ย ย ย ย  ย ย  4mks

 

  1. Two unbiased dice are thrown. Find the probability of obtaining (4mks)
  2. A product of 6
  3. A sum of 8

iii. The same number showing ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)

 

 

 

 

 

 

 

 

 

 

Two pulley wheels centers A and B are joined by a rubber band C D E F G H C round them.ย  Given that larger wheel has radius of 12 cm and AB = 20 cm, CD and GF are tangentsย  common toย  bothย  wheels and that CBA = 60o), Find

  1. BD (Length)
  2. CD

iii.ย  Arc length CHG and DEF, hence find the length of the rubber.

 

  1. V A B C D is a right pyramid with a square base A B C D of side 5 cm. Each of its four triangular

faces is inclined at 750 to the base. Calculate

  1. The perpendicular height of the pyramid
  2. The length of the slant edge VA
  3. The angle between edge VA and base A B C D
  4. The area of the face VAB

 

  1. Plot the graphs of y = sin xo and y = cos 2xo on the same axes for โ€“180 ยฃ x ยฃ180o.

Use your graphs to solve the equation 2 sin x = cosย  2x

 

  1. The depth of the water in a rectangular swimming pool increases uniformly from 1M at the shallow

end to 3.5m at the deep end.ย  The pool isย  25m longย  andย  12mย  wide. Calculate the volume of the pool

in cubic meters.

The pool is emptied by a cylindrical pipe of internal radius 9cm. The water flows through the pipe at speed of 3 metres per second.ย  Calculate the number of litres emptied from the pool in two minutes to the nearest 10 litres.ย ย ย ย ย ย ย ย ย  (Take II = 3.142)

 

 

 

  1. A rectangle A B C D is such that A and C lie on the line y = 3x. The images of B and D under a

reflection in the line y = x are B1 (-1, -3) and D1 (1,3) respectively.

  1. Draw on a cartesian plane, the line y = xย  and mark points B1 and D1
  2. Mark the points B and D before reflection
  3. Draw the line y = 3x hence mark the points A and C to complete and draw the rectangle ABCD.

State its co-ordinates, and these of A1 and C1.

  1. Find the image of D under a rotation, through – 900, Center the origin.

 

 

MATHEMATICS I

PART II

MARKING SCHEME.

  1. 1.43 X 0.091 X 5.04 Xย  100000 ย ย ย ย ย ย  91 X 504 ย ย ย ย ย ย ย ย  ย =ย ย ย ย ย ย ย  7/103

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  2.86 X 2.8 X 11.7ย ย ย ย ย ย  ย ย ย ย  ย 105 ย ย ย ย ย ย ย ย ย ย ย  2 x 28 x 117 x 103

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย = 0.007ย ย ย ย ย ย ย ย ย ย ย  (A)

  1. y = a/x + bx yx = a + bx2

Either

bx2 โ€“ yx + a = 0

 

x =ย ย ย ย  yย ย  ยฑย ย  v y2 –ย  4ab

2bย ย ย ย ย ย ย ย  ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

 

  1. 2x + 1ยฃย  10 โ€“ xย  ยฃย ย ย  bxย  -1

2x + 1 ยฃ 10 โ€“ x ย ย ย ย ย ย ย ย ย ย  10 โ€“x ยฃย  6x โ€“1

3x ยฃย ย  9 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  11ยฃย ย  7x

xย  ยฃย  3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย  xย ย  ยฃ 11/7 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

11/7 ยฃย  xย ย  ยฃย ย  3

 

  1. a = 4000 r = 110/100ย ย  =ย ย ย ย ย  1.1ย ย  ( 4000, 4000 + 4000, 4400 + 0/100 (4400——)

(a and r)

Snย  =ย  a(r n โ€“ 1)ย ย ย ย ย ย ย 

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Rย  -1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1.1 Logย  = 0.04139

ย ย ย ย ย Xย ย  5

0.20695

 

0.1ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4)

= 4000 (1.15 โ€“1) ย ย (any)

1.1 โ€“1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4000 (1.6 โ€“ 1)

0.1

Aย  =ย  4000 ( 0.6105)

0.1

= Sh. 2442ย ย ย ย ย ย  =ย ย ย  Sh. 24,420ย ย  ย ย ย  (A)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4)

0.1

 

  1. (i) b2=ย  a2 + b2 โ€“ 2ab Cos B

= 642 ย + 562โ€“ 2(64) (56) cos 78

= 4096 + 3136ย  – 7168 (0.2079)

= 7232ย  – km 1490.3

 

b2 ย = 5741.7ย  = 5.77 km ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย (5)

 

(ii)ย  ย ย ย ย ย  bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  a

ย ย ย ย ย ย ย ย ย ย ย  Sin Bย ย ย  ย ย ย ย ย  Sin A

 

75.77ย  ย  =ย ย ย ย ย  64

Sin 78ย  ย ย ย ย ย ย  sin Aย ย ย ย ย ย ย ย  Sin A = 64 x 0.9781ย ย ย ย ย 

75.77ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย 

Sin A = 0.08262

Aย  = 55.70ย  (or B = 46.30)

 

Bearing = 90 โ€“ 28 โ€“ 55.7

= 0.06.30ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

  1. (x + y) 6 =ย  1 (x) 6 (y)0 + 6 (x)5 (y)1+15(x)4 (y)2 + 20x3y3 + 15x2y4 + 6xy5 + y6

(1.02)6 = (1 +0.02)6 x = 1

y = 0.02

 

(1.02)6 = 1+6 (0.02) + 15 (0.02)2 + 15(0.02) + 20(0.02)3 + 15 (0.02)4ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย 

=ย  1 + 0.12ย  + 0.006 + 0.00016

= 1.12616

= 1.126ย  (to 3 d.p) ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

 

  1. ย ย ย ย ย ย 3(1 +ย  3)ย ย ย ย  ย ย ย ย  ย ย ย ย ย ย  =ย  3ย  +ย  3ย ย ย ย ย ย ย ย ย  3 + v3

(1-ย  3)(1+ย  3) ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1-3ย ย ย ย ย ย ย ย ย ย  ย  ย ย ย ย ย ย ย ย ย  ย ย ย 2

 

  1. Moving averages of order 4

M1ย ย ย ย ย ย ย  =ย  52 + 41 + 43 + 48 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  184ย ย ย ย ย ย  = 146

4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4

M2ย ย ย ย ย ย ย  =ย  ย ย ย ย 184 โ€“ 52 + 40ย ย  = 172ย  = 43ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย  for 7

4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  for > 4

M3ย ย ย ย ย ย ย ย ย ย ย ย  = 172โ€“ 40 + 38 = 170ย ย ย  = 42.5

4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4

M4ย ย ย ย ย ย ย ย ย ย ย ย  =ย  170 โ€“ 38+36ย  = 168ย ย  = 42

4ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย 4

M5ย ย ย ย ย ย ย  = 168 โ€“ 36 + 40 = 173ย ย ย  = 43ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

4ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  4

M6ย ย ย ย ย ย ย ย ย ย ย ย  = 172 โ€“ 40 + 44 = 176ย ย ย  = 44

4ย ย ย ย ย ย ย ย ย ย ย ย ย  4

M7 ย ย ย ย ย ย ย ย ย ย ย  = 176 โ€“ 44 + 45 = 177ย ย ย  = 44.25

4ย ย ย ย ย ย ย ย ย ย ย ย  4

 

  1. y = 3x + 4

A(0,4) B (1,7) Object points

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Aย ย ย ย ย ย ย ย  Bย  ย ย ย ย ย ย ย  Aย ย ย ย ย ย ย ย  B

2ย ย ย ย ย ย ย ย ย  1ย ย ย ย ย ย ย ย ย  0ย ย ย ย ย ย ย ย ย  1ย ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย ย  9

=

-1ย ย ย ย ย ย ย ย  2ย ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย ย  7ย ย ย ย ย ย ย ย ย  8ย ย ย ย ย ย ย ย ย  13

Y =ย  Mx + C

M = 13 โ€“ 8ย  =ย  5ย  = 1

9-4ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย  5ย ย ย ย  1

 

y = x+cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  y = x + 4

8 = 4 + cย ย ย  cย  = 4

 

  1. AB = 8ย ย ย ย  -4ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  BC =ย ย  16ย ย ย ย ย  – 8ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  -8ย ย ย ย  for either

=

7ย ย ย ย  -8 ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย -1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  5ย ย ย ย ย ย ย  – 7ย ย ย ย ย ย ย ย ย ย ย ย  -2

 

 

AB = ยฝย ย  BCย  and AB and BC share point B.

A,B,Cย  are collinear.ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

 

  1. 2ย ย ย ย ย ย ย ย ย  -3

 

4ย ย ย ย ย ย ย ย ย  3ย ย ย ย ย ย ย ย ย  det. = 6 + 12 = 18

Inv.=ย ย ย ย  1ย ย ย ย ย  ย ย  3ย ย ย ย ย ย ย ย ย  3

18

-4ย ย ย ย ย ย ย ย  2

1ย ย ย ย ย ย ย ย  3ย ย ย ย ย  3ย ย ย ย  2ย ย ย ย  -3ย ย  xย ย ย ย ย ย  1ย ย ย  ย ย ย ย ย  ย 3ย ย  3ย ย ย ย ย ย  7

18ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  18

-4ย ย ย  2ย ย ย  ย  4ย ย ย ย ย ย  2ย  yย  ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย -4ย  2ย ย ย ย ย ย  5

xย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย  36

1

yย ย ย ย ย ย ย ย ย  18ย ย ย ย ย ย ย  -18ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

x = 2, y = -1ย ย ย ย ย  (A)

 

  1. 1/7 + 3/12.4 + 7/0.103

1/7 + 3/1.24 x 10-1 + 7/1.03 x 10-1

ย 

ย  0.1429 + 3(0.8064) + 7 x 10 (0.9709)

10

= 0.1429 + 0.2419 + 67.96ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3)

=70.52ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. (i) ADC = 2×73

= 1460

ย 

(ii) OCB = x = 180 โ€“ 146 = 34

(iii) 360 โ€“ 107 โ€“ 146 โ€“ 34

= 73 0

 

  1. Tan 300 = y/x yย  =ย  x tanย  30

Tan 600 ย = 1000 + yย ย ย ย ย ย  ;ย ย ย ย  y = x tan 60 โ€“ 1000

X

X tan 300 ย = x tan 60 โ€“ 1000

0.5773 x = 1.732x โ€“ 1000

1.732x โ€“ 0.577 = 1000

1.155x = 1000

x = 1000

1.155ย ย ย ย ย ย ย ย ย ย  = 866.0 mย ย ย ย ย ย ย ย  (A)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4)

 

 

  1. 5 km Slower speed = x km/hr

Timeย ย ย  = 5/x

Faster = (x+20) k/h

Time = 5/x=20ย  ย ย ย ย ย ย ย ย ย  T1 โ€“ T2 = 5/xย  – 5/x+20 = 30/3600

5 (x+20) โ€“5xย ย ย ย  ย  1

x(x+20) 120

120 (5/x + 100 โ€“ 5x) = x2 + 20xย ย ย ย ย ย ย ย ย ย ย ย  (5)

x2 + 20x โ€“ 12000

x = –20 ย ย ย ย ย 400 + 48000

2

x = -20 ยฑย  220

2

Spd = 100 km/h

And x = 120 km/hย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

  1. Log x = a log y = b

Logย  x3 ย = Log x3ย  –ย  log y ยฝ

y

= 3 Log x โ€“ ยฝ Log y

= 8a โ€“ย  ยฝ ab

 

SECTION B

ย 

17.

Marks Mid point (x) d = x-44.5 F E = d/10 Ft T2 Ft2ย ย  v
0-9 4.5 -40 2 -4 -8 16 32
10-19 14.5 -30 4 -3 -12 9 36
20-29 24.5 -20 7 -2 -14 4 28
30-39 34.5 -10 19 -1 -19 1 19
40-49 44.5 -0 26 0 0 0 0
50-59 54.5 -10 15 1 15 1 15
60-69 64.5 20 12 2 24 4 48
70-79 74.5 30 5 3 15 9 45

=90ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =223

 

 

(a)ย ย  Mean = (1 / 90 x 10) + 44.5 = 44.5 + 0.111

= 44.610

 

(b)ย ย  Standard deviation = 10ย  233/90ย  – (1/90)2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  10ย  2.478ย  – 0.0001ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8)

10ย ย  2.478

10 x 1.574ย  = 15. 74ย ย ย  (A)

(c)ย ย ย  Median 45.5th valueย  = 39.5ย  + (13.5 x 10/ 26)

39.5 + 5.192ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

44.69

 

(a)ย  ย ย ย The probabilityย  = Shaded area

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Large circle area

Shaded area = ะŸR2 โ€“ ะŸ r2

= 22/7 (42 – 32) vย  = 22/7 x 7ย  = 22

ย ย ย ย ย ย ย ย ย ย ย  Large areaย  = 22/7 x4x4 = 352/7 (A)

Probability = 22 ย ย ย ย ย ย ย ย = 22ย  xย  7 =ย ย ย  7

352/7ย ย ย ย ย ย ย ย ย ย ย  352ย ย ย ย ย  16

 

(b)

  1 2 3 4 5 6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6

(M)

 

(i)ย ย ย  P(Product of 6) = P((1,6) or (2,3) or (3,2) or (6,1))

= 4/36ย ย  =ย  1/9

(4)

(ii)ย ย  P (sum of 8)ย ย  = P( (2,6) or (3,5) or (4,4) or (5,3) or (6,2) )

= 5/36ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

(iii)ย  P (same number)ย  = P (1,1) or (2,2) or (3,3) or (4,4) or (5,5) or (6,6)

6/36ย  = 1/6ย ย  (A)

 

 

 

 

 

 

 

 

 

 

(i)ย ย ย ย ย ย ย ย  Cos 60ย ย  = x/20 x = 20 x 0.5ย  = 10 cm

BD = 12 โ€“ 10 = 2 cm

 

(ii)ย ย ย ย ย ย ย ย ย  CD = yย  Sin 60ย  = y/20ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  y = 20x 0.8666

CD = 17.32 cm

 

 

 

 

(iii)ย ย ย ย ย ย ย  CHGย  = 120ย ย ย ย ย ย ย  reflexย  = 2400

CHG = 240/360 x 2 x p x r

= 50.27

DBF = 1200/360ย  x 2 x ย ะŸ xย  rย  =ย  1/3 x 2 x 3.142 x 2

=ย  4.189 ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

Length C D E f G H Cย  =ย ย ย ย ย ย ย ย ย  50.27 + 2(17.32) + 4.189

= 89.189ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

  1. (a) From the diagram, XO = 5/2 = 2.5

Tan 750 = VO/2.5ย ย ย ย ย ย ย ย ย  v m

VOย  =ย  2.5 x 3.732

 

Perpendicular heightย  = VOย  = 9.33 cm

2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  (A)

  1. Diagonal of base = 52 + 52 ย = 50
ย ย ย ย ย ย ย  Length of diag.ย ย  50ย ย ย ย ย ย  = 7.071ย ย ย  = 5.536

VA2 = AO2 + VO2ย ย ย ย  (m)

3.5362ย  + 9.32

12.50 + 87.05

= 99.55 = 9.98 cm2ย ย ย ย ย ย ย  (A)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8)

 

 

(c )ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = VAOย  Tan =ย ย ย ย ย  9.33ย ย  ย ย = 2.639

3.536

VAO = 69.240ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

 

(d) ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Cos VBA = = 2.5 /9.98ย ย  = 0.2505

VBA = 75.490

Area VBA = ยฝย  x 5ย  x 4.99 x sin 75.45ย ย ย ย ย ย ย ย ย ย ย ย  mย  ( or other perimeter)

= 5 x 4.99 xย ย  0.9681

= 24.15 cm2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

  1. Volume = cross โ€“ section Area x L

X-sec Area = (1 x 25)ย  +ย  (ยฝย  x 25 x 2.5)

=ย  25 + 31.25ย  =ย  56. M

Volumeย  = 56.25 x 12

= 675 m3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

ย ย ย ย ย ย ย ย ย ย ย  Volume passed / secย  = cross section area x speed

= ะŸ r2 x lย ย ย ย ย ย ย ย ย ย  = 3.14ย  xย  9/100 xย  9/100ย  x 3 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8)

= 0.07635ย  m3 /secย ย ย ย ย ย ย ย  v (M)

Volume emptied in 2 minutes

= 0.07635 x 60 x 2

= 9.162 m2 ย ย ย ย ย ย  ย ย ย ย ย ย ย ย (A)

1 m3 ย = 1000 l

= 9.162 litres

= 9160 litresย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (A)

 

 

 

 

 

24.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MATHEMATICS II

PART I

 

SECTION A (52 MARKS)

ย 

  1. Use tables to evaluate

3ร– 0.09122 + ร– 3.152 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (5mks)

0.1279 x 25.71

  1. Simplify ย (a โ€“ b)2

a2 โ€“ b2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

 

 

  1. The gradient function of a curve that passes through point: (-1, -1) is 2x + 3.

Find the equation of the curve. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (3mks)

  1. Find the value of k for which the matrix kย ย ย ย  3

has no inverse.ย ย  (2mks) ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย 3ย ย ย ย  k

  1. Without using tables, evaluate ย ย ย ย ย  log 128 โ€“ log 18

log 16 โ€“ log 6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

  1. Find the equation of the locus of points equidistant from point L(6,0) and N(-8,4). (3mks)
  2. The value of a machine is shs. 415,000. The machine depreciates at a rate of 15% p.a. Find how many years it will take for the value of the machine to be half of the original value. (4mks)
  3. Use reciprocal tables to evaluate to 3 d.p. 2ย ย ย ย ย ย ย ย ย  ย ย ย 1ย ย ย 

0.321ย ย ย ย ย ย ย ย ย ย  n2.2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (4mks)

  1. Using the trapezium rule, estimate the area bounded by the curve y = x2, the x โ€“ axis and the co-ordinates x = 2 and x = 5 using six strips. (4mks)
  2. Solve the equation for 00 ยฃ q ยฃ 3600 and Cos2q + ยฝ Cosq = 0 (3mks)
  3. Point P divides line MK in the ratio 4:5. Find the co-ordinates of point P if K is point (-6,10) and M is

point (3,-8) ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย (3mks)

  1. How many multiples of 3 are there between 28 and 300 inclusive. (3mks)
  2. The line y = mx โ€“ 1, where m is a constant , passes through point (3,1). Find the angle the line makes with the x โ€“ axis. (3mks)
  3. In the figure below, AF is a tangent to the circle at point A. Given that FK = 3cm, AX = 3cm, KX = 1.5cm and AF = 5cm, find CX and XN. (3mks)

 

 

 

 

 

 

 

 

 

 

 

  1. Make X the subject of the formula (3mks)

V = 3ร– k + x

sk – x

 

 

 

 

 

 

 

 

  1. Write down the inequalities that describe the unshaded region below. (4mks)

y

 

 

0.5ย ย ย ย ย ย ย ย  2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  x

 

-1.5

-2

 

ย ย ย ย ย  SECTION B (48 MARKS)

ย ย ย ย ย 

  1. Draw the graph of y = -x2 + 3x + 2 for โ€“4 ยฃ x ยฃ 4. Use your graph to solve the equations

(i.) 3x + 2 โ€“ x2 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (ii) โ€“x2 โ€“ x = -2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)

 

  1. The marks obtained by Form 4 students in Examination were as follows:

 

 

Marks 0-9 10-19 20-29 30-39 40-49 50-59
No. of students 2 8 6 7 8 10
Marks 60-69 70-79 80-89 90-99  
No. of Students 9 6 3    

ย ย ย ย ย  Using 74.5 as the Assumed mean, calculate:

(i) The mean mark

(ii) The standard deviationย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)

  1. In the figure below, a and b are the position vectors of points A and B respectively. K is a point on

AB such that the AK:KB = 1:1. The point R divides line OB in the ratio 3:2 and point S divides OK in

the ratio 3:1.

 

B

R

Bย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  ย ย ย ย  ย ย ย ย ย ย ย ย K

 

0 ย ย ย ย ย ย ย ย  ย ย ย ย ย aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  A

(a) Express in terms of a and b

(i) OKย  ย ย ย ย  (iii) RS

(iii) OS ย ย ย ย  (iv) RA

(b) Hence show that R,S and A are collinear.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)

 

  1. The figure below is the roof of a building. ABCD is a rectangle and the ridge XY is centrally placed.

 

 

 

 

 

 

 

 

 

 

 

Calculate:

(i) The angle between planes BXC and ABCD.

(ii) The angle between planes ABXY and ABCD.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (8mks)

  1. On the same axis, draw the graph of y = 2cosx and y = sin ยฝx for 00 ยฃ x ยฃ 1800, taking intervals of 150

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (6mks)

From the graph, find:

(a) The value of x for which 2cosx = sin ยฝ xย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (1mk)

(b) The range of values of x for which โ€“1.5 ยฃ 2cos x ยฃ 1.5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (1mk)

  1. Two towns T and S are 300km apart. Two buses A and B started from T at the same time travelling towards S. Bus B travelled at an average speed of 10km/hr greater than that of A and reached S 1 ยผ hrs earlier.

(a) Find the average speed of A.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (6mks)

(b) How far was A from T when B reached S.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (2mks)

  1. P and Q are two ports 200km apart. The bearing of Q from P is 0400. A ship leaves port Q on a bearing of 1500 at a speed of 40km/hr to arrive at port R 7 ยฝ hrs later. Calculate:

(a) The distance between ports Q and R.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (2mks)

(b) The distance between ports P and R. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

(c) The bearing of port R from port P. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

  1. A farmer has 15 hectares of land on which he can grow maize and beans only. In a year he grows maize on more land than beans. It costs him shs. 4400 to grow maize per hectare and shs 10,800 to grow beans per hectare. He is prepared to spend at most shs 90,000 per year to grow the crops. He makes a profit of shs 2400 from one hectare of maize and shs 3200 from one hectare of beans. If x hectares are planted with maize and y hectares are planted with beans.

(a) Write down all the inequalities describing this information.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (13mks)

(b) Graph the inequalities and find the maximum profit he makes from the crops in a year.ย ย  ย ย ย ย ย  ย (5mks)

 

ย 

MATHEMATICS II

PART II

 

  1. Use logarithm tables to Evaluate

3ร– 36.5 x 0.02573

1.938 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

  1. The cost of 5 shirts and 3 blouses is sh 1750. Martha bought 3 shirts and one blouse for shillings 850. Find the cost of each shirt and each blouse. ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  2. If K = ( y-cย  )1/2

4p

  1. a) Make y the subject of the formula. ย ย ย ย ย  (2mks)
  2. b) Evaluate y, when K = 5, p = 2 and c = 2ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (2mks)
  3. Factorise the equation:

x + 1/x = 10/3ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

  1. DA is the tangent to the circle centre O and Radius 10cm. If OD = 16cm, Calculate the area of the shaded Region. ย ย ย ย ย  (3mks)

 

 

 

 

 

 

 

 

 

 

 

  1. Construct the locus of points P such that the points X and Y are fixed points 6cm apart and

รXPY = ย ย ย  600.ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (2mks)

  1. In the figure below, ABCD is cyclic quadrilateral and BD is diagonal. EADF is a straight line,

CDF = 680, BDC = 450 and BAE = 980.

 

 

 

 

 

 

 

 

Calculate the size of:ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

  1. a) รABD ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  b) รCBD
  2. Otieno bought a shirt and paid sh 320 after getting a discount of 10%. The shopkeeper made a profit of 20% on the sale. Find the percentage profit the shopkeeper would have made if no discount was allowed? ย ย ย ย ย  (2mks)
  3. Calculate the distance:
  4. i) In nautical miles (nm)
  5. ii) In kilometres (km)

Between the two places along the circle of Latitude:

  1. a) A(300N, 200E) and B(300N, 800E) (Take Radius of Earth = 6371Km).ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
  2. b) X(500S, 600W) and Y(500S, 200E) (Take Radius of Earth = 6371Km).ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
  3. A rectangular tank of base 2.4m by 2.8m and height 3m contains 3,600 litres of water initially. Water flows into the tank at the rate of 0.5m/s. Calculate the time in hours and minutes required to fill the tank. (4mks)
  4. Expand (1 + a)5 up to the term of a power 4. Use your expansion to Estimate (0.8)5 correct to 4 decimal places. (4mks) ย 
  5. A pipe is made of metal 2cm thick. The external Radius of the pipe is 21cm. What volume of metal is there in a 34m length of pipe (p = 3.14). ย ย ย ย ย  (4mks)
  6. If two dice are thrown, find the probability of getting: a sum of an odd number and a sum of scoring more than 7 but less than 10. (4mks)
  7. Find the following indefinite integral รฒ 8x5 โ€“ 3x dxย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (4mks)

x3

  1. The figure below represents a circle of radius 14cm with a sector subtending an angle of 600 at the centre.

 

 

.

 

 

 

 

 

 

 

Find the area of the shaded segment.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (3mks)

 

 

 

 

 

 

 

 

  1. Use the data below to find the standard deviation of the marks.

 

Marks (x ) Frequency (f)
5

6

7

8

9

3

8

9

6

4

(4mks)

 

SECTION II (48MKS)

 

  1. The figure below shows a cube of side 5cm.

 

 

 

 

 

 

 

 

 

 

 

 

 

Calculate:

  1. a) Length FC ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)
  2. b) Length HB ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)
  3. c) Angle between GB and the plane ABCD. (1mk)
  4. d) Angle between AG and the Base. ย ย ย ย ย  (1mk)
  5. e) Angle between planes AFC and ABCD. (2mks)
  6. f) If X is mid-point of the face ABCD, Find angle AGX. (2mks)
  7. Draw on the same axes the graphs of y = Sin x0 and y = 2Sin (x0 + 100) in the domain 00 ยฃ x0 ยฃ 1800
  8. i) Use the graph to find amplitudes of the functions.
  9. ii) What transformation maps the graph of y = Sin x0 onto the graph of : y = 2Sin (x0 +100).
  10. The table below shows the masses to the nearest gram of 150 eggs produced at a farm in Busiro

country.

Mass(g) 44 45 46 47 48 49 50 51 52 53 54 55
Freq. ย 1 ย 2 ย 2 ย 1 ย 6 ย 11 ย 9 ย 7 ย 10 ย 12 ย 16 ย 16
Mass(g) 56 57 58 59 60 61 62 63 64 65 70  
Freq. ย 10 ย 11 ย 9 ย 7 ย  5 ย 3 ย 4 ย 3 ย 3 ย 1 ย 1  

 

Make a frequency Table with class-interval of 5g. Using 52g as a working mean, calculate the mean mass. Also calculate the median mass using ogive curve.

  1. A shopkeeper stores two brands of drinks called soft and bitter drinks, both produced in cans of same

size. He wishes to order from supplies and find that he has room for 1000 cans. He knows that bitter

drinks has higher demand and so proposes to order at least twice as many cans of bitter as soft. He

wishes however to have at least 90cans of soft and not more than 720 cans of bitter. Taking x to be

the number of cans of soft and y to be the number of cans of bitter which he orders. Write down the

four inequalities involving x and y which satisfy these conditions. Construct and indicate clearly by

shading the unwanted regions.

 

 

 

 

  1. Two aeroplanes, A and B leave airport x at the same time. A flies on a bearing 0600 at 750km/h and B flies on bearing of 2100 at 900km/h:
  2. a) Using a suitable scale draw a diagram to show the positions of Aeroplanes after 2hrs.
  3. b) Use your graph to determine:
  4. i) The actual distance between the two aeroplanes.
  5. ii) The bearing of B from A.

iii) The bearing of A from B.

  1. The Probabilities that it will either rain or not in 30days from now are 0.5 and 0.6 respectively. Find the probability that in 30 days time.
  2. a) it will either rain and not.
  3. b) Neither will not take place.
  4. c) One Event will take place.
  5. Calculate the Area of each of the two segments of y = x(x+1)(x-2) cut off by the x axis. (8mks)
  6. Find the co-ordinates of the turning point on the curve of y = x3 โ€“ 3x2 and distinguish between them.

 

MATHEMATICS II

PART I

MARKING SCHEME:

 

  1. 0.09122 = (9.12 x 10-2)2 = 0.008317

ร– 3.152 = 1.776

3ร– 1.776 + 0.008317

0.1279 x 25.91

= 3ร– 1.784317ย ย ย ย ย ย ย ย ย ย ย ย ย  No.ย ย ย ย ย ย ย  ย ย ย ย  logย  ย ย ย ย 

0.1279 x 25.91ย ย ย ย ย ย ย ย ย ย  1.784ย ย ย ย ย ย ย ย  0.2514

0.1279 ย ย  -1.1069

25.71ย ย ย ย ย ย ย ย ย ย  1.4101 +

0.5170

-1.7344

x 1/3

10-1 x 8.155(6)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1-1.9115

Or 0.8155(6)

ย 

  1. (a โ€“ b)(a โ€“ b) =ย  a โ€“ b

(a โ€“ b)(a + b)ย ย ย ย ย ย  a + b

 

  1. dy = 2x + 3

dx

y = x2 + 3x + c

-1 = 1 โ€“ 3 + c

c = 1ย ย ย ย  ;ย ย ย ย  E.gย  y = x2 + 3x + 1

 

  1. K2 โ€“ 9 = 0

K = ยฑ 3

 

  1. log 128ย ย ย  =ย  logย ย ย ย ย ย  64

18ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 9

 

logย ย  16ย ย ย ย  ย ย  logย ย ย  ย 8ย 

6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

=ย  2 log (8/3)

log (8/3)

= 2

 

  1. Midpoint -8 + 6, 4 + 0ย ย ย ย ย  ย ย  (-1, 2)

2ย ย ย ย ย ย ย ย  2

Gradient of LN = 4/-14 = -2/7

Gradient of ^ bisector = 7/2

y โ€“ 2 ย = 7/2

x + 1

y = 7/2X + 11/2

 

  1. 207,500 = 415,000(1 โ€“ 15 )n

100

0.5 = ( 85 )n

100

0.5 = 0.85n

log 0.5 = n log 0.85

log 0.5 ย = n

log 0.85

n = –1.6990ย ย  =ย ย ย  -0.3010 = 4.264yrs

-1.9294ย ย ย ย ย  -0.0706

 

  1. 2 x ย ย ย ย ย 1ย ย ย ย  ย ย  = ย ย 1ย  . x 20 = 0.3115 xย  20 = 6.230

3.21 x 10-1ย ย ย  3.21

ย ย ย 1 ย ย ย ย = ย ย ย ย ย ย ย ย 1 ย ย ย ย ย =ย  0.5807 = 0.005807

172.2ย ย ย  1.722 x 102ย ย ย ย ย ย ย ย ย ย  100

6.230 โ€“ 0.005807 = 6.224193

= 6. 224(3d.p)

 

X 2 2.5 3 3.5 4 4.5 5
y 4 6.25 9 12.25 16 20.25 25

h = ยฝ

Area= ยฝ x ยฝ[29+2(6.25+9+12.25+16+20.25+25)]

= ยผ [29 + 127.5]

= ยผย  x 156.5ย  =ย  39.125ย  sq. units.

 

  1. Cos q (cos q + ยฝ ) = 0

cos q = 0ย ย ย ย ย ย ย  cos q = -0.5

q = 900, 2700ย ย ย  q = 1200, 2400ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

\ q = 900, 1200, 2400, 2700

 

  1. MP = 4 MK MK =ย ย ย ย ย  -9

9ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  -18

MP = 4 ( -9ย  ) = ( -4 )

9ย  -18ย ย ย ย ย ย ย ย ย  8

\ P is ( -1,0 )

 

  1. a = 30 d = 3ย ย  l = 300

300 = 30 + 3 (n โ€“ 1 )

300 = 30 + 3n โ€“ 3

300 โ€“ 27 = 3n

273 = 3n

91 = n ย 

 

 

 

 

  1. y = mx โ€“ 1

1 = 3m โ€“ 1

m = 2/3 = 0.6667

tan q = 0.6667ย  ;ย ย ย ย  q = 33.690ย ย ย ย 

 

  1. FK x FC = FA2

FC = 25/3 = 8 1/3 cm

CX = 81/3 โ€“ 9/2 = 23/6 = 35/6 cm

CX x XK = XA x XN

33/6 x 3/2 = 3 x XN

\ XN = 111/12 cm

 

  1. V3 = k + x

k โ€“ x

V3k โ€“ V3x = k + x

V3k โ€“ k = x + V3x

V3k โ€“ k = x( 1 + v3)

V3k โ€“ k ย = x

1 + V3

 

  1. (i.) x = 2 รž x ยฃ 2

(ii) y = -2 รž y > -2

(iii)pts. (0.5,0)

(0,-1.5)

m = -1.5 โ€“ 0 ย = 3

0 – 0.5

Eq. Y = 3x โ€“ 1.5ย ย ย  y < 3x โ€“ 1.5

 

ย ย ย ย ย 

SECTION B

ย 

X -4 -3 -2 -1 0 1 2 3 4
Y -26 -16 -8 -2 2 4 4 2 -2

(i) Roots are x = -0.5ย ย  x = 3.6

 

(ii)ย  y = -x2 + 3x + 2

0 = -x2 โ€“ x + 2ย 

y = 4xย ย ย ย  (-2, -8) (1, 4)

Roots are x = -2, x = 1

 

  1. class x fย ย ย ย ย ย  d=x-74.5ย ย ย ย ย ย  fdย ย ย ย ย ย ย ย ย ย ย ย  d2ย ย ย ย ย ย  fd2ย ย ย ย 

0 โ€“ 9ย ย ย ย ย ย ย  4.5ย ย ย  2ย ย ย ย ย ย ย ย  – 70ย ย ย ย ย ย ย ย  – 140ย ย ย ย ย ย  4900ย ย ย ย ย ย ย  9800

10 โ€“ 19ย ย ย  14.5ย ย ย ย  8ย ย ย ย ย ย ย ย  – 60 ย ย ย ย ย ย ย ย – 480ย ย ย ย ย ย  3600ย ย ย ย  28,800

20 โ€“ 29ย ย ย  24.5ย ย ย ย  6ย ย ย ย ย ย ย ย  – 50ย ย ย ย ย ย ย ย  – 300ย ย ย ย ย ย  2500ย ย ย ย  15,000

30 โ€“ 39ย ย ย  34.5ย ย ย ย  7ย ย ย ย ย ย ย ย  – 40ย ย ย ย ย ย ย ย  – 280ย ย ย ย ย ย  1600ย ย ย ย  11,200

40 โ€“ 49ย ย ย  44.5ย ย ย ย  8ย ย ย ย ย ย ย ย  – 30ย ย ย ย ย ย ย ย  – 240ย ย ย ย ย ย ย ย  900ย ย ย ย ย ย  7,200

50 โ€“ 59ย ย ย  54.5ย ย ย  10ย ย ย ย ย ย ย  – 20ย ย ย ย ย ย ย ย  – 200ย ย ย ย ย ย ย ย  400ย ย ย ย ย ย  4,000

60 โ€“ 69ย ย ย  64.5ย ย ย ย  9ย ย ย ย ย ย ย ย  – 10ย ย ย ย ย ย ย ย ย ย  – 90ย ย ย ย ย ย ย ย  100ย ย ย ย ย ย ย ย ย  900

70 โ€“ 79ย ย ย  74.5ย ย ย ย  6ย ย ย ย ย ย ย ย ย ย ย  0ย ย ย ย ย ย ย ย ย ย ย ย ย ย  0ย ย ย ย ย ย ย ย ย ย ย ย ย  0ย ย ย ย ย ย ย ย ย ย ย ย ย  0

80 โ€“ 89ย ย ย  84.5ย ย ย ย  3ย ย ย ย ย ย ย ย ย  10ย ย ย ย ย ย ย ย ย ย ย ย ย  30ย ย ย ย ย ย ย ย  100ย ย ย ย ย ย ย ย ย  300

90 โ€“ 99ย ย ย  94.5ย  ย ย ย 1ย ย ย  ย ย ย ย ย ย 20ย ย ย ย  ย ย ย ย ย ย ย 20ย ย ย  ย ย ย ย ย 400ย ย ย  ย ย ย ย ย ย 400ย ย ย 

Sf =ย ย ย ย ย ย  Sfd =ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Sfd2 =ย ย ย ย  77,600

60ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  -1680

(i) Mean = 74.5 + -1680

60

= 74.5 โ€“ 28ย  =ย ย ย  46.5

(ii) Standard deviation = ร– 77600 โ€“ ( –1680 )2

60ย ย ย ย ย ย ย ย ย ย ย  60

= ร– 1283.3 โ€“ 784

= ร– 499.3 = 22.35

 

  1. a (i.) OK = OA + AK = ยฝ a + ยฝ b

(ii) OS = ยพ OK = 3/8 a + 3/8 b

(iii)RS = RO + OS = 3/8 a โ€“ 9/40 b

(iv) RA = RO + OA = – 3/5 b + a

 

  1. RA = a โ€“ 3/5 bย ย  RS = 3/8 a + 9/40 b

= 3/8( a โ€“ 3/5 b)

\ RS = 3/8 RA

The vectors are parallel and they have a common

point Rย  \ point R, S and A are collinear

 

 

 

 

 

 

 

 

 

 

 

 

 

KB = 3mย ย  NK = 1.5mย ย  XB = 5m

(i) ย XK = ร– 52 โ€“ 32 ย = ร– 16 = 4m

let รXKN = q

cos q = 1.5 ย = 0.375

4

q = 67.97(8)0

 

(ii) In DXNK

XN = ร– 42 โ€“ 1.52 = ร– 13.75 = 3.708

In D SMR; MR = KB = 3m

SM = XN = 3.708m

Let รSRM = a

tan a = 3.708 ย =1.236

3

a = 51.02(3)0

 

 

 

 

 

 

 

 

 

21.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

21.

 

  0 150 300 450 600 750 900 1050 1200 1350 1500 1650 1800
Y =2cosX 2.00 1.93 1.73 1.41 1.00 0.52 0.00 -0.52 -1 -1.41 -1.73 -1.93 -2.00
Y = sin ยฝ X 0.00 0.13 0.26 0.38 0.50 0.61 0.71 0.79 0.87 0.92 0.97 ย 0.99 1.00

(a) X = 730 ยฑ 10

(b) Between 40.50 and 139.50

 

 

  1. 300km

Tย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย  S

Let the speed of A be X km/hr

Speed of B = (X + 10) km/hr

Time taken by A = 300 hrs

X

Time taken by B = 300 hrs

X + 10

300 โ€“ 300 ย =ย  5

xย ย ย  x + 10ย ย ย  4

300(x + 10) โ€“ 300x ย = 5

x(x + 10) ย ย  4

300x + 300 โ€“ 300x = 5

x2 + 10x

x2 + 10x โ€“ 2400 = 0.

x = 44.25

X = -54.25 N/A

(b) Distance covered by A in 1 ยผ hrsย  = 44.25 x 5/4ย  = 55.3 km

Distance of A from T is 300 โ€“ 55.3 = 244.7 km

 

 

 

 

 

 

 

 

 

  1. (a) Distance = 15 x 40 = 300km

2

(b)

 

 

 

 

 

 

 

 

 

 

 

PR2 = 2002 + 3002 โ€“2x 200 x 300 cos700

= 130,000 โ€“ 41040ย ย  =ย ย  88,960

PR = 298.3 km

 

(c) 298.3 ย = 300

sin 700ย ย ย  sin a

sin a = 300 sin 700

298.3

= 0.9344

a = 69.10

 

Bearing of R from P is

40 + 69.1 = 109.10

 

  1. (i.) X > y

(ii) 4,400X + 10,800Y ยฃ 90,000

Simplifies to 11X + 27y ยฃ 225

(iii) X + y ยฃ 15

X > 0;ย  y > 0

Boundaries

x = y pts (6,6) (12,12)

11x + 27y = 225 pts (13,3) (1,8)

X + y = 15 pts (0,15) (8,7)

Objective function

2400 x 3200y

(pt (2,1)

2400X + 3200y = 8000

Search line ยฎ 3X + 4y = 10

Point that give maximum profit is (12,3)

\ maximum profit

= 2400 x 12 + 3200 x 3 = 38,400 shs.

 

 

 

 

 

 

 

 

 

MATHEMATICSย  II

PART II

MARKING SCHEME

ย 

  1. No log.

36.5ย ย ย ย ย ย ย  1.5623

0.02573ย ย  –2.4104 +

-1.9727

1.938ย ย ย ย ย ย ย ย  0.2874 โ€“

-1.6853

 

-3 ย + 2.6853ย 

3ย ย ย ย ย ย ย ย  3

-1 + 0.8951

1.273(4) ยฌ 0.1049

= 1.273(4)

 

  1. Let shirt be sh x,

let blouse be sh. y.

5x + 3y =1750 (i.)

3x + y = 850ย ย ย  (ii)

mult (ii) by 3

9x + 3y = 2550 (iii)

Subtractย  (iii) โ€“ (i.)

– 4x = -800

Subt for x

  1. = 250

Shirt = sh 200ย  ;ย ย  Blouse = sh 250

ย ย ย ย ย 

  1. (a) K2 = y โ€“ c

4p

y โ€“ c = 4pK2

y = 4pK2 + c

(b)ย ย ย  y = 4 x 2 x 25 + 2ย ย  ;ย ย ย ย ย  y = 202

 

  1. x2 + 1 โ€“ 10x = 0

3

3x2 โ€“ 10x + 3 = 0

3x (x โ€“ 3) โ€“ 1(x โ€“ 3) = 0

(3x โ€“ 1) (x โ€“ 3 ) = 0

x = 1/3ย  or x = 3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

 

  1. Area D OAD pyth theorem AD =12.49cm

ยฝย  x 12.49 x 10ย  =ย ย  62.45cm2

Cos q = 10/16 = 0.625

q = 51.30ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 62.5

Sector 57.30 ย x 3.14 x 100ย ย ย  40.2 –

360ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 22.3

 

 

 

 

 

 

 

 

  1. รXPY = 600

\รXC1Y = 1200

ย ย ย ย ย ย ย ย ย ย ย ย ย  B1ย ย ย ย ย ย ย ย ย ย ย ย  \รC1XY = รC1YX

= 1800 โ€“ 1200 ย = 300

2

 

 

 

 

Construct 300 ย angles

at XY to get centres

B1ย  ย ย ย ย ย ย ย ย ย C1 and C2 ย mojar arcs drawn

2ย ย ย ย ย ย ย ย ย ย ย  on both sides with C1X and C2X

as centres.

 

 

 

 

 

 

 

 

 

 

  1. DAB = 1800 โ€“ 980 ย = 820

ADB = 180 โ€“ (68 + 45 ) = 670

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ABD = 180 โ€“ (67 + 82)

= 310

 

(a) 1800 โ€“ (67 + 82)0 = 310

ย  ย ย ย ย ย รABD = 310ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย Opp = 1800

(b) (180 โ€“ 82)0 = 980ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  82 + 98 = 1800

ย ย ย ย ย  ย ย 1800 ย – (980 โ€“ 450) =

รCBD = 370ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย 180 โ€“ (98 + 45)0ย 

= 370

  1. 10 x 320

100ย ย ย ย  Discount = sh 32

Sold atย ย ย ย ย  sh 288

If no Discount = ( 320 x 20 ) % = 22.7%

288

ย 

  1. (a) Dist along circle of lat.

Long diff x 60 x cos q nm

100 x 60 x Cos 500

100 x 60 x 0.866

5196nm =ย ย ย ย  ย 100 x 2pR Cos 500

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  360

100 ย x 2 x 3.14 x 6371

360ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  5780Km

 

 

 

 

 

 

(b) 80 x 60 Cos 50ย  = 3895 Km

 

  1. Vol =2.8 x 2.4 x 3 = 20.16m3

ย ย ย ย ย ย ย ย ย  1m3 = 1000 L

20.16m3 = 20160 L

20160

ย ย ย ย 3600 ย ย ย ย ย ย 

16560 L to fill

0.5 L โ€“ 1 sec

16560 L – ?

ย 165600

5 x 3600

33120 ย hr

3600ย ย ย ย ย ย ย ย ย ย ย ย  @ 9.41 hrsย ย ย ย  ;ย ย ย ย  @ 564.6 min.

 

  1. 15 + 5.14a + 10.13.a2 + 10.12a3 + 5.1.a4

a = -0.2

1 + 5(-0.2) + 10(-0.2)2 + 10(-0.2)3ย  + 5 (-0.2)4

1 โ€“ 1.0 + 0.4 โ€“ 0.08 + 0.008ย  =ย ย  0.3277 (4d.p)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

 

  1. Area of metal : Material โ€“ Cross section.

p(R2 โ€“ r2)

3.14 (21 โ€“19)

Volย  6.28cm2 x 3400cm

= 215.52m3ย ย ย ย ย ย ย ย 

ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

  1. Possibility space:

 

.ย ย ย ย ย ย ย ย ย ย ย  1ย  2ย  3ย  4ย  5ย  6ย 

1ย ย ย ย  2ย  3ย  4ย  5ย  6ย  7

2ย ย ย ย  3ย  4ย  5ย  6ย  7ย  8

3ย ย ย ย  4ย  5ย  6ย  7ย  8ย  9

4ย ย ย ย  5ย  6ย  7ย  8ย  9ย  10

5ย ย ย ย  6ย  7ย  8ย  9 10 11

6ย ย ย ย  7 8ย  9 10 11 12

 

P(odd) = 3/6 = ยฝ

P(Sum > 7 but < 10)ย ย  =ย ย  9 /36

\ P(odd) and P(sum > 7 but < 10 )

= ยฝย  x 9/36 = 9/72ย ย ย ย  =ย  1/8

ย 

  1. รฒ( 8x5/x3 โ€“ 3x/x3) d4

รฒ( 8x2 โ€“ 3x-2) d4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย 

16x3/3 + 6x-3/-3ย  + Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย 

16x3/3 โ€“ 2/x3ย  + C

 

  1. Area of DAOB

ยฝย  x 14 x 14 x 0.866ย  =ย  84.866cm2

Area of sectorย  =ย  60 ย x3.14 x 14 x14 = 10.257

360

Shaded Area

84.666ย  –ย  10.257 = 74.409cm2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

 

 

 

 

 

Marks F Fx fx2
5 3 15 75
6 8 48 288
7 9 63 441
8 6 48 384
9 4 36 324

 

รฅx =ย ย ย  รฅf=30ย ย  รฅfx=210ย ย  1512

S.d =ย  ร– รฅfx2 ย –ย  ( รฅfx )2

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย รฅfย ย ย ย ย ย ย ย ย ย ย  รฅf

= ร– 1512ย  ย –ย  (210)2ย 

30ย ย ย ย ย ย ย ย ย ย ย  30

=ย  ร– 50.4 โ€“ 49

=ย ย  ร– 1.4ย  = 1,183ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย 

ย 

ย ย ย ย ย ย  SECTION IIย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  .

 

  1. (a) FC = ร– 52 + 7.072 = ร– 50 = 7.071

(b) HB = ร– 52 + 7.072ย ย ย  = ร– 75 = 8.660

(c) q = Tan-1 5/5 = Tan-1ย  ย = 450ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

(d)ย  b = Tan-1 5/7.071 = Tan-1 0.7071ย  =ย  35.30ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

(e)ย  y = Tan-1 5/3.535ย ย  = Tan-1ย ย ย  = 54.70ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

(f) รAGX = 19.40

 

 

  1. y = Sin x
ย ย ย ย ย  x0 00 300 600 900 1200 1500 1800
sin x0 0 0.50 0.66 1.00 0.866 0.500 0

 

y = 2 Sin (x0 + 100)

ย ย ย ย ย  X0 00 300 600 900 1200 1500 700
2 Sin(x +100) 0.3472 1.286 1.8794 1.286 0.3472 -0.3472 -1.8794

Amplitudes for y = Sin x0 is 1

For

y = Sin(x+100) is 2.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c.f X F
61 53 12
16   54
93 55 16
103 56 10
11   57
123 58 9
130 59 7
135 60 5
138 61 3
142 62 4
145 63 3
148 64 3
149 65 1
150 70 1

 

Mean =ย  xย ย  ย + 52ย  + -4

150

52 –ย  0.02

=ย ย ย ย  51.08

Medianย  =ย ย ย ย  51.4g.

 

class interval 59

Class interval mid point Freg. c.f
44-48 46 12 12
49-53 51 49 61
54-58 56 64 125
59-63 69 22 147
64-68 66 3 130
69-73 71 1 150

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. X + Y ยฃ 1000

X ยฃ 2Y

Y < 720

X > 90

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

21.(a)ย ย ย  1cm = 200Km/h

A = 200 x 7.5ย  =ย  1500 Km

B =ย  200 x 9ย  = 1800Km.

 

(b) (i.) 15.8cm x 200ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (ii) Bearing 2240

= 3160 Km.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  (iii) Bearing 0490

 

  1. (a) P(R) x P(R)1 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (b) P(R)ยข x P(R) ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (c) P(R) x P(Rโ€™)

= 0.5 x 0.6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  0.5 x 0.4 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย P(R)โ€™ x P(R)

= 0.3ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย =ย  0.2ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 0.5 x 0.6 = 0.3

0.5 x 0.4 = 0. 2= 0.5

  1. y = x(x + 1)(x โ€“ 2)

= x3 โ€“ x2 โ€“ 2x

A1 = รฒ(x3 โ€“ x2 โ€“2x) d4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

-1[ยผ x4 โ€“ย  1/3 x2]-1

= 0 โ€“ ( ยผ + 1/3 โ€“ 1)ย ย ย  =ย  5/12

A2 = 2รฒ(x3 โ€“ x2 โ€“2x) d4

=ย  0รฒ ยผ x4 โ€“ 1/3 x3 โ€“ x2)-20ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

= ( ยผ .16 โ€“ 1/3 .8 โ€“ 8 )

= 4-0 โ€“ 8/3 โ€“ 4ย  =ย ย  – 8/3

ย ย ย ย ย ย ย ย ย ย ย ย ย  A1 = 5/12= A2 = 2 2/3ย ย  ย ย ย ย ย ย 

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

  1. y = x3 โ€“ 3x2

dy ย = 3x2 โ€“ 6x

At stationary

Pointsย ย ย ย ย  dy = 0

dx

i.eย ย  3x2 โ€“ 6x = 0

3x(x โ€“ 2) = 0

x = 0 or 2

Distinguish

dy = 3x2 โ€“ 6x

dx

d2y ย =ย  6x โ€“ 6

dx2

ย ย  ย (i)ย ย ย  x = 0ย  dy2 = 6x โ€“ 6 = -6ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  (ii)ย ย ย ย ย ย  x = 2

dx2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  d2y ย =ย  6

-6 < 0 โ€“ maximum.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  dx2

\ (0,0) Max Pt. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  6 > 0 hence

Minimum Pt.

x = 2,ย  y = 8 โ€“ 12 = -4

(2, -4)ย ย ย ย  minimum point.

ย 

MATHEMATICS II

PART I

 

SECTION 1 (52 Marks)

  1. Without using tables evaluate:

 

ร–7.5625 x 3ร–3.375

15ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (5 mks)

 

  1. Make k the subject of the formula.

y = 1ย  ร–k + yย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 

T2ย ย ย ย ย  kย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

 

  1. If A = (x, 2) and xBย ย ย ย  =ย ย ย ย  xย ย ย ย  and if AB = (8), find the possible values of x.

-2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. Simplify completely. (3 mks)

rx4 – r

2xr – 2r

 

  1. Solve the equation. (3 mks)

Log 3 (8-x)ย  –ย  log 3 (1+x) = 1

 

  1. Under an enlargement scale factor -1, A(4,3) maps onto A1 (4,-5). Find the co-ordinates of the centre of enlargement. (3 mks)

 

  1. Find the equation of the line perpendicular to the line 4x-y = -5 and passing through the point (-3,-2). ย ย ย ย ย  (2 mks)
  2. Find the standard deviation of the data below:

3,5,2,1,2,4,6,5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4 mks)

 

  1. What is the sum of all multiples of 7 between 200 and 300? (4 mks)

 

  1. Solve the equation.

ยฝ tan xย  =ย  sin x for -1800 ย ยฃย  xย  ยฃย  3600.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks).

 

  1. Expand (1-2x)4. Hence evaluate (0.82)4 correct to 5d.p. (4 mks)

 

  1. The line y = mx – 3 passes through point (5,2). Find the angle that the line makes with the x-axis. (2 mrks)
  2. A two digit number is such that 3 times the units digit exceed the tens digit by 14. If the digits are reversed, the value of the number increases by 36. Find the number (4 mks)

 

 

 

 

 

 

  1. In the figure below, O is the centre of the circle, OA = 7 cm and minor arc AB is 11 cm long. Taking P = 22/7, find the area shaded. (3 mks)

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. A box contains 36 balls, all identical except for colour. 15 of the balls are black, 15 are brown and the rest are white. Three balls are drawn from the box at random, one at a time, without replacement. Find the probability that the balls picked are white, black and brown in that order. (2 mks)

 

  1. Find the inequalities that describe the unshaded region R below. (4 mks)

y

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SECTIONย  2 (48 Marks)

ย 

  1. Draw the graph of y = x2 + x – 6 for -4 ยฃ x ยฃ

Use your graph to solve the equations.

(i)ย  x2 + x – 6 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (ii) x2 + 2x – 8 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

 

  1. The diagram below represents a bucket that has been placed upside down. The radius of the top surface is 15cm and that of the bottom is 40cm. The vertical height of the bucket is 50cm.

 

 

 

 

 

 

 

 

 

 

 

 

 

Determine:-

  • The volume of the bucket.
  • The curved surface area of the bucket. (leave your answers in terms of p)

 

  1. Draw, on the same axes, the graphs of y = cos q and y = 5 sin q for – 1800 ยฃ q ยฃ 1800
  • From your graph, determine the amplitude of each wave.
  • For what value(s) of q is cosq – 5 sin q = 0 (8 mks)

 

  1. A point P lies on a coast which runs from West to East. A ship sails from P on a bearing of 0320. When it reaches Q, 7km from P, a distress signal is observed coming from another ship at R. Given that R is N.E of P and on a bearing of 0660 from Q, calculate:
  • ร
  • The distance QR, between the two ships.
  • The shortest distance from R to the shore. (8 mks)

 

  1. A bag contains x red balls and y yellow balls. Four times the number of red balls is equal to nine times the number of yellow balls and twice the total number of balls exceeds the number of yellow balls by 44.
  • How many balls of each colour are three in the bag?
  • If two balls are drawn out of the bag at random one at a time with replacement what is the probability that the two balls are red? (8 mks)

 

  1. A Kenyan businessman goes on a trip to West Germany through Italy and back to Kenya. In Kenya he is allowed to take Ksh. 67,000 for sales promotion abroad. He converts the Kenya currency into US dollars. While in Italy, he converts 2/5 of his dollars into Italian lire, which he spends in Italy. While in West Germany, he converts 5/8 of the remaining dollars into Deutsche marks which he uses up before coming to Kenya. Using the conversion rates 1 US dollar = 1.8 Deutsche marks = 16.75

Ksh = 1340 Italian lire. Answer the following questions:

  • How many US dollars did he take out of Kenya?
  • How many Italian lire did he spend in Italy?
  • How much money, in Deutsche marks did he spend in West Germany?
  • How much money in Ksh. did he have on his return to Kenya? (8 mks)

 

  1. PQRS is a parallelogram in which PQ = r and PS = h. Point A is the midpoint of QR and B is a point on PS such that PS : PB = 4:3. PA and QB intersect at M.

 

 

 

 

 

 

 

 

 

 

Given that PM = kPA and BM = tBQ where k and t are scalars, express PM in two different ways and hence find the values of k and t.

Express PM in terms of r and h only.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

 

 

 

 

 

 

 

 

  1. Two variables T and X are connected by the equation T = abx where a and b are constants. The values of T and X are given in the table below:

 

T 6.56 17.7 47.8 129 349 941 2540 6860
X 2 3 4 5 6 7 8 9

 

 

Draw a suitable straight line graph and use it to estimate the values of a and b.ย ย ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

 

ย 

MATHEMATICS III

PART II

ย 

Section I:ย ย  (52 Marks)

 

  1. Use mathematical tables to evaluate:

 

8.67ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

ร– 0.786 x (21.72)3

 

  1. Simplify completely. (3 mks)

4ย ย ย ย ย  –ย ย ย  1

x2 – 4ย ย ย ย ย ย ย  x-2

 

  1. An Indian on landing at Wilson Airport changes Re 6000 into Kenya shillings when the exchange rate is Re = Ksh. 1.25. He spent Ksh. 5000 when in Kenya and converted the remaining amount to Rupees at the same rate as before. Find out how much the Indian is left with in Rupees. (3mks)

 

  1. The last of three consecutive odd numbers is (2x+3). If their sum is 105, find the value of x. (4 mks)

 

  1. a Sย  b is defined by:ย ย ย ย ย ย ย ย ย ย  a S bย  =ย  (a + b)

ab

If B Sย ย  (2ย  Sย ย  3)ย  =ย  4ย  Sย ย  1, Find B.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. Find the value of M. (3 mks)

 

 

M

 

850

 

1600

 

 

  1. (a) Expand (1+2x)6 upto the term containing x3 .ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

 

(b)ย  By putting x = 0.01, find the approximate value of (1.02)6 correct to 4 S.F. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

 

 

  1. Show that x is the inverse of : Y =ย ย ย  3ย ย ย ย ย ย ย ย ย  -3ย ย ย ย ย  1ย ย ย ย ย ย ย ย ย ย  X =ย ย ย ย ย ย  2ย ย ย ย ย  1ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย  (3 mks)

-5ย ย ย ย ย ย ย  2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  5ย ย ย ย ย  3

 

 

 

 

 

 

  1. The probabilities of three candidates K, M and N passing an examination is 2/3, ยพ and 4/5 Find the probability that :

(a)ย  All pass:ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1 mk)

(b)ย  At least one fails:ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

 

  1. In the figure, PR is tangent to the circle centre O. If รBQR=300, รQBC=270,and รOBA=370, find รBAC and ร

 

Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A

 

 

 

 

Bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  Pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย  R

  1. A frustrum of height 10cm is cut off from a cone of height 30cm. If the volume of the cone before cutting is 270cm3 , find the volume of the frustrum. (3 mks)

 

  1. Evaluate 0 (2 mks)

( 3x2 –ย  1 ) dx

4 x 2

1

  1. If one litre of water has a mass of 1000g, calculate the mass of water that can be held in a rectangular tank measuring 2m by 3m by 1.5m. (give your answer in tonnes). (2 mks)
  2. Write down the three inequalities which define the shaded region. (3 mks)

 

 

 

(3,2)

 

 

 

 

 

 

(2,1)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4,1)

 

 

 

 

  1. The depth of sea in metres was recorded on monthly basis as follows:

 

Month March April May June July
Depth (m) 5.1 4.9 4.7 4.5 4.0

Calculate the three monthly moving averages.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

  1. A number of women decided to raise sh. 6300 towards a rural project for bee keeping. Each woman had to contribute the same amount. Before the contribution, seven of them withdrew from the project. This meant the remaining had to pay more. If n stands for original number of women, show that the increase in contribution per woman was: 44100 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3 mks)

n(n-7)

 

 

 

 

 

SECTION II:ย ย  (48 Marks)

 

  1. Find the distance between points A(500 S, 250 E) and B(500 S, 1400 E) in:

(i)ย ย  Kmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (ii)ย ย  nmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

(take radius of earth to be 6400km, P =ย  3.14)

 

  1. The distance S in metres, covered by a moving particle after time t in seconds, is given by :

Sย  =ย  2t3 + 4t3– 8t + 3.

Find:

(a)ย  The velocity at :ย ย ย ย ย ย ย ย ย ย ย  (i)ย  tย  =ย  2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (ii)ย  tย  =ย  3

  • The instant at which the particle is at rest. (8 mks)

 

  1. A car starts from rest and its velocity is measured every second for six seconds. (see table below).
Time (t) 0 1 2 3 4 5 6
Velocity v(ms -1) 0 12 24 35 41 45 47

 

Use trapezium rule to calculate the distance travelled between t = 1 and t = 6.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

 

  1. Using a pair of compass and ruler only, construct triangle ABC such that AB=9cm, BC=14cm and รBAC = 1200 . Draw a circle such that AB, BC and AC are tangents. What is the radius of this circle?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (8 mks)
  2. The marks scored by 100 students in mathematics test is given in the table below:
Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79
No. of students 8 15 15 20 15 14 13

 

(a)ย  Estimate the median mark.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

(b) Using 44.5 as the assumed mean, calculate:-

(i)ย ย ย ย ย ย ย ย  The mean mark:ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

(ii)ย ย ย ย ย ย ย  The variance:ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

(iii) ย ย ย ย ย  The standard deviation:ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

 

  1. (a) On the same axes, draw the graphs of : yย  =ย  sin xย  ;ย  yย  =ย  cos x

yย  =ย  cosxย  +ย  sin X for 00 ร X ร 3600 .

(b)ย  Use your graph to deduce

(i) The amplitude

(ii) The period of the wave y = cos x + sin x.

(c) Use your graph to solve:

Cos xย  = – sin x for 00 ร X ร 3600 .

 

  1. Given a circle of radius 3 units as shown in the diagram below with its centre at O(-1, 6). If BE and DE are tangents to the circle where E (8,2). Given further that ร DAB = 800.

B

 

 

Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  E

C

 

 

D

(a)ย  Write down the equation of the circle in the form ax2 + bx + cy2 + dy + e = 0 where a, b, c, ย ย ย ย ย ย ย ย ย ย ย  d, e are constants.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

(b)ย  Calculate the length DE.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

(c)ย  Calculate the value of angle BED.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

(d)ย  Calculate the value of angle DCB.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2 mks)

 

  1. A building contractor has to move 150 tonnes of cement to a site 30km away. He has at his disposal 5 lorries. Two of the lorries have a carrying capacity of 12 tonnes each while each of the remaining can carry 7 tonnes. The cost of operating a 7 tonne lorry is sh. 15 per km and that of operating a 12 tonne lorry is sh. 25 per km. The number of trips by the bigger lorries should be more than twice that made by smaller lorries. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

 

(a)ย  Represent all the information above as inequalities.

  • How should the contractor deploy his fleet in order to minimise the cost of moving the cement?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (8 mks)

 

 

MATHEMATICS III

PART I

MARKING SCHEME

ย 

ย 

ย 

ย 

ย 

ย 

SOLUTION MRK AWARDING  
1. ร–7.5625 = 2.75

 

3ร–3.375 = 3ร–3375 X 3ร–10-3

 

=ย  3 ร–33 x 53 x 10-1 = 3 x 5 x 10-1 = 1.5

 

= 2.75 x 1.5ย  =ย  2.75ย  =ย  0.275

1.5 x 10ย ย ย ย ย ย ย ย ย  10

 

1

 

1

 

1

1

1

 

ย 

Method for ร–7.5625

Square root

 

Method for 3ร–

3ร–

Answer

 
    5    
2. T2yย  =ย  ร– k+y

K

T4y2k =ย  k+y

T4y2k – kย  =ย  y

K(T4y2-1) =ย  y

Kย  =ย  y

T4y2 โ€“ 1

 

 

1

 

 

1

 

1

ย 

 

Removal of square root

 

Rearrangement of terms

Answer

 
    3    
3. (x 2)ย ย ย ย ย ย ย ย  xย ย ย ย ย  =ย  (8)

-2

 

x2 – 4ย  =ย  8

 

xย  =ย  +ร–12 = + 2ร–3 = + 3.464

 

1

 

 

1

 

1

 

ย 

Matrix equation

 

 

Quadratic equation

Answers in any form

 
  ย  3    
4. r(x2 โ€“ 1)

2r(x โ€“ 1)

 

r(x2 โ€“ 1)(x2 + 1)

2r (x – 1)

 

r(x – 1)(x + 1)( x2 + 1)

2r (x – 1)

 

=ย ย  (x + 1)( x2ย  + 1)

2

 

 

 

 

1

 

 

 

1

 

 

1

 

 

 

 

Complete factorisation of numerator

 

Factorisation of denominator

 

Answer

 
    3    
5. ย ย ย ย ย  1ย  =ย  log3 3

8 – xย ย ย  =ย ย  3

1+x

 

-4xย  =ย  -5

 

x = 5

4

1

 

 

 

1

 

1

 

ย 

ย 

Logarithic expression.

 

 

Equation

 

Answer

 

 

 
    3    
6. Let the centre be (a,b)

 

4-9ย ย ย ย  ย ย ย =ย  -1ย ย ย ย ย  4-a

-5-bย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย 3-b

 

4-aย  =ย  -4+9ย ย ย ย ย ย ย ย ย ย  -5-bย  =ย  -3+b

aย  =ย  4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  bย  =ย  -1

centre is (4,-1)

 

 

 

1

 

 

1

 

1

 

ย 

 

 

 

Equation

 

 

Linear equations

 

Centre

 

 
    3    
7. Yย  =ย  4x + 5

Gradient = 4

Gradient of ^ line – ยผ

y + 2ย  =ย  – 1

x + 3ย ย ย ย ย ย ย  4

4y + xย  =ย  -11

 

 

1

1

ย 

 

Gradient of ^ line.

Equation.

 

 
    2    

8.

Xย  = 28 ย =ย  3.5

8

 

 

standard deviation = ร– 22 = ร–2.75ย  =ย  1.658

8

 

 

1

 

 

1

1

 

1

ย 

Mean

 

 

d values

d2 values

 

Answer

 
    4

 

   
9. a = 203ย ย ย  d = 7ย ย  L = 294

 

294ย  =ย  203 + 7(n-1)

nย  =ย  14

 

S 14ย  =ย  14 (203 +ย  294)

2

 

=ย  7 x 497

=ย  3479

 

1

 

1

 

 

1

 

 

 

1

ย 

For both a and b

Equation

 

 

For n

 

 

 

Sum

 

 
  ย  4    
10. Sin xย  =ย  2 sin x

Cos x

 

Sin xย  =ย  2 cosx

Sin x

 

2 cos xย  =ย  1

cos xย  =ย  0.5

 

xย  =ย  600, 3000, -600

 

 

 

1

 

 

 

1

 

1

ย 

 

 

 

Simplification

 

 

 

Equation

 

All 3 values

 
    3    
11. (1 +-2x)4ย  =ย  1-8x + 24x2 – 32x3 + 16x4

 

(0.82)4ย  =ย  (1 + -2 x 0.09)4

xย ย ย ย  =ย  0.09

(0.82)4ย  = 1 – 0.72 + 0.1944 โ€“ 0.023328 + 0.00119376

= 0.35226576

@ย  0.35227 (5 d..p)

1

 

 

1

1

 

1

 

Expansion

 

 

Value of x

All terms

 

 

Rounded

 
    4    
12. ย  2ย  =ย  5m โ€“ 3

m =ย  1

tan qย  =ย  1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  qย  =ย  450

 

1

1

ย 

 

Value of m.

Angle

 
    2    
13. ย Let the number be xy

3yย  =ย  x + 14

10y + xย  =ย  10x + y + 36ย  =ย  9y – 9xย  รžย  36

3y โ€“ xย  =ย  14

9y โ€“ 9xย  =ย  36

yย  =ย  5

xย  =ย  1

the number is 15.

 

1

1

 

1

 

 

1

ย 

 

1st equation

2nd equation

 

method of solving

 

Answer

 

 
   

 

S

4    
14. Let รAOBย  =ย  q

ย  qย  xย  2ย  xย ย  22ย  xย  7ย  =ย  11

360ย ย ย ย ย ย ย ย ย ย ย ย ย  7

qย  =ย  900

 

Area shadedย  =ย ย  90 x 22 x 7 x 7 – 1 x 7 x 7

360ย ย ย  7ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

=ย  77 โ€“ 49

2ย ย ย ย  2

= 28ย  =ย  14cm2

2

 

 

 

1

 

1

 

 

1

ย 

 

 

 

Value of q

 

Substitution

 

 

Answer

 
    3    
15. P(WBb)ย  =ย  6 x 15 x 15

36ย ย ย  35ย ย  34

 

=ย ย  15

476

1

 

 

1

 

Method

 

 

Answer

 
    2    
16. Equationย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  inequality

L1 ย ย ย y =ย  xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  yย  ยฃย  x

L2ย ย ย  y = -2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  yย  ยณ -2

L3 ย ย ย 2y + 5x = 21ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2y + 5x < 21

1

1

1

1

ย 

1 mark for each inequality.

Method for obtaining L3

 

 
 

 

 

 

 

 

 

 

 

 

 

 

(i)ย  roots are x = -3

x = 2

(ii)ย  y = x2 + x-6

0 = x2 + 2x-8

y = -x + 2

roots are x = -4

x =ย  2

4 2

 

 

 

 

 

 

 

1

 

1

 

 

1

 

1

 

1

 

For all correct points.

1 for atleast five correct points.

 

 

 

Correct plotting.

 

Scale

 

 

Smoothness of

curve

 

Both roots

 

 

Linear equation

 

 

Both roots

 

   

 

  8  
         
18. ย ย  hย ย ย ย  =ย  15

h+50ย ย ย ย  40

 

hย ย  = 30cm

Hย  =ย  80cm

 

(a)ย  Volumeย  =ย  1/3 p x 40 x 40 x 80 – 1/3 ย p x 15 x 15 x 30

 

=ย  128000 pย  –ย  6750 p

3ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3
=ย ย  121,250p cm3

3

 

(b)ย ย  L2ย  =ย  802 + 402ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Lย ย ย  =ย  152 + 302

= 6400 + 1600ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 225 + 900

=ย  8000ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 1125

Lย ย ย  =ย  89.44 cmย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Lย ย ย  =ย  33.54 cm

Curved surface area of bucket = p x 40 x 89.44

p x15x33.54

= 3577.6p – 503.1p

=ย  3074.5cm2

1

 

 

1

 

 

1

 

 

 

 

1

 

1

 

1

 

1

 

1

ย 

Expression

 

 

Value of H

 

 

Substitution

 

 

 

 

Volume

 

L

 

L

 

 

Substitution

 

Area

 

 
    8    
 

 

19.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

     
         
 

20.

 

 

 

(i)ย  รRPQย  =ย  130

ย ย ย ย ย ย ย  รPQRย  =ย  320+900+240 =ย  1460

รPRQย  =ย  1800 – (1460 + 130)

=ย  210

 

(ii)ย ย ย  Pย ย ย ย ย  =ย ย ย ย ย ย ย  7

sin130 ย ย ย ย ย ย ย ย sin 210

Pย ย ย  =ย ย  7 sin 130

Sin 210

=ย  4.394km

 

 

 

 

 

 

 

 

 

 

 

Pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  T

 

(iii)ย ย ย  Let PRย  =ย  q

 

qย ย ย ย ย ย  =ย ย ย ย ย ย  7

sin 1460ย ย ย ย ย  sin 210

 

qย ย ย ย  =ย  7 sin 1460

sin 210ย 

qย ย ย ย ย ย  =ย  10.92 km

 

sin 450ย  =ย ย ย  RT

10.92

 

RTย  =ย  10.92 sin 450

 

= 7.72 km (2 d..p)

 

 

 

 

1

 

 

 

 

 

 

1

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

1

 

 

 

 

 

1

 

 

1

 

1

ย 

 

 

Fair sketch

 

 

 

 

 

 

รPRQ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Equation

 

Method

 

 

 

Equation

 

 

 

 

 

 

Distance PR

 

 

Equation

 

RT

 

 

 

 

 
  8    
21. (a)ย  4xย  =ย  9y

2(x+y)ย  =ย  y+44ย  รžย  2x + yย  =ย  44

 

4x – 9y = 0

4x + 2y = 88

11y = 88

yย ย  =ย  8

 

xย  =ย  18

(b)ย  P(RR)ย  =ย ย  18ย  xย  18ย ย  =ย  81

26ย ย ย ย ย  26ย ย ย ย ย  169

 

1

 

1

 

2

 

1

1

1

1

 

ย 

 

Equation

 

Equation

 

Method of solving

Value y

Value x

Method

Answer

 
    8    
22. (a)ย  67,000 Kshย  =ย  67,000 US dollars

16.75

= 4,000 dollars

 

(b)ย  2 x 4,000ย  =ย ย  1600 US dollars

5

1600 US dollarsย  =ย  1600 x 1340

=ย  2,144,000 Italian lire

(c)ย  Remainderย  =ย  2400 US dollars

5ย  xย  2400ย ย  =ย  1500 US dollars

8

1500 US dollars = 1500 x 1.8

= 2700 Deutche marks

(d)ย  Remainderย  =ย  900 US Dollars

900 US Dollars = 900 x 16.75 Ksh.

=ย  15,075 Ksh.

 

1

 

1

 

1

 

1

1

 

 

1

1

1

 

 

Method

 

Answer

 

Method

 

Answer

 

For 1500

 

 

Answer

 

Method

Ksh.

 
    8    
23. PMย  =ย  kPA

=ย  k(r + 1h)

2

=ย  kr + 1kh

2

PMย  =ย  PB +ย  BM

3h + t BQ

4

=ย ย  3h + t(-3h + r)

4ย ย ย ย ย ย ย ย ย  4

 

=ย  3h – 3t h + tr

4ย ย ย ย  4

=ย  3 –ย ย  3tย ย ย  h + tr

4ย ย ย ย  4

 

t = kย  ย ย ย ย ย ย ย ย ย 33tย  =ย  1k

4ย ย  4ย ย ย ย ย ย  2

33t = 1 t

4ย ย ย  4ย ย ย ย  2

5tย  =ย ย  3

4ย ย ย ย ย ย  4

tย  =ย  3 + 4

4ย ย ย  5

= 3

5

\ย ย  k = 3

5

\ย ย  PMย  =ย  3rย  +ย  3h

5ย ย ย ย ย ย  10

 

 

 

1

 

1

 

 

1

 

 

1

 

1

 

 

1

 

1

1

 

ย 

 

 

 

PM

 

PM

 

 

PM simplified

 

 

 

 

 

 

 

Both equations

 

method

 

 

 

 

Value of k

 

Value k

PM

 

 
    8    
         
 

 

 

24.

Y

LogT

 

 

 

Log Tย  =ย  log a + x log b

Log Tย  รžย  0.82, 1.25, 1.68, 2.11, 2.54, 2.97, 3.40, 3.84

 

y โ€“ intercept = log a = 0

a = 1

gradientย  =ย  3.84 – 0.82ย  =ย ย  3.02

9 – 2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  7

= 0.4315

 

log b = 0.4315ย ย  =ย  0.4315

b = antilog 0.4315

bย  =ย  2.7

 

1
1

 

 

 

 

 

 

 

1

2

 

1

 

1

 

 

 

 

1

8

Plotting
Labeling of axis

 

 

 

 

Linear

All correct logs

 

Value of a

Method of gradient

 

Value ofย  b

 

MATHEMATICS III

PART II

MARKING SCHEME

ย 

  1. SOLUTION MARKSย ย ย  AWARDING
1. ย ย  Noย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  log

 

8.69ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  0.9390

0.786ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1.8954

21.72ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1.3369

1.2323

1.7067 – 2

 

2 +ย  1.7067

2ย ย ย ย ย ย ย ย ย ย  2

– 1ย  +ย  0.8533

0.7134 x 10 -1ย ย ย ย  =ย  0.07134

 

 

 

 

 

M1

 

 

M1

 

 

 

A1

 

 

รผ reading to 4 s.f

 

 

 

 

 

 

Rearranging

    3  
2.  

ย 4ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย –ย ย ย ย ย ย ย ย  1

(x-2)(x+2)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (x-2)

 

ย – x+2

(x-2(x+2)

– (x-2)

(x-2(x+2)

 

-1

x+2

 

 

 

M1

 

 

M1

 

 

A1

 

ย 

 
    3  
3.  

Re6000ย  =ย  Ksh. 75000

Spent 5000 Rem 2500

Remย ย ย  2500

1.25

Re 2000

M1

 

 

M1

 

A1

ย 

 
    3  
4. 2x – 1ย  ,ย  2z + 1ย  ,ย  2x + 3

6x +ย  3ย  =ย  105

6xย  =ย  102

xย  =ย  17

M1

M1

A1

A1

ย 

Allow M1 for us of different variable.
    4  
5.  

4 * 1ย  =ย  5

4

2 * 3ย  =ย  5

6

A * 5ย  =ย  5

6ย ย ย ย ย  4

A + 5ย  =ย  5ย  xย  5A

6ย ย ย ย ย  4ย ย ย ย ย ย  6

A +ย  5ย  =ย  25 A

6ย ย ย ย ย ย  24

Aย ย  =ย  20

 

 

M1

 

 

 

 

M1

 

 

A1

3

 
6.  

 

 

 

 

180 โ€“ M + 20 + 95ย  =ย  180

295ย  –ย  Mย  =ย  180

– Mย  =ย  – 115

Mย  =ย  115

 

 

 

B1

 

 

B1

 

 

A1

ย 

 
    3  
 

7.

 

1 + 2x + 60x2 + 160x3 +

1 + 0.2 + 0.006 + 0.00016

=ย  1.20616

=ย  1.206

 

M1

M1

M1

A1

4

 

Only upto term in x3.

Correct substitution

 

Only 4 s.f.

 

8.  

3ย ย  -1ย ย ย ย ย  2ย ย ย  1ย ย ย  =ย ย ย  I

-5ย ย  2ย ย ย ย ย ย  5ย ย ย  3

 

6ย ย  -5ย ย ย ย ย ย ย ย ย ย ย ย  3ย ย ย  -3

-10 +10ย ย ย ย ย ย ย ย  -5 + 6

 

1ย ย ย ย ย  0

0ย ย ย ย  ย ย 1

 

 

M1

 

M1

 

 

A1

ย 

 

Matrix multiplication gives :

 

Iย ย ย ย ย ย  1ย ย  0

0ย ย  1

  3  
9. (a)ย ย  2ย  xย  3ย  xย  4ย ย ย ย ย  =ย  2

3ย ย ย ย ย  4ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย  5

(b)

2ย  xย  3ย  x 1ย ย ย ย  +ย ย ย ย  2ย  xย  1ย  xย  4ย ย ย ย  +ย ย ย ย  1ย  xย  3ย  xย  4
3ย ย ย ย ย  4ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย ย  3ย ย ย ย ย  2ย ย ย ย ย  5ย ย ย ย ย ย ย ย ย ย ย ย  3ย ย ย ย ย  4ย ย ย ย ย  5

 

1ย  +ย  4ย  +ย  1

10ย ย ย ย  15ย ย ย ย  5

 

=ย ย ย ย  17

10

M1

 

 

M1

 

 

 

 

A1

 

ย 

 
    3  
10. รQCBย  =ย  300

180 – (27 + 30)ย  =ย  1230

\ย ย ย ย  BACย  =ย  570.

 

 

 

 

OBAย  =ย  370

OABย  =ย  370

 

 

AOBย  =ย  1060

\ ACBย  =ย  530

 

 

 

M1

 

 

 

 

 

M1

 

 

A1

ย 

 

 

 

 

 

 

 

 

Isosceles triangle.

 

Angle at centre is twice angle at circumference.

    3  
11. Vย  =ย  1ย  xย  3.14ย  xย  r 2 ย x 10ย  =ย  270

L.S.F.ย ย ย ย ย  20ย ย  =ย  2

30ย ย ย ย ย ย  3

V.S.Fย  =ย ย ย  2ย ย  3ย ย ย ย ย ย ย  =ย ย ย ย  8

3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  27

Vol. of coneย  =ย  8ย  xย  270

27ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย ย ย ย ย  80cm3

\ Vol. Of frusturmย  = (270 – 80)ย  =ย  190cm3

 

 

 

M1

 

M1

 

 

A1

ย 

 
    2  
12.

 

 

 

 

 

 

 

 

3x 3ย  –ย  xย  -1ย ย ย ย ย ย ย ย ย  2

3ย ย ย ย ย ย  -1ย ย ย ย ย ย ย ย  1

 

x 3ย  +ย  1ย ย ย ย  2

xย ย ย ย  1

 

8ย  +ย  1ย ย ย ย  –ย ย  ( 1ย  –ย  1)

2

8 1ย  –ย  2ย ย ย ย  =ย ย ย ย ย ย ย ย  6ย  1

2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

 

 

 

 

 

 

M1

 

 

 

A1

2

 
13. (2 x 3 x 1.5)ย  volume

9 m3

1Lย  ยบย  1000 cm3

1000 Lย  =ย  1 m3

9000 Lย  =ย  9 m3

1000 Lย  =ย  1 tonne

9000 Lย  =ย  9 tonnes.

 

 

M1

 

 

 

A1

ย 

 
    2  
14. ย ย ย ย  yย ย  ยณ 1ย ย ย ย ย ย ย ย ย ย ย  (i)

yย ย  <ย  x – 1ย ย ย ย  (ii)

yย ย  <ย  5 – xย ย ย ย  (iii)

 

B1

B1

ย 

 
    3  
15. M1 ย =ย  5.1ย  +ย  4.9ย  +ย  4.7ย  =ย  4.9

3

M2ย  =ย  4.9 + 4.7 + 4.5ย  =ย  4.7

3

M3 ย =ย  4.7 + 4.5 + 4.0ย  =ย  4.4

3

M1

M1

M1

 

ย 

 
    3  
16. Original contribution per womanย  =ย  6300

N

Contribution when 7 withdrawย  =ย  6300

(n-7)

Increaseย ย  –ย  Diff.

6300ย ย  –ย ย  6300

n-7ย ย ย ย ย ย ย ย ย  n

6300nย  –ย  6300(n-7)

n(n-7)

6300n โ€“ 6300 + 44100

n(n-7)

44100

n(n-7)

 

 

 

M1

 

 

M1

 

1

3

 
SECTION II (48 Marks)

ย 

17. (i)

1150

 

Aย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย B

 

Centre of circles of latitude 500 S.ย  R Cos 500

ABย  =ย  115ย  xย  2p R Cos 50o

115ย  xย  40192ย  xย  0.6428

360

=ย  8252.98ย  km

 

(ii)ย ย  Arc AB 60 x 115ย  Cos 50 nm

60 x 115 x 0.6428 nm

4435 nm

 

 

 

 

 

M1

M1

 

 

M1

 

A1

 

M1

M1

M1

A1

ย 

 

 

 

 

 

 

No.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  log

60ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1.7782

1+5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2.0607

0.6428ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1.8080

4435nmย ย ย ย ย ย ย ย ย ย ย ย  3.6469

    8  
18. (a)ย  Vย  =ย  dsย  =ย  6t2 + 8t – 8

dt

(i)ย  tย  =ย  2

Vย  =ย  6×4 + 8×2 – 8

= 32 ms-1

(ii)ย  tย  =ย  3

V =ย  6×9 + 8×3 – 8

= 70ms-1

 

(b)ย  Particle is at rest when V = 0

6t2 + 8t – 8 = 0

2(3t – 2) (t+2) = 0

tย  =ย  2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  tย  =ย  -2

3

particle is at rest at t = 2 seconds

3

ย   

 

 

 

 

 

 

 

 

 

 

 

Do not accept t = -2. Must be stated.

    8  
19. Area under velocity – time.

graphย  gives distance.

 

Aย  = { h ยฝย  (y1 + y6 ) + y2 + y3 + y4 + y5 )}

 

= 1 { ยฝ ( 12+47) + 24 + 35 + 41 + 45)}

=ย  29.5 + 14.5

=ย  174.5m

 

 

B1

B1

M1

M1

B1

B1

A1

ย 

Trapezium rule only accepted.

Formula.

 

Substitution into formular.

    8  
20. ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย Drawing actual

Scale 1cmย  =ย  2cm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Radiusย ย ย ย ย  1cm

=ย  2cm

 

M1

 

M1

 

M1

 

M1

 

M1

M1

 

M1

M1

ย 

 

Bisect รA

 

Bisect ร B

 

Intersection at centre of inscribed circle.

Draw circle.

 

Measure radius.

Arcs must be clearly shown.

  8  
 

 

 

21.

 

 

 

 

mean = 44.5 +ย  130

100

=ย  44.5ย  +ย  1.3

=ย  45.8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b)ย  Varianceย  S (x – A) 2ย  =ย  2800

Sfย ย ย ย ย ย ย ย ย ย ย ย ย ย  100

= 28

S.D.ย  =ย  ร– 28ย  =ย  5.292

 

 

 

 

M1

 

 

 

 

 

A1

 

M1

 

A1

M1

A1

ย 

 
    8  
 

 

 

 

 

22.

y = sin x

xย ย ย  0ย ย ย ย ย ย ย  60ย ย ย ย ย ย ย  120ย ย ย ย ย ย ย  180ย ย ย ย  240ย ย ย ย ย  30ย ย ย ย ย  360

sin x 0ย ย ย  0.866ย ย ย ย  0.866ย ย ย ย ย  0ย ย ย ย  -0.866ย ย  -0.866ย ย ย  0

y = cos x

xย ย ย ย  qย ย  ย ย ย ย ย 60ย ย ย ย ย ย ย  120ย ย ย ย ย ย ย  180ย ย ย ย  240ย ย ย  300ย  360

cos x 1ย ย ย ย  0.5ย ย ย ย ย ย  -0.5ย ย ย ย ย ย  -1.0ย ย ย ย  -0.5ย ย ย ย  0.5ย ย  1.0

y = cosx + sinx

xย ย ย ย ย ย ย ย ย ย ย  qย ย ย ย ย ย ย  60ย ย ย ย ย ย  120ย ย ย ย ย ย ย  180ย ย ย ย  240ย ย ย ย ย  30ย ย ย ย  360

cosx + sinx 1ย  1.366ย ย  0.366ย ย ย ย ย ย  -1ย ย  -1.366ย  -0.366 1.0

(c)ย ย ย ย ย  Cos x = – sin x

xย  =ย  450 , 2250

   
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

23.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(i)ย  amplitudeย ย  =ย  1.366

(ii)ย  Periodย  =ย  3000

 

 

 

(a)ย  (x+1) 2ย  +ย  (y-6)2ย  =ย  32

x2 + 2x + 1 + y2 – 12y + 36ย  =ย  9

x2 + 2x + y2 โ€“ 12y + 28ย  =ย  0

 

(b)ย  cos 10ย  =ย  ODย ย ย ย ย ย ย ย ย ย ย ย  DEย  =ย  3

DEย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  0.9848

DEย  =ย  3.046

 

(c)ย  Twice รOED

100 x 2ย  =ย  200

 

(d)ย  DABย  =ย  800

\ DCBย  =ย  1000

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

A1

 

M1

A1

 

 

M1

A1

 

M1

A1

ย 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Formular

(x-a)2 + (y-b)2 = r2

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

Cyclic quad.

ย 

 

    8  
24. Let number of trips by 12 tonne lorry be x.

Let number of trips by 7 tonne lorry be y.

 

(a)ย ย  x > 0ย  ;ย  y > 0

24x + 21yย  ยฃย  150

 

12 x 25 x X + 15 x 7 x y ยฃ 1200

300x + 105yย  ยฃย  1200

x > 2y

 

(b)ย  Ref. Graph paper.

Minimising:

3 – 12 tonne lorry and 2 – 7 tonne lorries should be deployed.

 

 

 

B1

 

 

 

B1

B1

 

 

 

 

 

 

 

MATHEMATICS IV

PART I

ย 

SECTION 1 (52MKS)

 

  1. Evaluate using logarithms 3ร–7.673 โ€“ 15.612

12.3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)

 

  1. Solve xย ย  –ย  3xย  –ย  7ย ย ย  =ย  x โ€“ 2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

3 ย ย ย ย ย ย ย ย ย ย ย 5ย ย ย ย ย ย ย ย ย ย ย ย  5

 

  1. In the given figure CD is parallel to BAC, calculate the values of x and y. (3mks)

 

 

Cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  D

 

 

 

 

 

 

Bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A

 

  1. The surface area and volume of a sphere are given by the formulars S = 4pr2 and V= 4/3 pr3.

Express V in terms of S only.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (3mks)

 

  1. A line perpendicular to y = 3-4x passes through (5,2) and intercepts y axis at (0,k)

Find the value of K.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (3mks)

 

  1. An alloy is made up of metals P,Q,R, mixed in the ratio 4:1: 5: A blacksmith wants to make 800g of the

alloy. He can only get metal P from a metallic ore which contains 20% of it. How many Kgs of the ore

does he need.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

 

  1. The co-ordinate of point Aย  is (2,8) vector AB =ย ย  5ย ย ย  and vector BCย  =ย  4ย ย  Find the

-2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย 3

co-ordinate of point C.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  3mks)

 

  1. Two buildings are on a flat horizontal ground. The angle of elevation from the top of the shorter building to the top of the taller is 200 and the angle of depression from the top of the top of the shorter building to the bottom of the taller is 300. If the taller building is 80m, how far apart are they

(4mks)

  1. The given figure is a quadrant of a piece of paper from a circle of radius 50cm. It is folded along AB

and AC to form a cone . Calculate the height of the cone formed.

(4mks)

 

 

 

 

5Ocm

 

 

50cm

 

 

  1. Express 3.023 as a fraction ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (2mks)
  2. Point A (1,9), Point B(3,5) and C (7,-3). Prove vectorically that A,B and C are collinear.ย ย ย ย ย ย  (4mks)
  3. A salesman gets a commission of 4% on sales of upto shs 200,000 and an additional 2% on

sales above this. If in January he got shs 12,200 as commission, what were his total salesย ย ย  (4mks)

  1. Water flows through a cylindrical pipe of diameter 3.5cm at the rate of 2m/s. How long to the nearest minute does it take to fill a spherical tank of radius 1.4m to the nearest minute? (4mks)
  2. Rationalize the denominator in ร–3

ร– 7 – 2

Leaving your answer in the form ร–a + ร–b

C

Where a ,b, and c are integersย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

  1. For positive values of x, write the integral solutions of 3ยฃ x2ย  ยฃย  35ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย (4mks)
  2. 8 girls working 5 hours a day take 12 days to drain a pool. How long will 6 girls working 8 hours a day take to drain the pool?( Rate of work is equal) (2mks)

 

SECTION IIย  (48 mks)

 

  1. In the given circle centre O , A,E,F, is target to the circle at E. Angle FED = 300ย  <DEC = 200 andย  <BC0ย  = 150

 

 

 

 

Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  F

 

 

 

 

Calculateย ย  (i) <CBEย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (3mks)

(ii)ย  <BEAย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

(iii) <EABย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (3mks)

 

  1. The sum of the 2nd and third terms of a G.P is 9/4 If the first term isย  3,

(a) Write down the first 4 terms of the sequence .ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย (5mks)

(b) Find the sum of the first 5 terms using positive values of the common ratio (r)

(3mks)

  1. E and F are quantities related by a law of the form E = KFn Where k and n are

constants. In an experiment , the following values of E and F were obtained .

 

E 2 4 6 8
F 16.1 127.8 431.9 1024

 

Use graphical method to determine the value of k and n (Graph paper provided) ย ย ย ย  (8mks)

 

  1. In the domain โ€“2 ยฃ x ยฃ 4 draw the graph of y = 3x2 + 1 โ€“2x .Useย  your graph to solve the equation.ย  6x2 4x + 4 = 0 (graph paper provided)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)
  2. A solid sphere of radius 18cm is to be made from a melted copper wire of radius 0.4mm . Calculate the length of wire in metres required to make the sphere.ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  (5mks)

(b) If the density of the wire is 5g/cm3. Calculate the mass of the sphere in kg.ย ย ย ย ย ย ย  (3mks)

 

  1. A right cone with slantย  height of 15cm and base radius 9cm has a smaller cone of height 6cm chopped off to form a frustum. Find the volume of the frustum formedย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  ย  (8mks)

 

 

 

 

 

 

 

 

9cm

 

  1. PQRS are vertices of a rectangle centre. Given that P(5,0) and Q and R lie on the line x+5 = 2y, determine

(a) The co-ordinates of Q,R,S,ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (6mks)

(b) Find the equation of the diagonal SQย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย ย (2mks)

  1. A tap A takes 3 hours to fill a tank. Tap B takes 5 hours to fill the same tank. A drain tap C takes 4 hours to drain the tank. The three taps were turned on when the tank was empty for 1ยฝ hours. Tap A is then closed. Find how long it takes to drain the tank.

(8mks)

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

MATHEMATICS IV

PART II

ย 

SECTIONย ย  Iย  (52MKS)

ย 

  1. Without using mathematical tables, evaluate ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (3mks)

 

ร– 0.0784 x 0.27 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (leave your answer in standard form)

0.1875

 

  1. A father is three times as old as his son. In ten years time , the son will be half as old as the father . How

old are they now?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. A,B,C,D, is a parallelogram diagram. ADE is an equilateral triangle. AB and CD are 3cm apart.

AB = 5cm. Calculate the perimeter of the trapezium ABCEย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

Eย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Dย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย C

Aย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  B

  1. Given that a = -2, b = 3 and c = -1, Find the value ofย ย  a3 โ€“ b โ€“ 2c2ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  (2mks)

2b2 โ€“ 3a2c

 

  1. The exchange rate in January 2000 was US $ 1 = Ksh 75.60. and UK ยฃ1 = Ksh 115.80.ย ย ย  A touristย  came to Kenya with US $ 5000 and out of it spent ksh.189,000. He changedย ย  the balance in UK ยฃ . How many pounds did he receive?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (4mks)

 

  1. ABC is a cross โ€“ section of a metal bar of uniform cross section 3m long. AB = 8cm andย  AC = 5cm.

Angle BAC = 600 . Calculate the total surface area of the bar in M2. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (4mks)

 

  1. The bearing of a school chapel C, from administration block A, is 2500 and 200mย  apart.

School flag F is 150m away from C and on a bearing of 0200. Calculate the distance and

bearing of A from F.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (5mks)

  1. A box has 9 black balls and some white balls identical except in colour. The probability of picking a white ball is 2/3

(i) Find the number of red ballsย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย  ย ย ย ย ย ย  (2mks)

(ii) Ifย  2ย  balls are chosen at random without replacement, find the probability that they are of different colour.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

  1. Under an enlargement of linear scale factor 7, the area of a circle becomes 441.p

Determine the radius of the original circle.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  (3mks)

  1. A circle has radius 14cm to the nearest cm . Determine the limits of its area. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ( 3mks)
  2. Expand (1 + 2x)5 up to the term with x3. Hence evaluate 2.045 to the nearest 3 s.f. (4mks)
  3. The nth term of aย  G.P is given byย  5 x 2 n-2

(i) Writeย  down the first 3 terms of the G.Pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  ย ย  (1mk)

(ii) Calculate the sum of the first 5 termsย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (2mks)

  1. 3 bells ring at intervals of 12min, 18min and 30min respectively. If they rang together at 11.55am, when will they ring together again. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย  ย (3mks)
  2. On a map scale 1:20,000 a rectangular piece of land measures 5cm by 8cm. Calculate its actual area in hectares.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  3. It costs Maina shs. 13 to buy 3 pencils and 2 rubbers; while Mutiso spent shs.9 to buy one pencil and 2 rubbers. Calculate the cost of a pencil and one rubber ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. Three angles of a pentagon are 1100, 1000 and 1300. The other two are 2x and 3x respectively. Find their values . ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย (2mks)

 

SECTION II (48MKS)

ย 

  1. Members of a youth club decided to contribute shs 180,000 to start a company. Two members withdrew their membership and each of the remaining member had to pay shs. 24,000 more to meet the same expense. How many members remained? (8mks)
  2. A box contains 5 blue and 8 white balls all similar . 3 balls are picked at once. What is the probability that

(a)ย  The three are whiteย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  (2mks)

(b)ย  At least two are blueย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (3mks)

(c) Two are white and one is blueย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย (3mks)

 

  1. A rectangular tennis court is 10.5m long and 6m wide. Square tiles of 30cm are fitted on the floor.

(a) Calculate the number of tiles needed.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (2mks)

(b) Tiles needed for 15 such rooms are packed in cartons containing 20 tiles. How many cartons are

there in total?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย (2mks)

(c)ย  Each carton costs shs. 800. He spends shs. 100 to transportย  each 5 cartons. Howย  much would one

sell each carton to make 20% profit ?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (4mks)

  1. The following was Kenya`s income tax table in 1988.

Income in Kยฃ P.aย ย ย ย ย ย ย ย ย ย ย ย  Rate (Ksh) ยฃ

1ย ย ย ย ย ย ย ย ย  –ย ย  2100ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

2101ย ย ย  –ย ย  4200ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

4201ย ย ย ย  –ย  6303ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  5

6301ย ย ย ย  –ย  8400ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  7

 

(a) Maina earns ยฃ 1800 P.a. How much tax does he pay?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (2mks)

(b) Okoth is housed by his employer and therefore 15% is added to salary to makeย  taxable income. He

pays nominal rent of Sh.100 p.m His total tax relief is Shs.450. If he earns Kยฃ3600 P.a, how much

tax does he pay?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (6mks)

  1. In the given figure, OA = a , OB =b,ย  OP: PA =3:2,ย  OQ:QB = 3:2

Q

B
R

Oย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย A

(a) Write in terms of a and b vector PQย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (2mks)

(b) Given that AR = hAB where h is a scalar, write OR in terms h, a. and bย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย  (2mks)

(c) PRย  =ย  K PQ Where K is a scalar, write OR in termsย  of k, a and bย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (1mk)

(d) Calculate the value of k and hย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย (3mks)

 

  1. A transformation P = and maps A(1,3) B(4,1) and C(3,3) onto A1B1C1. Find the

 

 

co-ordinates of A1B1C1 and plot ABC and A1B1C1 on the given grid.

Transformation Q maps A1B1C1ย  onto A11 (-6,2) B11(-2,3) and C11(-6,6). Find the matrix Q and plot

A11B11C11on the same grid. Describe Q fully.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)

 

  1. By use of a ruler and pair of compasses only, construct triangle ABC in which AB = 6cm,

BC = 3.5cm and AC = 4.5cm. Escribe circleย  centre 0 on BC to touch AB and

AC produced at P and Q respectively. Calculate the area of the circle.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย  ย (8mks)

  1. The following were marks scored by 40 students in an examination

330ย ย ย ย ย ย  334ย ย ย ย ย  354ย ย ย ย  348ย ย ย ย  337ย ย ย ย  349ย ย ย ย  343ย ย ย  335ย ย ย  344ย ย ย  355

392ย ย ย ย ย ย  341ย ย ย ย ย  358ย ย ย  ย 375ย ย ย ย  353ย ย ย ย  369ย ย ย ย  353ย ย ย  355ย ย ย  352ย ย ย  362

340ย ย ย ย ย ย  384ย ย ย ย ย  316ย ย ย ย  386ย ย ย ย  361ย ย ย ย  323ย ย ย ย  362ย ย ย  350ย ย ย  390ย ย ย  334

338ย ย ย ย ย ย  355ย ย ย ย ย  326ย ย ย ย  379ย ย ย ย  349ย ย ย ย  328ย ย ย ย  347ย ย ย  321ย ย ย  354ย ย ย  367

 

(i) Make a frequency table with intervals of 10 with the lowest class starting at 31ย ย  ย ย ย ย ย  ย (2mks)

(ii) State the modal and median classย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  ย (2mks)

(iii) Calculate the mean mark using an assumed mean of 355.5ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย (4mks)

 

 

MATHEMATICS IV

PART 1

MARKING SCHEME

 

1.  

ร– –ย  7.939

12.3

 

=ย ย ย  ย ย Noย ย ย ย ย ย ย ย ย ย ย ย  log

7.939ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  0.8998

12.3 ย ย ย ย ย ย ย ย ย ย ย ย ย 1.0899

T.8099ย ย  1/3 = 3 + 2.8099ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  T.9363ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3

 

=ย  -0.8635

B1

 

 

 

 

B

 

M1

 

A1

4

 

ย Subtraction

 

 

 

 

Logs

 

Divide by 3

 

Ans

2. 5x โ€“ 3 (3x โ€“7 )ย ย ย  =ย  3(x โ€“ 2 )

5x โ€“ 9x + 21ย ย ย  =ย ย  3x โ€“ 6

-7xย ย ย ย ย ย ย ย ย ย ย ย  = -27

xย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  36/7

 

M1

M1

 

A1

3

Multiplication

Removal ( )

 

Ans

3. 3x +5y + x =ย  180

9xย ย  =ย  180

xย ย ย  =ย ย  20

yย ย  =ย ย ย  60

M1

A1

B1

3

Eqn

X

B

 

 

4.  

.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  rย ย  =ย ย ย ย ย ย  3vย ย ย ย ย  1/3

4P

 

.ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย rย ย  =ย ย ย ย ย ย  ย Sย  ย ย ย ย ย ยฝ

4P

 

\ 3Vย ย ย ย ย  1/3ย ย  ย ย ย ย ย ย ย ย ย ย ย =ย ย ย ย ย ย ย ย  Sย  ย ย ย ยฝ

4Pย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4P

 

3V ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย =ย ย ย ย ย ย  Sย ย ย ย ย ย  3/2

4Pย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 4P

 

Vย ย ย ย ย ย ย ย ย ย ย ย  =ย ย ย ย ย ย  4P ย ย ย ย ย S ย ย ย ย 3/2

3ย ย ย ย ย ย ย ย ย ย ย  4P

 

 

 

B1

 

 

 

 

 

 

 

M1

 

 

 

A1

3

 

 

 

Value r

 

 

 

 

 

 

 

Equation

 

 

 

Expression

5.

 

 

 

 

6.

Gradย  lineย ย ย ย ย ย ย ย ย  = ยผ

y โ€“ 2ย ย ย ย ย ย ย  = ยผ

x โ€“ 5

yย ย ย ย ย ย ย ย ย  ย ย =ย  ยผ x + ยพ

kย ย ย ย ย ย ย ย ย ย ย ย  =ย ย ย ย ย  ยพ

P in Alloyย ย ย ย ย ย ย ย  = 4/10ย  x 800

= 320g

=ย  100 x 320

20

=ย  3.2 kg

 

M 1

 

A1

A 1

3

 

B1

 

M1

 

A 1

 

Equation

 

Equation

K

 

 

P in alloy

 

Expression

 

Ans

 

 

 

 

7.

 

 

 

 

B (a,b) ,ย ย ย ย ย ย ย ย ย ย ย  C (x ,y)

.a โ€“ 2ย ย ย ย ย ย ย ย ย  =ย ย ย  5

.b โ€“ 8ย ย ย ย ย ย ย ย ย ย ย ย ย ย  -2

.aย  = 8ย ย ย ย  b = 6ย ย ย ย ย  B(8, 6 )

x โ€“ 8ย ย ย ย ย ย ย ย ย  =ย ย  3

y โ€“ 6ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4

x = 11,ย  y = 10 c(11,10)

 

 

 

 

 

B1

 

M1

 

 

A1

3

 

 

 

 

B conduct

 

Formular

 

 

C

8.  

 

 

 

 

 

80 – x

 

 

 

 

 

.h = x tan 70

h = (80 โ€“ x ) tan 60

\ย ย  x tan 70 = 80 tan 60-x tan 60

2.7475x + 1.732x = 138.6

4.4796 xย ย ย ย ย ย  =ย ย  138.6

.hย ย ย ย  =ย ย ย  138.6 x tan 60

4.4796

 

= 53.59

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

M1

 

 

 

M1

 

A1

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Expression forย  h both

Equation

 

 

 

Expression for h

 

Ans

9. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2prย ย ย  =ย  90ย  x 2p x 50

360

rย ย ย  =ย  12.5

hย ย ย ย  =ย  ร–2500 –ย  156.25

=ย ย  ร–2343.75

=ย ย  48.41 cm

 

M1

P

A1

M1

 

A1

4

Equation

 

.r

expression for h

 

ans

 

 

10.

 

100 nย ย ย ย ย  =ย ย  302.323

ย ย ย ย ย nย ย ย ย ย  =ย ย ย ย ย  3.023ย ย ย 

99nย ย ย ย ย ย  =ย ย  299.3

nย ย ย ย ย  =ย ย ย  2993

990

=ย ย ย  323/990

 

M1

 

 

A1

4

 

 

Equation

 

 

Ans

 

11. ABย ย ย ย ย ย ย  =ย ย ย ย  3-1

5-9

=ย ย ย ย  2

-4

BCย ย ย ย ย ย ย ย  =ย ย ย ย  4

-8

ABย ย ย ย ย ย ย ย  = ยฝย ย  BC

\ AB // BC

But B is common

\ A,B,C are collinear.

 

 

 

 

B1

 

 

 

 

 

B1

 

 

B1

3

 

 

A B &ย  BC

 

 

 

 

 

Both

 

 

Both

 

12. ย ย ย ย ย  4% of 200,000ย  = 8000/=

balanceย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 4200/=

6% ofย  xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 4200/=

xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 4200 x 100

6

=ย  70,000

salesย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  sh. 270,000

B1

 

 

M1

A1

B1

4

 

 

Both

 

 

Expression

Extra sales

Ans

 

 

 

 

 

 

13 .

 

 

 

 

 

Timeย ย ย ย ย ย ย ย ย  =ย ย  22/7 x 3.5/2x 3.5/2 x 200ย ย  hrs

22/7x 140x140x 140x 3600

 

=ย  8960

3600

= 2 hrs 29min

 

 

 

 

 

M1

M1

 

M1

 

A1

4

 

 

 

 

 

 

Vol tank

Vol tank

 

Div x 3600

 

Tank

 

 

14.

 

 

 

 

 

 

 

ย ย  ย ร–3ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย =ย ย ย  ย ร–3ย ย ย ย ย ย ย ย ย ย  ร–7 + ร–2

ร–7ร–2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ร–2ร–2ย ย ย ย ย ย ย ย  ร–7+ ร–2

 

= ร–3 ร–7 + ร–2

5

 

= ร–21 + ร–6

5

M1

 

M1

 

A1

3

Multi

 

Expression

 

 

 

Ans

15. ย ย ย ย ย ย ย ย ย  3 ยฃ x 2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  x2 ยฃ 35

ยฑ1.732 ยฃxย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  x ยฃ ยฑ 5.916

1.732 ยฃ xย ย ย ย ย ย ย ย ย ย  ยฃ 5.916

integral x : 2, 3, 4, 5

 

B1

B1

B1

B1

4

Lower limit

Upper limit

Range

Integral values

 

16. ย No of daysย ย  =ย  8/6 x 5/8ย  x 12

=ย ย  10 days

M1

A1

2

Expression

days

17. (i)ย  รCEDย ย ย ย ย  =ย  รECDย ย  = 30

ร CDEย ย ย ย  =ย  180 โ€“ 60

=ย  120

ร CBEย ย ย  =ย  180-120

=60

(ii) ร AECย  = 90+30

= 120

ร EABย  = 180-(120+45)

= 150

(iii) รBEOย  = 90-45

= 45

B1

B1

B1

B1

 

B1

 

B1

B1

 

B1

8

 

 

 

 

 

 

 

รA EB = 450

 

รBEO

18. ย  .ar + ar2ย ย ย  =ย  9/4

3r + 3r2ย ย  =ย  9/4

12r2ย  + 12r โ€“ 9 = 0

4r2 ย + 3r โ€“ 3ย ย  = 0

4r2 + 6r โ€“ 2r โ€“3 = 0

(2r – 1) (2r + 3)ย  = 0

rย  = ยฝย  or rย ย  = -11/2

 

Ssย ย ย ย ย  = 3(1- (1/2 )5)

1 โ€“ ยฝ

 

= 3 (1-12/3 2)

ยฝ

= 6 ( 31/32)

= 6 31/32

 

B1

B1

 

B1

 

M1

A1

 

M1

 

 

 

M1

 

 

A1

8

 

 
19.

LOGย  E.ย ย ย  0.3010ย ย  0.6021ย ย ย ย  0.7782ย ย ย ย  0.9031

LOGย  Fย ย ย ย ย  1.2068ย ย  2.1065ย ย ย ย  2.6354ย ย ย ย  3.0103

 

Log E =n log Fย  + Log K

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.nย  = gradientย ย ย  = 2ย ย ย ย ย ย ย  2.4 โ€“ 1.4ย ย  =ย  12 ย =ย  3

Log k.ย ย ย ย ย ย ย ย ย ย ย ย  =ย  0.3ย ย ย ย ย ย  0.7 โ€“ 0.3ย ย ย ย ย ย  4

.kย ย ย ย ย ย ย ย ย ย ย ย ย  = 1.995

ยพ 2

ย‹ย ย ย ย ย ย ย ย  Eย ย ย ย  =ย  2F 3

B1

B1

 

 

S1

 

 

P1

 

 

L1

 

 

M1

A1

 

B1

8

 

Log E

Log F

 

 

Scale

 

 

Plotting

 

 

Line

 

 

Gradient

 

 

K

 

 

 

20 ย 

.xย ย ย ย ย ย  -2ย ย ย ย  -1ย ย ย ย  0ย ย ย  1ย ย ย ย  2ย ย ย  3ย ย ย ย ย  4

.yย ย ย ย ย  17ย ย ย ย ย  6ย ย ย ย ย  1ย ย ย  6ย ย ย ย  9ย  22ย ย ย ย  41

 

.yย  =ย  3x 2ย  – 2x + 1ย ย ย  –

0ย ย ย ย ย ย  =ย  3x 2 โ€“ 3x โ€“ 2

yย ย  =ย  xย ย ย ย  +ย  3

 

 

 

 

 

 

 

B2

 

B1

 

B1

 

S1

P1

C1

 

L1

 

B1

 

8

 

 

 

All values

 

At leastย  5

 

Line

 

Scale

Plotting

Smooth curve

 

Line drawn

 

Value of r

 

 

21. .hย ย ย ย ย ย ย ย ย  = ยพ p x 18 x 18x 18

p x 0.04 x 0.04

= 24 x 18x 18x 18

0.04ย ย  x 0.04 x 100

 

=ย  48,600m

 

density ย = 4/3 x 22/7 x 18 x 18x 18x 15 kg

1000

= 122.2kg

M1

M1

M1

M1

 

A1

 

M1

M1

A1

8

N of wire

ยธ to length in cm

ยธ for length

conversing to metres

 

length

 

expression for density

conversion to kg

ans

 

 

22.  

H = ร–152 โ€“ 92

= ร–144

= 12

 

X/6ย  = 9/12

Xย ย ย  = 4.5

Volumeย ย  = 1/3 x 22/7x (81 x 12 โ€“20.25×6 )

 

= 22/21ย  (972 โ€“ 121 -5)

 

=ย ย  891ย  cm3

 

 

M1

 

 

A1

 

M1

A1

M1

M1

M1

 

A1

8

Method

 

 

 

 

Method

Radius

Small vd

Large vol

Subtraction of vol.

 

Ans

23. R(-a , b) , Q (c,d), S(x , y) ,P (5,0)

PR isย  diagonal

(a)ย ย ย  Mid pointย  PRย  (0,0)

a + 5ย ย ย  = 0

2

.aย ย ย ย ย ย ย ย  =ย ย  -5

b- 0ย ย ย ย  =ย ย  0

2
b = 0

R (-5,0)

Gradย  PQย ย  = -2

Grad RSย ย  = -2

.d โ€“ 0ย ย  =ย  -2

c โ€“5

.d โ€“ 0 ย ย ย ย ย = ยฝ

c+5

.d+ 2cย ย ย ย  = 10

2d โ€“ cย ย ย ย  = 5×2ย ย ย  ย ย ย ย ย –

4d โ€“ 2cย ย  = 10

5dย ย ย ย ย ย ย ย  = 20

dย ย ย ย ย ย ย ย  = 4

cย ย ย ย ย ย ย ย  = 3

Q (3, 4)

x + 3ย  ,ย ย ย  y+4ย ย  =ย  (0,0)

2ย ย ย ย ย ย ย ย ย ย  2

xย  =ย  -3 , y = -4ย ย  \ s(-3 -4)

 

(b) y โ€“ 4ย ย  =ย ย  8

x โ€“ 3ย ย ย ย ย ย ย  6

3yย  = 8x โ€“ 12

 

 

 

 

 

 

 

B1

 

 

 

M1

 

 

 

 

M1

 

 

 

 

A1

 

M1

A1

 

M1

 

A1

8

 

 

 

 

 

 

 

Ans .

 

 

 

Expression both correct

 

 

 

Equation

 

 

 

 

Ans

 

 

 

 

Expression

 

Equation

 

       

MATHEMATICS IV

PART II

MARKING SCHEME

ย 

 

1. ย ย ย ย ย ย ย ย ย ย ย ย ย ย  784 X 27ย ย ย ย ย ย ย  =

187500

ร– 784 x 9ย ย ย ย ย ย ย ย ย ย  =ย ย  ย 4 x 7x 3

62500ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  250

=ย ย ย ย ย ย  42

125

=ย ย ย ย ย ย  0.336

 

 

 

M1

 

M1

 

 

A1

 

 

 

Factors for

Fraction or equivalent

 

C.A.O

    3  
2. ย ย ย ย  Father 3x ,ย  r sonย  = x

2(x +10)ย ย ย ย ย ย ย  = 3x + 10

2x +20ย ย ย ย ย ย  =ย  3x + 10

xย ย ย ย ย ย ย  = 10

fatherย ย ย ย ย ย ย ย ย ย ย  = 30

M1

 

 

A1

B1

 

Expression

 

 

 

 

 

 

    3  
3. 3ย ย  = sinย ย  60

AE

AE ย = 3

Sin 60

= 3.464

perimeterย  = 5×2 + 3.464 x 3

= 10+10.393

= 20.39

M1

 

 

 

A1

 

 

B1

Side of a triangle

 

 

 

 

 

 

Perimeter

    3  
4. ย ย  .a3 โ€“ b-2c2ย  =ย  (-2)3 โ€“ 3 โ€“2(-1)2

2b2 โ€“ 3a2cย ย ย  ย ย 2(3)2 โ€“3(-2)2(-1)

= -8 โ€“3-2

18 + 12

= -13

30

M1

 

 

M1

 

A1

Substitution

 

 

Signs

 

C.A.O

    3  
5. ย ย ย ย ย ย  Kshย  189,000ย ย ย ย ย ย ย ย ย  =ย ย  $ 189,000

75.6

= $ 2500

balanceย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = $ 2500

=ย  Kshs. 189,000

Kshs. 189,000ย ย ย ย ย ย ย ย ย  =ย ย ย ย ย ย ย ย ย ย ย ย  189,000

115.8

Ukย ย ย  โ‚ค1632

M1

 

A1

 

M1

A1

 

A1

4

 

Conversion

 

 

 

Conversion

 

6. Area of 2 trianglesย  =ย ย  2 (ยฝ x 8x 5 sin 60)

=ย ย  40 sin 60

=ย ย  40x 0.8660

= 34.64 cm2

Area of rectangleย ย ย  = 300 x 8 + 300 x 5 +300 x BC

BCย ย ย ย ย ย ย ย ย ย ย ย ย  = ร–64 +25 โ€“ 2 x 40cos 60

= ร–89 โ€“ 80 x 0.5

= ร–89 โ€“ 40

= ร–49

= 7

Totalย ย  S.A.ย ย ย ย ย ย ย ย ย ย ย ย ย  = 300 (8+5+7) + 34.64 cm2

= 6000 + 34.64

= 6034.64 cm2

M1

 

 

 

 

M1

 

 

 

 

M1

 

A1

Areas of D

 

 

 

 

B.C. expression

 

 

 

 

Area

 

    4  
7. ย ย  AF2ย ย ย  = 32+42+-2+12x cos 50

= 25 โ€“ 24 x 0.6428

= 25-15.43

= 9.57

AFย ย ย ย ย  =ย  3.094 x 50

AFย ย ย ย ย  =ย  154.7m

Sin Qย  =ย  200 sin 50o

154.7

= 0.9904

Qย ย  = 82.040

Bearing = 117.960ย 

M1

 

 

 

 

A1

M1

 

 

A1

B1

 

 

 

 

 

 

 

 

 

 

 

Bearing

    5  
8. (i)ย  No. of whiteย  = w

w ย ย ย ย ย ย = 2

w+9ย ย ย ย  ย ย ย ย 3

3wย ย ย ย ย ย  = 2w + 18

wย ย ย ย ย  =ย  18

(ii)ย  p(different colour )ย  = p(WB Nย  BW)

= 2 ย ย xย ย  9ย ย  + 9 xย  18

3ย ย ย ย ย  25ย ย ย ย  27ย ย ย  25

= 12/25

M1

 

 

 

 

A1

M1

 

A1

 
    4  
9. A.sfย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  1

49

smaller areaย ย ย ย ย ย  = 1 x 441 p

49

=ย  9p

pr2ย ย ย ย ย ย ย ย ย  = 9p

r2ย ย ย ย ย ย ย ย  =ย  9

rย ย ย ย ย ย ย ย ย ย  = 3

 

 

 

M1

 

M1

 

 

A1

 

 
    3  
10. ย Largest areaย ย ย ย ย ย ย ย  = 22 x (14.5)2

7

=ย  660.8 cm 2

smallest areaย ย ย ย ย ย ย ย ย  =ย  22/7 x (13.5)2

= 572.8

572.8ย ย ย  ยฃ Aย  ยฃ 660.81

M1

 

 

M1

 

A1

 
    3  
11. (1 +2 x)5ย  =ย  1 + 5 (2x) + 10 (2x)2 + 10 (2x)3

=ย  1 + 10xย ย  + 40x2ย  + 80x3

2.0455ย ย ย  =ย ย  1+2 (0.52)5

= 1+10 (0.52)+ 40(0.52)2+80(0.52)3

= 1+5.2 + 10.82 + 11.25

= 28.27

M1

A1

 

M1

 

A1

 
    4  
12. ย ย ย ย ย ย ย ย  Tnย ย ย ย ย ย ย ย ย ย  =ย  5x 2n โ€“2

(i)ย ย ย ย ย ย ย ย ย ย ย ย ย ย  T1 , T2, T3 = 2.5, 5, 10

(ii)ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย S5ย ย ย ย ย  =ย  2.5(25-1)

2-1

= 2.5 (31)

= 77.5

 

B1

M1

 

 

A1

 

All terms

 

    3  
13. 12ย ย ย ย ย ย ย ย  = 22 x 3

18ย ย ย ย ย ย ย ย  = 2 x 32

30ย ย ย ย ย ย ย ย  = 2x3x5

Lcmย ย ย ย ย ย ย ย  = 22 x 32x 5 = 180 min

=ย  3hrs

time they ring together =11.55 +3 = 2.55 p.m

M1

 

 

 

A1

B1

 
    3  
14. ย Map areaย ย ย ย ย  = 40cm 2

Actual areaย ย  =ย  200x200x40m2

= 200x200x40ha

100×100

= 320ha

M1

M1

 

 

A1

Area in m2

Area in ha

 

 

CAO

    3  
15. ย ย ย  3p + 2rย ย ย  = 13

p + 2rย ย ย  =ย ย  9ย  –

2pย ย ย ย ย ย ย ย ย ย  =ย ย  4

pย ย ย ย  = sh 2

rย ย ย ย  = 3.50

M1

 

 

A1

B1

 
    3  
16. 110 + 100+130+2x +3x = 540

5xย  = 200

xย  = 400

2x , 3xย ย ย ย  = 80 and 1200 res

M1

 

A A1

2

 
17. Contribution / personย ย ย  = 180,000

X

New contributionย ย ย  = 180,000

x โ€“ 2

180,000ย ย  – 180,000ย  = 24,000

x โ€“2ย ย ย ย ย ย ย ย ย ย ย ย ย ย  x

180,000x โ€“ 180,000x +360,000 = 24,000(x-2)x

24,000x2 ย –ย  48,000x โ€“ 360,000 =0

x2 ย – 2x โ€“ 15 = 0

x2 โ€“ 5x + 3x โ€“ 15 = 0

x (x โ€“ 5)+ 3 (x โ€“ 5) = 0

(x + 3 )(x – 5)ย  = 0

xย ย ย ย  = -3

orย ย ย ย  = 5

remaining membersย ย ย ย ย ย ย ย ย ย ย  = 5-2

= 3

B1

 

B1

 

M1

M1

 

 

A1

M1

 

 

A1

 

B1

 

โ€˜Cโ€™

 

 

 

eqn

mult

 

 

eqn

factor

 

 

both ans

 

remaining members

    8  
18. (a) P (3 white)ย ย ย ย ย ย ย ย  =ย  8ย ย  xย  7ย  xย ย  6ย  =ย  28

13ย ย ย ย ย  12ย ย ย ย  11ย ย ย  143

(b) P(at least 2 blue)=p(WBBorBBWorBWB)orBBB

= 8ย  xย ย  5ย  xย ย  4 ย ย +ย  5ย  xย ย  4ย  xย  8

13ย ย ย ย  12ย ย ย ย  11ย ย ย ย ย  13ย ย ย ย  12ย ย ย  11

+ 5 ย xย ย  8 ย xย ย  4 +ย ย  8 xย ย  7 xย ย  6

13ย ย ย ย  12ย ย ย ย  11ย ย ย  13ย ย ย ย  12ย ย ย  11

= 204

429

= 68

143

(c) p(2 white and one blue )= p(WWB or WBW or BWW)

= 8 ย xย  7ย  xย  5 ย +ย  8ย  xย  5 xย  7ย  +ย  5 xย  8 xย  7

13ย ย ย ย  12ย ย ย  11ย ย  13ย ย ย ย  12ย ย  11ย ย  13ย ย ย  12ย ย  11

= 3 x 8 x 7 x 5

13 x 12 x 11

 

= ย 70

143

M1

A1

 

 

M1

 

M1

 

 

 

A1

 

 

 

M1

M1

 

 

 

A1

 

 
    8  
19. (a) recourt areaย ย ย  =ย  10.5 x 6ย  m2

titleย  areaย ย ย ย ย ย  =ย ย ย  0.3 x 0.3 m2

No of tilesย ย ย ย  =ย ย ย  10.5 x 6

0.3 x 0.3

=ย  700

(b) No of cartons = 700 x 15

20

= 52.5

 

(c) Cost of 525 cartonsย  =ย ย  525 x 100 + 800 x 525

+ transportย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย 5

=ย  10,500+420,000

=ย ย  430,500

sale priceย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  120 x 4.30,500

100

=ย  shย ย ย  516,600

s.p of a cartonย ย ย ย ย ย ย ย ย ย ย  =ย  516,600

525

= sh. 984

 

 

M1

A1

 

M1

 

A1

 

 

 

B1

 

M1

 

 

M1

 

A1

 

 

 
    8  
20. (a) Maina`s tax duesย ย ย ย ย ย  = 1800 x 10

100

=ย ย ย ย ย ย ย  180

(b) Taxable incomeย ย ย ย ย ย ย  = 3600 x 115 โ€“ n rent

100

= 36 x 115 โ€“ 100 x 12

20

= 4140 โ€“ 60

=ย ย ย ย ย ย ย ย  4080

Tax duesย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 10ย  ย ย x 2100ย  + 15ย  x 1980

100ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  100

= 210 + 297

=ย ย ย ย ย ย ย  507

Taxย  reliefย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย ย ย ย ย ย ย  270-

Taxย  paidย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย ย ย ย ย ย ย  237

M1

 

A1

 

 

M1

 

 

A1

M1

M1

 

A1

 

B1

 

 

 

 

 

 

 

 

 

 

1st slab

2nd slab

    8  
21. ย (a)ย ย ย ย ย ย ย ย ย ย ย  PQย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  –3/5 aย ย  +ย  3/1b

= ย 31/2 โ€“ 3/5 a

(b)ย ย ย ย ย ย ย ย ย ย ย ย  ORย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย ย  h a + h b

=ย ย  a โ€“ ha + hb

=ย  (1-h) a + h b

(c)ย ย ย ย ย ย ย ย ย ย ย ย ย  ORย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  =ย  3/5 aย ย  + k (31/2 b โ€“ 3/5a)

=ย  (3/5 โ€“ 3/5k)a +3k b

(d)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  1 โ€“ hย ย ย ย  =ย  3/5 โ€“ 3/5kย ย ย  (i)

3kย ย ย  =ย  hย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (ii)

Sub (i)ย ย ย ย ย ย ย ย ย ย ย ย ย  1 โ€“ 3kย ย ย  =ย  3/5 โ€“ 3/5k

5- 15kย ย ย  = ย 3-3k

12kย ย ย  =ย  2

kย ย ย  =ย ย  1/6

hย ย ย ย  =ย  ยฝ

 

 

B1

 

M1

A1

M1

A1

 

 

M1

 

 

A1

B1

 
     

8

 
 

22.

 

P(ABC) =ย ย ย ย  0ย  – 1ย ย ย ย ย  1ย  4ย  3ย ย ย ย ย  =ย  -3 ย -1ย  -3

1ย ย ย  0ย ย ย ย ย  3ย  1ย  3ย ย ย ย ย ย ย ย ย ย ย  1ย ย  4ย ย  3

A1 (-3,1)B1 (-1,4)C1(-3,3)

Q(A1B1C1) =ย  aย  bย ย ย  -3 โ€“1 -3ย ย ย  =ย ย ย ย ย ย ย  -6 โ€“2 โ€“6

c dย ย ย ย ย ย  1ย ย  4ย  3ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2ย ย  8ย ย  6

 

=> -3a + b =ย  -6ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย -3c + d = 2

-a + 4bย ย  =ย  -2 x 3ย ย ย ย ย ย ย ย  -c + 4d = 8 x 3

– 3aย  + 12b = -6ย ย ย ย ย ย ย ย ย ย ย ย ย  – 3c + 12d = 24

11bย  = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  -11dย  = -22

b = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  d = 2

a = 2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  d = 2

cย  = 0

Q =ย ย ย  2ย ย ย ย  0

0ย ย ย ย ย ย  2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

A1

 

 

M1

 

 

M1

 

 

 

 

 

A1

 

 

 

B1

 

 

 

 

B1

 

 

 

B1

 

 

A1 B1 C1

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

L Q

 

 

 

A1 B1 C1 drawn

 

 

 

 

All BII CII

Ploted

 

 

 

 

Destruction

 

 

 

    8  
23.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

24.

Rย ย ย ย  = 2.2CM ยฑ 0.1

Area = 22 xย  2.2 x 2-2

7

= 15.21cm2

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

ย 

 

 

Ef =40ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  efd = -80

(ii) model classย ย ย  = 351- 360

modern classย  = 341 โ€“ 350

(iii) meanย ย ย ย ย ย ย ย ย ย ย ย  = 355.5ย  – 80

40

=ย  355.5 โ€“ 2

= ย 353.5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

 

B1

 

B1

 

B1

 

B1

 

B1

 

M1

 

 

 

1

1

 

 

8

 

B1

B1

 

M1

 

 

A1

 

B1

 

B1

B1

B1

 

 
    A1  
    8  

 

ย 

ย 

MATHEMATICS V

PART I

ย 

SECTION 1 (52 MARKS)

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  1. Use logarithms to evaluate 6 Cos 40ย ย  0.25
    63.4ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
  2. Solve for x in the equation (x + 3) 2 โ€“ 5 (x + 3) = 0 (2mks)
  3. In the triangle ABC, AB = C cm. AC = bcm. รBAD = 30o and รACD = 25o. Express BC in terms of b and c.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  4. Find the equation of the normal to the curve y = 5 + 3x โ€“ x3 when x = 2 in the form
    ay + bx = cย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
  5. Quantity P is partly constant and partly varies inversely as the square of q. q= 10 and p = 5 ยฝย  when q =20. Write down the law relating p and q hence find p when qs is 5.ย ย ย ย ย ย ย ย ย ย ย  (4mks)
  6. Solve the simultaneous equation below in the domain 0ย  ยฃ x ยฃย  360 and Oยฃย  y ยฃ 360
    2 Sin x + Cos y = 3
    3 Sin x โ€“ 2 Cos y = 1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
  7. Express as single factor 2ย ย ย ย  –ย ย ย ย  x + 2ย ย ย ย ย ย ย ย  +ย ย ย ย ย ย  1
    x + 2ย ย ย  x2 + 3x + 2ย ย ย ย ย ย ย ย  x + 1ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  8. By use of binomial theorem, expand (2 โ€“ ยฝ x )5 up to the third term, hence evaluate (1.96)5
    correct to 4 sf.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
  9. Points A(1,4) and B (3,0) form the diameter of a circle. Determine the equation of the circle and write it in the form ay2 + bx2 + cy + dy = p where a, b, c, d and p are constants.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
  10. The third term of a GP is 2 and the sixth term is 16. Find the sum of the first 5 terms of the GP. (4mks)
  11. Make T the subject of the formulae 1ย ย ย ย ย ย  –ย  3mย ย  +ย  2
    T2ย ย ย ย ย ย ย ย  Rย ย ย ย ย ย ย ย  Nย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  12. Vectors, a =ย ย  2ย ย ย ย  b =ย ย  2ย ย  andย ย  c โ€“ย ย  6
    2ย ย ย ย ย ย ย ย ย ย ย ย ย  0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  4
  13. By expressing a in terms of b and c show that the three vectors are linearly dependent.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
    A cylindrical tank of base radius 2.1 m and height is a quarter full. Water starts flowing into this tank at 8.30 a.m at the rate of 0.5 litres per second. When will the tank fill up? (3mks)
  14. A piece of wood of volume 90cm3 weighs 54g. Calculate the mass in kilograms of 1.2 m3 of the wood. ย ย ย ย  (2mks)
  15. The value of a plot is now Sh 200,000. It has been appreciating at 10% p.a. Find its value 4 years ago.
    (3mks)
  16. 12 men working 8 hours a day take 10 days to pack 25 cartons. For how many hours should 8 men be

working in a day to pack 20 cartons in 18 days?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

SECTION II (48MARKS)

  1. The tax slab given below was applicable in Kenya in 1990.
    Income in p.a.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  rate in sh
    1ย  โ€“ 1980ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2
    1981 โ€“ 3960ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  3
    3961 โ€“ 5940ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  5
    5941 โ€“ 7920ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  7
    Maina earns Sh. 8100 per month and a house allowance of Sh. 2400. He is entitled to a tax relief of Sh.

800 p.m. He pays service charge of Sh 150 and contributes Sh 730 to welfare. Calculate Mwangis net

salary per month.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)

  1. OAB is a triangle with OA = a , OB = b. R is a point of AB. 2AR = RB. P is on OB such that
    3OP = 2PB. OR and AP intersect at Y, OY = m OR and AY = nAP. Where m and n are scalars.ย ย ย  Express in terms of a and b.
    (i) ORย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)
    (ii)APย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)

    (b) Find the ratio in whichย  Y divides APย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (6mks)

  2. The table below gives related values of x and y for the equation y = axn where a and n are constants
X 0.5 1 2 3   10
Y 2 8 32   200 800

By plotting a suitable straight line graph on the graph provided, determine the values of a and n.

20.ย ย ย ย ย ย  Chalk box x has 2 red and 3 blue chalk pieces. Box Y has same number of red and blue

pieces. A teacher picks 2 pieces from each box. What is the probability that
(a)ย ย ย ย ย ย ย  They are ofย  the same colour.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
(b)ย ย ย ย ย ย ย  At least one is blueย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(c)ย ย ย ย ย ย ย  At most 2 are redย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

21.ย  Point P(50oN, 10oW) are on the earthโ€™s surface. A plane flies from P due east on a parallel of

latitude for 6 hours at 300 knots to port Q.
(a) Determine the position of Q to the nearest degree.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(b)ย  If the time at Q when the plane lands is 11.20am what time is it in P.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(c) The plane leaves Q at the same speed and flies due north for 9 hours along a longitude to

airport R. Determine the position of R.ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย (3mks)
22.ย ย ย ย ย ย  Using a ruler a pair of compasses only, construct :
(a) ย ย ย ย ย ย  Triangle ABC in which AB = 6cm, AC = 4cm and ร ABC = 37.5o.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(b)ย ย ย ย ย ย ย  Construct a circle which passes through C and has line AB as tangent to the circle at A. ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(c)ย ย ย ย ย ย ย  One side of AB opposite to C, construct the locus of point P such thatย  รAPB = 90o.ย  ย ย ย ย ย ย ย ย ย ย ย  (2mks)
23. ย ย ย ย ย  A particle moves in a straight line and its distance is given by S = 10t2 โ€“ t3 + 8t where S is

distance in metres at time t in seconds.
Calculate:
(i) Maximum velocity of the motion.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
(ii) The acceleration when t = 3 sec.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(iii) The time when acceleration is zero.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (2mks)

 

 

 

  1. A rectangle ABCD has vertices A(1,1) B(3,1), C(3,2) and D(1,2). Under transformation

matrix M =ย ย  2ย  2ย ย  ABCD is mapped onto A1B1C1D1

1ย ย  3
under transformation M =ย ย  -1ย  0ย ย ย  A1B1C1D1 is mapped ontoย  A11B11C11D11. Draw on the given grid
0 โ€“2

(a)ย ย ย ย ย ย  ABCD, A1B1C1D1 and A11B11C11D11ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)
(b)ย ย ย ย ย ย ย  If area of ABCD is 8 square units, find area of A11B11C11D11.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(c)ย ย ย ย ย ย ย  What single transformation matrix maps A11B11C11D11 onto A1B1C1D1ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)

MATHEMATICS V

PART II

ย 

SECTION 1 (52 Marks)

 

  1. Evaluate without using mathematical tables (2.744 x 15 5/8)1/3 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  2. If 4 ยฃ x ยฃ 10 and 6 ยฃ y ยฃ5, calculate the difference between highest and least
    (i) xyย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย ย ย ย ย ย ย  (2mks)
    (ii)ย  y/xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
  3. A 0.21 m pendulum bob swings in such a way that it is 4cm higher at the top of the swing than at the bottom. Find the length of the arc it forms. ย ย ย ย ย  (4mks)
  4. Matrix 1ย ย ย ย ย ย ย  2xย ย  has on inverse, determine xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
    x +3ย ย ย ย ย  x2
  5. The school globe has radius of 28cm. An insect crawls along a latitude towards the east from A(50o, 155oE) to a point B 8cm away. Determine the position of B to the nearest degree. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย ย ย ย ย  (4mks)
  6. The diagonals of triangle ABCD intersect at M. AM = BM and CM = DM. Prove that triangles ABM and CDM are Similar. ย ย ย ย ย  (3mks)
  7. Given that tan x = 5/12, find the value of 1ย  –ย ย  sinx
    ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Sin x + 2Cos x,ย  ย for 0 ยฃ x ยฃ 90ย ย ย ย ย ย ย ย ย ย  (3mks)

 

  1. Estimate by MID ORDINATE rule the area bounded by the curve y = x2 + 2, the x axis and the lines x = O and x = 5 taking intervals of 1 unit in the x. (3mks)
  2. MTX is tangent to the circle at T. AT is parallel to BC. ร MTC = 55o and ร XTA = 62o. Calculate ร (3mks)
  3. Clothing index for the years 1994 to 1998 is given below.
Year 1994 1995 1996 1997 1998
Index 125 150 175 185 200

Calculate clothing index using 1995 as base year.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)

  1. A2 digit number is such that the tens digit exceeds the unit by two . If the digits are reversed, the number formed is smaller than the original by 18. Find the original number. (4mks)
  2. Without using logarithm tables, evaluate log5 (2x-1) โ€“2 + log5 4 = log5 20ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
  3. Mumiaโ€™s sugar costs Sh 52 per kg while imported sugar costs Sh. 40 per kg. In what ratio should I mix the sugar, so that a kilogram sold at Sh. 49.50 gives a profit of 10%. (4mks)
  4. The interior angles of a regular polygon are each 172o. Find the number of sides y lie polygon. ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
  5. Evaluate 2xย ย  =ย ย ย ย ย ย  2ย ย ย  +ย ย ย ย ย ย ย  3
    341ย ย ย ย ย ย  9.222ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
  6. A water current of 20 knots is flowing towards 060o. A ship captain from port A intends to go to port

Bย ย  at a final speed of 40 knots. If to achieve his own aim, he has to steer his ship at a course of 350o.

Find the bearing of A from B.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

SECTION IIย  (48 MARKS)

  1. 3 taps, A, B and C can each fill a tank in 50 hrs, 25 hours and 20 hours respectively. The three taps are turned on at 7.30 a.m when the tank is empty for 6 hrs then C is turned off. Tap A is turned off after four hours and 10 minutes, later. When will tap B fill the tank? (8mks)
  2. In the domain โ€“5 ยฃ x ยฃ 4, draw the graph of y = x2 + x โ€“ 8. On the same axis, draw the graph of y + 2x = -2. Write down the values of x where the two graphs intersect. Write down an equation in x whose roots are the points of intersection of the above graphs. Use your graph to solve. 2x2 + 3x โ€“ 6 = 0.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (8mks)
  3. The average weight of school girls was tabulated as below:
Weight in Kg 30 โ€“ 34 35 โ€“ 39 40 โ€“ 44 45 โ€“ 49 50 โ€“ 54 55-59 60-64
No. of Girls 4 10 8 11 8 6 3

(a) State the modal class.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)
(b) Using an assumed mean of 47,
(i) Estimate the mean weightย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(ii) Calculate the standard deviation.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (4mks)

 

 

 

 

 

 

 

 

 

 

  1. The table below shows values of y = a Cos (x โ€“ 15) and y = b sin (x + 30)
X 0 15 30 45 60 75 90 105 120 135 150
a Cos(x-5) 0.97       0.71 0.5       -0.5 -0.71
b sin(x+3) 1.00       2.00       1.00   0.00

(a) Determine the values of a and bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(b) Complete the tableย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(c) On the same axes draw the graphs of y = across(x โ€“ 15) and y = b sin(x + 30)ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(d) Use your graph to solve ยฝ cos (x โ€“ 15) = sin(x + 30)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (1mk)

21.ย ย ย  The diagram below is a clothing workshop. ร ECJ = 30o AD, BC, HE, GF are vertical

walls. ABHG is horizontal floor. AB = 50m, BH = 20m,ย  AD=3m

 

 

 

(a) Calculate DEย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(b) The angle line BF makes with plane ABHGย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(c) If one person requires minimum 6m3 of air, how many people can fit in the workshopย ย ย ย ย ย ย ย  (3mks)

  1. To transport 100 people and 3500 kg to a wedding a company has type A vehicles which take ย ย ย ย ย ย ย ย  10 people and 200kg each and type B which take 6 people and 300kg each. They must not use more

than 16 vehicles all together.
(a)ย ย ย ย  Write down 3 inequalities in A and B which are the number of vehicles used and plot them

in a graph. ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(b)ย ย ย ย  What is the smallest number of vehicles he could use.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

(c)ย ย ย ย  Hire charge for type A is Sh.1000 while hire for type B is Sh.1200 per vehicle. Find the cheapest

hire charge for the whole functionย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)

A circle centre A has radius 8cm and circle centre B has radius 3cm. The two centres are

12cm apart. A thinย  tight string is tied all round the circles to form interior common tangent. The tangents CD and EF intersect at X.

(a) Calculate AXย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)
(b) Calculate the length of the string which goes all round the circles and forms the tangent.
(6mks)

 

  1. Airport A is 600km away form airport B and on a bearing of 330o. Wind is blowing at a speed of

40km/h from 200o. A pilot navigates his plane at an air speed of 200km/h from B to A.
(a)ย ย ย ย  Calculate the actual speed of the plane.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(b)ย ย ย ย  What course does the pilot take to reach B?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (3mks)
(c)ย ย ย ย  How long does the whole journey take?ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  (2mks)

 

MATHEMATICS V

PART I

MARKING SCHEME

 

1 SOLUTION MKS AWARDING
  Noย ย ย ย ย ย ย ย  Log

13.6ย ย ย ย ย ย ย  1.1335ย ย  +

Cos 40ย ย ย  1.8842

1.0177ย ย  –

63.4ย ย ย ย ย ย  1.8021

1. 2156

(4 + 3.2156) 1/4

1.8039

Antilogย ย ย  0.6366

 

B1

 

M1

 

 

M1

 

A1

 

Log

 

+

 

 

divide by 4

 

C.A.O

    4  
2. (x + 3) (x + 3 โ€“ 5) = 0

(x +3)b (x โ€“ 2) = 0

x = -3 or x = 2

M1

 

A1

 

Factors

 

Both answers

3 BD = C Sin 30ย  = 0.05

CD = b Cos 25

= 0.9063b

ย‹ BC = 0.9063b + 0.5 C

B1

 

B1

B1

 

BD in ratio from

 

CD in ratio form

Addition

    3  
4 ย Dyย  = 3 โ€“ 3x2
dx
x = 2, grad = 1
9
Point (2,3)
y โ€“ 3ย  = 1
x โ€“ 2ย ย ย ย  9

9y โ€“ 27ย  = x โ€“ 2
9y โ€“ xย ย  =ย  25

B1

 

B1

 

M1

 

 

A1

 

Grad equ

 

Grad of normal

 

Eqn

 

 

Eqn

 

    4  
5 ย  700 = 100 + n
2200 = 400 + n

1500 = 300m

m = 5

n = 200

P = 5 + 200
q2
When q = 5 P = 13

M1

 

 

A1

 

 

B1

B2

Equan

 

 

Both ans

 

 

Eqn (law)

Ans (P)

    4  
 

6

 

4 Sin x + 2 cos y = 6

3 Sin x โ€“ 2 Cos y = 1
7 sin xย ย  ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย = 7

Sin xย ย ย ย ย ย ย ย ย ย ย  = 1

Xย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  = 90

Cos yย ย ย ย ย ย ย ย ย  = 1

Yย ย ย ย ย ย ย  = 0o

 

M1

M1

 

 

A1

 

B1

 

Elim

Sub

 

 

 

 

 

7 2(x +1) โ€“ 1(x + 2) + x + 2

(x+2) (x +1)
= 2x +2 โ€“ x โ€“ 2 + x = 2

(x +2) (x + 1)

=ย ย ย ย  2x + 2

(x + 2) ย (x + 1)

=ย ย ย ย  2
x + 2

M1

M1

 

 

 

A1

Use of ccm

Substitution

 

 

 

Ans

8 (-2 โ€“ ยฝ x)5ย  = 25ย  – 5 (2)4 ( ยฝ x) + 10(2)3( ยฝ x)2

=ย  32 โ€“ 40x + 20x2

= 32 โ€“ 4 (0.08) + 20 (0.08)2

= 32 โ€“ 0.32 + 0.128
= 3

M1

A1

 

M1

A1

 

 

 

 

 

  ย  4  
9. Circle centre C = (3 +1,ย ย  0 + 4)

2ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

C( 2, 2)

R =ร– (2 โ€“ 0)2 + (2 โ€“ 3)2

=ร– 5

(y โ€“ 2)2 + (x โ€“ 2)2 = ร–5

y2 + x2 โ€“ 4y โ€“ 4x =ย  8 + ร–5

B1

 

B1

 

M1

 

A1

Centre

 

Radius

 

 

 

 

    4  
10 ย ar2 =2,ย  ar5 = 16

aย  = 2ย  \ 2 r5 = 16

r2ย ย ย ย ย ย  r2

2r3 = 16

r3 = 8

r = 2, a = ยฝ

 

S5= ยฝ (1 โ€“ ( ยฝ )5)

ยฝ

= 1 โ€“ 1/32

= 31/32

M1

 

 

 

 

A1

 

M1

 

 

A1

 

 

 

 

 

Both

 

Sub

 

 

CAO

    4  
11 NR โ€“ 3MT2ย  = 2RT2

T2(2R + 3M) = NR

T2ย ย  =ย ย  NR

2R + 3m

T = ย ! ร–ย  NR
2R + 3m

M1

 

M1

 

A1

X mult

 

72

 

ans

    3  
12 ย 2ย  = mย ย  2ย ย  + nย ย ย  6

2ย ย ย ย ย ย ย ย ย ย ย  0ย ย ย ย ย ย ย ย ย ย  4

2 = 2m + 6n

2 = 0 + 4n

n = ยฝ

m = – ยฝ

\a = – ยฝ b + ยฝ c

\a b c are linearly dep

M1

 

 

 

 

A1

 

B1

 
    3  
13 Volume = 22 x 2.1 x 2.1 x 2 x ยพ m3

7

Time = 11 x 0.3 x 2.1 x 3 x 1,000,000

500 x 3600

= 11.55

= 11.33 hrs

time to fill = 8.03 pm

M1

 

 

M1

 

 

 

A1

 
    3  
14 Mass = 54ย ย  xย  1.2 x 1,000,000

90ย ย ย ย ย ย ย ย ย ย ย ย ย  1000

= 720kg

M1

 

A1

 
    2  
15 V3 = P

P(0.9)3ย ย ย ย  = 200,000

P = 200,000

0.93

= 200,000

0.729

= Sh 274,348

M1

 

M1

 

 

 

A3

 
    3  
16 No of hours = 8 x 12 x10 x 20

8 x 18 x 25

= 19200

3600

= 5hrs, 20 min

M1

 

 

 

A1

 
    2  
17 ย Taxable income = 8100 + 2400

= sh. 10,500

=ย ย  โ‚ค6300

Tax duesย ย ย ย ย  = Sh 1980 x 2 + 1980 x 3 + 1980 x 5 + 3670 x 7

12

= 22320

12

= Sh 1860

net tax = 1860 โ€“ 800 p.m.

= Sh 1060

Total deduction = 1060 + 150 + 730

= 1940

Net salary = 10,500 โ€“ 1940

= Sh 8560 p.m.

B1

 

 

M1

M1

 

A1

 

B1

 

B1

 

M1

A1

Tax inc

 

 

2

2

 

 

 

net tax

 

total dedu.

    8  
18 OR = 2/3 a + 1/3b or (1/3 (2a + b)

AP = 2/5 b โ€“ a

OY = m OR = A + n (2/5b โ€“ a)

2/5m b + ma = (1 โ€“ n)a + 2/5 n b

2/5m = 2/5n
m = n

\m = 1 โ€“ m

2m = 1

mย  = ยฝ = n

ยฝ AP = Ay

AY:AP = 1:1

B1

B1

 

B1

M1

M1

A1

A1

 

 

B1

 

 

 

EXP, OY

Eqn

M = n

Sub

CAO

 

 

Ratio

    8  
 

19

 

 

 

 

Log y = n log x + log a

Log a = 0.9031

A = 8

Grad = 1.75 โ€“ 0.5

0.4 + 0.2

= 1.25
0.6

= 2.08

n = 2

\y = 8x2

x = 3ย  y = 8 x 32ย ย  = 72

y = 200ย ย ย ย ย ย ย ย ย ย  x = 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

B1

 

 

 

B1

 

B1

B1

S1

P1

L1

 

 

 

 

Log x

Log y

 

 

 

A

 

N

Missing x and y

Scale

Points

Line

    8  
 

 

 

20

 

 

 

P (same colour) = P (XRRrr orXBB or YXX or YBB)

= ยฝ (2/5 x ยผ + 3/5 x 2/4)ย  x 2

=ย  2ย  +ย  6
20ย ย ย ย  20

=ย ย ย  8
20

=ย  2/5

(b) P(at least 1B) = 1 โ€“ P(non blue)

= 1 โ€“ P (XRR or YRR)

= 1 โ€“ ยฝ (2/5 x ยผ) x 2

= 1 โ€“ 1/10

= 9/10

(c) P(at most 2 Red) = 1 โ€“ P (BB)

= 1 โ€“ ยฝ (3/5 x 2/4)2

= 1 โ€“ 6/20

= 14/20 or 7/10

 

 

 

M1

M1

 

M1

 

A1

 

 

M1

 

A1

M1

 

 

A1

 

 

 

Any 2

Any 2

 

Fraction

 

 

    8  
21 (a) PQย  = 1800nm

qย ย ย ย  =ย ย ย ย  1800

60 x 0.6428

= 46.67

= 47o

Q (50oN, 37oE)

 

(b) Time diff = 47 x 4
60

= 3.08

Time at P = 9.12am

(c) QR = 2700 nm

x oย ย  = 2700

60

= 45o

R (85oN, 133oW)

M1

 

 

 

A1

 

 

M1

 

A1

 

M1

 

 

A1

B1

 
    8  
 

 

22

   

 

 

 

B1

B1

 

B1

B1

B1

B1

B1

B1

 

 

 

 

 

Bisector of 150

Bisector 75

 

ABย  AC

^ at A

Bisector AC

Circle

ร AB

Locus P with Aย  B excluded

    8  
24 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  A1B1 C1D1

2ย  2ย  1 3 3 1ย ย  =ย  4ย  8 10 6

1ย  3ย  1 1 2 2ย ย ย ย ย ย  4ย  6ย  9ย  7

 

A11 B11 C11ย  D11

-1ย ย  0ย ย ย ย  4ย  8 10ย  6ย ย ย ย ย ย  =ย ย  -4ย  โ€“8ย ย  -10ย ย  -6

0 โ€“2ย ย ย ย  4ย  6ย  9ย ย  7ย ย ย ย ย ย ย ย ย ย ย  -8ย ย  -12ย  -18ย  -14

 

NM =ย ย  -1ย  0ย ย ย ย ย ย ย  2ย  2

0 โ€“2ย ย ย ย ย ย  1ย  3

 

=ย  -2ย  -2

-2ย ย  -6

 

 

(b)ย ย ย ย ย  detย  = Asfย  =ย  12 โ€“ 4ย ย ย  = 8

Area A11 B11 C11 D11ย  = 8 x 8

= 64ย  U2

(c) Single matrix = Inv N
= ยฝย ย ย  -2 โ€“ย  0

0ย ย ย ย ย ย  โ€“1

 

=ย ย ย ย  -1ย ย ย ย  0

0ย ย ย ย ย ย  – ยฝ

 

 

B1

 

 

B1

 

 

 

 

 

 

 

 

 

 

B1

M1

A1

 

 

 

 

B1

 

 

Product

 

 

Product

 

 

 

 

 

 

 

 

 

 

Det

 

 

 

 

 

 

Inverse

    6  
23  

Dsย  = 20tย  – 3t2 + 8 =0

Dtย ย ย ย  3t2 โ€“ 20t โ€“ 8 = 0

T =ย  20 !ย  ร–400 + 4 x 3 x 8

6

t = 7.045 sec

max velย ย ย ย ย ย ย ย ย  = 148.9 โ€“ 140.9 โ€“ 8

= 0.9 m/s


d2 s
ย  = 6t โ€“ 20

dt2

when t = 3ย ย  a = -2m/s2

6t โ€“ 20 = 0

6tย  = 20

t = 3 2/3 sec

 

 

M1

 

A1

M1

A1

M1

 

A1

M1

 

A1

 
    8  
       

 

 

 

 

 

 

 

 

MATHEMATICS V

PART II

MARKING SCHEME

 

No Solution Mks Awarding
1 ย 2744 x 125ย  ย 1/3

1000ย ย ย ย ย ย ย ย ย ย ย  8

 

2744ย  1/3ย  xย ย  53ย ย ย ย  1/3

1000ย ย ย ย ย ย ย ย ย ย ย  23

 

23 x 73ย  1/3 ย xย ย  5

103ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  2

 

2 x 7 xย  5ย ย  = 3.5

10ย ย ย ย ย  2

 

 

 

M1

 

 

 

 

M1

A1

 

 

 

Factor

 

 

 

 

Cube root

 

    3  
2 (i) Highest โ€“ 10 x 7.5 = 75

Lowestย  โ€“ 6 x 4 =ย  24 โ€“

51

(ii) Highest = 7.5 = 1.875

4

Lowest = 6ย ย  = 0.600

10ย ย  1.275

M1

A1

 

M1

 

A1

Highest

 

 

Fraction

 

 

    4  
3 Cos qย  =ย  17ย  = 0.8095

21

 

q = Cos 0.8095

= 36.03o

 

Arc length = 72. 06 x 2 x 22 x 21

360ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  7

= 26.422cm

M1

 

 

A1

 

 

M1

 

A1

 

 

 

q

    4  
4 ย x2 โ€“ 2x(x +3) = 0

x2 โ€“ 2x2 โ€“ 6x = 0

-x2 โ€“ 6x = 0

either x = 0

orย  x = 6

M1

 

M1

 

A1

Equ

 

Factor

 

Both A

    3  
 

 

5

 

8ย  = xย  x 2 x 22 x 28 Cos 60o

360 ย ย ย ย ย ย ย ย ย ย ย 7

 

8 =ย  xย ย ย  x 44 x 28 x 0.5

360ย ย ย ย ย  ย ย ย 7

x = ย ย 8 x 360 x 7
ย ย ย ย ย ย ย  44 x 28 x 0.5

= 32.73o

= 33o

 

 

M1

 

 

 

 

M1

 

A1

B1

 

 

 

 

 

 

 

x exp

 

 

 

6

 

 

 

รDMC = ร AMB vert. Opp = q

รMABย  = ร MDC = 180 – q BASE Ls of an isosc. <

2
ร MBA = ร MACย ย  180 – q base angles of isos <

2

<โ€™s AMC and < CDM are equiangle

 

\ Similar proved

 

 

 

 

B1

 

 

 

 

 

B1

 

B1

 
    3  
7 Tan x = 5/12

h = ร– b2 + 122

= ร–25 + 144

= ร–169

= 13

 

1 โ€“ Sinx ย ย ย ย ย ย ย ย ย ย ย ย ย ย =ย ย ย ย ย ย  1 โ€“ 5

sin x + 2 Cos xย ย ย ย ย  5/13 + 2 x 12/13

 

12/13ย ย ย ย ย  = 12 x 13ย  =ย  12

29/13ย ย ย ย ย ย ย ย ย  13ย ย  29ย ย ย ย ย  29

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

M1

 

A1

 

 

 

 

 

 

 

 

 

 

 

 

 

Hypo

Sub

 

    3  
8 Y = x 2 + 2

 

 

 

 

 

Area = h (y1, = y2 +โ€ฆโ€ฆ..yn)

= 1(2.225 + 4.25 + 8.25 +14.25 + 22.25)

= 51.25 sq units

 

 

 

B1

 

 

M1

 

A1

 

 

 

Ordinals

    3  
 

9

รCBA = 117o

ร ACD = 55

ร BAC = 180 โ€“ (117 + 55) = 8o

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

B1

B1

 

3

 
10  

 

 

 

B1

B1

B1

B1

1994

1996

1997

1998

    4  
11. Xy = 35

y = 35/x

9x โ€“ 9y = -18

Sub x2 + 2x โ€“ 35 = 0

x2 + 7x โ€“ 5x โ€“ 35 = 0

x (x + 7) โ€“ 5(x + 7) = 0

(x โ€“ 5) (x + 7) = 0

xย  = -7

x = +5

y = 7

Smaller No.

= 57

= 75

B1

 

M1

 

 

 

 

A1

 

 

 

B1

 

 
    3  
12 Log5 (2x โ€“ 1 )4ย  = log552

20

4(2x โ€“ 1)ย  = 52

20

2x โ€“ 1 = 25

5

2x โ€“ 1 = 125

2x = 126

x = 63

M1

 

M1

 

 

 

 

 

A1

 
    3  
13 C.P = 100 x 49.50

110

= 45/-

52x + 40y = 45

x + y

45x + 45yย  = 52x + 40

-7xย  = -54

x/yย  = 5/7

x : y = 5 : 7

 

 

B1

M1

 

 

M1

 

A1

 
    4  
14 ย 

2n โ€“ 4 it angle = 172

n

(2n โ€“ 4) x 90 = 172n

n

90 (2n โ€“ 4) x 90 = 172

n

180 n โ€“ 360 = 172n

 

180n โ€“ 172n = 360

8n = 360

n = 45

 

M1

 

A1

 

M1

 
  ย  2  
15 2 x = 2.ย ย ย  1ย ย ย  +ย ย ย  3.ย ย ย  1

6.341ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย 9.22

2x = 2 x 0. 1578 + 3 x 0.1085

= 0.3154 + 0.3254

= 0.6408

x = 0.3204

 

 

B1

 

 

A1

 

 

Tables

    2  
16 Bearing 140o

Sin q = 20 Sin 110

40

= 0.4698

= 228.02

Bearing of A from B = 198.42

 

M1

 

 

A1

B1

 
    3  
17 Points that each tap fills in one hour

 

A =ย  1ย ย  Bย  = 1 ย ย ย ย ย ย C โ€“ 1
ย ย ย ย ย ย ย ย ย  50ย ย ย ย ย ย ย ย  25ย ย ย ย ย  ย ย ย ย ย ย 20

In one hour all taps can fill = 1ย  +ย  1 ย ย +ย  1ย ย  =ย  11

50ย ย ย  25ย ย ย ย ย  20ย ย ย ย  100

In 6hrs all can fill =ย  11ย  x 6 = 33 parts

100ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย 50

taps A and B can fill =ย  = 1 ย +ย  1ย  = 3 part in 1 hr

50ย ย ย  25ย ย ย  50

In 4 1 hrs, A and B =ย  25 x 3ย  +ย  1

6ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  6ย ย ย ย  50ย ย ย ย  4

Parts remaining for B to fill = 1 โ€“ 33ย  + 1ย  ย = 1ย  – 91ย ย  = 9 parts

50ย ย ย ย ย ย ย  ย 4ย ย ย ย ย ย ย ย ย ย  100ย ย ย  100

Timeย  taken =ย  9ย  xย  25ย  hrs = 2 ยผ hrs

100ย ย ย ย ย ย ย ย ย  1

7.30 am

6.ย ย ย ย  hrs

13.30

ย  4.10

5.40pm

ย  2.15

ย  7.55 pm

 

 

 

M1

 

 

 

B1

 

 

 

 

 

B1

 

B1

 

 

 

 

 

M1

 

A1

 

 
 

 

18

 

 

 

 

 

 

 

 

 

x2 + x โ€“ 8 = -2 โ€“ 2x

y = x2 + 3x โ€“ 6

Points of intersection (-4, 1.4)

y = x2 + x โ€“ 8 = 2x2 + 3x โ€“ 6

x2 + 2x + 2

y = x2 + x โ€“ 8 x 2

2y = 2x2 + 2x โ€“ 16

0 = 2x2 + 3x โ€“ 6

2y = -xย  – 10

y = – 2.6

Ny = 1.2

 

8

 

 

 

 

 

 

 

 

 

B1

B1

 

 

 

 

 

B1

 

B1

 

 

 

 

 

 

 

 

 

 

Eqn

Point of inter

 

 

 

 

 

Line eqn

 

Both

 

 

19

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a)ย ย ย  Modal class = 45 โ€“ 49

(i)ย ย ย ย ย ย ย ย ย ย ย ย ย ย  Mean = 47 + -55

50

= 47 โ€“ 1.1

= 45.9

 

(ii) Standard deviation = ร– 3575 –ย  –55 2
50ย ย ย ย ย ย ย ย  50

=ย  ร–71.5ย ย  – 1.21

=ร– 70.29

= 8.3839

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

B1

 

 

 

 

M1

 

 

A1

B1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

fd

fd2

    8  
20  

 

 

 

 

 

 

 

 

(a)ย ย ย  a =ย ย  1
b = 2

ยฝ cos (x โ€“ 15) = Sin (x + 30)

has no solution in the domain

 

 

 

 

 

 

 

 

B1

B1

B1

 

B1

 

 

 

 

 

 

 

 

All

All

A & b

 

 

    8  
21 (a)ย ย ย ย ย ย  O Cos 30 = 20

X

X =ย  20

0.866

= 23.09

 

DE = ร– 502ย ย  + 23.092

= ร– 2500 + 533.36

= ร– 3033.36

= 55.076m

 

(b)ย ย ย ย ย ย  GB =ย  ร– 202ย  + 502

= 53.85

Tan q = 14.55
53.85

=ย  0.27019

qย ย ย  = 15.12o

 

 

 

 

B1

 

M1

 

 

 

A1

 

 

M1

 

 

A1

 
  8  
  (c)ย ย ย ย ย ย  Volume of air = 50 x 20 x 3 + ยฝ x 20 x 11.55 x 50

= 3000 + 5775

= 8775

No. of peopleย  =ย ย  8775
ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  6

= 1462.5

j 1462

 

M1

 

M1

 

 

A1

 
    8  
22 (a)ย ย ย  A + B [ 16

5A + 3B ยณ 50

2A + 3B [ 35

 

 

(b)ย ย  14 vehicles

 

(c)ย ย ย  A – 6 vehicles

B –ย  8

Cost = 6 x 1000 + 8 x 1200

= 6000 + 9600

= 15,600/=

 

 

B1

 

 

B1

 

B1

 

M1

 

A1

 

 

 

In equation 3

 

 

Vehicles

    8  

23

 

 

 

 

 

 

 

 

 

 

 

 

 

x ย ย ย ย ย ย ย =ย ย ย ย ย  8

12 โ€“ xย ย ย ย ย ย ย ย ย ย  3

 

= 8.727

FBX =ย ย ย  3ย ย ย  =ย  0.9166ย ย  = 23.57
3.273

 

3FBX = 47.13

 

Reflexย  ร FBD = 312.87

 

Reflex arc FD = 312.87ย ย  x 22ย  x 6
360ย ย ย ย ย ย  ย ย ย ย 7

 

= 16.39cm

Reflex Arc CE = 312.87 x 22 x 16
360ย ย ย ย ย ย ย ย  7

 

=ย  43.7cm

 

FE (tangent) =ย  ร–144 โ€“ 121

= ร– 23

= 4.796cm

2 FEย ย ย ย ย ย ย ย ย ย ย  =ย  9.592

 

Total length = 9.592 + 4.796 + 43.7 + 16.39

= 74.48 cm2

 

 

 

 

 

 

 

 

 

M1

 

 

A1

 

 

 

 

 

 

 

M1

 

 

A1

 

M1

 

 

A1

 

 

 

 

 

 

M1

A1

 
    8  

24

 

 

 

 

 

 

 

 

 

 

 

 

(b)ย ย ย ย ย ย ย ย  200ย ย ย ย ย  =ย ย ย  40

Sin 50ย ย ย ย ย ย  Sin q

 

Sin q =ย  40Sin 50
ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  200
= 0.7660
5
=0.1532

qย ย ย ย ย ย ย ย  = 8.81o

ร ACB = 180 โ€“ (50 + 8.81)o

= 121.19o

ย ย ย  xย ย ย ย  ย ย ย ย ย ย ย ย =ย ย  200
Sin 121.19ย ย ย ย  Sin 50

 

xย  = 200 x Sin 121.19
Sin 50

= 200 x 0.855645
0.7660

= 223.36Km/h

 

(b)ย  Course = 330o โ€“ 8.81o

= 321.19o

 

(c) Timeย  =ย ย ย  600
321.19o

 

= 2.686 hrs

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

 

 

 

 

 

A1

 

 

 

M1

 

 

M1

 

 

A1

 

B1

 

 

 

M1

 

A1

 

 

8

 

 

 

One thought on “KCSE MATHEMATICS REVISION SERIES (QUESTIONS & ANSWERS)”
  1. I was wondering if you ever thought of changing the layout of your blog? Its very well written; I love what youve got to say. But maybe you could a little more in the way of content so people could connect with it better. Youve got an awful lot of text for only having 1 or 2 pictures. Maybe you could space it out better?

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