MATHEMATICS I
PART I
SECTION I (50 MARKS)
- Evaluate without mathematical tables leaving your answer in standard form
0.01712 X 3
855 X 0.531 (2 Mks)
- 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 days (3 Mks)
- M In quadrilateral OABC, OA = 4i – 3j. OC = 2i + 7j
AB = 3OC. cm: mB = 2:3. Find in terms of i and j
C vector Om (3 Mks)
O A
- By matrix method, solve the equations
5x + 5y = 1
4y + 3x = 5 (3 Mks)
- In the given circle centre O, ÐABC = 1260.
Calculate ÐOAC (3 Mks)
A C
B
- Solve the equation
2(3x – 1)2 – 9 (3x – 1) + 7 = 0 (4 Mks)
- 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’s (3 Mks)
- Expand (2 + 1/2x)6 to the third term. Use your expression to evaluate 2.46 correct to 3 s.f (3 Mks)
- The probability of failing an examination is 0.35 at any attempt. Find the probability that
(i) You will fail in two attempts (1 Mk)
(ii) In three attempts, you will at least fail once (3 Mks)
- 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 line (2 Mks)
- During a rainfall of 25mm, how many litres collect on 2 hectares? (3 Mks)
- Solve the equation a – 3a – 7 = a – 2 (3 Mks)
3 5 6
- The sum of the first 13 terms of an arithmetic progression is 13 and the sum of the first 5 terms is
–25. Find the sum of the first 21 terms (5 Mks)
- The curved surface of a core is made from the shaded sector on the circle. Calculate the height of
the cone. (4 Mks)
O
20cm 1250 20 cm
- Simplify (wx – xy – wz + yz) (w + z) (3 Mks)
z2 – w2
- 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. (3 Mks)
SECTION II (50 MARKS)
- In the given circle centre O, ÐQTP = 460, ÐRQT = 740 and ÐURT = 390
U T P
Q
S 390
Calculate R
(a) ÐRST (1 Mk)
(b) ÐSUT (3 Mks)
(c) Obtuse angle ROT (2 Mks)
(d) ÐPST (2 Mks)
- The exchange rate on March 17th 2000, was as follows: –
1 US$ = Kshs.74.75
1 French Franc (Fr) = Kshs.11.04
A Kenyan tourist had Kshs.350,000 and decided to proceed to America
(a) How much in dollars did he receive from his Kshs.350,000 in 4 s.f? (2 Mks)
(b) The tourist spend ¼ of the amount in America and proceeded to France where he spend Fr
16,200. Calculate his balance in French Francs to 4 s.f (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? (3 Mks)
- On the provided grid, draw the graph of y = 5 + 2x – 3x2 in the domain -2 £ x £ 3 (4 Mks)
(a) Draw a line through points (0,2) and (1,0) and extend it to intersect with curve y = 5 + 2x – 3 x 2
read the values of x where the curve intersects with the line (2 Mks)
(b) Find the equation whose solution is the values of x in (a) above (2 Mks)
- (a) Using a ruler and compass only, construct triangle PQR in which PQ = 3.5 cm, QR = 7 cm
and angle PQR = 300 (2 Mks)
(b) Construct a circle passing through points P, Q and R (2 Mks)
(c) Calculate the difference between area of the circle formed and triangle PQR (4 Mks)
- The given Region below (unshaded R) is defined by a set of inequalities. Determine the inequalities (8 Mks)
Y
4
2 R (3,3)
X
-3 5
- The table below shows the mass of 60 women working in hotels
| Mass (Kg) |
60 – 64 |
65 – 69 |
70 – 74 |
75 – 79 |
80 – 84 |
85 – 89 |
| No. of women |
8 |
14 |
18 |
15 |
3 |
2 |
(a) State (i) The modal class (1 Mk)
(ii) The median class (1 Mk)
(b) Estimate the mean mark (4 Mks)
(c) Draw a histogram for the data (2 Mks)
- 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. (2 MKS)
Z
C
.. B
X
A Y
(a) Calculate, length XA (2 Mks)
(b) The shaded area (6 Mks)
- Maina bought a car at Kshs.650,000. The value depreciated annually at 15%
(a) After how long to the nearest 1 decimal place will the value of the car be Kshs.130,000 (4 Mks)
(b) Calculate the rate of depreciation to the nearest one decimal place which would make the value of
the car be half of its original value in 5 years (4 Mks)
MATHEMATICS I
PART II
SECTION 1 (50 MARKS)
- Simplify 32a10 -2/5 ÷ 9b4 11/2
b15 4a6 (2 Mks)
- Use logarithm tables to evaluate
Ö0.375 cos 75
tan 85.6 (4 Mks)
- 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)
- 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)
- In the circle centre O, chords AB and CD intersect at X. XD = 5 cm
XC = 1/4 r where r is radius. AX:XO = 1:2 Calculate radius of the circle. (3 mks)
A 5cm D
C O
B
- Simplify 2 – 1 (3 mks)
5 – 2Ö3 5 + 2Ö3
- 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 (4 mks)
- The figure on the side is a tent of uniform cross-section A F
ABC. AC = 8m, BC = 8m, BD = 10m and (ACB = 1200. 8m
If a scout needs 2.5 m3 of air, how many scouts can fit 120o C E
in the tent. 8m (4 mks)
B D
10m
- The length of a rectangle is given as 8 cm and its width given as 5 cm. Calculate its maximum % error in its perimeter (3 mks)
- 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
- Expand (1 + 2x)6 to the 3rd term hence evaluate (1.04)6 (4 mks)
- 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 (4 mks)
- 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. (3 mks)
- 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 train (3 mks)
- 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 speed (3 mks)
- 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)
- The income tax in 1995 was collected as follows:
Income in Kshs. p.a rate of tax %
1 – 39,600 10
39,601 – 79,200 15
79,201 – 118,800 25
118,801 – 158,400 35
158,401 – 198,000 45
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 income (8 Mks)
- 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) All the three are correct (2 Mks)
(b) Two out of three are correct (3 Mks)
(c) At least two are correct (3 Mks)
- 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 prices (8 mks)
- In D OAB, OA = a, OB = b
OPAQ is a parallelogram.
ON:NB = 5:-2, AP:PB = 1:3
Determine in terms of a and b vectors
(a) OP (2 Mks)
(b) PQ (2 Mks)
(c) QN (2 Mks)
(d) PN (2 mks)
- 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) Calculate the volume in M3 added in 2 days (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 tank (4 mks)
- 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 get (5 mks)
(b) If the sponsor’s money was shared according to the schools needs while the rest was in the ratio of
students, how much does each school get if school A has 780 students and school B 220
students (3 mks)
- 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
V
|
0.35 |
0.49 |
0.72 |
0.98 |
1.11 |
| E |
0.45 |
0.61 |
0.89 |
1.17 |
1.35 |
By drawing a suitable linear graph, determine values of k and n hence V when E = 0.75(8mks)
- The vertices of triangle P,Q,R are P(-3,1), Q (-1,-2), R (-2,-4)
(a) Draw triangle PQR and its image PIQIRI of PQR under translation T = 3 on the provided grid 4 (2 Mks)
(b) Under transformation matrix m = 4 3 , PIQIRI is mapped on to PIIQIIRII. Find the
co-ordinates of PIIQIIRII and plot it 1 2 on the given grid (4 Mks)
(c) If area of D PIQIRI is 3.5 cm2, find area of the images PIIQIIRII (2 Mks)
MATHEMATICS I
PART 1
MARKING SCHEME
- 171 X 171 X 3 X 10-5 M1
855 X 531
= 2 X 10-6 A1
2
- No. of men = 6 X 14 X 8 X 2500 M1
2 X 5 X 2240
= 75 A1
Extra men = 75 – 6 = 69 B1
3
- OM = 2i + 7j + 2/5 (4i – 3j + 6i + 21j – 2i – 7j) M1
= 2i + 7j + 2/5 (8i + 11j) M1
= 26 i + 57 j
5 5 A1
3
- 2 5 x = 1
3 4 y 5 M1
x -1/7 5/7 1
y = 3/7 -2/7 5 M1
x = 3
y -1
x, 3, y = -1 A1
3
- Reflex ÐAOC = 126 x 2 = 2520 B1
Obtuse ÐAOC = 360 – 252 = 1080 B1
= 1/2 (180 – 108)0
= 360 B1
3
- 18x2 – 39x + 18 = 0
6x2 – 13x + 6 = 0 B1Ö equation
6x2 – 9x – 4x + 6 = 0
3x(2x – 3) (3x – 2) = 0 M1
x = 2/3 or A1
x =1 ½ B1
4
- M : K : O = 4 : 2 : 3 B1Ö ratio
Maina’s = 4/9 X 180
= 80/- B1Ö Omondi’s
Omondi’s = 60/- and Maina’s
Difference = Shs.20/- B1 difference
3
- (2 + 1/2x)6 = 26 + 6(25) (1/2x + 15 (24) (1/2 x)2 M1
= 64 + 96x + 60x2 A1
2.46 = (2 + 1/2 (0.8))6
= 64 + 96 (0.8) + 60 (0.64) M1
= 179.2
@179 to 3 s.f A1
4
- P (FF) = 7/20 X 7/20
= 49/100 B1
P (at least one fail) = 1 – P (FI FI FI)
= 1- 13/20 3 M1
= 1 – 2197 M1
8000
= 5803
8000 A1
4
- grad = term 135
= -1 B1
y = mx + c
y = -x + 5 B1
2
- Volume = 2 x 10,000 x 10,000 x 25 M1Ö x section area
1000 10 M1Ö conv. to litres
= 500,000 Lts A1
3
- 10a – 6(3a – 7) = 5(a -2) M1
10a – 18a + 42 = 5a – 10
– 13a = -52 M1
a = 4 A1
3
- 2a + 12d = 2
2a + 4d = -10 M1
8d = 12
d = 11/2 A1
a = -8 B1
S21 = 21/2 (-16 + 20 X 3/2) M1
= 147 A1
5
- 2 p r = 120 x p x 40 M1
360
r = 6.667 cm A1
h = Ö 400 – 44.44 M1
= 18.86 cm A1
4
- = (w (x – z) – y (x – z)) (w + z) M1Ö factor
(z – w) (z + w)
= (w – y) (x – z) (w + z) M1Ö grouping
(z – w) (z + w)
= (w – y) (x – z)
z – w A1
3
R
250 B1Ö sketch
- 550
Q 125 PR = 4 sin 125 M1
Sin 25
A1
30
P 3
- (a) <RST = 1800 – 740 = 1060 B1
(b) < RTQ = 900– 740 = 160 B1
< PTR = 460 + 160 = 620 B1
< SUT = 620 – 390 = 230 B1
(c) Reflex ÐRQT = 180 – 2 x 16
= 180 – 32 = 1480 B1
Obtuse ROT = 360 – 148 = 2120 B1
(d) < PTS = 46 + 180 – 129 = 970 B1
< PST = 180 – (97 + 39) = 440 B1
8
(a) Kshs.350,000 = $ 350,000 M1
74.75
= $ 4682 A1
(b) Balance = 3/4 x 4682
= $ 3511.5 B1
$3511.5 = Fr 3511.5 x 74.75 M1
11.04
= Fr 23780 A1
Expenditure = Fr 16 200
Balance = Fr 7580
(c) Value on arrival = Kshs.7580 X 12.80
= Kshs.97,024
Value on departure = Kshs.7580 X 11.04 B1 bothÖ
= Kshs.83 683.2
Difference = Kshs.97,024 – 83683.2 M1
= Kshs.13,340.80 A1
8
-
| X |
-2 |
-1 |
0 |
1 |
2 |
3 |
| Y |
-11 |
0 |
5 |
4 |
-3 |
-16 |
B1Övalues
y
S1Ö scale
8 — P1Ö plotting
6 — C1 Ö curve
4 —
2
-2 — 1 2 3 x
-4 —
-6 —
-8 — y=2x=2
-10 —
-12 —
-14 — x =-0.53 + 0.1 BI
-16 — Nx = 1.87+ 0.1
y = 5+2x-3x2 =2-2x MI for equation
3x2-4x-4x-3=0 AI equation
8
x = -0.53 ± 0.1 B1
mx = 1.87 ± 0.1
y = 5 + 2x – 3x2 = 2 – 2x M1 Ö for equation
\ 3×2 – 4x – 3 = 0 MA1 Ö equation
8
20.
B1 Ö 300
R B1 Ö 2 ^ PQ, QR
B1 Ö 2 ^ bisectors
B1 Ö circle
9 Q
Radius = 4.2 ± 0.1 B1Ö radius
Area of circle = 22/7 x 4.22
= 55.44 ± 3 cm2
Area of D PQR = 1/2 x 3.5 x 7.5 sin 30 M1Ö D and circle
= 6.5625 cm2
Difference = 55.44 – 6.5625 M1Ö sub
= 48.88 cm2 A1
8
- Line (i) y/2 + x/5 = 1
5y + 2x = 10 B1Öequation
5y + 2x = 10 B1Ö inequality
Line (ii) y/4 + x/-3 = 1
3y = 4x + 12 B1Ö equation
3y < 4x + 12 or 3y – 4x < 12 B1Ö inequality
Line (iii) grad = -1/3 y inter = 4
3y + x = 12 or 3y = -x + 12 B1Ö equation
3y + x < 12 B1Ö inequality
Line (iv) y – 3 = -3
x – 3 2
2y + 3x = 15 B1Ö equation
\ 2y + 3x £ 15 B1Ö equation
8
-
CLASS
|
F |
x |
Fx |
Cf |
| 60 – 64
65 – 69
70 – 79
75 – 79
80 – 84
85 – 89 |
8
14
18
15
3
2 |
62
67
72
77
82
87 |
496
938
1296
1155
246
174 |
8
28
40
55
58
60 |
| |
Sf = 60 |
|
Sfx 3809 |
|
B1Ö x column
B1Ö f column
(a) (i) Modal class = 70 – 74 B1Ö model class
(ii) Median class = 70 – 74 B1Ö median
(b) Mean = 3809
60 M1
= 63.48 A1
S1Ö scale
B1 Ö blocks
59.5 – 64.5
64.5 – 69.5 e.t.c.
8
(c)
Histogram
20 —
15 —
10 – –
5 —
55 60 65 70 75 80 85 90
- (a) XA = a, YA = 10 – a, YB = 10 – a, CZ = 10 – a = ZB
YZ = 10 – a + 12 – a = 8 M1
2a = 14
a = 7 cm A1
Cos X = 100 + 144 – 64
240 M1Ö any angle of the D
= 0.75
X = 41.410
1/2 X = 20.700 A1Ö 1/2 of the angle
r = OA = 7tan 20.7 B1 Ö radius
= 2.645 cm
Shaded area = 1/2 X 10 X 12 sin 41.41 – 22/7 X 2.6452 M1 Ö D & circle
= 39.69 – 21.99
= 17.7 cm2 A1Ö
8
- (a) 650,000 (0.85)n = 130,000 M1Ö formula
1.15n = 0.2
n = log 0.2 M1Ö
log 0.85
= 1.3010
1.9294
= – 0.6990 M1
– 0.0706
= 9.9 years A1
(b) 650,000 (1 – r/100) 5 = 325,000 M1
(1 – r/100) 5 = 0.5
1 – r/100 = 0.5 1/5 M1
= 0.8706
r/100 = 0.1294 A1
r = 12.9 % B1
8
MATHEMATICS I
PART II
MARKING SCHEME
SECTION I (50 MARKS)
- = b15 2/5 X 4a6 3/2
32a10 9b4 M1Ö reciprocal
= 2a5 A1
27 2
-
No. Log.
0.375 1.5740 +
cos 75 1.4130
2.9870 _
tan 85.6 1.1138
3.8732 = 4 + 1.8732
2 2
2.9366
0.0864
- S. Price = 80 X 600
100
= Shs.480 B1
Cost Price = x
96x = 480 M1
100
x = Shs.500 A1
3
- r2 + hr = A/2p M1
r2 + hr + (h/2)2 = A/2A + h/4 M1
(r + h/2)2 = Ö 2A + h2
4p M1
r = -h/2 ± Ö2A + h2 A1
4p 4
- (12/3r) (1/3 r) = (1/4 r) (5) M1
4r2 – qr = 0
r(4r – q) = 0 M1
r = 0
or r = 2.25 A1
3
- = 2 (5 + 2Ö3) – 1 (5 – 2Ö3) M1
(5 – 2Ö3) (5 + 2Ö3)
= 10 + 4Ö3 – 5 +2Ö3 M1
13
= 5 + 6Ö3 A1
13 3
- P = Kq2 + c
6 = 4k + c
16 = 9k + c M1 Ö subtraction
5k = 10
k = 2
c = -2 A1 Ö k and c
P = 64 2q2 = 66
q = Ö33
= ± 5.745 A1
4
- Volume = 1/2 X 8 X 8 sin 120 X 10 M1 Ö area of x-section
No. of scouts = 32 sin 60 X 10 M1 Ö volume
2.5 M1
= 110.8
= 110 A1
3
- Max. error = 2(8.5 + 5.5) – 2(7.5 + 4.5)
2
= 2 B1
% error = 2/26 X 100 M1
= 7.692% A1
G 3
- B1 Ö net
H D G H B1 Ö dimen. FE must be 10cm
4cm 4cm
B1 Ö labelling
E 4cm A 12cm F 10cm E 3
4cm 12cm
E
F
- (1 + 2x)6 = 1 + 6(2x) + 15 (2x)2 M1
= 1 + 12x + 60x2 A1
(1.04)6 = (1 + 2(0.02))6
= 1 + 12 (0.02) + 60(0.02)2 M1
= 1.264 A1
4
- BT = 10 cm B1
CT = 10 sin 40 M1
= 6.428 m A1
A1 10cm B C h = 6.428 + 1.5
1-5 = 7.928 B1
4
- A.S.F = 405 2/3 = 27 2/3 = 9 B1
960 64 10
smaller area = 29 X 120 M1
164
= 67.5 cm2 A1
3
- Relative speed = (x – 30)km/h B1
2 km = 2 hrs
(x – 30)km/h 60 M1
2x – 60 = 120
x = 90 km/h A1
3
- 16 Km/h 5 Km/hr
x Km (22 – x) Km
x + 22 – x = 11
16 5 4 M1
5x + 352 – 16x = 220 M1Ö x-multiplication
11x = 132
x = 12 km A1
3
- AP = 2/9 x 12 = 22/3 cm B1 Ö both AP & AQ
AQ = 5/9 x 12 = 62/3 cm
\ PQ = 62/3 – 22/3 = 4 cm B1 Ö C.A.O
2
- Taxable income = 115/100 x 8000 M1
= Shs.9200 p. m
= Shs.110,400 p.a A1
Tax dues = 10/100 x 39600 + 15/100 x 39600 + 25/100 x 31200 M1 Ö first 2 slabs
= 3960 + 5940 + 7800 M1 Ö last slab
= Shs.17,700 p.a
= 1475 p.m A1
net tax = 1475 – 400
= Shs.1075 B1 Ö net tax
Total deductions = 1075 + 130
= Shs.1205
net income = 8000 – 1205 M1
= Shs.6795 A1
8
-
(a) P (all correct) = 2/3 x 3/5 x 2/5 M1
= 12/125 A1
(b) 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
M1
= 18/125 + 4/125 + 24/125 M1
= 46/125 A1
(c) P (at least 2 correct)
= P(2 correct or 3 correct)
= 46/125 + 12/125 M1
= 46 + 12 M1
125
= 58
125 A1
8
- Old price/pen = 440
x
New price/pen = 494 B1Öboth expressions
x + 3
440 – 494 = 1.50
x x + 3 M1 Ö expression
440(x + 3) – 494x = 1.5x2 + 4.5x M1Ö x-multiplication
x2 + 39x – 880 = 0 A1 Ö solvable quad. Eqn
x2 + 55x – 16x – 880 = 0 M1 Ö factors or equivalent
(x – 16) (x + 55) = 0
x = -55
or x = 16 A1 Ö both values
\ x = 16
difference in purchase = 19 X 1.50 M1
= Shs.28.50 A1
8
- (a) OP = a + 1/4 (b – a) M1
= 3/4 a + 1/4 b A1
(b) PQ = PO + OQ
= –3/4 a – 1/4 b + 1/4 (a – b) M1
= –1/2 a – 1/2 b A1
(c) QN = QO + ON
= 1/4 (b – a) + 5/3 b M1
= 23/12 b – 1/4 a A1
(d) PN = PB + BN
= 3/4 (b – a) + 2/3 b M1
= 17/12 b – 3/4 a A1
8
- (a) Volume in 2 days = 22 x 3.5 x 3.5 x 150 x 20 x 3600 M1 Ö area of x-section
7 2 2 1,000,000 M1 Ö volume in cm3
= 103.95 m3 M1 Ö volume in m3
(b) 22 X r2 x 8 = 103.95 x 15 x 7 M1
7 2
r2 = 103.95 x 15 x 7 M1
2 x 2 2x 8
= 31.01 M1
r = 5.568 m A1
8
- (a) Ration of needs for A:B = 5:8
A’s share = 5/13 x 16.9 + 1/2 x 13.5 M1
= 13.25 Million A1
B’s share = (13.5 + 16.9) – 13.25 M1
= 13.25 M1
- A’s share 5/13 x 16.9 + 39/50 x 13.5
6.5 + 10.53
= 17.03 m A1
B’s share = 30.4 – 17.03 M1
= 13.37 Million A1
8
- 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 |
B1Ö log V all points
B1Ö log E all points
S1 Ö scale
P1Ö plotting
Log V = n log E + log K L1 Ö line
Log K = 0.08
K = 1.2 ± 0.01 B1 Ö K
N = 0.06/0.06 B1 Ö n
= 1 ± 0.1
\ v = 1.2E B1Ö v
when E = 0.75, V = 0.9 ± 0.1 8
- (a) T 3 PQR ® PIQIRI
4 PI (0,5), QI (2,2) RI (1,0)
PI QI RI PII QII RII
(b) 4 3 0 2 1 = 15 14 4 M1 Ö
1 2 5 2 0 10 6 1 A1 Ö
PII (15,10), QII (14,6), RII (4,1) B1Ö
(c) Area s.f = det M
= 5
area of PII QII RII = 5 (area PIQIRI)
= 5 X 3.5 M1Ö
= 16.5 cm2 A1
8
MATHEMATICS 2
PART I
SECTION A:
- Use logarithm tables to evaluate (4 mks)
0.0368 x 43.92
361.8
- Solve for x by completing the square (3mks)
2x2 – 5x + 1 = 0
- 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 (3mks)
- 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. (3mks)
- 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 (3mks)
- Find the value of θ within θ < θ < 360O if Cos (2 θ + 120) = γ3 (3mks)
2
- A quantity P varies inversely as Q2 Given that P = 4 When Q = 2. , write the equation joining P and Q
hence find P when Q = 4 a (3mks)
- A rectangle measures 3.6 cm by 2.8 cm. Find the percentage error in calculating its perimeter. (3mks)
- Evaluate: 11/6 x ¾ – 11/12 (3mks)
½ of 5/6
- 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 ( 3mks)
- A regular octagon has sides of 8 cm. Calculate its area to 3 s.f. (4mks)
- Find the values of x and y if ( 2 mks)
3 x 1 = 2
2 1 -1 y
- An equation of a circle is given by x2 + y2 – 6x + 8y – 11 = 0 (3mks)
Find its centre and radius
- In the figure given AB is parallel to DE. Find the value of x and y
- A line pass through A (4,3) and B(8,13). Find (6 mks)
(i) Gradient of the line
(ii) The magnitude of AB
(iii) The equation of the perpendicular bisector of AB.
- 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:
- Construct without using a protractor the triangle ABC so that BC=10cm, angle ABC = 600 and
BCA = 450
- On the diagram , measure length of AC
- Draw the circumference of triangle ABC
- Construct the locus of a set of points which are equidistant from A and B.
- Hence mark a point P such that APB = 450 and AP = PB
- Mark a point Q such that angle AQB = 450 and AB = AQ
- (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 matrix -2 0 to obtain its image AI B I CI DI. under a second transformation
0 – 2
which has a rotation centre (0,0) through –900 , the image AII BII CII DII of AI BI CI DI is
obtained. Plot the three figures on a cartesian plane (6mks)
(b) Find the matrix of transformation that maps the triangle ABC where A (2,2) B (3,4) C (5,2)
onto A B C where A( 6,10) B (10,19 ) C ( 12, 13). ( 2mks)
19.
In the triangle OAB, OA = 3a , OB = 4b and OC = 5/3 OA. M divides OB in the ratio 5:3
- Express AB and MC in terms of a and b
- By writing MN in two ways, find the ratio in which N divides
- AB
- MC
- In the figure below, SP = 13.2 cm, PQ = 12 cm, angle PSR = 80O and angle PQR = 900. S and Q are the centres (8mks)
Calculate:
The area of the intersection of the two circles
The area of the quadrilateral S P Q R
The area of the shaded region
- 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 |
- By plotting 1/y against x, confirm that y is related to x by an equation of the form
Y = q
P + x
where p and q are constants. (3mks)
(b) Use your graph to determine p and q (3mks)
(c ) Estimate the value of (i) y when x = 14
(ii) x when y = 0.46 (2mks)
- 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
- Find the times taken to complete the uphill and downhill sections of the course.
- Calculate the cyclists average speed over the 150km.
- 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
- Find the length of XB as a fraction
- Show that XAD is similar to XBC
- Given that the area of AXD = 6cm2, find the area of BXC
- Find the value of the ratio
Area of AXB
Area of DXC
- 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. (3mks)
(scale 1cm rep 10 km)
From this find,
(a) The distance of AD and DC in km (2mks)
(b) The distance and bearing of B from D (2mks)
(c) The bearing of C from A (1mk)
MATHEMATICS 2
PART 1
MARKING SCHEME (100MKS)
- No. Log
= 3.6502
0.3681 2.5660
0.3682 1.6427 + -4 = 1.6502 = 2.8251
0.2087 Logs 2
361.8 2.5585 + – v ans (4) 6.6850 x 10 -2
3.6502 = 0.06685
- 2 x2 – 5x + 1 = 0
x2 – 5 x + ½ = 0
2
x2 – 5 x = – ½
2
x – 5x + –5 2 = –½ + –5 2 (m)
2 4 4
= x – 5 = – ½ + 25 = 17 (3)
4 16 16
= x – 5/4 = 17/16 = 1.0625
x – 5/4 ± 1.031
X1 = -1.031 = 1.25 = 0.2192
X2 = 1.031 + 1.25 = 1.281
- A1 = P(1 + R/100)2 = 6000 x 113/100 x 113/100 = Sh. 7661.40
A2 = P + PRT/100 = 6000 + 15 X 2 = 6000 + 1800
100
= Shs. 7800
Amount by simple interest is more by Shs. (7800 – 7661. 40)
Shs. 138.60
- Let a plate be p and a cup c.
3p + 4c = 380 x 5 15p + 20c = 1900
4p + 5c = 490 x 4 16p + 20c = 1960
-p -60 (m)
p = Shs 60
3(60) + 4 c = 380
4c = 380 –180 = 2000 (3)
c= Shs. 50
Plate = Shs. 60 , Cup = Shs. 50 (A both)
- Ratio of juice to water = 1 : 7
In 1 glass = 1/8 x 200 = Sh 25
3 litres = 300 ml (undiluted concentrate) (3)
No. of glasses =v 3000 = 120 glasses
25
- Cos (2 θ + 120) = 3/2 = 0.866
Cos 30 , 330, 390, 690, 750 ….
2 θ + 120 = 330
2 θ = 210 , = 1050 (3)
2 θ = 390 – 120 = 2700 , θ2 1350
2 θ = 690 – 120 = 5700 , θ3 2850 (for 4 ans)
θ4= 315o ( for >2)
2 θ = 750 – 120 = 6300 ,
- P = k 4 = K/4 (substitution)
Q2 9
K = 4 X 4 = 16
9 9
P = 16 v when Q = 4
9Q2
P = 16 = 1/9 (A) (3)
9x4x4
- The perimeter = (3.6 + 2.8 ) x 2 = 12.8 cm
Max perimeter = (3.65 + 2.85) x 2 = 23 cm Expressions
% error = 13 –12.8 x 100 m = 0.2 x 100 (3)
12.8 12.8
= 1.5620% (A)
- 1 1/6 x ¾ – 11/12 = (7/6 x ¾ ) -11/12 = 7/8 – 11/12 = 21-22
½ of 5/6 ½ of 5/6 5/12 5/12
= -1/24 = -1 x 12 = -1
5/12 24 5 10 (3)
- Volume of rod = П r2h = 22/7 x 4 x 14 = 704cm3 (m)
Volume of each cube = 2x2x2 = 8 cm3 A
No. of cubes = 704 /8 = 88 cm3 A
- < AOB = 360 = 450
8
Tan 67.5 = h
4
h = 4 x 2.414 A
= 9.650cm
Area of 1 triangle = ½ x 8 x 9.656 x 8 cm = 38.628 x 8 vm
Octagon area = 38.628 x 8 m
= 309.0 cm2 (A)
- 3 2 -1 2
=
2 1 -1 y
3 – x = 2 (1) x = 1 (2)
2 – 1 = y y = 1 (A)
- x 2 + y2 – 6x + 8y – 11 = 0
x2 – 6x + (-3)2 + y2 + 8y + (4)2 = 11 + (-3)2 + (4)2 (completing the square)
(x – 3)2 + (y+4)2 = 11 + 9 + 16 = 36
(x – 3)2 + (y + 4)2 = 62
Centre is (3, -4)
Radius = 6 units As (3)
14.
Figs A C B and D C E are similar
AB = AC = and AB = BC
DE DC DE CE
10 = 6 + x
3 6
= 10 = 15 + y, m
3 y 60 = 18 + 3x
10y = 15 + 3y 3x = 42
7y = 15 x = 14
y = 15/7 (A) (3)
A (4 , 3) B(8,13)
- (i) gdt = change in y = 13-3 = 10 = 5
change in x 8-4 4 2
(ii) Mag AB = 8 -4 4 =
13 -3 10
Length = Ö42 + 102 = Ö116 = 10.77 units
(iii) Mid point = 4 +8 , 3 + 3
2 2
= (6, 8) (mid point) (5 mks)
gdt of perpendicular to AB = -ve rec. of 5/2
-2/5
Eqn is y = -2/5 x + c
8 = -2/5 x 6 + c = 40 = -12 + 5c
= c = 52/5
y = -2/5 x + 52/5 (A)
- Acceleration = Change in velocity
Time
= (34 – 10) m/s = 24 m/s
60 x 10 600
= 0.04m/s2- (2)
17.
Triangle (8)
AC = 9cm
Circumference Centre
Circle
Perpendicular bisector of AB
P
Q
- (b) a b 2 3 5 6 10 12
c d 2 4 2 10 19 13
2a +2b = 6 x 2 = 49 + 4b = 12
3a + 4b = 10 3a + 4b = 10
a = 2 4 + 2b = b
2c + 2d = 10×2 = 4c + 4d = 20 2 b = 2 b = 1
3c + 4d = 19 3c + 4d = 19
c = 1
2 (1) + 2d = 10
2d = 8 Matrix is 2 1 (A)
d = 4 1 4
OC = 5/3 (31) = 5A
19.
(a) = AO + OB MC = MO + OC
= -3a = 4b = -5/8 (4b) + 5
= 5A – 5/2 b
(b) MN = 5 Mc = 3(5a – 5/2 b)
= 5 s a – 5/2 s b
MN = BN + BN
= 3/8 (4 b) + (1 – t) (-BA)
= 3/8 (4 b) + (1 – t)(3a – 4 b)
= 3/2 b + 3 ta –4b + 4tb
= (3-3t) a (4t – 5/2)b
MN = MN
= 5 s a – 5/2 sb = (3-3t)a + (4t – 5/2 )b
= 5 a = 3 – 3t = 5s + 3t =3
= -5/2 s = 4t –5/2 v 5s + 8t = 5
-5t = -2 t = 2/5
5 s = 3 – 3(2/5)
= 3 – 6/5 = 9/5
= 3 – 6/5 = 9/5
s = 9/25
(i) AN : NB = 2 : 3
(ii) MN : 9 : 16
20.
θ x pr2
360
- Area of sector SPR = 80/360 x 13.2 x 13.2 x 3.142
= 121.6
Area of triangle SPR ½ x 13.2 x 13.2 x sin 80
= 85.8 cm2
(m of area of ) A (at least one)
(m of area) A(at least one)
Area of segment = 121.6 – 85.8
= 35.8 cm2
Area of sector QPR = 90/360 x 3.142 x 12 x12
Area of PQR = ½ x 12 x 12 = 722
Area of segment = 113.1 – 72
= 41.1cm2
Area of intersection = (35.8 + 41.1) = 76.9 cm2
b). Area of quadrilateral = Area of PQR + SPR
= 85.8 + 72 = 157.8cm2
Area of shaded region = Area of Quadrilateral – Area of sector SPR
= 157.8 – 121.6
= 36.2 cm2
- y = q p + x = q 1 = x + p
p + x y y q q
Gradient = 1/q at (0, 0.95) (8,2.0) (8,2.0) gradient = 2.0 – 0.95 = 1.05
8 8
1 = 0.1312
q
q = 1 = 7.619
0.1312
q = 7.62.
y(1/y) Intercept p = 0.95 P = 0.95
q 7.62
p = 7.62 x 095 = 7.27
at x = 14, y = 2.7
at y = 0.46, 1/y = 2.174
x = 9.6.
- a) Distance = 75km uphill speed = vkm/h
uphill Time = 75/v hrs
Downhill speed = ( + 20) km/h
Downhill Time = 75 hrs.
v + 20
Takes larger uphill
75 – 75 = 1
v v+20
75 (v+20) – 75v = 1
v(v + 20) 1
75v + 1500 – 75v = v(v + 20) = v2 + 20v.
v2 + 20v – 1500 = 0
v = – 20 + 202 – 4(1) (-1500)
2(1)
v = –20 + 400 + 6000 = –20 + v6400
2 2.
V1 = –20 + 80 = 30km/hr
2
V2 = – 20 – 80 X impossible
2
speed uphill = 30 km /hr, T = 75 time = 2 ½ hrs
30
speed downhill = 50 km /hr Time = 75 Time = 2 ½ hr
50
Average speed = Total distance = 150km = 37.5 km/ hr
Total time 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 |
- A B
D C
A x X x C = BX . XD
8 x 4 = 6BX
BX = 8 x 142 = 16
6 3
X AD = XBC
XA = 8 = 24 = 3
XB 16 16 2
XD = 6 = 3
XC 4 2
< AXD = BXC (vertically opposite <s))
SAS holds : they are similar.
LSF = 3/2 ASF = (3/2)2 = 9/4
Area A x A = 6cm2 Area B x C = 6 x 9 = 27 = 13.5cm2
4
24.
- a) AD = 50km
DC = 35km
BD = 90km
Bearing is 0200
Bearing is 134o (8mks)
MATHEMATICS I
PART II
SECTION (52 MARKS)
- Without using tables, simplify
1.43 x 0.091 x 5.04
2.86 x 2.8 x 11.7 (3mks)
- Make x the subject of the formula if
y = a/x + bx (3mks)
- Give the combined solution for the range of x values satisfying the inequality
2x + 1< 10 – x < 6x – 1 (3mks)
- 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 years (4mks)
- A town A is 56 km from B on a bearing 0620. A third town C is 64 km from B on the bearing of 140o. Find
(i) The distance of A to C (2mks)
(ii) The bearing of A from C (3mks)
- Expand (x + y)6 hence evaluate (1.02) to 3d.p. (3mks)
- Rationalise the denominator in (2mks)
Ö 3
1 – v3
- The table below shows daily sales of sodas in a canteen for 10 days.
| 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 data (3mks)
- Find the image of the line y = 3x = 4 under the transformation whose matrix is.
3mks
2 1
-1 2
- Three points are such that A (4 , 8), B(8,7), C (16, 5). Show that the three points are collinear (3mks)
- Write down the inverse of the matrix 2 – 3 hence solve for x and y if
4 3
2x – 3y = 7
4x + 3y +5 (3mks)
- Use the table reciprocals to evaluate to 3 s.f. 3mks
1/7 + 3/12 + 7/0.103
-
Given that O is the centre of the circle and OA is parallel to CB, and that angle
ABC = 1070, find
(i) Angles AOC, (ii) OCB (iii) OAB (3mks)
- 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 (4 mks)
- 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. (5mks)
- If log x = a and log y = b, express in terms of a and b
Log x 3
VY (2mks)
SECTION B:
- 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
- The mean
- The standard deviation
- Find the median mark
- Draw the graph of y = 2x2 – x – 4 for the range of x -3 = x = 3. From your graph
State the minimum co-ordinates
- Solve the equations
- 2x2 – x – 4 = 0
- 2x2 – 3x – 4 = 0
-
- 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 region 4mks
- Two unbiased dice are thrown. Find the probability of obtaining (4mks)
- A product of 6
- A sum of 8
iii. The same number showing (4mks)
-
Two pulley wheels centers A and B are joined by a rubber band C D E F G H C round them. Given that larger wheel has radius of 12 cm and AB = 20 cm, CD and GF are tangents common to both wheels and that CBA = 60o), Find
- BD (Length)
- CD
iii. Arc length CHG and DEF, hence find the length of the rubber.
- 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
- The perpendicular height of the pyramid
- The length of the slant edge VA
- The angle between edge VA and base A B C D
- The area of the face VAB
- Plot the graphs of y = sin xo and y = cos 2xo on the same axes for –180 £ x £180o.
Use your graphs to solve the equation 2 sin x = cos 2x
- 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. The pool is 25m long and 12m 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. Calculate the number of litres emptied from the pool in two minutes to the nearest 10 litres. (Take II = 3.142)
- 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.
- Draw on a cartesian plane, the line y = x and mark points B1 and D1
- Mark the points B and D before reflection
- 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.
- Find the image of D under a rotation, through – 900, Center the origin.
MATHEMATICS I
PART II
MARKING SCHEME.
- 1.43 X 0.091 X 5.04 X 100000 91 X 504 = 7/103
2.86 X 2.8 X 11.7 105 2 x 28 x 117 x 103
(3)
= 0.007 (A)
- y = a/x + bx yx = a + bx2
Either
bx2 – yx + a = 0
x = y ± v y2 – 4ab
2b (3)
- 2x + 1£ 10 – x £ bx -1
2x + 1 £ 10 – x 10 –x £ 6x –1
3x £ 9 11£ 7x
x £ 3 x £ 11/7 (3)
11/7 £ x £ 3
- a = 4000 r = 110/100 = 1.1 ( 4000, 4000 + 4000, 4400 + 0/100 (4400——)
(a and r)
Sn = a(r n – 1)
R -1 1.1 Log = 0.04139
X 5
0.20695
0.1 (4)
= 4000 (1.15 –1) (any)
1.1 –1 4000 (1.6 – 1)
0.1
A = 4000 ( 0.6105)
0.1
= Sh. 2442 = Sh. 24,420 (A) (4)
0.1
- (i) b2= a2 + b2 – 2ab Cos B
= 642 + 562– 2(64) (56) cos 78
= 4096 + 3136 – 7168 (0.2079)
= 7232 – km 1490.3
b2 = 5741.7 = 5.77 km (5)
(ii) b a
Sin B Sin A
75.77 = 64
Sin 78 sin A Sin A = 64 x 0.9781
75.77
Sin A = 0.08262
A = 55.70 (or B = 46.30)
Bearing = 90 – 28 – 55.7
= 0.06.30 (A)
- (x + y) 6 = 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)4
= 1 + 0.12 + 0.006 + 0.00016
= 1.12616
= 1.126 (to 3 d.p) (3)
- 3(1 + 3) = 3 + 3 3 + v3
(1- 3)(1+ 3) 1-3 2
- Moving averages of order 4
M1 = 52 + 41 + 43 + 48 184 = 146
4 4
M2 = 184 – 52 + 40 = 172 = 43 for 7
4 4 for > 4
M3 = 172– 40 + 38 = 170 = 42.5
4 4
M4 = 170 – 38+36 = 168 = 42
4 4
M5 = 168 – 36 + 40 = 173 = 43 (3)
4 4
M6 = 172 – 40 + 44 = 176 = 44
4 4
M7 = 176 – 44 + 45 = 177 = 44.25
4 4
- y = 3x + 4
A(0,4) B (1,7) Object points
A B A B
2 1 0 1 4 9
=
-1 2 4 7 8 13
Y = Mx + C
M = 13 – 8 = 5 = 1
9-4 5 1
y = x+c y = x + 4
8 = 4 + c c = 4
- AB = 8 -4 4 BC = 16 – 8 -8 for either
=
7 -8 -1 5 – 7 -2
AB = ½ BC and AB and BC share point B.
A,B,C are collinear. (3)
- 2 -3
4 3 det. = 6 + 12 = 18
Inv.= 1 3 3
18
-4 2
1 3 3 2 -3 x 1 3 3 7
18 18
-4 2 4 2 y -4 2 5
x 36
1
y 18 -18 (3)
x = 2, y = -1 (A)
- 1/7 + 3/12.4 + 7/0.103
1/7 + 3/1.24 x 10-1 + 7/1.03 x 10-1
0.1429 + 3(0.8064) + 7 x 10 (0.9709)
10
= 0.1429 + 0.2419 + 67.96 (3)
=70.52 (A)
- (i) ADC = 2×73
= 1460
(ii) OCB = x = 180 – 146 = 34
(iii) 360 – 107 – 146 – 34
= 73 0
- Tan 300 = y/x y = x tan 30
Tan 600 = 1000 + y ; y = x tan 60 – 1000
X
X tan 300 = x tan 60 – 1000
0.5773 x = 1.732x – 1000
1.732x – 0.577 = 1000
1.155x = 1000
x = 1000
1.155 = 866.0 m (A) (4)
- 5 km Slower speed = x km/hr
Time = 5/x
Faster = (x+20) k/h
Time = 5/x=20 T1 – T2 = 5/x – 5/x+20 = 30/3600
5 (x+20) –5x 1
x(x+20) 120
120 (5/x + 100 – 5x) = x2 + 20x (5)
x2 + 20x – 12000
x = –20 400 + 48000
2
x = -20 ± 220
2
Spd = 100 km/h
And x = 120 km/h (A)
- Log x = a log y = b
Log x3 = Log x3 – log y ½
y
= 3 Log x – ½ Log y
= 8a – ½ ab
SECTION B
17.
| Marks |
Mid point (x) |
d = x-44.5 |
F |
E = d/10 |
Ft |
T2 |
Ft2 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 |
=90 =1 =223
(a) Mean = (1 / 90 x 10) + 44.5 = 44.5 + 0.111
= 44.610
(b) Standard deviation = 10 233/90 – (1/90)2
10 2.478 – 0.0001 (8)
10 2.478
10 x 1.574 = 15. 74 (A)
(c) Median 45.5th value = 39.5 + (13.5 x 10/ 26)
39.5 + 5.192 (A)
44.69
(a) The probability = Shaded area
Large circle area
Shaded area = ПR2 – П r2
= 22/7 (42 – 32) v = 22/7 x 7 = 22
Large area = 22/7 x4x4 = 352/7 (A)
Probability = 22 = 22 x 7 = 7
352/7 352 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) P(Product of 6) = P((1,6) or (2,3) or (3,2) or (6,1))
= 4/36 = 1/9
(4)
(ii) P (sum of 8) = P( (2,6) or (3,5) or (4,4) or (5,3) or (6,2) )
= 5/36 (A)
(iii) P (same number) = P (1,1) or (2,2) or (3,3) or (4,4) or (5,5) or (6,6)
6/36 = 1/6 (A)
-
(i) Cos 60 = x/20 x = 20 x 0.5 = 10 cm
BD = 12 – 10 = 2 cm
(ii) CD = y Sin 60 = y/20 y = 20x 0.8666
CD = 17.32 cm
(iii) CHG = 120 reflex = 2400
CHG = 240/360 x 2 x p x r
= 50.27
DBF = 1200/360 x 2 x П x r = 1/3 x 2 x 3.142 x 2
= 4.189 (A)
Length C D E f G H C = 50.27 + 2(17.32) + 4.189
= 89.189 (A)
- (a) From the diagram, XO = 5/2 = 2.5
Tan 750 = VO/2.5 v m
VO = 2.5 x 3.732
Perpendicular height = VO = 9.33 cm
2 (A)
- Diagonal of base = 52 + 52 = 50
Length of diag. 50 = 7.071 = 5.536
VA2 = AO2 + VO2 (m)
3.5362 + 9.32
12.50 + 87.05
= 99.55 = 9.98 cm2 (A) (8)
(c ) = VAO Tan = 9.33 = 2.639
3.536
VAO = 69.240 (A)
(d) Cos VBA = = 2.5 /9.98 = 0.2505
VBA = 75.490
Area VBA = ½ x 5 x 4.99 x sin 75.45 m ( or other perimeter)
= 5 x 4.99 x 0.9681
= 24.15 cm2 (A)
- Volume = cross – section Area x L
X-sec Area = (1 x 25) + (½ x 25 x 2.5)
= 25 + 31.25 = 56. M
Volume = 56.25 x 12
= 675 m3
Volume passed / sec = cross section area x speed
= П r2 x l = 3.14 x 9/100 x 9/100 x 3 (8)
= 0.07635 m3 /sec v (M)
Volume emptied in 2 minutes
= 0.07635 x 60 x 2
= 9.162 m2 (A)
1 m3 = 1000 l
= 9.162 litres
= 9160 litres (A)
24.
MATHEMATICS II
PART I
SECTION A (52 MARKS)
- Use tables to evaluate
3Ö 0.09122 + Ö 3.152 (5mks)
0.1279 x 25.71
- Simplify (a – b)2
a2 – b2 (2mks)
- The gradient function of a curve that passes through point: (-1, -1) is 2x + 3.
Find the equation of the curve. (3mks)
- Find the value of k for which the matrix k 3
has no inverse. (2mks) 3 k
- Without using tables, evaluate log 128 – log 18
log 16 – log 6 (3mks)
- Find the equation of the locus of points equidistant from point L(6,0) and N(-8,4). (3mks)
- 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)
- Use reciprocal tables to evaluate to 3 d.p. 2 – 1
0.321 n2.2 (4mks)
- Using the trapezium rule, estimate the area bounded by the curve y = x2, the x – axis and the co-ordinates x = 2 and x = 5 using six strips. (4mks)
- Solve the equation for 00 £ q £ 3600 and Cos2q + ½ Cosq = 0 (3mks)
- 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) (3mks)
- How many multiples of 3 are there between 28 and 300 inclusive. (3mks)
- The line y = mx – 1, where m is a constant , passes through point (3,1). Find the angle the line makes with the x – axis. (3mks)
- 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)
- Make X the subject of the formula (3mks)
V = 3Ö k + x
sk – x
- Write down the inequalities that describe the unshaded region below. (4mks)
y
0.5 2 x
-1.5
-2
SECTION B (48 MARKS)
- Draw the graph of y = -x2 + 3x + 2 for –4 £ x £ 4. Use your graph to solve the equations
(i.) 3x + 2 – x2 = 0 (ii) –x2 – x = -2 (8mks)
- 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 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Using 74.5 as the Assumed mean, calculate:
(i) The mean mark
(ii) The standard deviation (8mks)
- 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
B K
0 a A
(a) Express in terms of a and b
(i) OK (iii) RS
(iii) OS (iv) RA
(b) Hence show that R,S and A are collinear. (8mks)
- 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. (8mks)
- On the same axis, draw the graph of y = 2cosx and y = sin ½x for 00 £ x £ 1800, taking intervals of 150
(6mks)
From the graph, find:
(a) The value of x for which 2cosx = sin ½ x (1mk)
(b) The range of values of x for which –1.5 £ 2cos x £ 1.5 (1mk)
- 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 ¼ hrs earlier.
(a) Find the average speed of A. (6mks)
(b) How far was A from T when B reached S. (2mks)
- 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 ½ hrs later. Calculate:
(a) The distance between ports Q and R. (2mks)
(b) The distance between ports P and R. (3mks)
(c) The bearing of port R from port P. (3mks)
- 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. (13mks)
(b) Graph the inequalities and find the maximum profit he makes from the crops in a year. (5mks)
MATHEMATICS II
PART II
- Use logarithm tables to Evaluate
3Ö 36.5 x 0.02573
1.938 (3mks)
- 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. (3mks)
- If K = ( y-c )1/2
4p
- a) Make y the subject of the formula. (2mks)
- b) Evaluate y, when K = 5, p = 2 and c = 2 (2mks)
- Factorise the equation:
x + 1/x = 10/3 (3mks)
- DA is the tangent to the circle centre O and Radius 10cm. If OD = 16cm, Calculate the area of the shaded Region. (3mks)
- Construct the locus of points P such that the points X and Y are fixed points 6cm apart and
ÐXPY = 600. (2mks)
- 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: (2mks)
- a) ÐABD b) ÐCBD
- 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? (2mks)
- Calculate the distance:
- i) In nautical miles (nm)
- ii) In kilometres (km)
Between the two places along the circle of Latitude:
- a) A(300N, 200E) and B(300N, 800E) (Take Radius of Earth = 6371Km). (2mks)
- b) X(500S, 600W) and Y(500S, 200E) (Take Radius of Earth = 6371Km). (2mks)
- 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)
- 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)
- 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). (4mks)
- 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)
- Find the following indefinite integral ò 8x5 – 3x dx (4mks)
x3
- 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. (3mks)
- 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)
- The figure below shows a cube of side 5cm.
Calculate:
- a) Length FC (1mk)
- b) Length HB (1mk)
- c) Angle between GB and the plane ABCD. (1mk)
- d) Angle between AG and the Base. (1mk)
- e) Angle between planes AFC and ABCD. (2mks)
- f) If X is mid-point of the face ABCD, Find angle AGX. (2mks)
- Draw on the same axes the graphs of y = Sin x0 and y = 2Sin (x0 + 100) in the domain 00 £ x0 £ 1800
- i) Use the graph to find amplitudes of the functions.
- ii) What transformation maps the graph of y = Sin x0 onto the graph of : y = 2Sin (x0 +100).
- 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. |
1 |
2 |
2 |
1 |
6 |
11 |
9 |
7 |
10 |
12 |
16 |
16 |
| Mass(g) |
56 |
57 |
58 |
59 |
60 |
61 |
62 |
63 |
64 |
65 |
70 |
|
| Freq. |
10 |
11 |
9 |
7 |
5 |
3 |
4 |
3 |
3 |
1 |
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.
- 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.
- 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:
- a) Using a suitable scale draw a diagram to show the positions of Aeroplanes after 2hrs.
- b) Use your graph to determine:
- i) The actual distance between the two aeroplanes.
- ii) The bearing of B from A.
iii) The bearing of A from B.
- 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.
- a) it will either rain and not.
- b) Neither will not take place.
- c) One Event will take place.
- Calculate the Area of each of the two segments of y = x(x+1)(x-2) cut off by the x axis. (8mks)
- Find the co-ordinates of the turning point on the curve of y = x3 – 3x2 and distinguish between them.
MATHEMATICS II
PART I
MARKING SCHEME:
- 0.09122 = (9.12 x 10-2)2 = 0.008317
Ö 3.152 = 1.776
3Ö 1.776 + 0.008317
0.1279 x 25.91
= 3Ö 1.784317 No. log
0.1279 x 25.91 1.784 0.2514
0.1279 -1.1069
25.71 1.4101 +
0.5170
-1.7344
x 1/3
10-1 x 8.155(6) 1-1.9115
Or 0.8155(6)
- (a – b)(a – b) = a – b
(a – b)(a + b) a + b
- dy = 2x + 3
dx
y = x2 + 3x + c
-1 = 1 – 3 + c
c = 1 ; E.g y = x2 + 3x + 1
- K2 – 9 = 0
K = ± 3
- log 128 = log 64
18 9
log 16 log 8
6 3
= 2 log (8/3)
log (8/3)
= 2
- Midpoint -8 + 6, 4 + 0 (-1, 2)
2 2
Gradient of LN = 4/-14 = -2/7
Gradient of ^ bisector = 7/2
y – 2 = 7/2
x + 1
y = 7/2X + 11/2
- 207,500 = 415,000(1 – 15 )n
100
0.5 = ( 85 )n
100
0.5 = 0.85n
log 0.5 = n log 0.85
log 0.5 = n
log 0.85
n = –1.6990 = -0.3010 = 4.264yrs
-1.9294 -0.0706
- 2 x 1 = 1 . x 20 = 0.3115 x 20 = 6.230
3.21 x 10-1 3.21
1 = 1 = 0.5807 = 0.005807
172.2 1.722 x 102 100
6.230 – 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 = ½
Area= ½ x ½[29+2(6.25+9+12.25+16+20.25+25)]
= ¼ [29 + 127.5]
= ¼ x 156.5 = 39.125 sq. units.
- Cos q (cos q + ½ ) = 0
cos q = 0 cos q = -0.5
q = 900, 2700 q = 1200, 2400
\ q = 900, 1200, 2400, 2700
- MP = 4 MK MK = -9
9 -18
MP = 4 ( -9 ) = ( -4 )
9 -18 8
\ P is ( -1,0 )
- a = 30 d = 3 l = 300
300 = 30 + 3 (n – 1 )
300 = 30 + 3n – 3
300 – 27 = 3n
273 = 3n
91 = n
- y = mx – 1
1 = 3m – 1
m = 2/3 = 0.6667
tan q = 0.6667 ; q = 33.690
- FK x FC = FA2
FC = 25/3 = 8 1/3 cm
CX = 81/3 – 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
- V3 = k + x
k – x
V3k – V3x = k + x
V3k – k = x + V3x
V3k – k = x( 1 + v3)
V3k – k = x
1 + V3
- (i.) x = 2 Þ x £ 2
(ii) y = -2 Þ y > -2
(iii)pts. (0.5,0)
(0,-1.5)
m = -1.5 – 0 = 3
0 – 0.5
Eq. Y = 3x – 1.5 y < 3x – 1.5
SECTION B
-
| X |
-4 |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
4 |
| Y |
-26 |
-16 |
-8 |
-2 |
2 |
4 |
4 |
2 |
-2 |
(i) Roots are x = -0.5 x = 3.6
(ii) y = -x2 + 3x + 2
0 = -x2 – x + 2
y = 4x (-2, -8) (1, 4)
Roots are x = -2, x = 1
- class x f d=x-74.5 fd d2 fd2
0 – 9 4.5 2 – 70 – 140 4900 9800
10 – 19 14.5 8 – 60 – 480 3600 28,800
20 – 29 24.5 6 – 50 – 300 2500 15,000
30 – 39 34.5 7 – 40 – 280 1600 11,200
40 – 49 44.5 8 – 30 – 240 900 7,200
50 – 59 54.5 10 – 20 – 200 400 4,000
60 – 69 64.5 9 – 10 – 90 100 900
70 – 79 74.5 6 0 0 0 0
80 – 89 84.5 3 10 30 100 300
90 – 99 94.5 1 20 20 400 400
Sf = Sfd = Sfd2 = 77,600
60 -1680
(i) Mean = 74.5 + -1680
60
= 74.5 – 28 = 46.5
(ii) Standard deviation = Ö 77600 – ( –1680 )2
60 60
= Ö 1283.3 – 784
= Ö 499.3 = 22.35
- a (i.) OK = OA + AK = ½ a + ½ b
(ii) OS = ¾ OK = 3/8 a + 3/8 b
(iii)RS = RO + OS = 3/8 a – 9/40 b
(iv) RA = RO + OA = – 3/5 b + a
- RA = a – 3/5 b RS = 3/8 a + 9/40 b
= 3/8( a – 3/5 b)
\ RS = 3/8 RA
The vectors are parallel and they have a common
point R \ point R, S and A are collinear
-
KB = 3m NK = 1.5m XB = 5m
(i) XK = Ö 52 – 32 = Ö 16 = 4m
let ÐXKN = q
cos q = 1.5 = 0.375
4
q = 67.97(8)0
(ii) In DXNK
XN = Ö 42 – 1.52 = Ö 13.75 = 3.708
In D SMR; MR = KB = 3m
SM = XN = 3.708m
Let ÐSRM = a
tan a = 3.708 =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 ½ X |
0.00 |
0.13 |
0.26 |
0.38 |
0.50 |
0.61 |
0.71 |
0.79 |
0.87 |
0.92 |
0.97 |
0.99 |
1.00 |
(a) X = 730 ± 10
(b) Between 40.50 and 139.50
- 300km
T 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 – 300 = 5
x x + 10 4
300(x + 10) – 300x = 5
x(x + 10) 4
300x + 300 – 300x = 5
x2 + 10x
x2 + 10x – 2400 = 0.
x = 44.25
X = -54.25 N/A
(b) Distance covered by A in 1 ¼ hrs = 44.25 x 5/4 = 55.3 km
Distance of A from T is 300 – 55.3 = 244.7 km
- (a) Distance = 15 x 40 = 300km
2
(b)
PR2 = 2002 + 3002 –2x 200 x 300 cos700
= 130,000 – 41040 = 88,960
PR = 298.3 km
(c) 298.3 = 300
sin 700 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
- (i.) X > y
(ii) 4,400X + 10,800Y £ 90,000
Simplifies to 11X + 27y £ 225
(iii) X + y £ 15
X > 0; 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 ® 3X + 4y = 10
Point that give maximum profit is (12,3)
\ maximum profit
= 2400 x 12 + 3200 x 3 = 38,400 shs.
MATHEMATICS II
PART II
MARKING SCHEME
- No log.
36.5 1.5623
0.02573 –2.4104 +
-1.9727
1.938 0.2874 –
-1.6853
-3 + 2.6853
3 3
-1 + 0.8951
1.273(4) ¬ 0.1049
= 1.273(4)
- Let shirt be sh x,
let blouse be sh. y.
5x + 3y =1750 (i.)
3x + y = 850 (ii)
mult (ii) by 3
9x + 3y = 2550 (iii)
Subtract (iii) – (i.)
– 4x = -800
Subt for x
- = 250
Shirt = sh 200 ; Blouse = sh 250
- (a) K2 = y – c
4p
y – c = 4pK2
y = 4pK2 + c
(b) y = 4 x 2 x 25 + 2 ; y = 202
- x2 + 1 – 10x = 0
3
3x2 – 10x + 3 = 0
3x (x – 3) – 1(x – 3) = 0
(3x – 1) (x – 3 ) = 0
x = 1/3 or x = 3
- Area D OAD pyth theorem AD =12.49cm
½ x 12.49 x 10 = 62.45cm2
Cos q = 10/16 = 0.625
q = 51.30 62.5
Sector 57.30 x 3.14 x 100 40.2 –
360 = 22.3
- ÐXPY = 600
\ÐXC1Y = 1200
B1 \ÐC1XY = ÐC1YX
= 1800 – 1200 = 300
2
Construct 300 angles
at XY to get centres
B1 C1 and C2 mojar arcs drawn
2 on both sides with C1X and C2X
as centres.
- DAB = 1800 – 980 = 820
ADB = 180 – (68 + 45 ) = 670
ABD = 180 – (67 + 82)
= 310
(a) 1800 – (67 + 82)0 = 310
ÐABD = 310 Opp = 1800
(b) (180 – 82)0 = 980 82 + 98 = 1800
1800 – (980 – 450) =
ÐCBD = 370 180 – (98 + 45)0
= 370
- 10 x 320
100 Discount = sh 32
Sold at sh 288
If no Discount = ( 320 x 20 ) % = 22.7%
288
- (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 = 100 x 2pR Cos 500
360
100 x 2 x 3.14 x 6371
360 = 5780Km
(b) 80 x 60 Cos 50 = 3895 Km
- Vol =2.8 x 2.4 x 3 = 20.16m3
1m3 = 1000 L
20.16m3 = 20160 L
20160
3600
16560 L to fill
0.5 L – 1 sec
16560 L – ?
165600
5 x 3600
33120 hr
3600 @ 9.41 hrs ; @ 564.6 min.
- 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)3 + 5 (-0.2)4
1 – 1.0 + 0.4 – 0.08 + 0.008 = 0.3277 (4d.p)
- Area of metal : Material – Cross section.
p(R2 – r2)
3.14 (21 –19)
Vol 6.28cm2 x 3400cm
= 215.52m3
- Possibility space:
. 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
P(odd) = 3/6 = ½
P(Sum > 7 but < 10) = 9 /36
\ P(odd) and P(sum > 7 but < 10 )
= ½ x 9/36 = 9/72 = 1/8
- ò( 8x5/x3 – 3x/x3) d4
ò( 8x2 – 3x-2) d4
16x3/3 + 6x-3/-3 + C
16x3/3 – 2/x3 + C
- Area of DAOB
½ x 14 x 14 x 0.866 = 84.866cm2
Area of sector = 60 x3.14 x 14 x14 = 10.257
360
Shaded Area
84.666 – 10.257 = 74.409cm2
-
| Marks |
F |
Fx |
fx2 |
| 5 |
3 |
15 |
75 |
| 6 |
8 |
48 |
288 |
| 7 |
9 |
63 |
441 |
| 8 |
6 |
48 |
384 |
| 9 |
4 |
36 |
324 |
åx = åf=30 åfx=210 1512
S.d = Ö åfx2 – ( åfx )2
åf åf
= Ö 1512 – (210)2
30 30
= Ö 50.4 – 49
= Ö 1.4 = 1,183
SECTION II .
- (a) FC = Ö 52 + 7.072 = Ö 50 = 7.071
(b) HB = Ö 52 + 7.072 = Ö 75 = 8.660
(c) q = Tan-1 5/5 = Tan-1 = 450
(d) b = Tan-1 5/7.071 = Tan-1 0.7071 = 35.30
(e) y = Tan-1 5/3.535 = Tan-1 = 54.70
(f) ÐAGX = 19.40
- y = Sin x
| 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)
| 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 = x + 52 + -4
150
52 – 0.02
= 51.08
Median = 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 |
- X + Y £ 1000
X £ 2Y
Y < 720
X > 90
21.(a) 1cm = 200Km/h
A = 200 x 7.5 = 1500 Km
B = 200 x 9 = 1800Km.
(b) (i.) 15.8cm x 200 (ii) Bearing 2240
= 3160 Km. (iii) Bearing 0490
- (a) P(R) x P(R)1 (b) P(R)¢ x P(R) (c) P(R) x P(R’)
= 0.5 x 0.6 0.5 x 0.4 P(R)’ x P(R)
= 0.3 = 0.2 0.5 x 0.6 = 0.3
0.5 x 0.4 = 0. 2= 0.5
- y = x(x + 1)(x – 2)
= x3 – x2 – 2x
A1 = ò(x3 – x2 –2x) d4
-1[¼ x4 – 1/3 x2]-1
= 0 – ( ¼ + 1/3 – 1) = 5/12
A2 = 2ò(x3 – x2 –2x) d4
= 0ò ¼ x4 – 1/3 x3 – x2)-20
= ( ¼ .16 – 1/3 .8 – 8 )
= 4-0 – 8/3 – 4 = – 8/3
A1 = 5/12= A2 = 2 2/3
- y = x3 – 3x2
dy = 3x2 – 6x
At stationary
Points dy = 0
dx
i.e 3x2 – 6x = 0
3x(x – 2) = 0
x = 0 or 2
Distinguish
dy = 3x2 – 6x
dx
d2y = 6x – 6
dx2
(i) x = 0 dy2 = 6x – 6 = -6 (ii) x = 2
dx2 d2y = 6
-6 < 0 – maximum. dx2
\ (0,0) Max Pt. 6 > 0 hence
Minimum Pt.
x = 2, y = 8 – 12 = -4
(2, -4) minimum point.
MATHEMATICS II
PART I
SECTION 1 (52 Marks)
- Without using tables evaluate:
Ö7.5625 x 3Ö3.375
15 (5 mks)
- Make k the subject of the formula.
y = 1 Ök + y
T2 k (3 mks)
- If A = (x, 2) and xB = x and if AB = (8), find the possible values of x.
-2 (3 mks)
-
Simplify completely. (3 mks)
rx4 – r
2xr – 2r
- Solve the equation. (3 mks)
Log 3 (8-x) – log 3 (1+x) = 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)
- Find the equation of the line perpendicular to the line 4x-y = -5 and passing through the point (-3,-2). (2 mks)
- Find the standard deviation of the data below:
3,5,2,1,2,4,6,5 (4 mks)
- What is the sum of all multiples of 7 between 200 and 300? (4 mks)
- Solve the equation.
½ tan x = sin x for -1800 £ x £ 3600. (3 mks).
- Expand (1-2x)4. Hence evaluate (0.82)4 correct to 5d.p. (4 mks)
- The line y = mx – 3 passes through point (5,2). Find the angle that the line makes with the x-axis. (2 mrks)
- 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)
- 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)
- 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)
- Find the inequalities that describe the unshaded region R below. (4 mks)
y
SECTION 2 (48 Marks)
- Draw the graph of y = x2 + x – 6 for -4 £ x £
Use your graph to solve the equations.
(i) x2 + x – 6 = 0 (ii) x2 + 2x – 8 = 0 (8 mks)
- 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)
- Draw, on the same axes, the graphs of y = cos q and y = 5 sin q for – 1800 £ q £ 1800
- From your graph, determine the amplitude of each wave.
- For what value(s) of q is cosq – 5 sin q = 0 (8 mks)
- 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:
- Ð
- The distance QR, between the two ships.
- The shortest distance from R to the shore. (8 mks)
- 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)
- 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)
- 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. (8 mks)
- 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. (8 mks)
MATHEMATICS III
PART II
Section I: (52 Marks)
- Use mathematical tables to evaluate:
8.67 (3 mks)
Ö 0.786 x (21.72)3
- Simplify completely. (3 mks)
4 – 1
x2 – 4 x-2
- 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)
- The last of three consecutive odd numbers is (2x+3). If their sum is 105, find the value of x. (4 mks)
- a S b is defined by: a S b = (a + b)
ab
If B S (2 S 3) = 4 S 1, Find B. (3 mks)
- Find the value of M. (3 mks)
M
850
1600
- (a) Expand (1+2x)6 upto the term containing x3 . (2 mks)
(b) By putting x = 0.01, find the approximate value of (1.02)6 correct to 4 S.F. (2 mks)
- Show that x is the inverse of : Y = 3 -3 1 X = 2 1 (3 mks)
-5 2 5 3
- The probabilities of three candidates K, M and N passing an examination is 2/3, ¾ and 4/5 Find the probability that :
(a) All pass: (1 mk)
(b) At least one fails: (2 mks)
- In the figure, PR is tangent to the circle centre O. If ÐBQR=300, ÐQBC=270,and ÐOBA=370, find ÐBAC and Ð
C A
B P R
- 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)
- Evaluate 0 (2 mks)
( 3x2 – 1 ) dx
4 x 2
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)
- Write down the three inequalities which define the shaded region. (3 mks)
(3,2)
(2,1) (4,1)
- 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. (3 mks)
- 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 (3 mks)
n(n-7)
SECTION II: (48 Marks)
- Find the distance between points A(500 S, 250 E) and B(500 S, 1400 E) in:
(i) Km (ii) nm (8 mks)
(take radius of earth to be 6400km, P = 3.14)
- The distance S in metres, covered by a moving particle after time t in seconds, is given by :
S = 2t3 + 4t3– 8t + 3.
Find:
(a) The velocity at : (i) t = 2 (ii) t = 3
- The instant at which the particle is at rest. (8 mks)
- 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. (8 mks)
- Using a pair of compass and ruler only, construct triangle ABC such that AB=9cm, BC=14cm and ÐBAC = 1200 . Draw a circle such that AB, BC and AC are tangents. What is the radius of this circle? (8 mks)
- 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) Estimate the median mark. (2 mks)
(b) Using 44.5 as the assumed mean, calculate:-
(i) The mean mark: (2 mks)
(ii) The variance: (2 mks)
(iii) The standard deviation: (2 mks)
- (a) On the same axes, draw the graphs of : y = sin x ; y = cos x
y = cosx + sin X for 00 Ð X Ð 3600 .
(b) 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 x = – sin x for 00 Ð X Ð 3600 .
- 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 Ð DAB = 800.
B
A E
C
D
(a) Write down the equation of the circle in the form ax2 + bx + cy2 + dy + e = 0 where a, b, c, d, e are constants. (2 mks)
(b) Calculate the length DE. (2 mks)
(c) Calculate the value of angle BED. (2 mks)
(d) Calculate the value of angle DCB. (2 mks)
- 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. (8 mks)
(a) Represent all the information above as inequalities.
- How should the contractor deploy his fleet in order to minimise the cost of moving the cement? (8 mks)
MATHEMATICS III
PART I
MARKING SCHEME
|
|
SOLUTION |
MRK |
AWARDING |
|
| 1. |
Ö7.5625 = 2.75
3Ö3.375 = 3Ö3375 X 3Ö10-3
= 3 Ö33 x 53 x 10-1 = 3 x 5 x 10-1 = 1.5
= 2.75 x 1.5 = 2.75 = 0.275
1.5 x 10 10
|
1
1
1
1
1
|
Method for Ö7.5625
Square root
Method for 3Ö
3Ö
Answer |
|
| |
|
5 |
|
|
| 2. |
T2y = Ö k+y
K
T4y2k = k+y
T4y2k – k = y
K(T4y2-1) = y
K = y
T4y2 – 1
|
1
1
1
|
Removal of square root
Rearrangement of terms
Answer |
|
| |
|
3 |
|
|
| 3. |
(x 2) x = (8)
-2
x2 – 4 = 8
x = +Ö12 = + 2Ö3 = + 3.464
|
1
1
1
|
Matrix equation
Quadratic equation
Answers in any form |
|
| |
|
3 |
|
|
| 4. |
r(x2 – 1)
2r(x – 1)
r(x2 – 1)(x2 + 1)
2r (x – 1)
r(x – 1)(x + 1)( x2 + 1)
2r (x – 1)
= (x + 1)( x2 + 1)
2
|
1
1
1
|
Complete factorisation of numerator
Factorisation of denominator
Answer |
|
| |
|
3 |
|
|
| 5. |
1 = log3 3
8 – x = 3
1+x
-4x = -5
x = 5
4 |
1
1
1
|
Logarithic expression.
Equation
Answer
|
|
| |
|
3 |
|
|
| 6. |
Let the centre be (a,b)
4-9 = -1 4-a
-5-b 3-b
4-a = -4+9 -5-b = -3+b
a = 4 b = -1
centre is (4,-1) |
1
1
1
|
Equation
Linear equations
Centre
|
|
| |
|
3 |
|
|
| 7. |
Y = 4x + 5
Gradient = 4
Gradient of ^ line – ¼
y + 2 = – 1
x + 3 4
4y + x = -11
|
1
1
|
Gradient of ^ line.
Equation.
|
|
| |
|
2 |
|
|
|
8. |
X = 28 = 3.5
8
standard deviation = Ö 22 = Ö2.75 = 1.658
8 |
1
1
1
1
|
Mean
d values
d2 values
Answer |
|
| |
|
4
|
|
|
| 9. |
a = 203 d = 7 L = 294
294 = 203 + 7(n-1)
n = 14
S 14 = 14 (203 + 294)
2
= 7 x 497
= 3479
|
1
1
1
1
|
For both a and b
Equation
For n
Sum
|
|
| |
|
4 |
|
|
| 10. |
Sin x = 2 sin x
Cos x
Sin x = 2 cosx
Sin x
2 cos x = 1
cos x = 0.5
x = 600, 3000, -600 |
1
1
1
|
Simplification
Equation
All 3 values |
|
| |
|
3 |
|
|
| 11. |
(1 +-2x)4 = 1-8x + 24x2 – 32x3 + 16x4
(0.82)4 = (1 + -2 x 0.09)4
x = 0.09
(0.82)4 = 1 – 0.72 + 0.1944 – 0.023328 + 0.00119376
= 0.35226576
@ 0.35227 (5 d..p) |
1
1
1
1
|
Expansion
Value of x
All terms
Rounded |
|
| |
|
4 |
|
|
| 12. |
2 = 5m – 3
m = 1
tan q = 1 q = 450 |
1
1
|
Value of m.
Angle |
|
| |
|
2 |
|
|
| 13. |
Let the number be xy
3y = x + 14
10y + x = 10x + y + 36 = 9y – 9x Þ 36
3y – x = 14
9y – 9x = 36
y = 5
x = 1
the number is 15. |
1
1
1
1
|
1st equation
2nd equation
method of solving
Answer
|
|
| |
S |
4 |
|
|
| 14. |
Let ÐAOB = q
q x 2 x 22 x 7 = 11
360 7
q = 900
Area shaded = 90 x 22 x 7 x 7 – 1 x 7 x 7
360 7 2
= 77 – 49
2 2
= 28 = 14cm2
2 |
1
1
1
|
Value of q
Substitution
Answer |
|
| |
|
3 |
|
|
| 15. |
P(WBb) = 6 x 15 x 15
36 35 34
= 15
476 |
1
1
|
Method
Answer |
|
| |
|
2 |
|
|
| 16. |
Equation inequality
L1 y = x y £ x
L2 y = -2 y ³ -2
L3 2y + 5x = 21 2y + 5x < 21 |
1
1
1
1
|
1 mark for each inequality.
Method for obtaining L3
|
|
|
(i) roots are x = -3
x = 2
(ii) y = x2 + x-6
0 = x2 + 2x-8
y = -x + 2
roots are x = -4
x = 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. |
h = 15
h+50 40
h = 30cm
H = 80cm
(a) Volume = 1/3 p x 40 x 40 x 80 – 1/3 p x 15 x 15 x 30
= 128000 p – 6750 p
3 3
= 121,250p cm3
3
(b) L2 = 802 + 402 L = 152 + 302
= 6400 + 1600 = 225 + 900
= 8000 = 1125
L = 89.44 cm L = 33.54 cm
Curved surface area of bucket = p x 40 x 89.44
p x15x33.54
= 3577.6p – 503.1p
= 3074.5cm2 |
1
1
1
1
1
1
1
1
|
Expression
Value of H
Substitution
Volume
L
L
Substitution
Area
|
|
| |
|
8 |
|
|
|
19. |
|
|
|
|
| |
|
|
|
|
|
20. |
(i) ÐRPQ = 130
ÐPQR = 320+900+240 = 1460
ÐPRQ = 1800 – (1460 + 130)
= 210
(ii) P = 7
sin130 sin 210
P = 7 sin 130
Sin 210
= 4.394km
P T
(iii) Let PR = q
q = 7
sin 1460 sin 210
q = 7 sin 1460
sin 210
q = 10.92 km
sin 450 = RT
10.92
RT = 10.92 sin 450
= 7.72 km (2 d..p)
|
1
1
1
1
1
1
1
1
|
Fair sketch
ÐPRQ
Equation
Method
Equation
Distance PR
Equation
RT
|
|
|
|
8 |
|
|
| 21. |
(a) 4x = 9y
2(x+y) = y+44 Þ 2x + y = 44
4x – 9y = 0
4x + 2y = 88
11y = 88
y = 8
x = 18
(b) P(RR) = 18 x 18 = 81
26 26 169
|
1
1
2
1
1
1
1
|
Equation
Equation
Method of solving
Value y
Value x
Method
Answer |
|
| |
|
8 |
|
|
| 22. |
(a) 67,000 Ksh = 67,000 US dollars
16.75
= 4,000 dollars
(b) 2 x 4,000 = 1600 US dollars
5
1600 US dollars = 1600 x 1340
= 2,144,000 Italian lire
(c) Remainder = 2400 US dollars
5 x 2400 = 1500 US dollars
8
1500 US dollars = 1500 x 1.8
= 2700 Deutche marks
(d) Remainder = 900 US Dollars
900 US Dollars = 900 x 16.75 Ksh.
= 15,075 Ksh.
|
1
1
1
1
1
1
1
1
|
Method
Answer
Method
Answer
For 1500
Answer
Method
Ksh. |
|
| |
|
8 |
|
|
| 23. |
PM = kPA
= k(r + 1h)
2
= kr + 1kh
2
PM = PB + BM
3h + t BQ
4
= 3h + t(-3h + r)
4 4
= 3h – 3t h + tr
4 4
= 3 – 3t h + tr
4 4
t = k 3 – 3t = 1k
4 4 2
3 – 3t = 1 t
4 4 2
5t = 3
4 4
t = 3 + 4
4 5
= 3
5
\ k = 3
5
\ PM = 3r + 3h
5 10 |
1
1
1
1
1
1
1
1
|
PM
PM
PM simplified
Both equations
method
Value of k
Value k
PM
|
|
| |
|
8 |
|
|
| |
|
|
|
|
|
24. |
Y
LogT
Log T = log a + x log b
Log T Þ 0.82, 1.25, 1.68, 2.11, 2.54, 2.97, 3.40, 3.84
y – intercept = log a = 0
a = 1
gradient = 3.84 – 0.82 = 3.02
9 – 2 7
= 0.4315
log b = 0.4315 = 0.4315
b = antilog 0.4315
b = 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 of b |
|
MATHEMATICS III
PART II
MARKING SCHEME
- SOLUTION MARKS AWARDING
| 1. |
No log
8.69 0.9390
0.786 1.8954
21.72 1.3369
1.2323
1.7067 – 2
2 + 1.7067
2 2
– 1 + 0.8533
0.7134 x 10 -1 = 0.07134 |
M1
M1
A1
|
ü reading to 4 s.f
Rearranging |
| |
|
3 |
|
| 2. |
4 – 1
(x-2)(x+2) (x-2)
– x+2
(x-2(x+2)
– (x-2)
(x-2(x+2)
-1
x+2 |
M1
M1
A1
|
|
| |
|
3 |
|
|
3. |
Re6000 = Ksh. 75000
Spent 5000 Rem 2500
Rem 2500
1.25
Re 2000 |
M1
M1
A1
|
|
| |
|
3 |
|
| 4. |
2x – 1 , 2z + 1 , 2x + 3
6x + 3 = 105
6x = 102
x = 17 |
M1
M1
A1
A1
|
Allow M1 for us of different variable. |
| |
|
4 |
|
| 5. |
4 * 1 = 5
4
2 * 3 = 5
6
A * 5 = 5
6 4
A + 5 = 5 x 5A
6 4 6
A + 5 = 25 A
6 24
A = 20 |
M1
M1
A1
3 |
|
| 6. |
180 – M + 20 + 95 = 180
295 – M = 180
– M = – 115
M = 115
|
B1
B1
A1
|
|
| |
|
3 |
|
|
7. |
1 + 2x + 60x2 + 160x3 +
1 + 0.2 + 0.006 + 0.00016
= 1.20616
= 1.206 |
M1
M1
M1
A1
4 |
Only upto term in x3.
Correct substitution
Only 4 s.f.
|
| 8. |
3 -1 2 1 = I
-5 2 5 3
6 -5 3 -3
-10 +10 -5 + 6
1 0
0 1
|
M1
M1
A1
|
Matrix multiplication gives :
I 1 0
0 1 |
|
|
3 |
|
|
9. |
(a) 2 x 3 x 4 = 2
3 4 5 5
(b)
2 x 3 x 1 + 2 x 1 x 4 + 1 x 3 x 4
3 4 5 3 2 5 3 4 5
1 + 4 + 1
10 15 5
= 17
10 |
M1
M1
A1
|
|
| |
|
3 |
|
| 10. |
ÐQCB = 300
180 – (27 + 30) = 1230
\ BAC = 570.
OBA = 370
OAB = 370
AOB = 1060
\ ACB = 530
|
M1
M1
A1
|
Isosceles triangle.
Angle at centre is twice angle at circumference. |
| |
|
3 |
|
| 11. |
V = 1 x 3.14 x r 2 x 10 = 270
L.S.F. 20 = 2
30 3
V.S.F = 2 3 = 8
3 27
Vol. of cone = 8 x 270
27 = 80cm3
\ Vol. Of frusturm = (270 – 80) = 190cm3 |
M1
M1
A1
|
|
| |
|
2 |
|
| 12.
|
3x 3 – x -1 2
3 -1 1
x 3 + 1 2
x 1
8 + 1 – ( 1 – 1)
2
8 1 – 2 = 6 1
2 2 |
M1
A1
2 |
|
| 13. |
(2 x 3 x 1.5) volume
9 m3
1L º 1000 cm3
1000 L = 1 m3
9000 L = 9 m3
1000 L = 1 tonne
9000 L = 9 tonnes. |
M1
A1
|
|
| |
|
2 |
|
| 14. |
y ³ 1 (i)
y < x – 1 (ii)
y < 5 – x (iii) |
B1
B1
|
|
| |
|
3 |
|
| 15. |
M1 = 5.1 + 4.9 + 4.7 = 4.9
3
M2 = 4.9 + 4.7 + 4.5 = 4.7
3
M3 = 4.7 + 4.5 + 4.0 = 4.4
3 |
M1
M1
M1
|
|
| |
|
3 |
|
| 16. |
Original contribution per woman = 6300
N
Contribution when 7 withdraw = 6300
(n-7)
Increase – Diff.
6300 – 6300
n-7 n
6300n – 6300(n-7)
n(n-7)
6300n – 6300 + 44100
n(n-7)
44100
n(n-7) |
M1
M1
1
3 |
|
| SECTION II (48 Marks)
|
| 17. |
(i)
1150
A B
Centre of circles of latitude 500 S. R Cos 500
AB = 115 x 2p R Cos 50o
115 x 40192 x 0.6428
360
= 8252.98 km
(ii) Arc AB 60 x 115 Cos 50 nm
60 x 115 x 0.6428 nm
4435 nm
|
M1
M1
M1
A1
M1
M1
M1
A1
|
No. log
60 1.7782
1+5 2.0607
0.6428 1.8080
4435nm 3.6469 |
| |
|
8 |
|
|
18. |
(a) V = ds = 6t2 + 8t – 8
dt
(i) t = 2
V = 6×4 + 8×2 – 8
= 32 ms-1
(ii) t = 3
V = 6×9 + 8×3 – 8
= 70ms-1
(b) Particle is at rest when V = 0
6t2 + 8t – 8 = 0
2(3t – 2) (t+2) = 0
t = 2 t = -2
3
particle is at rest at t = 2 seconds
3 |
|
Do not accept t = -2. Must be stated. |
| |
|
8 |
|
| 19. |
Area under velocity – time.
graph gives distance.
A = { h ½ (y1 + y6 ) + y2 + y3 + y4 + y5 )}
= 1 { ½ ( 12+47) + 24 + 35 + 41 + 45)}
= 29.5 + 14.5
= 174.5m |
B1
B1
M1
M1
B1
B1
A1
|
Trapezium rule only accepted.
Formula.
Substitution into formular. |
| |
|
8 |
|
| 20. |
Drawing actual
Scale 1cm = 2cm
Radius 1cm
= 2cm |
M1
M1
M1
M1
M1
M1
M1
M1
|
Bisect ÐA
Bisect Ð B
Intersection at centre of inscribed circle.
Draw circle.
Measure radius.
Arcs must be clearly shown. |
|
|
8 |
|
|
21. |
mean = 44.5 + 130
100
= 44.5 + 1.3
= 45.8
(b) Variance S (x – A) 2 = 2800
Sf 100
= 28
S.D. = Ö 28 = 5.292
|
M1
A1
M1
A1
M1
A1
|
|
| |
|
8 |
|
|
22. |
y = sin x
x 0 60 120 180 240 30 360
sin x 0 0.866 0.866 0 -0.866 -0.866 0
y = cos x
x q 60 120 180 240 300 360
cos x 1 0.5 -0.5 -1.0 -0.5 0.5 1.0
y = cosx + sinx
x q 60 120 180 240 30 360
cosx + sinx 1 1.366 0.366 -1 -1.366 -0.366 1.0
(c) Cos x = – sin x
x = 450 , 2250
|
|
|
|
23. |
(i) amplitude = 1.366
(ii) Period = 3000
(a) (x+1) 2 + (y-6)2 = 32
x2 + 2x + 1 + y2 – 12y + 36 = 9
x2 + 2x + y2 – 12y + 28 = 0
(b) cos 10 = OD DE = 3
DE 0.9848
DE = 3.046
(c) Twice ÐOED
100 x 2 = 200
(d) DAB = 800
\ DCB = 1000 |
M1
A1
M1
A1
M1
A1
M1
A1
|
Formular
(x-a)2 + (y-b)2 = r2
Cyclic quad.
|
| |
|
8 |
|
| 24. |
Let number of trips by 12 tonne lorry be x.
Let number of trips by 7 tonne lorry be y.
(a) x > 0 ; y > 0
24x + 21y £ 150
12 x 25 x X + 15 x 7 x y £ 1200
300x + 105y £ 1200
x > 2y
(b) Ref. Graph paper.
Minimising:
3 – 12 tonne lorry and 2 – 7 tonne lorries should be deployed. |
B1
B1
B1
|
|
MATHEMATICS IV
PART I
SECTION 1 (52MKS)
- Evaluate using logarithms 3Ö7.673 – 15.612
12.3 (4mks)
- Solve x – 3x – 7 = x – 2 (3mks)
3 5 5
- In the given figure CD is parallel to BAC, calculate the values of x and y. (3mks)
C D
B A
- 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. (3mks)
- A line perpendicular to y = 3-4x passes through (5,2) and intercepts y axis at (0,k)
Find the value of K. (3mks)
- 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. (3mks)
- The co-ordinate of point A is (2,8) vector AB = 5 and vector BC = 4 Find the
-2 3
co-ordinate of point C. 3mks)
- 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)
- 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
- Express 3.023 as a fraction (2mks)
- Point A (1,9), Point B(3,5) and C (7,-3). Prove vectorically that A,B and C are collinear. (4mks)
- 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 sales (4mks)
- 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)
- Rationalize the denominator in Ö3
Ö 7 – 2
Leaving your answer in the form Öa + Öb
C
Where a ,b, and c are integers (3mks)
- For positive values of x, write the integral solutions of 3£ x2 £ 35 (4mks)
- 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 II (48 mks)
- In the given circle centre O , A,E,F, is target to the circle at E. Angle FED = 300 <DEC = 200 and <BC0 = 150
A F
Calculate (i) <CBE (3mks)
(ii) <BEA (2mks)
(iii) <EAB (3mks)
- The sum of the 2nd and third terms of a G.P is 9/4 If the first term is 3,
(a) Write down the first 4 terms of the sequence . (5mks)
(b) Find the sum of the first 5 terms using positive values of the common ratio (r)
(3mks)
- 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) (8mks)
- In the domain –2 £ x £ 4 draw the graph of y = 3x2 + 1 –2x .Use your graph to solve the equation. 6x2 4x + 4 = 0 (graph paper provided) (8mks)
- 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. (5mks)
(b) If the density of the wire is 5g/cm3. Calculate the mass of the sphere in kg. (3mks)
- A right cone with slant 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 formed (8mks)
9cm
- 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, (6mks)
(b) Find the equation of the diagonal SQ (2mks)
- 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 1½ hours. Tap A is then closed. Find how long it takes to drain the tank.
(8mks)
MATHEMATICS IV
PART II
SECTION I (52MKS)
- Without using mathematical tables, evaluate (3mks)
Ö 0.0784 x 0.27 (leave your answer in standard form)
0.1875
- 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? (3mks)
- 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 ABCE (3mks)
E D C
A B
- Given that a = -2, b = 3 and c = -1, Find the value of a3 – b – 2c2 (2mks)
2b2 – 3a2c
- The exchange rate in January 2000 was US $ 1 = Ksh 75.60. and UK £1 = Ksh 115.80. A tourist came to Kenya with US $ 5000 and out of it spent ksh.189,000. He changed the balance in UK £ . How many pounds did he receive? (4mks)
- ABC is a cross – section of a metal bar of uniform cross section 3m long. AB = 8cm and AC = 5cm.
Angle BAC = 600 . Calculate the total surface area of the bar in M2. (4mks)
- The bearing of a school chapel C, from administration block A, is 2500 and 200m apart.
School flag F is 150m away from C and on a bearing of 0200. Calculate the distance and
bearing of A from F. (5mks)
- 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 balls (2mks)
(ii) If 2 balls are chosen at random without replacement, find the probability that they are of different colour. (2mks)
- Under an enlargement of linear scale factor 7, the area of a circle becomes 441.p
Determine the radius of the original circle. (3mks)
- A circle has radius 14cm to the nearest cm . Determine the limits of its area. ( 3mks)
- Expand (1 + 2x)5 up to the term with x3. Hence evaluate 2.045 to the nearest 3 s.f. (4mks)
- The nth term of a G.P is given by 5 x 2 n-2
(i) Write down the first 3 terms of the G.P (1mk)
(ii) Calculate the sum of the first 5 terms (2mks)
- 3 bells ring at intervals of 12min, 18min and 30min respectively. If they rang together at 11.55am, when will they ring together again. (3mks)
- On a map scale 1:20,000 a rectangular piece of land measures 5cm by 8cm. Calculate its actual area in hectares. (3mks)
- 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 (3mks)
- Three angles of a pentagon are 1100, 1000 and 1300. The other two are 2x and 3x respectively. Find their values . (2mks)
SECTION II (48MKS)
- 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)
- A box contains 5 blue and 8 white balls all similar . 3 balls are picked at once. What is the probability that
(a) The three are white (2mks)
(b) At least two are blue (3mks)
(c) Two are white and one is blue (3mks)
- 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. (2mks)
(b) Tiles needed for 15 such rooms are packed in cartons containing 20 tiles. How many cartons are
there in total? (2mks)
(c) Each carton costs shs. 800. He spends shs. 100 to transport each 5 cartons. How much would one
sell each carton to make 20% profit ? (4mks)
- The following was Kenya`s income tax table in 1988.
Income in K£ P.a Rate (Ksh) £
1 – 2100 2
2101 – 4200 3
4201 – 6303 5
6301 – 8400 7
(a) Maina earns £ 1800 P.a. How much tax does he pay? (2mks)
(b) Okoth is housed by his employer and therefore 15% is added to salary to make taxable income. He
pays nominal rent of Sh.100 p.m His total tax relief is Shs.450. If he earns K£3600 P.a, how much
tax does he pay? (6mks)
- In the given figure, OA = a , OB =b, OP: PA =3:2, OQ:QB = 3:2
Q
B
R
O A
(a) Write in terms of a and b vector PQ (2mks)
(b) Given that AR = hAB where h is a scalar, write OR in terms h, a. and b (2mks)
(c) PR = K PQ Where K is a scalar, write OR in terms of k, a and b (1mk)
(d) Calculate the value of k and h (3mks)
- 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 A1B1C1 onto A11 (-6,2) B11(-2,3) and C11(-6,6). Find the matrix Q and plot
A11B11C11on the same grid. Describe Q fully. (8mks)
- 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 circle centre 0 on BC to touch AB and
AC produced at P and Q respectively. Calculate the area of the circle. (8mks)
- The following were marks scored by 40 students in an examination
330 334 354 348 337 349 343 335 344 355
392 341 358 375 353 369 353 355 352 362
340 384 316 386 361 323 362 350 390 334
338 355 326 379 349 328 347 321 354 367
(i) Make a frequency table with intervals of 10 with the lowest class starting at 31 (2mks)
(ii) State the modal and median class (2mks)
(iii) Calculate the mean mark using an assumed mean of 355.5 (4mks)
MATHEMATICS IV
PART 1
MARKING SCHEME
| 1. |
Ö – 7.939
12.3
= No log
7.939 0.8998
12.3 1.0899
T.8099 1/3 = 3 + 2.8099 T.9363 3
= -0.8635 |
B1
B
M1
A1
4
|
Subtraction
Logs
Divide by 3
Ans |
| 2. |
5x – 3 (3x –7 ) = 3(x – 2 )
5x – 9x + 21 = 3x – 6
-7x = -27
x = 36/7
|
M1
M1
A1
3 |
Multiplication
Removal ( )
Ans |
| 3. |
3x +5y + x = 180
9x = 180
x = 20
y = 60 |
M1
A1
B1
3 |
Eqn
X
B
|
| 4. |
. r = 3v 1/3
4P
. r = S ½
4P
\ 3V 1/3 = S ½
4P 4P
3V = S 3/2
4P 4P
V = 4P S 3/2
3 4P
|
B1
M1
A1
3
|
Value r
Equation
Expression |
|
5.
6. |
Grad line = ¼
y – 2 = ¼
x – 5
y = ¼ x + ¾
k = ¾
P in Alloy = 4/10 x 800
= 320g
= 100 x 320
20
= 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) , C (x ,y)
.a – 2 = 5
.b – 8 -2
.a = 8 b = 6 B(8, 6 )
x – 8 = 3
y – 6 4
x = 11, y = 10 c(11,10)
|
B1
M1
A1
3 |
B conduct
Formular
C |
| 8. |
80 – x
.h = x tan 70
h = (80 – x ) tan 60
\ x tan 70 = 80 tan 60-x tan 60
2.7475x + 1.732x = 138.6
4.4796 x = 138.6
.h = 138.6 x tan 60
4.4796
= 53.59 |
B1
M1
M1
A1
4 |
Expression for h both
Equation
Expression for h
Ans |
| 9. |
2pr = 90 x 2p x 50
360
r = 12.5
h = Ö2500 – 156.25
= Ö2343.75
= 48.41 cm
|
M1
P
A1
M1
A1
4 |
Equation
.r
expression for h
ans
|
|
10. |
100 n = 302.323
n = 3.023
99n = 299.3
n = 2993
990
= 323/990 |
M1
A1
4
|
Equation
Ans
|
| 11. |
AB = 3-1
5-9
= 2
-4
BC = 4
-8
AB = ½ BC
\ AB // BC
But B is common
\ A,B,C are collinear.
|
B1
B1
B1
3 |
A B & BC
Both
Both
|
| 12. |
4% of 200,000 = 8000/=
balance = 4200/=
6% of x = 4200/=
x = 4200 x 100
6
= 70,000
sales = sh. 270,000 |
B1
M1
A1
B1
4
|
Both
Expression
Extra sales
Ans
|
|
13 . |
Time = 22/7 x 3.5/2x 3.5/2 x 200 hrs
22/7x 140x140x 140x 3600
= 8960
3600
= 2 hrs 29min |
M1
M1
M1
A1
4
|
Vol tank
Vol tank
Div x 3600
Tank
|
| 14.
|
Ö3 = Ö3 Ö7 + Ö2
Ö7Ö2 Ö2Ö2 Ö7+ Ö2
= Ö3 Ö7 + Ö2
5
= Ö21 + Ö6
5 |
M1
M1
A1
3 |
Multi
Expression
Ans |
| 15. |
3 £ x 2 x2 £ 35
±1.732 £x x £ ± 5.916
1.732 £ x £ 5.916
integral x : 2, 3, 4, 5
|
B1
B1
B1
B1
4 |
Lower limit
Upper limit
Range
Integral values
|
| 16. |
No of days = 8/6 x 5/8 x 12
= 10 days |
M1
A1
2 |
Expression
days |
| 17. |
(i) ÐCED = ÐECD = 30
Ð CDE = 180 – 60
= 120
Ð CBE = 180-120
=60
(ii) Ð AEC = 90+30
= 120
Ð EAB = 180-(120+45)
= 150
(iii) ÐBEO = 90-45
= 45 |
B1
B1
B1
B1
B1
B1
B1
B1
8 |
ÐA EB = 450
ÐBEO |
| 18. |
.ar + ar2 = 9/4
3r + 3r2 = 9/4
12r2 + 12r – 9 = 0
4r2 + 3r – 3 = 0
4r2 + 6r – 2r –3 = 0
(2r – 1) (2r + 3) = 0
r = ½ or r = -11/2
Ss = 3(1- (1/2 )5)
1 – ½
= 3 (1-12/3 2)
½
= 6 ( 31/32)
= 6 31/32
|
B1
B1
B1
M1
A1
M1
M1
A1
8
|
|
| 19. |
LOG E. 0.3010 0.6021 0.7782 0.9031
LOG F 1.2068 2.1065 2.6354 3.0103
Log E =n log F + Log K
.n = gradient = 2 2.4 – 1.4 = 12 = 3
Log k. = 0.3 0.7 – 0.3 4
.k = 1.995
¾ 2
E = 2F 3 |
B1
B1
S1
P1
L1
M1
A1
B1
8
|
Log E
Log F
Scale
Plotting
Line
Gradient
K
|
| 20 |
.x -2 -1 0 1 2 3 4
.y 17 6 1 6 9 22 41
.y = 3x 2 – 2x + 1 –
0 = 3x 2 – 3x – 2
y = x + 3
|
B2
B1
B1
S1
P1
C1
L1
B1
8
|
All values
At least 5
Line
Scale
Plotting
Smooth curve
Line drawn
Value of r
|
| 21. |
.h = ¾ p x 18 x 18x 18
p x 0.04 x 0.04
= 24 x 18x 18x 18
0.04 x 0.04 x 100
= 48,600m
density = 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
¸ to length in cm
¸ for length
conversing to metres
length
expression for density
conversion to kg
ans
|
| 22. |
H = Ö152 – 92
= Ö144
= 12
X/6 = 9/12
X = 4.5
Volume = 1/3 x 22/7x (81 x 12 –20.25×6 )
= 22/21 (972 – 121 -5)
= 891 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 is diagonal
(a) Mid point PR (0,0)
a + 5 = 0
2
.a = -5
b- 0 = 0
2
b = 0
R (-5,0)
Grad PQ = -2
Grad RS = -2
.d – 0 = -2
c –5
.d – 0 = ½
c+5
.d+ 2c = 10
2d – c = 5×2 –
4d – 2c = 10
5d = 20
d = 4
c = 3
Q (3, 4)
x + 3 , y+4 = (0,0)
2 2
x = -3 , y = -4 \ s(-3 -4)
(b) y – 4 = 8
x – 3 6
3y = 8x – 12 |
B1
M1
M1
A1
M1
A1
M1
A1
8 |
Ans .
Expression both correct
Equation
Ans
Expression
Equation
|
| |
|
|
|
MATHEMATICS IV
PART II
MARKING SCHEME
| 1. |
784 X 27 =
187500
Ö 784 x 9 = 4 x 7x 3
62500 250
= 42
125
= 0.336
|
M1
M1
A1 |
Factors for
Fraction or equivalent
C.A.O |
| |
|
3 |
|
| 2. |
Father 3x , r son = x
2(x +10) = 3x + 10
2x +20 = 3x + 10
x = 10
father = 30 |
M1
A1
B1
|
Expression
|
| |
|
3 |
|
| 3. |
3 = sin 60
AE
AE = 3
Sin 60
= 3.464
perimeter = 5×2 + 3.464 x 3
= 10+10.393
= 20.39 |
M1
A1
B1 |
Side of a triangle
Perimeter |
| |
|
3 |
|
| 4. |
.a3 – b-2c2 = (-2)3 – 3 –2(-1)2
2b2 – 3a2c 2(3)2 –3(-2)2(-1)
= -8 –3-2
18 + 12
= -13
30 |
M1
M1
A1 |
Substitution
Signs
C.A.O |
| |
|
3 |
|
| 5. |
Ksh 189,000 = $ 189,000
75.6
= $ 2500
balance = $ 2500
= Kshs. 189,000
Kshs. 189,000 = 189,000
115.8
Uk ₤1632 |
M1
A1
M1
A1
A1
4
|
Conversion
Conversion
|
| 6. |
Area of 2 triangles = 2 (½ x 8x 5 sin 60)
= 40 sin 60
= 40x 0.8660
= 34.64 cm2
Area of rectangle = 300 x 8 + 300 x 5 +300 x BC
BC = Ö64 +25 – 2 x 40cos 60
= Ö89 – 80 x 0.5
= Ö89 – 40
= Ö49
= 7
Total S.A. = 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. |
AF2 = 32+42+-2+12x cos 50
= 25 – 24 x 0.6428
= 25-15.43
= 9.57
AF = 3.094 x 50
AF = 154.7m
Sin Q = 200 sin 50o
154.7
= 0.9904
Q = 82.040
Bearing = 117.960 |
M1
A1
M1
A1
B1
|
Bearing |
| |
|
5 |
|
| 8. |
(i) No. of white = w
w = 2
w+9 3
3w = 2w + 18
w = 18
(ii) p(different colour ) = p(WB N BW)
= 2 x 9 + 9 x 18
3 25 27 25
= 12/25 |
M1
A1
M1
A1 |
|
| |
|
4 |
|
| 9. |
A.sf = 1
49
smaller area = 1 x 441 p
49
= 9p
pr2 = 9p
r2 = 9
r = 3 |
M1
M1
A1
|
|
| |
|
3 |
|
| 10. |
Largest area = 22 x (14.5)2
7
= 660.8 cm 2
smallest area = 22/7 x (13.5)2
= 572.8
572.8 £ A £ 660.81 |
M1
M1
A1 |
|
| |
|
3 |
|
| 11. |
(1 +2 x)5 = 1 + 5 (2x) + 10 (2x)2 + 10 (2x)3
= 1 + 10x + 40x2 + 80x3
2.0455 = 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. |
Tn = 5x 2n –2
(i) T1 , T2, T3 = 2.5, 5, 10
(ii) S5 = 2.5(25-1)
2-1
= 2.5 (31)
= 77.5 |
B1
M1
A1 |
All terms
|
| |
|
3 |
|
| 13. |
12 = 22 x 3
18 = 2 x 32
30 = 2x3x5
Lcm = 22 x 32x 5 = 180 min
= 3hrs
time they ring together =11.55 +3 = 2.55 p.m |
M1
A1
B1 |
|
| |
|
3 |
|
| 14. |
Map area = 40cm 2
Actual area = 200x200x40m2
= 200x200x40ha
100×100
= 320ha |
M1
M1
A1 |
Area in m2
Area in ha
CAO |
| |
|
3 |
|
| 15. |
3p + 2r = 13
p + 2r = 9 –
2p = 4
p = sh 2
r = 3.50 |
M1
A1
B1 |
|
| |
|
3 |
|
| 16. |
110 + 100+130+2x +3x = 540
5x = 200
x = 400
2x , 3x = 80 and 1200 res |
M1
A A1
2 |
|
| 17. |
Contribution / person = 180,000
X
New contribution = 180,000
x – 2
180,000 – 180,000 = 24,000
x –2 x
180,000x – 180,000x +360,000 = 24,000(x-2)x
24,000x2 – 48,000x – 360,000 =0
x2 – 2x – 15 = 0
x2 – 5x + 3x – 15 = 0
x (x – 5)+ 3 (x – 5) = 0
(x + 3 )(x – 5) = 0
x = -3
or = 5
remaining members = 5-2
= 3 |
B1
B1
M1
M1
A1
M1
A1
B1
|
‘C’
eqn
mult
eqn
factor
both ans
remaining members |
| |
|
8 |
|
| 18. |
(a) P (3 white) = 8 x 7 x 6 = 28
13 12 11 143
(b) P(at least 2 blue)=p(WBBorBBWorBWB)orBBB
= 8 x 5 x 4 + 5 x 4 x 8
13 12 11 13 12 11
+ 5 x 8 x 4 + 8 x 7 x 6
13 12 11 13 12 11
= 204
429
= 68
143
(c) p(2 white and one blue )= p(WWB or WBW or BWW)
= 8 x 7 x 5 + 8 x 5 x 7 + 5 x 8 x 7
13 12 11 13 12 11 13 12 11
= 3 x 8 x 7 x 5
13 x 12 x 11
= 70
143 |
M1
A1
M1
M1
A1
M1
M1
A1
|
|
| |
|
8 |
|
| 19. |
(a) recourt area = 10.5 x 6 m2
title area = 0.3 x 0.3 m2
No of tiles = 10.5 x 6
0.3 x 0.3
= 700
(b) No of cartons = 700 x 15
20
= 52.5
(c) Cost of 525 cartons = 525 x 100 + 800 x 525
+ transport 5
= 10,500+420,000
= 430,500
sale price = 120 x 4.30,500
100
= sh 516,600
s.p of a carton = 516,600
525
= sh. 984 |
M1
A1
M1
A1
B1
M1
M1
A1
|
|
| |
|
8 |
|
| 20. |
(a) Maina`s tax dues = 1800 x 10
100
= 180
(b) Taxable income = 3600 x 115 – n rent
100
= 36 x 115 – 100 x 12
20
= 4140 – 60
= 4080
Tax dues = 10 x 2100 + 15 x 1980
100 100
= 210 + 297
= 507
Tax relief = 270-
Tax paid = 237 |
M1
A1
M1
A1
M1
M1
A1
B1
|
1st slab
2nd slab |
| |
|
8 |
|
| 21. |
(a) PQ = –3/5 a + 3/1b
= 31/2 – 3/5 a
(b) OR = h a + h b
= a – ha + hb
= (1-h) a + h b
(c) OR = 3/5 a + k (31/2 b – 3/5a)
= (3/5 – 3/5k)a +3k b
(d) 1 – h = 3/5 – 3/5k (i)
3k = h (ii)
Sub (i) 1 – 3k = 3/5 – 3/5k
5- 15k = 3-3k
12k = 2
k = 1/6
h = ½
|
B1
M1
A1
M1
A1
M1
A1
B1 |
|
| |
|
8 |
|
|
22. |
P(ABC) = 0 – 1 1 4 3 = -3 -1 -3
1 0 3 1 3 1 4 3
A1 (-3,1)B1 (-1,4)C1(-3,3)
Q(A1B1C1) = a b -3 –1 -3 = -6 –2 –6
c d 1 4 3 2 8 6
=> -3a + b = -6 -3c + d = 2
-a + 4b = -2 x 3 -c + 4d = 8 x 3
– 3a + 12b = -6 – 3c + 12d = 24
11b = 0 -11d = -22
b = 0 d = 2
a = 2 d = 2
c = 0
Q = 2 0
0 2
|
M1
A1
M1
M1
A1
B1
B1
B1
|
A1 B1 C1
L Q
A1 B1 C1 drawn
All BII CII
Ploted
Destruction
|
| |
|
8 |
|
| 23.
24. |
R = 2.2CM ± 0.1
Area = 22 x 2.2 x 2-2
7
= 15.21cm2
Ef =40 efd = -80
(ii) model class = 351- 360
modern class = 341 – 350
(iii) mean = 355.5 – 80
40
= 355.5 – 2
= 353.5
|
B1
B1
B1
B1
B1
B1
M1
1
1
8
B1
B1
M1
A1
B1
B1
B1
B1
|
|
| |
|
A1 |
|
| |
|
8 |
|
MATHEMATICS V
PART I
SECTION 1 (52 MARKS)
- Use logarithms to evaluate 6 Cos 40 0.25
63.4 (4mks)
- Solve for x in the equation (x + 3) 2 – 5 (x + 3) = 0 (2mks)
- In the triangle ABC, AB = C cm. AC = bcm. ÐBAD = 30o and ÐACD = 25o. Express BC in terms of b and c. (3mks)
- Find the equation of the normal to the curve y = 5 + 3x – x3 when x = 2 in the form
ay + bx = c (4mks)
- Quantity P is partly constant and partly varies inversely as the square of q. q= 10 and p = 5 ½ when q =20. Write down the law relating p and q hence find p when qs is 5. (4mks)
- Solve the simultaneous equation below in the domain 0 £ x £ 360 and O£ y £ 360
2 Sin x + Cos y = 3
3 Sin x – 2 Cos y = 1 (4mks)
- Express as single factor 2 – x + 2 + 1
x + 2 x2 + 3x + 2 x + 1 (3mks)
- By use of binomial theorem, expand (2 – ½ x )5 up to the third term, hence evaluate (1.96)5
correct to 4 sf. (4mks)
- 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. (4mks)
- 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)
- Make T the subject of the formulae 1 – 3m + 2
T2 R N (3mks)
- Vectors, a = 2 b = 2 and c – 6
2 0 4
- By expressing a in terms of b and c show that the three vectors are linearly dependent. (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)
- A piece of wood of volume 90cm3 weighs 54g. Calculate the mass in kilograms of 1.2 m3 of the wood. (2mks)
- 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)
- 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? (2mks)
SECTION II (48MARKS)
- The tax slab given below was applicable in Kenya in 1990.
Income in p.a. rate in sh
1 – 1980 2
1981 – 3960 3
3961 – 5940 5
5941 – 7920 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. (8mks)
- 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. Express in terms of a and b.
(i) OR (1mk)
(ii)AP (1mk)
(b) Find the ratio in which Y divides AP (6mks)
- 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. 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) They are of the same colour. (4mks)
(b) At least one is blue (2mks)
(c) At most 2 are red (2mks)
21. Point P(50oN, 10oW) are on the earth’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. (3mks)
(b) If the time at Q when the plane lands is 11.20am what time is it in P. (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. (3mks)
22. Using a ruler a pair of compasses only, construct :
(a) Triangle ABC in which AB = 6cm, AC = 4cm and Ð ABC = 37.5o. (3mks)
(b) Construct a circle which passes through C and has line AB as tangent to the circle at A. (3mks)
(c) One side of AB opposite to C, construct the locus of point P such that ÐAPB = 90o. (2mks)
23. A particle moves in a straight line and its distance is given by S = 10t2 – t3 + 8t where S is
distance in metres at time t in seconds.
Calculate:
(i) Maximum velocity of the motion. (4mks)
(ii) The acceleration when t = 3 sec. (2mks)
(iii) The time when acceleration is zero. (2mks)
-
A rectangle ABCD has vertices A(1,1) B(3,1), C(3,2) and D(1,2). Under transformation
matrix M = 2 2 ABCD is mapped onto A1B1C1D1
1 3
under transformation M = -1 0 A1B1C1D1 is mapped onto A11B11C11D11. Draw on the given grid
0 –2
(a) ABCD, A1B1C1D1 and A11B11C11D11 (4mks)
(b) If area of ABCD is 8 square units, find area of A11B11C11D11. (3mks)
(c) What single transformation matrix maps A11B11C11D11 onto A1B1C1D1 (1mk)
MATHEMATICS V
PART II
SECTION 1 (52 Marks)
- Evaluate without using mathematical tables (2.744 x 15 5/8)1/3 (3mks)
- If 4 £ x £ 10 and 6 £ y £5, calculate the difference between highest and least
(i) xy (2mks)
(ii) y/x (2mks)
- 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. (4mks)
- Matrix 1 2x has on inverse, determine x (3mks)
x +3 x2
- 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. (4mks)
- The diagonals of triangle ABCD intersect at M. AM = BM and CM = DM. Prove that triangles ABM and CDM are Similar. (3mks)
- Given that tan x = 5/12, find the value of 1 – sinx
Sin x + 2Cos x, for 0 £ x £ 90 (3mks)
- 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)
- MTX is tangent to the circle at T. AT is parallel to BC. Ð MTC = 55o and Ð XTA = 62o. Calculate Ð (3mks)
- 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. (4mks)
- 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)
- Without using logarithm tables, evaluate log5 (2x-1) –2 + log5 4 = log5 20 (3mks)
- Mumia’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)
- The interior angles of a regular polygon are each 172o. Find the number of sides y lie polygon. (2mks)
- Evaluate 2x = 2 + 3
341 9.222 (2mks)
- A water current of 20 knots is flowing towards 060o. A ship captain from port A intends to go to port
B 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. (3mks)
SECTION II (48 MARKS)
- 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)
- In the domain –5 £ x £ 4, draw the graph of y = x2 + x – 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 – 6 = 0. (8mks)
- The average weight of school girls was tabulated as below:
| Weight in Kg |
30 – 34 |
35 – 39 |
40 – 44 |
45 – 49 |
50 – 54 |
55-59 |
60-64 |
| No. of Girls |
4 |
10 |
8 |
11 |
8 |
6 |
3 |
(a) State the modal class. (1mk)
(b) Using an assumed mean of 47,
(i) Estimate the mean weight (3mks)
(ii) Calculate the standard deviation. (4mks)
-
The table below shows values of y = a Cos (x – 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 b (2mks)
(b) Complete the table (2mks)
(c) On the same axes draw the graphs of y = across(x – 15) and y = b sin(x + 30) (3mks)
(d) Use your graph to solve ½ cos (x – 15) = sin(x + 30) (1mk)
21. The diagram below is a clothing workshop. Ð ECJ = 30o AD, BC, HE, GF are vertical
walls. ABHG is horizontal floor. AB = 50m, BH = 20m, AD=3m
(a) Calculate DE (3mks)
(b) The angle line BF makes with plane ABHG (2mks)
(c) If one person requires minimum 6m3 of air, how many people can fit in the workshop (3mks)
- To transport 100 people and 3500 kg to a wedding a company has type A vehicles which take 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) Write down 3 inequalities in A and B which are the number of vehicles used and plot them
in a graph. (3mks)
(b) What is the smallest number of vehicles he could use. (2mks)
(c) 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 function (3mks)
A circle centre A has radius 8cm and circle centre B has radius 3cm. The two centres are
12cm apart. A thin tight string is tied all round the circles to form interior common tangent. The tangents CD and EF intersect at X.
(a) Calculate AX (2mks)
(b) Calculate the length of the string which goes all round the circles and forms the tangent.
(6mks)
- 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) Calculate the actual speed of the plane. (3mks)
(b) What course does the pilot take to reach B? (3mks)
(c) How long does the whole journey take? (2mks)
MATHEMATICS V
PART I
MARKING SCHEME
| 1 |
SOLUTION |
MKS |
AWARDING |
| |
No Log
13.6 1.1335 +
Cos 40 1.8842
1.0177 –
63.4 1.8021
1. 2156
(4 + 3.2156) 1/4
1.8039
Antilog 0.6366 |
B1
M1
M1
A1 |
Log
+
divide by 4
C.A.O |
| |
|
4 |
|
| 2. |
(x + 3) (x + 3 – 5) = 0
(x +3)b (x – 2) = 0
x = -3 or x = 2 |
M1
A1
|
Factors
Both answers |
| 3 |
BD = C Sin 30 = 0.05
CD = b Cos 25
= 0.9063b
BC = 0.9063b + 0.5 C |
B1
B1
B1
|
BD in ratio from
CD in ratio form
Addition |
| |
|
3 |
|
| 4 |
Dy = 3 – 3x2
dx
x = 2, grad = 1
9
Point (2,3)
y – 3 = 1
x – 2 9
9y – 27 = x – 2
9y – x = 25 |
B1
B1
M1
A1
|
Grad equ
Grad of normal
Eqn
Eqn
|
| |
|
4 |
|
| 5 |
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 – 2 Cos y = 1
7 sin x = 7
Sin x = 1
X = 90
Cos y = 1
Y = 0o |
M1
M1
A1
B1 |
Elim
Sub
|
| 7 |
2(x +1) – 1(x + 2) + x + 2
(x+2) (x +1)
= 2x +2 – x – 2 + x = 2
(x +2) (x + 1)
= 2x + 2
(x + 2) (x + 1)
= 2
x + 2 |
M1
M1
A1 |
Use of ccm
Substitution
Ans |
| 8 |
(-2 – ½ x)5 = 25 – 5 (2)4 ( ½ x) + 10(2)3( ½ x)2
= 32 – 40x + 20x2
= 32 – 4 (0.08) + 20 (0.08)2
= 32 – 0.32 + 0.128
= 3 |
M1
A1
M1
A1 |
|
| |
|
4 |
|
| 9. |
Circle centre C = (3 +1, 0 + 4)
2 2
C( 2, 2)
R =Ö (2 – 0)2 + (2 – 3)2
=Ö 5
(y – 2)2 + (x – 2)2 = Ö5
y2 + x2 – 4y – 4x = 8 + Ö5 |
B1
B1
M1
A1 |
Centre
Radius
|
| |
|
4 |
|
| 10 |
ar2 =2, ar5 = 16
a = 2 \ 2 r5 = 16
r2 r2
2r3 = 16
r3 = 8
r = 2, a = ½
S5= ½ (1 – ( ½ )5)
½
= 1 – 1/32
= 31/32 |
M1
A1
M1
A1 |
Both
Sub
CAO |
| |
|
4 |
|
| 11 |
NR – 3MT2 = 2RT2
T2(2R + 3M) = NR
T2 = NR
2R + 3m
T = ! Ö NR
2R + 3m |
M1
M1
A1 |
X mult
72
ans |
| |
|
3 |
|
| 12 |
2 = m 2 + n 6
2 0 4
2 = 2m + 6n
2 = 0 + 4n
n = ½
m = – ½
\a = – ½ b + ½ c
\a b c are linearly dep |
M1
A1
B1 |
|
| |
|
3 |
|
| 13 |
Volume = 22 x 2.1 x 2.1 x 2 x ¾ 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 = 54 x 1.2 x 1,000,000
90 1000
= 720kg |
M1
A1 |
|
| |
|
2 |
|
| 15 |
V3 = P
P(0.9)3 = 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 |
Taxable income = 8100 + 2400
= sh. 10,500
= ₤6300
Tax dues = Sh 1980 x 2 + 1980 x 3 + 1980 x 5 + 3670 x 7
12
= 22320
12
= Sh 1860
net tax = 1860 – 800 p.m.
= Sh 1060
Total deduction = 1060 + 150 + 730
= 1940
Net salary = 10,500 – 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 – a
OY = m OR = A + n (2/5b – a)
2/5m b + ma = (1 – n)a + 2/5 n b
2/5m = 2/5n
m = n
\m = 1 – m
2m = 1
m = ½ = n
½ 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 – 0.5
0.4 + 0.2
= 1.25
0.6
= 2.08
n = 2
\y = 8x2
x = 3 y = 8 x 32 = 72
y = 200 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)
= ½ (2/5 x ¼ + 3/5 x 2/4) x 2
= 2 + 6
20 20
= 8
20
= 2/5
(b) P(at least 1B) = 1 – P(non blue)
= 1 – P (XRR or YRR)
= 1 – ½ (2/5 x ¼) x 2
= 1 – 1/10
= 9/10
(c) P(at most 2 Red) = 1 – P (BB)
= 1 – ½ (3/5 x 2/4)2
= 1 – 6/20
= 14/20 or 7/10 |
M1
M1
M1
A1
M1
A1
M1
A1 |
Any 2
Any 2
Fraction
|
| |
|
8 |
|
| 21 |
(a) PQ = 1800nm
q = 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 o = 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
AB AC
^ at A
Bisector AC
Circle
Ð AB
Locus P with A B excluded |
| |
|
8 |
|
| 24 |
A1B1 C1D1
2 2 1 3 3 1 = 4 8 10 6
1 3 1 1 2 2 4 6 9 7
A11 B11 C11 D11
-1 0 4 8 10 6 = -4 –8 -10 -6
0 –2 4 6 9 7 -8 -12 -18 -14
NM = -1 0 2 2
0 –2 1 3
= -2 -2
-2 -6
(b) det = Asf = 12 – 4 = 8
Area A11 B11 C11 D11 = 8 x 8
= 64 U2
(c) Single matrix = Inv N
= ½ -2 – 0
0 –1
= -1 0
0 – ½
|
B1
B1
B1
M1
A1
B1
|
Product
Product
Det
Inverse |
| |
|
6 |
|
|
23 |
Ds = 20t – 3t2 + 8 =0
Dt 3t2 – 20t – 8 = 0
T = 20 ! Ö400 + 4 x 3 x 8
6
t = 7.045 sec
max vel = 148.9 – 140.9 – 8
= 0.9 m/s
d2 s = 6t – 20
dt2
when t = 3 a = -2m/s2
6t – 20 = 0
6t = 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 |
2744 x 125 1/3
1000 8
2744 1/3 x 53 1/3
1000 23
23 x 73 1/3 x 5
103 2
2 x 7 x 5 = 3.5
10 2 |
M1
M1
A1 |
Factor
Cube root
|
| |
|
3 |
|
| 2 |
(i) Highest – 10 x 7.5 = 75
Lowest – 6 x 4 = 24 –
51
(ii) Highest = 7.5 = 1.875
4
Lowest = 6 = 0.600
10 1.275 |
M1
A1
M1
A1 |
Highest
Fraction
|
| |
|
4 |
|
| 3 |
Cos q = 17 = 0.8095
21
q = Cos 0.8095
= 36.03o
Arc length = 72. 06 x 2 x 22 x 21
360 7
= 26.422cm |
M1
A1
M1
A1 |
q |
| |
|
4 |
|
| 4 |
x2 – 2x(x +3) = 0
x2 – 2x2 – 6x = 0
-x2 – 6x = 0
either x = 0
or x = 6 |
M1
M1
A1 |
Equ
Factor
Both A |
| |
|
3 |
|
|
5 |
8 = x x 2 x 22 x 28 Cos 60o
360 7
8 = x x 44 x 28 x 0.5
360 7
x = 8 x 360 x 7
44 x 28 x 0.5
= 32.73o
= 33o |
M1
M1
A1
B1 |
x exp |
|
6 |
ÐDMC = Ð AMB vert. Opp = q
ÐMAB = Ð MDC = 180 – q BASE Ls of an isosc. <
2
Ð MBA = Ð MAC 180 – q base angles of isos <
2
<’s AMC and < CDM are equiangle
\ Similar proved
|
B1
B1
B1 |
|
| |
|
3 |
|
| 7 |
Tan x = 5/12
h = Ö b2 + 122
= Ö25 + 144
= Ö169
= 13
1 – Sinx = 1 – 5
sin x + 2 Cos x 5/13 + 2 x 12/13
12/13 = 12 x 13 = 12
29/13 13 29 29 |
M1
M1
A1 |
Hypo
Sub
|
| |
|
3 |
|
| 8 |
Y = x 2 + 2
Area = h (y1, = y2 +……..yn)
= 1(2.225 + 4.25 + 8.25 +14.25 + 22.25)
= 51.25 sq units |
B1
M1
A1 |
Ordinals |
| |
|
3 |
|
|
9 |
ÐCBA = 117o
Ð ACD = 55
Ð BAC = 180 – (117 + 55) = 8o |
B1
B1
B1
3 |
|
| 10 |
|
B1
B1
B1
B1 |
1994
1996
1997
1998 |
| |
|
4 |
|
| 11. |
Xy = 35
y = 35/x
9x – 9y = -18
Sub x2 + 2x – 35 = 0
x2 + 7x – 5x – 35 = 0
x (x + 7) – 5(x + 7) = 0
(x – 5) (x + 7) = 0
x = -7
x = +5
y = 7
Smaller No.
= 57
= 75 |
B1
M1
A1
B1
|
|
| |
|
3 |
|
| 12 |
Log5 (2x – 1 )4 = log552
20
4(2x – 1) = 52
20
2x – 1 = 25
5
2x – 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 + 45y = 52x + 40
-7x = -54
x/y = 5/7
x : y = 5 : 7 |
B1
M1
M1
A1 |
|
| |
|
4 |
|
| 14 |
2n – 4 it angle = 172
n
(2n – 4) x 90 = 172n
n
90 (2n – 4) x 90 = 172
n
180 n – 360 = 172n
180n – 172n = 360
8n = 360
n = 45 |
M1
A1
M1 |
|
| |
|
2 |
|
| 15 |
2 x = 2. 1 + 3. 1
6.341 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 = 1 B = 1 C – 1
50 25 20
In one hour all taps can fill = 1 + 1 + 1 = 11
50 25 20 100
In 6hrs all can fill = 11 x 6 = 33 parts
100 50
taps A and B can fill = = 1 + 1 = 3 part in 1 hr
50 25 50
In 4 1 hrs, A and B = 25 x 3 + 1
6 6 50 4
Parts remaining for B to fill = 1 – 33 + 1 = 1 – 91 = 9 parts
50 4 100 100
Time taken = 9 x 25 hrs = 2 ¼ hrs
100 1
7.30 am
6. hrs
13.30
4.10
5.40pm
2.15
7.55 pm
|
M1
B1
B1
B1
M1
A1
|
|
|
18 |
x2 + x – 8 = -2 – 2x
y = x2 + 3x – 6
Points of intersection (-4, 1.4)
y = x2 + x – 8 = 2x2 + 3x – 6
x2 + 2x + 2
y = x2 + x – 8 x 2
2y = 2x2 + 2x – 16
0 = 2x2 + 3x – 6
2y = -x – 10
y = – 2.6
Ny = 1.2
|
8
B1
B1
B1
B1 |
Eqn
Point of inter
Line eqn
Both |
|
19 |
(a) Modal class = 45 – 49
(i) Mean = 47 + -55
50
= 47 – 1.1
= 45.9
(ii) Standard deviation = Ö 3575 – –55 2
50 50
= Ö71.5 – 1.21
=Ö 70.29
= 8.3839
|
4
B1
B1
M1
A1
B1
|
fd
fd2 |
| |
|
8 |
|
| 20 |
(a) a = 1
b = 2
½ cos (x – 15) = Sin (x + 30)
has no solution in the domain |
B1
B1
B1
B1 |
All
All
A & b
|
| |
|
8 |
|
| 21 |
(a) O Cos 30 = 20
X
X = 20
0.866
= 23.09
DE = Ö 502 + 23.092
= Ö 2500 + 533.36
= Ö 3033.36
= 55.076m
(b) GB = Ö 202 + 502
= 53.85
Tan q = 14.55
53.85
= 0.27019
q = 15.12o |
B1
M1
A1
M1
A1 |
|
|
|
8 |
|
| |
(c) Volume of air = 50 x 20 x 3 + ½ x 20 x 11.55 x 50
= 3000 + 5775
= 8775
No. of people = 8775
6
= 1462.5
j 1462 |
M1
M1
A1 |
|
| |
|
8 |
|
| 22 |
(a) A + B [ 16
5A + 3B ³ 50
2A + 3B [ 35
(b) 14 vehicles
(c) A – 6 vehicles
B – 8
Cost = 6 x 1000 + 8 x 1200
= 6000 + 9600
= 15,600/= |
B1
B1
B1
M1
A1
|
In equation 3
Vehicles |
| |
|
8 |
|
|
23 |
x = 8
12 – x 3
= 8.727
FBX = 3 = 0.9166 = 23.57
3.273
3FBX = 47.13
Reflex Ð FBD = 312.87
Reflex arc FD = 312.87 x 22 x 6
360 7
= 16.39cm
Reflex Arc CE = 312.87 x 22 x 16
360 7
= 43.7cm
FE (tangent) = Ö144 – 121
= Ö 23
= 4.796cm
2 FE = 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) 200 = 40
Sin 50 Sin q
Sin q = 40Sin 50
200
= 0.7660
5
=0.1532
q = 8.81o
Ð ACB = 180 – (50 + 8.81)o
= 121.19o
x = 200
Sin 121.19 Sin 50
x = 200 x Sin 121.19
Sin 50
= 200 x 0.855645
0.7660
= 223.36Km/h
(b) Course = 330o – 8.81o
= 321.19o
(c) Time = 600
321.19o
= 2.686 hrs
|
M1
A1
M1
M1
A1
B1
M1
A1
8
|
|
