Categories: Teachers' Resources

KCSE REVISION MATHEMATICS (QUESTIONS & ANSWERS)

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

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 135⁰with 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)

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

S 39⁰

Calculate R

(a) ÐRST (1 Mk)

(b) ÐSUT (3 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)

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)

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

2 R (3,3)

-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)

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)

.. B

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

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 120^o 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)

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)

(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 P^IQ^IR^I of PQR under translation T = 3 on the provided grid 4 (2 Mks)

(b) Under transformation matrix m = 4 3 , P^IQ^IR^I is mapped on to P^IIQ^IIR^II. Find the

co-ordinates of P^IIQ^IIR^II and plot it 1 2 on the given grid (4 Mks)

MATHEMATICS I

PART 1

MARKING SCHEME

171 X 171 X 3 X 10^-5 M1

855 X 531

= 2 X 10^-6 A1

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

2 X 5 X 2240

= 75 A1

Extra men = 75 – 6 = 69 B1

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

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

Reflex ÐAOC = 126 x 2 = 252⁰ B1

Obtuse ÐAOC = 360 – 252 = 108⁰ B1

= ¹/₂ (180 – 108)⁰

= 36⁰ B1

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

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

(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

P (FF) = ⁷/₂₀ X ⁷/₂₀

= ⁴⁹/₁₀₀ B1

P (at least one fail) = 1 – P (F^IF^I F^I)

= 1- ¹³/₂₀³ M1

= 1 – 2197 M1

8000

= 5803

8000 A1

grad = term 135

= -1 B1

y = mx + c

y = -x + 5 B1

Volume = 2 x 10,000 x 10,000 x 25 M1Ö x section area

1000 10 M1Ö conv. to litres

= 500,000 Lts A1

10a – 6(3a – 7) = 5(a -2) M1

10a – 18a + 42 = 5a – 10

– 13a = -52 M1

a = 4 A1

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

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

= (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

25⁰ B1Ö sketch

55⁰

Q 125 PR = 4 sin 125 M1

Sin 25

P 3

(a) <RST = 180⁰ – 74⁰ = 106⁰ B1

(b) < RTQ = 90⁰– 74⁰ = 16⁰ B1

< PTR = 46⁰ + 16⁰ = 62⁰ B1

< SUT = 62⁰ – 39⁰ = 23⁰ B1

= 180 – 32 = 148⁰ B1

Obtuse ROT = 360 – 148 = 212⁰ B1

(d) < PTS = 46 + 180 – 129 = 97⁰ B1

< PST = 180 – (97 + 39) = 44⁰ B1

(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

= 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

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

B1Övalues

S1Ö scale

8 — P1Ö plotting

6 — C1 Ö curve

4 —

-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

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

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

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

CLASS

60 – 64

65 – 69

70 – 79

75 – 79

80 – 84

85 – 89

496

938

1296

1155

246

174

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.

(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 = ¹/₂X 10 X 12 sin 41.41 – ²²/₇ X 2.645² M1 Ö D & circle

= 39.69 – 21.99

= 17.7 cm² A1Ö

(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

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

r² + hr = ^A/₂_p 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

= 2 (5 + 2Ö3) – 1 (5 – 2Ö3) M1

(5 – 2Ö3) (5 + 2Ö3)

= 10 + 4Ö3 – 5 +2Ö3 M1

= 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

Volume = ¹/₂X 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

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

= 2 B1

% error = ²/₂₆X 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

(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

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

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

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

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

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

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

(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

= P(2 correct or 3 correct)

= ⁴⁶/₁₂₅ + ¹²/₁₂₅ M1

= 46 + 12 M1

125

= 58

125 A1

Old price/pen = 440

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

(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

= 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

(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

(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

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 ® P^IQ^IR^I

4 P^I (0,5), Q^I (2,2) R^I (1,0)

P^I Q^I R^I P^II Q^II R^II

(b) 4 3 0 2 1 = 15 14 4 M1 Ö

1 2 5 2 0 10 6 1 A1 Ö

P^II (15,10), Q^II (14,6), R^II (4,1) B1Ö

= 5

area of P^II Q^II R^II = 5 (area P^IQ^IR^I)

= 5 X 3.5 M1Ö

= 16.5 cm² A1

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 < 360^Oif Cos (2 θ + 120) = γ3 (3mks)

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 = 45⁰and 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 A^I B ^I C^I D^I. under a second transformation

0 – 2

which has a rotation centre (0,0) through –90⁰ , the image A^II B^II C^II D^II of A^I B^I C^I D^I 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 = 80^O 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 255⁰from 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)

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

x² – 5 x = ^– ½

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

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= 315^o ( 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

Tan 67.5 = h

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/s_2-(2)

17.

Triangle (8)

AC = 9cm

Circumference Centre

Circle

Perpendicular bisector of AB

(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

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 = 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

V₂ = – 20 – 80 X impossible

speed uphill = 30 km /hr, T = 75 time = 2 ½ hrs

speed downhill = 50 km /hr Time = 75 Time = 2 ½ hr

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

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

24.

a) AD = 50km

DC = 35km

BD = 90km

Bearing is 020⁰

Bearing is 134^o (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 140^o. Find

(i) The distance of A to C (2mks)

(ii) The bearing of A from C (3mks)

Expand (x + y)⁶hence 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 60^o and 30^o 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 = 60^o), 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 x^o and y = cos 2x^o on the same axes for –180 £ x £180^o.

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

-4 2

1 3 3 2 -3 x 1 3 3 7

18 18

-4 2 4 2 y -4 2 5

x 36

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)

= 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 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

x = -20 ± 220

Spd = 100 km/h

And x = 120 km/h (A)

Log x = a log y = b

Log x³ = Log x³ – log y ^½

= 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	Ft²v
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)

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 r²x 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 £ 360⁰and 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)

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 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)

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

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 = 45⁰and BAE = 98^0.

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)

(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

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 = 8^1/3 – 9/2 = 23/6 = 3^5/6 cm

CX x XK = XA x XN

3^3/6 x 3/2 = 3 x XN

\ XN = 1^11/12 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

= 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

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

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

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

(b)

PR² = 200² + 300² –2x 200 x 300 cos70⁰

= 130,000 – 41040 = 88,960

PR = 298.3 km

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

y – c = 4pK²

y = 4pK² + c

(b) y = 4 x 2 x 25 + 2 ; y = 202

x² + 1 – 10x = 0

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

Construct 30⁰ angles

at XY to get centres

B1 C₁and 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 = 98⁰82 + 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

(d) b = Tan^-1 5/7.071 = Tan^-10.7071 = 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

i.e 3x²– 6x = 0

3x(x – 2) = 0

x = 0 or 2

Distinguish

dy = 3x² – 6x

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)

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 = ab^xwhere 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)

If B S (2 S 3) = 4 S 1, Find B. (3 mks)

Find the value of M. (3 mks)

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

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 = cos^x + 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.

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.

A E

(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)

(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 = ³Ö3³x 5³x 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 = log₃3 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	1^st equation 2^nd 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 50^o 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

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 2^nd and third terms of a G.P is ⁹/₄If 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 = KFⁿWhere 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 n^th 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)

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)

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

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)

(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 A¹B¹C¹on the given grid.

Transformation Q maps A¹B¹C¹onto 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

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

A1

Subtraction

Logs

Divide by 3

Ans

5x – 3 (3x –7 ) = 3(x – 2 )

5x – 9x + 21 = 3x – 6

-7x = -27

x = 3⁶/₇

A1

Multiplication

Removal ( )

Ans

3x +5y + x = 180

9x = 180

x = 20

y = 60

B1

Eqn

. r = 3v ¹/₃

. r = S ^½

\ 3V ¹/₃ = S ^½

4P 4P

3V = S ³/₂

4P 4P

V = 4P S ³/₂

3 4P

A1

Value r

Equation

Expression

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

= 3.2 kg

M 1

A 1

Equation

P in alloy

Expression

Ans

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)

A1

B conduct

Formular

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

A1

Expression for h both

Equation

Expression for h

Ans

2pr = 90 x 2p x 50

360

r = 12.5

h = Ö2500 – 156.25

= Ö2343.75

= 48.41 cm

A1

Equation

expression for h

ans

10.

100 n = 302.323

n = 3.023

99n = 299.3

n = 2993

990

= 3²³/₉₉₀

A1

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

A B & BC

Both

12.

4% of 200,000 = 8000/=

balance = 4200/=

6% of x = 4200/=

x = 4200 x 100

= 70,000

sales = sh. 270,000

B1

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

A1

Vol tank

Div x 3600

Tank

14.

Ö3 = Ö3 Ö7 + Ö2

Ö7Ö2 Ö2Ö2 Ö7+ Ö2

= Ö3 Ö7 + Ö2

= Ö21 + Ö6

A1

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

Lower limit

Upper limit

Range

Integral values

16.

No of days = ⁸/₆ x ⁵/₈ x 12

= 10 days

A1

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

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})

= 6 ( ³¹/₃₂)

= 6 ³¹/₃₂

A1

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

Log E

Log F

Scale

Plotting

Line

Gradient

.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

B1

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

A1

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

A1

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

.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

A1

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 50^o 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 2^{n –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	1^st slab 2^nd 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 A^ll B^II C^II 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 = 30^o and ÐACD = 25^o. 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(50^oN, 10^oW) 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 = 37.5^o.                                (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 = 90^o.             (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(50^o, 155^oE) 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 = 55^o and Ð XTA = 62^o. 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)

A²digit 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 172^o. 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 060^o. 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 350^o.

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 = 30^o 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)

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 330^o. Wind is blowing at a speed of

40km/h from 200^o. 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 = 0^o	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 = 47^o Q (50^oN, 37^oE) (b) Time diff = 47 x 4 60 = 3.08 Time at P = 9.12am (c) QR = 2700 nm x ^o = 2700 60 = 45^o R (85^oN, 133^oW)	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 5^{3 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 = 36.03^o 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 60^o 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 = 32.73^o = 33^o	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 +……..y_n) = 1(2.225 + 4.25 + 8.25 +14.25 + 22.25) = 51.25 sq units	B1 M1 A1	Ordinals
		3
9	ÐCBA = 117^o Ð ACD = 55 Ð BAC = 180 – (117 + 55) = 8^o	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 140^o 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 = 15.12^o	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 = 8.81^o Ð ACB = 180 – (50 + 8.81)^o = 121.19^o 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 = 330^o – 8.81^o = 321.19^o (c) Time = 600 321.19^o = 2.686 hrs	M1 A1 M1 M1 A1 B1 M1 A1 8

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	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

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

	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

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

KCSE REVISION MATHEMATICS (QUESTIONS & ANSWERS)

MATHEMATICS I

PART I

Calculate R

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

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

Income in Kshs. p.a rate of tax %

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)

ON:NB = 5:-2, AP:PB = 1:3

V

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

855 X 531

2 X 5 X 2240

Maina’s = 4/9 X 180

Omondi’s = 60/- and Maina’s

Difference = Shs.20/- B1 difference

8000 A1

Difference = 55.44 – 6.5625 M1Ö sub

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

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

Line (iv) y – 3 = -3

CLASS

60 M1

P = 64 2q2 = 66

No. of scouts = 32 sin 60 X 10 M1 Ö volume

AQ = 5/9 x 12 = 62/3 cm

M1

125 A1

2 x 2 2x 8

Log K = 0.08

2 θ + 120 = 330

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

DE DC DE CE

v + 20

SECTION B:

MATHEMATICS I

MARKING SCHEME.

R -1 1.1 Log = 0.04139

Sin B Sin A

A B A B

2 1 0 1 4 9

Large circle area

Large area = 22/7 x4x4 = 352/7 (A)

Length of diag. 50 = 7.071 = 5.536

Volume passed / sec = cross section area x speed

MATHEMATICS II

Using 74.5 as the Assumed mean, calculate:

MATHEMATICS II

MATHEMATICS IV

A1

A1

B1

A1

A 1

A 1

A1

A1

A1

A1

B1

B1

A1

A1

B1

A1

B1

A1

LOG E. 0.3010 0.6021 0.7782 0.9031

B1

B1

A1

A1

A1

MATHEMATICS IV

PART II

MARKING SCHEME

A1

A A1

1

MATHEMATICS V

Line (ii) ^y/₄ + ^x_/-3 = 1

Line (iii) grad = ^-1/₃ y inter = 4

P = 64 2q² = 66

AQ = ⁵/₉ x 12 = 6²/₃ cm

	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

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