# Mathematics Paper 1 Questions and Answers - Lainaku II Joint Mock Examination 2023

Instructions to candidates

• This paper consists of TWO sections I and II.
• Answer ALL questions in section I and any five from section II.
• Show all the steps in your calculations giving your answers at each stage in the spaces below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculation and KNEC Mathematical tables may be used.

SECTION I (50 Marks)
Answer all questions in this section in the spaces provided

1. Without using mathematical table or a calculator, evaluate: (3 mks)
36 – 8x -4 – 15 ÷ −3
3x -3 + -8 (6 - (-2)
3. Solve the inequality −3x + 2 < x + 6 ≤ 17 – 2x and write down the integral values satisfying the inequality. (3 mks)
4. A triangle whose area is 16cm2 has its shortest side as 12cm. Find the length of the shortest side of a similar triangle whose area is 36cm2 (3mks)
5. Solve for x given that 52x + 2 – 20 x 52x = 625 (3mks)
6. The line y = mx + 6 makes an angle of 780 581 with the x – axis. Find the co-ordinates of the point where the line cuts the X – axis. (3mks)
7. 3g of metal A of density 2.7g/cm3 is mixed with 1.6cm3 of metal B of density 3.2g/cm3
Determine the density of the mixture (3mks)
8. A two digit number is such that when the digits are reversed, the value of the number increases by 36. If the sum of the unit digit and twice the tens digit is 16, find the number. (3mks)
9. On Monday the currency exchange rate was
1 Euro (E) = Kshs.95.65
1 US dollar(\$) = Ksh.76.50
A gentle man Tourist decided to exchange half of his 2400E into Dollars.
Calculate to 2 decimal places the number of dollars he received. (3 marks)
10. In the figure given below, AC is an arc of a circle centre B. Angle ABD = 60°, AB = BC = 7cm and CD = 5 cm.

Calculate (3 d.p)
1. The area of triangle ADB (2mks)
2. The area of the shaded region. (2mks
11. If cos A = − 40 and A is obtuse, find without using tables or calculators
41
1. Sin A (2mks)
2. Tan A (1mk)
12. A matatu left town X at 9.37a.m. towards town Y at an average speed of 78km/h. At 9.47a.m. A car left the same venue at an average speed of 84.5km/h. Determine the time when the car caught up with the matatu . (3mks)
13. Using a set square, a ruler and a pair of compasses, divide the given line into five equal parts. Measure the length of one part. (3 marks)

14. Simplify the expression.    x2 + 14x + 49 (3mks)
x2 − 49
15. The acceleration of a particle in ms-2 is given by the expression Acc=3t –4
Given that t=0 sec V=3m/s and S=0 metres
Find:
1. an expression for velocity Vms-1 (2 mark)
2. An expression for distance S metres from a fixed point O. (2 marks)
16. A salesman earns a basic salary of Kshs.19, 000 per month. In addition, he earns a commission of 5% for all sales above ksh.20,000. In February 2023, he sold goods worth 115,000. Calculate his total earnings that month . (3mks)

SECTION II (50 MARKS)

1.  A triangle with A(-4, 2), B(-6, 6) and C(-6, 2) is enlarged by a scale factor -1 and centre (-2, 6) to produce triangle A1 B1 C1. Triangle A1 B1 C1 is then reflected in the line y = x to give triangle A11 B11 C11
1. Draw triangle ABC, A1 B1 C1 and A11 B11 C11 and state the co-ordinates of A1 B1 C1 and A11 B11 C11 (6mks)
2. If triangle A11 B11 C11 is mapped onto A111 B111 C111 whose co-ordinates are A111(0,−2), B111(4, −4) and C111 (0, −4) by a rotation. Find the centre and angle of rotation (4mks)
2. A cylindrical water tank of diameter 5m and height 1.8m is supplied with water by pipe P of internal radius 2.5cm. Water flows through this pipe at the rate of 50m per minute. A drainage pipe Q can empty the full tank in 8 hours.
1. Calculate the time in hours that pipe P alone would take to fill the empty tank (4mks
2. The tank is initially half full. Water flows through pipe P into the tank for 2 hours. The drainage pipe Q is then opened and both pipes left running.
Determine how long it will take to fill the tank (6 mks)
3. Ken and Jane cycle to school 20km away. Jane cycles at 2km/h faster than Ken and reaches there half an hour earlier. Given that the speed of Ken is xkm/hr. Find in terms of x
1. The time taken by Ken. (1 mk)
2. The time taken by Jane. (1 mk)
3. Form an equation in x. (1 mk)
4. Solve the equation in (a) above and find the speed of Ken and Jane. (7 mks)
4. The table below shows some paired values of X and Y for a known curve.
 X 0 0.2 0.4 0.6 0.8 1 Y 0 0.4 1.6 3.6 6.4 10
By establishing how x and y relates.
Estimate the area under the curve for the interval 0 < X< 1 using
1. The mid – ordinate rule with five mid – ordinates. (4mks)
2. The trapezium rule with five Trapezia. (2mks)
3. If the exact area is 10/3 square units.
Calculate the percentage error in the two estimates. (4mks)
5. The following are speeds in m/s of the first 50 vehicles at a police check point.
 Speed   in m/s                              x No. of  vehicles   f                             f(x) 10 - 19 20 – 29 30 - 39 40 - 49 50 - 59 60- 69 70 – 79 80 - 89 90- 99 100 -109 3 1 2 5 611 9 8 3 2

1. Calculate the mean speed (4mks)
2. Draw on the same axes using this information
1. The Histogram (3 mks)
2. The Frequency polygon (1 mk)
3. Calculate the median speed. (2mks)
6. A trader sold an article at sh.4800 after allowing his customer a 12% discount on the marked price of the article. In so doing he made a profit of 45%.
1. Calculate
1. The marked price of the article. (2 marks)
2. The price at which the trader had bought the article (2 mks)
2. If the trader had sold the same article without giving a discount. Calculate the percentage profit he would have made. (3 mks)
3. To clear his stock, the trader decided to sell the remaining articles at a loss of 12.5%. Calculate the price at which he sold each article.
(3 mks)
7. The diagram below shows a parallelogram OPQR. Point M divides line PQ in the ration 2:3 PR and OM intersect at N. Given that OP = p and OR = r

1. Express the following vectors in terms of   (2 mk)

2. Given that
By Expressing ON in two different ways find the ratio in which N divides PR. (8 mks)
8. The diagram below shows the graph of a moving matatu from Nakuru to Naivasha.

1. Find the acceleration of the matatu. (2mks)
2. Find the deceleration of the matatu (2mks)
3. Calculate the distance the matatu travelled while accelerating. (2mks)
4. Calculate the distance the matatu covered while traveling at an acceleration of 0m/s2 (2mks)
5. Find the distance between the two Towns. (2 mks)

MARKING SCHEME

SECTION A
1.    36 + 32 + 5

3x −3 + −8 (8)
73
− 73
= − 1

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A1

Simplification of num
Simplification of deno
03
2.

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Removal of fractional powers.
Removal of negative power and simplim
03
3.    −3x + 2 < x + 6
−4 x < 4
x > −1
x + 6 ≤ 17 – 2x
3x ≤ 11
X ≤ 11/3
−1< x ≤ 3 2/3
Integral values of x={ 0, 1, 2, 3}

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03
4.      A.S.F = 16:36
A.S.F = 4/9
L.S.F = √4/9
= 2/3

2/3 = 12/x
= 12 x 3
2
= 18 cm

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For  L.S.F

Equating the ratios
03
5.     Let U=52X

52X(52) - 20(52X) = 625
25U - 20U = 625
U=125
5U = 625
5        5
U = 125

52X = 53
2x = 3
X = 3/2
x = 1.5

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A 1

Expressing in the same base
04
6.  Gradient = Tan 78° 581
= Tan 78.97°
y = 5.13x + 6
At x –axis, y = 0
0 = 5.13x + 6
X = − 1.17
Co-ordinate (−1.17, 0)

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Substitution for y=0

Co- ordinates

04
7.   Metal A: Volume =   mass
density
=   3g
2.7 g/cm3
= 1.111cm3
Metal B: Mass = d x v
= 1.6 x 3.2
= 5.12g
Total mass = 5.12 + 3 (mixture)
= 8.12g
Total volume = 1.111 + 1.6
= 2.711
Density =   8.12
2.711
= 2.995g/cm3

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For both vol &mass
04
8.

Let no.be = xy
10x + y is the value of the number
⇒ reversing digits, the new value is 10y + x
∴ 10y + x – (10x + y) = 36
10y + x – 10x – y = 36
9y – 9x = 36
9(y – x) = 36

y – x = 4…..(i)
y + 2x = 16 …..(ii)
y – x = 4
y + 2x = 16
−3x = −12
x = 4
and y – 4 = 4
y = 8
∴ number is 48

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For both equations✔

✔elimination of one variable or equivalent

03
9.  ½ of 2400E = 1200E

In ksh. = 1200E x 95.65
= Ksh.114,780
Number of dollar = Kshs.114,780
76.50

= 1500.39 dollars ( to 2 d.p)

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For conversion.
03
10.
1.
Area of triangle ADB is ½ x 7 x 12 sin 60°
7 x 6 sin 60°
= 36.373cm2 (to 3 d.p)
2.
60 x 22 x 7 x 7 = 25.6667
360    7
Shaded area = 36.373 − 25.667
= 10.706cm2 (to 3 d.p)

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Exp for the area
04
11.

Sin A = 9/41
Tan A = − 9/40 (the ans must be –ve)

BI

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Sin Ratio
Accept 0.2195

Tan Ratio
Accept − 0.225

03
12.    9.47 – 9.37 = 10min
60min = 78km/hr
= 10 x 78
60
Dist. apart = 13km
R.S = (84.5 – 78) = 6.5km/hr
Time taken = 6.5
Time to
Catch up = 9.47a.m. + 2hrs
= 11.47 a.m.

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Exp for Dist apart
03
13.
Length of one part = 2.5cm

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Drawing a line at an acute angle

Subdividing the line

Accept 2.4cm or 2.6cm

03
14.   Num = (x + 7) (x +7)

Den = (x + 7) ( x – 7)
(x + 7) (x +7)
(x + 7) ( x – 7)
x + 7
x - 7

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Num simplified

Deno simplified

03
15.

dv = 3t – 4
dt

1. V= ∫3t – 4dt
V= 3t2 − 4t + c
2
Since V = 3 when t = 0
Then
3(0)2 − 4(0) + c = 3
2
c = 3
V= 3t2 − 4t +3
2
2. S =
= t3 – 2t2 +3t + k
2
= S = 0, when t = 0
k = 0
s = t3 – 2t2 + 3t
2

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04
16.   Extra sales = Kshs. 115,000

− 20,000
Kshs. 95,000
Commission earned = x 95,000 = 4,750
100
Total earning = 4,750 + 19,000
= Kshs. 23,750

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Exp for the commission
03
17.  (a)  B1 for object drawn
B2 1st image drawn
B1 coordinates of 1st image
B2 for 2nd image & its coordinates given

(b)  B1 for the 1st bisector
B1 for the 2nd bisector
B1 for the centre of rotation C(5,−1)
B1 for the angle of rotation =180° or equivalent

10
18.
1. Rate of water flow
= 22/7 x 2.5 x 2.5 x 5000
= 98,214.28571cm3
Volume of tank = 22/7 x 2.5 x 2.5 x 1.8
= 35.35714286 m3
35.35714286 x 1000000
98214.28571
= 360mins
= 6hrs
2. Half full ⇒ pipe has worked for 3hrs
Total hours = 2 + 3
= 5 hrs
Fraction of
Tank without water = 1 – 5/6
= 1/6 of tank
Both taps working together
= 1/61/8
= 1/24 of tank
1/24 of tank filled in 1hr
1/6 of tank = ?
= 1/6 x 24/1 x 1
= 4hrs

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Rate of flow

vol of tank

Division

NB
Follow through for other   possible alternative   methods

10
19.  Ken’s speed = x km/h.

Distance = 20km.

1. Time taken by Ken = (20/x)hrs
2. Time taken by Jane = (20/x+2)hrs
3. 20/x20/x+2 = ½
4. 20x2 (x+2) – 20 x 2 x x = x(x + 2)
20 x 2(x + 2) – 20 x 2 x X = x2 + 2x

40x + 80 – 40x = x2 + 2x
x2 + 2x – 80 = 0

x2 - 8x +10x – 80 = 0
(x + 10)(x – 8) = 0
Either x = -10 or x = 8
∴ Ken’s speed = 8km/h
and Jane’s speed=10km/h

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For the correct eqn.
Accept equivant

Removal of fractions and   partial factorisation

For the eqn

Removal or fractions

Partial simplification

Factorization equated to 0

Accept equivalent   methods

10
20.
1. x and y are related by the eqn y = 10x2
 X 0 0.2 0.4 0.6 0.8 1 Y 0 0.4 1.6 3.6 6.4 10
Area = 0.2(0.1 + 0.9 + 2.5 + 4.9 + 8.1)
= 3.3 square units
2. Trapezium rule
Area = ½ x 0.2 ( 0.0 + 10.0 ) + 2 ( 0.4 + 1.6 + 3.6 + 6.4 )
= 0.1 (10) + 2 (12)
= 0.1 x 34
= 3.4 square units

3.  % error in mid ordinate rule
10 − 33 x 100
3     10
10
3  = 1%
% error in trapezium rule
10 − 17 x 100
3     5
10
3   =  2%

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All mid values✔
B1 for any 3✔

✔substitution

10
21.
 Speed Mid/points(x) class f fx c.f 10-19 14.5 9.5-19.5 3 43.5 3 20-39 24.5 19.5-29.5 1 24.5 4 30-39 34.5 29.5-39.5 2 69 6 40-49 44.5 39.5-49.5 5 222,5 11 50-59 54.5 49.5-59.5 6 327 17 60-69 64.5 59.5-69.5 11 709.5 28 70-79 74.5 69.5-79.5 9 670.5 37 80-89 84.5 79.5-89.5 8 676 45 90-99 94.5 89.5-99.5 3 283.5 48 100-109 104.5 99.5-109.5 2 209 50 Σf= 50 Σfx= 3235

x = mean speed = Σfx
Σf
= 3235 = 64.7 m/s
50

(c) Median speed will be given by 25th vehicle.
Median = 59.5 + 8 x 10 =
11
= 66.7727m/s

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B2

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✔values for all   the mid pts

for all the fx   values✔

Correct linear   scale

All bars
Polygon

10
22.
1.
1.  S.P. = 88%

88% =shs.4800
M.P.= shs 4800/88 x 100
M.P. = sh.5454.54
Accept M.P= Shs.5454.55

2.  B. P.=145% = 4800
B.P. = 48000 x 100
145
= shs 3310.34
Shs 3310.35

2. Profit= Shs.5454.50- 3310.30
percentage Profit = 5454.54 – 3310.34 x 100
10
3310.34
= 0.6477 x 100
= 64.77%

3.  87.5 x 3310.30
100

= shs 2,896.55

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For the profit

10
23.
1.
1. PR = PO + OR

2. OM = OP + PM

2. ON = KOM

ON = OP + hPR

K = 1 – h ……………………….(i)
2/5K = h………………………...(ii)
Substituting for h in (i)
K = 1 – 2/5K
K = 5/7
h = 2/5 (5/7)
h = 2/7
PN:NR=2:5

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For equating   eqn1 &2

For extracting   the eqns.

10
24.
1.  ACC = 15 – 0

20
= 0.75m/s2

2. Dece = 0 – 15
20
= − 0.75m/s2

3. Area = 1 x 20 x 15
2
= 150m

4. Area = 20 x 15
= 300m

5. Area = 1 (20 + 60) X 15
2
= 600m

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10

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