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

Instructions to candidates.

• This paper consists of two sections: Section I and Section II.
• Answer all the questions in Section I and only five questions from Section II
• Show all the steps in your calculations, giving your answer at each stage in the spaces provided below each question
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.

SECTION 1 (50 MARKS)

Answer all questions in the spaces provided.

1. A sum of money is divided between three men x, y and z in the ratio 5:3:1. If y has Shs. 700/= more than z, calculate how much x has.
(3 marks)
2. Simplify the expression  12x2 + ax - 6ax2           (3 marks)
9x2 - 4a2
3. Find the integral values of x of which  (3 marks)
5 ≤ 3x + 2
3 x − 14 ≤ − 2
4. The figure below shows triangle PQR in which PQ = 7cm, angle QPR = 100° and angle PRQ = 35°. Calculate to 2 decimal places the length of PR hence the area of triangle PQR.

5. Evaluate without using a calculator  3 marks)
23.4 − 2(5.2 + 5.3)
3.2 x 1.2
6. The figure below represents a kite ABCD. Use a pair of compasses and ruler only to rotate it through -60° about point P.  (2 marks)

7. Find the inverse of the matrix (1/3 1/1) hence determine the point of intersection of the lines. (4 marks)
y + x = 7
3x + y = 15
8. A trader had a bag of rice, when he packed the rice in 6kg packets, he had 1 kg left over, when he packed the rice in 8kg packets, again he had 1kg left over. When he packed the rice in 9kg packets, he had 1kg left over. What is the smallest amount of rice that he must have had.
(3 marks)
9. A Histogram is drawn from the set of data given below. Complete the missing bars.  (3 marks)
 Marks 6-10 11-20 21-35 36-55 56-65 Frequency 8 14 18 24 10
10. Kinyua marked an article at Kshs. 1200/- and sold it to a customer at a discount of 15%. Find the percentage profit he made if he had bought the article at Kshs. 900/=.                (3 marks)
11. The diagram below represents a solid of a conical frustrum. (Use π 22/7) (4 marks)

Calculate the volume of the solid
12. Given that a = (−28), b = −64) and c = −4 2 and given that = 4a − 8b + 6c, find /P/  (3 marks)
13. Evaluate using tables of reciprocals  (3 marks)
5      +    1
807      0.0591
14. Solve for x in the following equation  (3 marks)

15. Each interior angle of a regular polygon is four times greater than the exterior angle. How many sides does this polygon have?  (3 marks)
16. A boat is at point P, a distance of 100km from the bottom of a hill. The angle of elevation of the top of the hill is 30° from P. The boat sails straight towards the hill to a point Q from where the angle of elevation to the top of the hill is now 60°. Calculate the distance PQ   (3 marks)

SECTION II (50 MARKS)

Answer only five question from this section in the spaces provided.

1. Given that a line L1 passes through the points A(−1, 5) and B (3, −1), find
1. The equation of line L1 in the form y = mx + c     (2 marks)
2. The equation of a line L2, which is a perpendicular bisector of L1. Leave your answer in the form
ax + by = c where a, b, c are integers  (3 marks)
3. Given that another line L3 is parallel to L2 and passes through point (−3, −5) and intersects lines L1 at point P. Find the equation of L3 in the form ax + by +c = 0  (2 marks)
4. The coordinates of the point of intersection of lines L1 and L3   (3 marks)
2. Four towns P, R, T and S are such that R is 70km directly to the north of P and T is on a bearing of 280° from P at a distance of 75km. S is on a bearing of 320° from T and a distance of 45km.
1. Using a scale of 1cm to represent 10km, make an accurate scale drawing to show the relative position of the towns. (4 marks)
2. Using your diagram above, find the distance in km of
1. R from T  (3 marks)
2. S from R
3. From your diagram, what is the compasses bearing of (3 marks)
1. R from T
2. P from S
3. Nyaugenya bus leaves Port Victoria for Eldoret at 7:00 a.m. at an average speed of 80km/h. Climax bus leaves Eldoret towards Port Victoria at 7:30 a.m on the same day using the same route at an average speed of 60km/h. The distance from Eldoret to Port Victoria is 450km. After travelling for one hour and a half, climax bus developed a mechanical problem which took 45 minutes to repair before continuing at its speed in the same direction.
1. Determine the time when the two buses met. (4 marks)
2. Calculate the distance from Eldoret at the time of the two buses meeting.  (3 marks)
3. For how long did the Nyaugenya bus stay in Eldoret before Climax bus arrived at Port Victoria.  (3 marks)
4.
1. On the grid provided draw triangle ABC with vertices A(0, -1), B(4, 3) and C(2, 2)   (1 mark)

2. Draw triangle A1B1C1, the image of ABC under a translation defined by the translation vector T = (1 −2)  (3 marks)
Write down the coordinates A1B1C1.
3. A11B11C11 is the image of ABC under an enlargment, scale factor −2, centre (3,1). On the same grid draw A11B11C11 and write down its coordinates.
4. Draw A11B11C11, the image of A11B11C11, 1 under reflection on the line x = 0
5. The motion of a particle P moving along a straight line is described by the equation S = (8+10t)t – t3
Calculate;
1. The distance when t = 3 sec.  (2 marks)
2. The maximum velocity of the motion. (4 marks)
3. The acceleration of motion after 4 seconds.  (2 marks)
4. The time at which the velocity is zero.  (2 marks)
1. Use the trapezium rule with 8 ordinates to approximate the area under the curve   (5 marks)
y = x+ x + 3 and the x-axis between lines x = −3 and x = 4
2. Use the mid-ordinate rule with 5 strips to calculate the area under the curve y = 3x2 + 8 and bounded by lines y = 0, x = 1 and x = 6.
(5 marks)
6. Three people; A, B and C work together to make a certain number of tins. It person C was to work alone he will take 44/hours to complete the job. If all working together they will take 1 hour 40 min. to complete the job. They all started working together however person B left after the first 40 minutes, while person C left 20 minutes later. Person A took a further 1hr 46mins. Calculate how long it would take if all the tins were made by;
1. Person A alone  (6 marks)
2. Person B alone (2 marks)
3. Person A and C alone  (2 marks)
7. In Busia County, a tailor bought a number of suits at a cost of Shs. 57,600/= from a wholesaler. Had he bought the same number of suits from a supermarket, it would have cost him Shs. 480/= less per unit. This would have enabled him to buy four extra suits for the same amount of money.
1. By letting the number of suits that the tailor bought to be n, write an expression in n for cost of a suit from;
1. The wholesaler (1 mark)
2. The supermarket (1 mark)
2. Find the number of suits that the trailer bought  (4 marks)
3. The tailor later sold each suit for Shs. 720/= more than he paid for it. Determine the percentage profit he made.    (4 marks)

MARKING SCHEME

No.  Working   Marks Allocation
1.   3 − 1 = 2
700 x 5
2
sh. 1750
B1 - difference in ratio
M1
A1
03
2.  12x2 + 9ax − 8ax − 6a2
3x(4x + 3a) −2a(4x + 3α)
(4x + 39) (3x − 2a)

(3x − 29) (3x + 2a)
(4x + 39)(3x − 29)
(3x − 29) (30 + 24)
4x + 39
3x + 29
B1

B1

A1

03
3.

3 ≤ x
1 ≤ x
or
x ≥ 1
3x ≤ 12
x ≤ 4
or
4 ≥ x
x = 1, 2, 3, 4

B1

B2

A1
03
4.

7       =  QR
Sin 35    sin 100
QR = 12.02 cm

½ x 7 x 12.02 Sin 45
=   29.75cm2

M1
A1  ( C.A.O)
M1
A1
04
5.      2.4
3.84
24 x 10
384
=  0.625
M1 - for simplifying
M1 - for cancellation
A1  ( C.A.O) DO NOT AWARD IF FRACTION
03
6.       B1 - for 60° measurement clockwise about                      0
B1 - for correct diagram
02
7.     (1 x 1) − (3 x 1)

M1 - For determinant
A1 - C.A.O
M1 - For premultipying by inverse
otherwise award 0
A1

04
8.
LCM = 23 x 32
= 72
Amount of rice = 72 + 1
= 73 kg
M1 - For finding lcm

A1
03
9.      B1 - Each correct bar B1
Three correct bars drawn
03
10  S.P = 85 x 1200
100
= 1020
profit = 1020 − 900
= 120
% profit = 120 x 100
900
= 13 1/3 or 13.3 or 13.33%
M1 - for selling price                        calculations

M1 - for % profit calculation
A1
03
11.

h    =    2.1
h + 4       4.9
h = 3
Vol = 1/3 x 22/7 ( 4.92 x 7 − 2.12 X 3)
= 162.2133 cm3

M1 - for finding height of the                   original cone
B1 - for height
M1 - for finding volume
A1
04
12.

P   = 4 (−2 8) −8 (–6 4) + 6(–4 2)
= (16 12)
/P/ = √162 + 122
= 20 units

M1 - for substitution
M1 - for calculation of magnitude
A1
03
13.   (5 X 0.1293 X 10−2)
0.006195
0.1692 X 102
16.92
0.006195 + 16.92 = 16.926195
B1 - for finding reciprical using                tables
B1 -  for finding reciprical using                tables
A1 - no mark for using calculators
C.A.O
03
14.  (2−2)(x − 2)  =  2(x + 2)
−2x + 4 = x + 2
x = 2/3
M1 - for expressing to similar                   base.
M1 - for interpretation of equation
A1
03
15.   x + 4x = 180
x = 36
n = 360
36
= 10
M1

M1
A1
03
16.    tan 30 = h/100
100tan30 = h
tan 60 =  100tan30
100 − x
x = 662/3km or 66.6km
M1 - for finding height
M1 - for substituting height
A1 - Accept 4.sf.
03
17.
1.  M1−1 − 5
3− (−1)
= −3/
y − 5  = − 3
x −(−)       2
y = −3/2x + 7/2

2. M2 = 2/3
Midpoint = (1,2)
y − 2 = 2
x − 1    3
2x − 3y = − 4 or −2x + 3y = 4

3. M3 = 2/3
y − (−5) =  2
x − (−3)     3
2x − 3y − 9 = 0 or −2x + 3y + 9 = 0

4. 2x − 3(−3x + 7) − 9 = 0
2      2
x = 3
y = −3 x 3 + 7
2           2
= −1
P(3, −1)

M1 - correct substitution
A1

B1 - for both M2 & midpoint                   correct
M1 - correct substitution
A1

M1 - correct substitution
A1

M1 - correct substitution

M1 - correct substitution

A1

10
18.
1.
2.
1.   9.4 x 10
94km ± 1km

2.  10.5 x 10
105km ± 1km
3.
1. N52°E ± 1°
2. S65°E ± 1°

M1
A1

M1
A1

B1
B1

19.
1.  Distance by Nyangenya = 80 x (½ + 1½ + ¾)
= 220km
Distance by Climax = 60 x 1½
= 90km
RD = 450 − (220 + 90)
= 140km
RS = 80 + 60
= 140km/h
T = 140
140
= 1hr
Time of meeting = 7:30
3:15
= 10.45 a.m
2. Time by Climax bus = 1.5 + 1
= 2.5h
Distance = 60 x 2.5
= 150km
3. Time taken by Nyangenya =  150
80
=  17/8 hr
Time taken by Climax =  300
60
= 5h
Time difference = 5h − 17/8
B1 - for distcance by both buses

B1 - for both RD & RS

M1

A1
M1

M1
A1(accept relevant alternative)
M1

M1

10
20.
1.
2. A1 (1, −3) , B1 (5, 1), C1(3, 0)

3. A11 (7, 9), B11 (−1, 1), C11 (3, 3)

B1 - for correct Δ ABC
B2 - for correct Δ A1B1C1

B3 - for correct Δ A11B11C11

B2 - for correct Δ A111B111C111

B1 - for correct co-ordinates

B1 - for correct co-ordinates

10
21.
1.  s = (8 + 10(3))3 − 33
= 87 m

2. v = 8 + 20t − 3t2
a = 20 − 6t
6t = 20
t = 10/3
v = 8 + 20 (10/3) −3(10/3)2
= 411/3 or 41.3
3. a = 20 − 6(4)
= 4m/s2
4. 8 + 20t − 3t2 = 0

= 7.045 sec 0r − 0.375
∴ t = 7.045 sec

M1 - correct substitution
A1

B1 - for correct a
M1 - for correct substitution
M1 - for correct substitution

A1
M1 - correct substitution
A1

M1

A1

10
22.
1.
 x − 3 − 2 − 1 0 1 2 3 4 y = x2 + x +3 9 5 3 3 5 9 15 23

A = ½ x 1{(9 + 23) + 2(5 + 3 + 3 +5 + 9 + 15)}
= 56 sp. units
2.
 x 1.5 2.5 35 4.5 5.5 y = 3x2 + 8 14.75 26.75 44.75 68.75 98..75

A = 1(14.75 + 26.75 + 44.75 + 68.75 + 98.75)
= 253.75 sq units
B3 - for all correct
B2 - for any 5 correct
B1 - for any 3 correct

M1
A1

B1 - for all correct
B2 - for all correct
B1 - for any 3 correct

M1
A1

10
23
1.  A + B + C ⇒ 1h 40 min = 5/3h
∴ Work done in 1h = 3/5
⇒ work done in 40min
40/60 x 3/5 = 2/5
Let work completion by A and B be a and b respectively.
C  takes 40/9 h
∴ When B leaves, work done by A and C = 20/60 (1/9 + 9/40)
= 1/3a + 3/40
When C leaves, work done by A = 1/a (53/30a
= 53/30a
∴ Completing the job
2/5 + 1/3a + 3/40 + 53/30a = 1
a = 4h
2. 1/4 + 1/b + 9/40 = 3/5
1/b = 1/8
b = 8h
3. 1/4 + 9/10 = 19/40

B1 - for work done in 40                minutes

M1 - for work done by
A & C
B1

M1 - for work done by all

A1

M1 - for work done by the              3
A1

M1 - for work done by A

24
1.
1.  57600
n
2. 57600
n + 4
2.
57600 − 57600 = 480
n          n + 4
n2 + 4n − 380 = 0

n = − 24 or 20
∴ No. of suits bought = 20
3. 720 x 20
1400

14400 x 100
57600
= 25%
B1

B1

M1
B1 - for correct quadratic                equation
M1 - accept any other                    method

A1

M1
B1
M1

A1 - acccept alternative

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