# Mathematics Questions and Answers - Form 2 End Term 3 Exams 2023

Instructions

• This paper consists of two sections; Section I and Section II.
• Answer all the questions in Section I and any five questions from Section II
• Show all the steps in your calculations, giving your answers 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 Kenya National Examinations Council Mathematical tables may be used, except where stated otherwise.

SECTION I (50 Marks)

Answer all the questions in this section

1. George spends five-eighths of his salary on food, one-sixth of the remainder on school fees, two-thirds of what remains on contingencies and saves Kshs. 6,000. Calculate George’s monthly salary. (3 marks)
2.
1. Find the GCD of 1240 and 860 (1 mark)
2. Hence find the number of square tiles that can be used to cover a floor of dimensions 12.4 m by 8.6 m (2 marks)
3. Without using a mathematical table or a calculator, find the values of a and b given that: (3 marks)
0.0168×2.46×7 = a
5.74×0.0112       b
4. Ten years ago, Mary was twice as old as her daughter Ruth. If the sum of their ages now is 56 years, calculate their present ages.
(3 marks)
5. Simplify the expression below (3 marks)
2x−3y+4x2−6xy
1−4x2
6. A rectangular plot of land measures 120 m by 50 m. One of the longer sides has a wall already erected along it. The plot is to be fenced using chain link. One roll of chain link measures is 16 m long. A gate 6 m wide is left on one of the shortest sides. Calculate the number of rolls of chain link required to completely fence the plot. (3 marks)
7. In the figure below, O is the centre of the circle. The reflex ∠AOD=244° and ∠BCD=480.

Calculate the size of ∠BDC (3 marks)
8. The figure below shows a face and an edge of a triangular prism ABCDEF.

Complete the prism, labeling it accordingly and showing the hidden lines as broken. (2 marks)
9. Use tables of logarithms only to evaluate (4 marks)
10. Evaluate without using mathematical tables (3 marks)
(125)2/3 × (625/64)−½
11. A tourist arrived in Kenya with Great Britain Pounds (GBP) 8,640 all of which she exchanged into Kenya Shillings. He spend Kshs. 517,906 while in Kenya and converted the rest of the money into US dollars (USD). Use the exchange rates below to calculate the amount of money he received to the nearest USD. (3 marks)
 Currency Buying (Kshs.) Selling (Kshs.) 1 GBP 145.40 154.29 1 USD 104.61 111.00
12. Two similar cups have capacities of 1 litre and 343cm3. The surface area of the smaller cup is 122.5cm2, calculate the surface area of the larger cup. (3 marks)
13. The figure below shows the cross-section of a solid equipment used in a construction firm. It comprises of a circular hole of diameter 56 cm drilled on to a square of side 70 cm. The solid is 1.5 m long.

Calculate the volume of the solid in m3. Use =  22/7 (4 marks)
14. Find the modulus of the AB given that a = −4i−7j and b = 3i−8j correct to 2 decimal places. (3 marks)
15. Use a ruler and pair of compasses only in this question.
Construct a trapezium PQRS such that PQ=7.4 cm, ∠PQR=105°, ∠SPQ=∠RSP=90°, PS=5.3 cm and PQ parallel to SR. (4 marks)
16. Show on a number line the range of all integral values of x which satisfy the following pair of inequalities. (3 marks)
3−x≤1−1/2x
1/2(x−5) < 7−x

SECTION II (50 Marks)

Answer any five questions in this section

1. Four schools, K, L, M and N are such that L is on a bearing of 055° and a distance of 8.5 km from K. M is 7.2 km from L on a bearing of 144°. N is 5 km due south of M. A campsite P is due west of N and on a bearing of 159° from K
1. Using a scale of 1: 100,000 make a scale drawing to show the relative positions of K, L, M, N and P (5 marks)
2. Use the scale drawing in (a) above to find:
1. The distance between K and M (2 marks)
2. The bearing on P from L. (1 mark)
3. Show the position T of a communication mast installed such that it is equidistant from K, N and P. (2 marks)
2.
1. The ratio a: b = 2: 3 and the ratio b: c = 2: 5. If the sum of the three quantities is 2,050, calculate the difference between quantities c and a. (4 marks)
2. Two business partners Eric and Bryan contributed Kshs. 100,000 and Kshs. 120,000 respectively to start a tech firm. They agreed that 30% of the proceeds from the firm would be shared in the ratio of their contributions and another 30% shared equally. At the end of the year, the business realized a profit of Kshs. 198,000
Calculate:
1. The amount of money they shared (2 marks)
2. The amount received by each. (4 marks)
3. The figure below shows a solid drum that is in the shape of a frustum of a cone with top and bottom radii as 30 cm and 20 cm respectively. The drum is 45 cm high.

1. Calculate the height of the original cone from which the drum was made. (2 marks)
2. Calculate to 2 decimal places, the surface area of the drum. Use π = 3.142 (8 marks)
4. Below are masses of students taken in a medical centre.
35    55    32    37    60    35    38
20    54    66    59    34    56    42
31    44    47    45    55    48    39
29    27    24    33    36    35    43
45    48    28    52    39    64    41
1. Starting with the class 20 – 29 and class size of 10, make a frequency distribution table for the data. (2 marks)
2. Calculate the mean mass (4 marks)
3.
1. On the grid provided, draw a histogram to represent the data above (2 marks)
2.  On the histogram, draw a line to estimate where the median mass lies. (2 marks)
5.
1. Calculate the time in seconds it takes a bus 5 m long and moving with a speed of 60 km/h to go past a truck 15 m long and heading in the opposite direction with a speed of 50 km/h if the distance between the bus and the truck is 200 m (3 marks)
2. A bus left Eldoret at 7.12 a.m. towards Nairobi at an average speed of 72 km/h. At 8.22 a.m., a car left Eldoret for Nairobi at an average speed of 100 km/h. The distance between Nairobi and Eldoret is 348 km.
Calculate:
1. The time of the day when the car caught up with the bus. (4 marks)
2. The distance from Nairobi where the car caught up with the bus. (3 marks)
6. A youth group plans to buy a branding machine at a cost of Kshs. 72,000 by contributing equal amount of money. Before they contribute, 3 new members join the group, and as such each of the members has to contribute Kshs. 1,200 less.
1. By taking x as the original number of members of the group, write an expression for:
1. The original amount to be contributed by each member (1 mark)
2. The amount contributed by each member after the new members joined (1 mark)
2. Find the total members who contributed. (6 marks)
3. At the end of the year, the machine realized a profit Kshs. 337,500. The profit was shared equally among the members. Calculate the amount of money received by each member of the group. (2 marks)
7. Triangle T has coordinates A(−2, −2), B(−5, −5) and C(0, −4). T1 is the image of T after a transformation M such that A'(2, 2), B'(5, 5) and C'(4, 0)
1.
1. On the same pair of axes, draw triangles T and T1 (2 marks)
2. Describe the transformation M fully (2 marks)
2. T2 is the image of T1after a rotation of +900 about (1, 1). Draw T2 on the same axes and state its coordinates (3 marks)
3. T3 is the image of T2 under a translation described by vector  .
1. Draw T3 (2 marks)
2. Which pair of the triangles exhibit opposite congruency? (1 mark)
8. A straight line L1 passes through the points (−2, 8) and (4,−4)
1.
1. Find the equation L1 in the form ax+by=c, where a, b and c are integral values. (3 marks)
2. L1 crosses the x-axis at point P, determine the coordinates of P. (2 marks)
2.
1. Another line L2 is perpendicular to L1 through point P. Determine the y-intercept of L2. (3 marks)
2. Determine the angle L2 makes with the x-axis correct to 2 decimal places. (2 marks)

### MARKING SCHEME

NO.  WORKING   MARKS   REMARKS
1.  5/8 → food
1/6 × 3/8 = 1/16 → fees
1 − (5/8+1/16) = 5/16
2/3 × 5/16 = 5/24 → contingencies
1 − (5/8 + 1/16 + 5/24) = 5/48
5/48 → 6,000
6,000 × 48/5
=Kshs. 57,600

M1

M1

A1
Fraction saved
3
2. (a) GCD

1240 = 23 × 5 × 31
860 = 22 × 5 × 43
GCD = 22 × 5 = 20
(b) Number of Tiles
12.4×8.6
0.2×0.2
= 2,666 tiles

B1

M1

A1

3
3.  0.0168×2.46×7×106
5.74×0.0112×106
168×246×7
574×112
23×3×7×2×3×41×7
2×7×41×24×7
243272×41
2572×41
32 = = 4.5
2     2
M1

M1

A1

3
4.
 10 years ago Now Mary 2x 2x + 10 Ruth x x + 10
2x+10+x+10=56
3x+20=56
3x=56-20=36
x=12
Mary→2×12+10=34 years
Ruth→12+10=24 years

M1

A1

B1

Equating sum of current ages to 56

Value of x

Both ages correct

3
5.

2x+4x2−3y−6xy = 2x1+2x−3y(1+2x)
1−4x2                  (1−2x)(1+2x)
(2x−3y)(1−2x)
(1−2x)(1+2x)
2x−3y
1+2x

M1

M1

A1
Expression for perimeter to be fenced

Expression of number of rolls

14 rolls seen
3
6. Perimeter = 2×50+120−6=214 metres
Number of rolls = 214
16
= 13.375
= 14 rolls
M1

M1
A1
Expression for perimeter to be fenced

Expression for number of rolls
14 rolls seen
3
7. ABD = ½ × 244° = 122°
CBD= 180° − 122° = 58°
BDC = 180° − (58°+48°) = 74°
B1
B1
B1

3
8.

B1

B1

B1

Completing the solid

Showing hidden lines

Correct labeling
3
9.

M1

M1

M1

A1
Reject all if logs from calculator used

All logs correct i.e. 1.3228, 1.9799 and 1.8727 seen.

_
Correct addition and subtraction of logs i.e. 1.3027 and 3.5573 seen
_                 _
Correct multiplication of logs by 2 and 1/3 i.e. 3.7454 and 1.1858 seen

0.1534 seen
4
10.
23 = 8
M1

M1
A1
Expression in index form

Simplification
3
11. 145.4×8,640
=1,256,256
1,256,256 − 517,906 = 738,350
738,350 = 6,651.802
111
=USD 6,651
M1

M1
A1
Converting to Kshs.

Converting to USD
USD 6,651
3
12. V.s.f = 1000
343
A.s.f =
100    x
49     122.5
x = 100×122.5 = 250cm2
49

M1

M1
A1

Area scale factor

3
13. Cross − sectional area = 0.7×0.7− 22/7×0.28×0.28
= 0.49 − 0.2464
= 0.2436
Volume   = 0.2436 × 1.5
Volume = 0.3654m3

M1
A1
M1
A1

4
14 AB = b−a = (3 − 8) − (−4− 7) = (7 − 1)
lABl = √(72+(−1)2)
= √50
=7.071
=7.07
B1

M1

A1
AB in column form or in terms of i and j

Expression for modulus

7.07 seen
4
15.  B1

B1

B1

B1
PQ = 7.4 cm ± 0.1 cm drawn

Construction of 90° at P and S

Construction of 105°

Locating R/completing trapezium
4
16. 3−x ≤ 1−½x
3−1 ≤ −½x+x
2 ≤ ½x → 4 ≤ x
−½(x−5) < 7− x
1/2x+5/2 < 7 − x
−½x + x < 7 − 5/2
½x < 9/2 → x < 9
4 ≤ x < 9
Integral values →4, 5, 6, 7 and 8

B1

B1
B1

3
17. (a) Scale drawing

(b) Using scale drawing
(i) Distance K and M
11.2 cm±0.1
11.2 km±0.1
(ii) Bearing P from L
204° ± 1°
(c) Locating T

S1

B1

B1

B1

B1

B1

B1
B1

B1
B1

Given scale

Locating L

Locating M

Locating N

Locating P

KM ,in cm

KM in km
Bearing (S24°W)

Bisecting PN and PK
Locating T
10
18 (a) a:b=2:3… ×2→4:6
b:c=2:5… ×3→6:15
a:b:c = 4: 6:15
15/25 × 2,050 − 4/25×2,050
1,230 − 328=902
(b)(i) Amount shared
30/100×198,000 + 30/100×198,000
59,400 + 59,400 = 118,800
59,400 = 29,700
2
Ratio = 100,000:120,000=5:6
Eric → 29,700 + 5/11 × 59,400 = 56,700
Bryan → 29,700 + 6/11 × 59,400 = 62,100
M1

A1
M1
A1

M1
A1

M1

M1
A1
B1
Multiplying a: b by 2 and b: c by 3

Ratio a: b: c simplified

Sum of shared equally and shared in ratio

Ratio of Eric: Bryan
Eric's total
Bryan's total
10
19. (a) Height
h   =  20
h+45    30
30h = 20h+900
30h − 20h = 900
10h = 900→h = 90 cm
Hence H = 90+45 = 135 cm
(b) Surface Area
L= √(1352+302) =15√85
l = √(902+202) = 10√85
CSA = 3.142[(15√85×30) − (10√85×20)]
CSA = 3.142[4148.795006 − 1843.908891]
CSA = 3.142 × 2304.886115
CSA = 7241.95
Top = 3.142 × 30 × 30 = 2,827.8
Bottom = 3.142 × 20 × 20 = 1256.8
Total = 7241.95 + 2,878 + 1256.8
Total = 11,326.55

M1

A1

M1
M1

M1

A1
M1, M1

M1
A1

10
20.  (a) Frequency distribution table
 Mass Tally f x fx 20 - 29 llll 5 24.5 122.5 30 - 39 llll  llll  ll 12 34.5 414 40 - 49 llll  llll 9 44.5 400.5 50 - 59 llll  l 6 54.5 327 60 - 69 lll 3 64.5 193.5 Σf = 35 Σfx = 1457.5
(b) Mean
Mean= Ʃfx =1457.5
Σf       35
Mean = 41.64
(c) Histogram

T.Area = 10×5+10×12+10+10×9+10×6+10×3
Total area = 350 hecatres
Median = 350÷2 = 175
50 + 120 = 170
175 − 170 = 5

B1

B1

B1

M1

A1

B1

B1

B1

M1

A1

10
21 (a) Total distance = 5+15+200 = 220 m
Relative speed = 60+50 = 110 km/h
Time =  0.22 km  × 3,600
110 km/h
= 7.2 seconds
(b) (i) 8.22 a.m. – 7.12 a.m. = 1 hr 10 minutes
Bus→72×116=84 km
Relative speed = 100 − 72 = 28 km/h
Time = 84/28 = 3 hours
Time of the day = 8.22 a.m. + 3 hours = 11.22 a.m.
(ii) Bus distance→3×100=300 km
384 − 300 = 84 km
M1

M1

A1

M1
M1
M1
A1
M1, A1

Total distance and relative speed

Expression for time

Time in seconds

Distance between them at 8.22

Relative speed
Duration to catch up

10
22. (a) (i) – Before
72,000
x
(ii) – After
72,000
x+3
(b) Value of x
72,00072,000 = 1,200
x            x+3
72000(x+3) − 72000x = 1,200
x(x+3)
72,000x+216,000−72,000 = 1,200
x2+3x
216,000 = 1,200(x2+3x)
x2+3x−180=0
x2+15x−12x−180 = 0
x(x+15) −12(x+15) = 0
(x−12)(x+15) = 0
Either x−12 = 0 hence x = 12
Or x+15=0 thus x=−15 (discriminate)
Hence number = 12+3=15
Profit
337,500
12+3
= 22,500

M1

M1

M1

M1

M1

A1
B1

M1

A1

Amount before

Amount after

Expression for difference

Factorization

Both values seen
15 seen

10
23. (a)(i) T and T1
(ii) M described
(b) Rotation
T2
A''0, 2, B''(−3, 5) and C''(2, 4)
Translation
(i) T3
(ii) T and T1, T1 and T2, T2 and T3
B1
B1
B1
B1
B1
B1
B1
B1
B1
B1

Drawing T
Drawing T1
Reflection
Along y + x = 0
Drawing T2
Coordinates of T2
Translation
Drawing T3
Pairs of opposite congruency

10
24 (a) Equation of L1
(i) Equation of L1
m1 = 8−(−4) = −2
−2−4
−2 = y−(−4)
x−4
−2 = y+4
x−4
−2 + 8 = y + 4
8−4 = y+2x
2x+y = 4
(ii) Coordinates of P
At P→y=0
2x=4
x=2
Hence P(2, 0)
(b)(i) y-intercept of L2
−2m2 = −1
m2 =1/2
Consider m2 = 1/2, P(2, 0)
y = mx+c
0 = 1/2 × 2+c
0 = 1+c→c = −1
Hence→y = 1/2x−1
y-intercept=−1
(ii) Angle of L2 with x-axis
tan θ = 1/2
θ = tan−1(1/2)
θ = 26.57°

M1

M1

A1

M1
A1

M1

M1

A1

M1

A1

2x + y = 4 seen

2x + 0 = 4
In coordianate form

10

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