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

17. 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)
18.  
    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)
19. 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)
20. 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)
21.  
    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)
22. 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)
23. 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)
24. 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 = 9  = 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 (ii) Amount received 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  Amount received equally  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 72,000 − 72,000 = 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  Quadratic equation formed  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  Rotation about (1, 1)  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	Gradient of L1  Equating m1 to gradient  2x + y = 4 seen  2x + 0 = 4  In coordianate form  Gradient of L2
		10