Mathematics Paper 1 Questions and Answers- Mokasa 1 Joint Pre Mock Exams 2023

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Mathematics Paper 2 Questions and Answers - Mokasa 1 Joint Pre Mocks Exams 2023

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Instructions to candidates

This paper consists of two sections: Section I and Section II.
Answer all questions in section I and only five questions from section II.
Show all the steps in your calculations, giving the 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 KNEC mathematical tables may be used, except where stated otherwise.
Candidates should answer the questions in English.

SECTION I (50 marks)

Answer all questions in this section in the spaces provided below each question

Solve without using tables or calculators. (3 marks)
²/₃(1 −²/₅ + ¹/₄)
⁷/₃₀ +¹/₃ of ³/₅
The present age of a father is 10 years more than twice the present age of his son. In how many years time will the age of the father be twice the age of the son? (3 marks)
Solve for y in the equation 36^(y−1) × 6^(2y−2) = 1 (3 marks)
Given that x=1.313313313…….., find the exact value of 999x . (3 marks)
The marked price of a radio which gives a profit of 20% on the cost price is ksh 3600. During the actual sale, the salesman allowed a discount of 5% off the marked price. Calculate the percentage profit made by the salesman after allowing the discount. (3 marks)
Use tables of squares and reciprocals to find to 3 significant figures; (3 marks)
0.3628² + 1
21.63
Simplify: 18t² − 50a² (3 marks)
6t² − 19at + 15a²
The boundaries PQ, QR, RP of a ranch are straight lines such that Q is 16 km on a bearing of 0400 from P. R is directly south of Q and east of P. Using a scale of 1 cm to represent 2 km, show the above information in a scale drawing. Hence find the distance of PR.
(4 marks)
The circle below circumscribes a triangle ABC where AB = 6.3cm, BC = 5.7cm and AC = 4.8cm. <BAC= 60°. Find correct to two decimal places, the area of the shaded part. (use π = 3.142). (4 marks)
Below is a Cartesian plane with an inequality represented graphically.
1. Determine a simple statement inequality that satisfy the inequality drawn on the Cartesian plane above. (2 marks)
2. Represent the following inequalities on the grid drawn above.
  1. y<1 (1 mark)
  2. y≤x+2 (1 mark)
In Equity bank Eldoret branch, customers may withdraw cash through one of the five tellers at the counter. On average, the 1^st teller takes 2 minutes, the 2n^d teller takes 3 minutes, the 3^rd teller takes 4 minutes, the 4^th one takes 5 minutes and the 5^th takes 6 minutes to serve a customer. If the five tellers start to serve the customers at the same time, find the shortest time it takes to serve 348 customers. (4 marks)
An athlete can run 100m in 10 seconds. He can also run 200m in 19 seconds after having a rest of 30 minutes between the two races. On a particular day, an athlete ran 100m race starting from 1:34:27 p.m. He rested for 30minutes before going in for a 200m race. At what time of the day, did he complete his two races? (2 marks)
A point A is mapped onto A¹ by a translation vector −7 3 . If point A¹ is −2, 4, find the distance between point A and point B(10,−13) correct to two decimal places. (3 marks)
Given that 3x is an acute angle and sin 3x − co(s x−50°) =0, find tan (x+35°) giving your answer to 4 significant figures. (3 marks)
Thirty two men working at the rate of 9 hours a day can complete a piece of work in 7 days. How many more men working at a rate of 8 hours a day would complete the same work in 6 days. (3 marks)
A triangular Prism ABCDEF, with the cross-section ABF is such that AB= 3cm, BF= 5.5 cm and AF= 4 cm. The length of the prism is 6cm. Draw the net of the solid. (3 marks)

SECTION II ( 50 marks)

Answer only five questions in this section in the spaces provided below each question

A triangle DEF is such that angle EDF = 90°. The coordinates of the vertices of the triangle are (−6, 2) , E(5, −8) and F(x,y).
1. Find equation of line DE in the double intercept form. (3 marks)
2. Find the coordinates of F(x,y) if the gradient of EF is −2²/₃₅ (5 marks)
3. A line l passes through (4 , 4) and is parallel to EF, find the equation of line l. (2 marks)
Maru Secondary School buys 40 bags of maize and 25 bags of beans per term and pays total of sh 380,000. While Kadu Secondary School buys 35 bags of maize and 18 bags of beans per term and pays a total of sh. 301,500.
1. Write down the two equations representing the cost of grains in the two schools. (1 mark)
2. Using matrix method, calculate the cost of each bag of maize and beans. (4 marks)
3. The following term the price of each bag of maize increased by 20% while that of beans decreased by 15%.
  1. How much did Maru pay for the grains? (3 marks)
  2. If the school enrolment increased by 10% and the food ratio also increased proportionally, how much would the school spend on the grains? (2 marks)
The diagram below represents a lampshade in the form of an open-ended frustum of a cone. Its bottom and the top diameters are 21cm and 7cm respectively. Its height is 10cm. (Use = ²²/₇)
1. Find;
  1. The height of the cone from which the frustum was cut. (2 marks)
  2. The curved surface area of the frustum. (3 marks)
  3. The volume of the frustum. (2 marks)
2. The material used for making a lampshade is sold at ksh 800 per square metre and the labour of making one lampshade is ksh150. Find the cost of making 20 similar lampshades. (3 marks)
Two express matatus P and Q are travelling towards a certain town R along the same route at 96 km/hr and 80 km/hr respectively. When P is 32 km from R, Q is 42 km from R. If P overtook Q at a place x km from R.
1. Find;
  1. the value of x. (6 marks)
  2. the time, in minutes, it takes matatu P to reach R after overtaking matatu Q. (2 marks)
2. If matatu P had a puncture which took 13 minutes to repair after overtaking Q, determine which matatu would have arrived at R first. (2 marks)
The number of students in 30 schools in Kikuyu sub-county were recorded as follows.
200 350 258 400 198 240
202 338 372 392 303 250
250 300 400 405 338 302
330 370 345 209 220 230
275 280 390 403 352 340
1. Using a class width of 50 and ending with the class 395-444, make a frequency distribution table for the data. (2 marks)
2. Calculate the median correct to a whole number. (3 marks)
3. On the grid provided, and on the same axes, draw a histogram and frequency polygon to represent the above information.
  (5 marks)
A group of farmers decided to raise Ksh 120 000 to start a rabbit rearing project. Before they could start contributions, 6 members opted out and as a result the remaining farmers had to contribute each Ksh 1000 more in order to raise the required amount of money.
1. Taking the original number of farmers of the group to be x, write expressions for:
  1. the amount of money each farmer would have contributed before the 6 farmers opted out. (1 mark)
  2. the amount of money that each of the remaining farmers contributed after the 6 farmers opted out. (1 mark)
2. Find the original number of farmers of the group. (6 marks)
3. Calculate the amount of money contributed by each of the remaining farmers. (2 marks)
The vertices of triangle ABC are A(−2,−2), B(0,−5) and C(1,−1). Triangle A′B′C′ is the image of triangle ABC under a reflection in the line
x + y = 3
1. Draw triangle ABC and A′B′C′ on the same axes. (3 marks)
2. A′′B′′C′′ is the image A′B′C′ under a translation T = . Draw triangle A″B″C″. State the coordinates of A′′B′′C′′ (3 marks)
3. Draw triangle A‴B‴C‴the image of A′′B′′C′′ under a positive quarter turn about (0,0). (2 marks)
4. State two pairs of the triangles which are directly congruent. (2 marks)
Use a ruler and a pair of compasses only for all constructions in this question.
1. Construct quadrilateral PQRS such that PQ = PS = 5 cm, SR = 4.5 cm, angle SPQ = 75° and angle PSR=90°. (4 marks)
2. Drop a perpendicular from S to meet line PQ at N. Measure SN and calculate the area of the triangle SPN. (4 marks)
3. Construct a circle passing through vertices P, Q and R of quadrilateral PQRS. Measure the radius of the circle. (3 marks)

MARKING SCHEME

²/₃(⁵/₄ −²/₅) M1 - numerator
⁷/₃₀ +¹/₅ M1 - denominator

²/₃(¹⁷/₂₀) × ³⁰/₁₃ = ¹⁷/₁₃ = 1⁴/₁₃ A1
Father Son
10+2x x → present ages
In t years time;
Father Son
(10+2x) + t x + t
(10+2x+t) − (x + t) = x + t
10 + 2x − x + t − t = x + t
t = 10 years
6^2(y−1) × 6^(2y−2) = 6⁰
2y − 2 + 2y − 2 = 0
4y = 4
y = 1
x = 1.313313313
1000x = 1313.313
999x = 1312
x = 1312
999
999 × 1312 = 1312
999
3600 = 20%
Cost price = 100%
100 × 3600 = Ksh 3000
120
S.P = ⁹⁵/₁₀₀ × 3600 = 3420
(3420 − 3000) × 100
3000
= 14%
3.628 × 10⁻¹ 2.163 × 10¹
13.162 × 10⁻² 0.4623 × 10⁻¹
(0.13162) + (0.04623)
= 0.17785
≅ 0.178
2(9t² − 25a²)
6t² − 10at − 9at + 15a²
2(3t + 5a) (3t − 5a)
2t(3t−5a) − 3a(3t − 5a)
2(3t + 5a) (3t − 5a)
(2t − 3a) (3t − 5a)
6t + 10a
2t − 3a
PR = 5.2cm × 2 = 10.4km Accept 10.2km and 10.6km
5.7 = 2R
Sin 60
R = 3.291cm
3.142 × (3.291)²
= 34.0299997cm²
½ × 6.3 × 4.8 sin 60
= 13.09430411
34.02999970
13.09430411−
20.93569559
≅ 20.94cm²
1. ^x/₃ + ^y/₄ = 1
  ^x/₃ + ^y/₄ ≤ 1
2² × 3 × 5 = 60min
1st 2nd 3rd 4th 5th
⁶⁰/₂ + ⁶⁰/₃ + ⁶⁰/₄ + ⁶⁰/₅ + ⁶⁰/₆
30 + 20 + 15 + 12 + 10 = 87
In 1 hour 87 people are served
∴ 348 people = ?
348 × 1 = 4 hours
87
1:34:27pm + 10sec + 19 sec + 30
= 1 : 35 : 26pm
= 14.87 units
Sin 3x = Cos (x − 50)
3x + (x − 50) = 90
x = 35
tan 70 = 2.747
Men Hours Days
32 9 7
? 8 6
⁷/₆ × ⁹/₈× 32 = 42 men
42 − 32 = 10 more men
3. Gradient = ⁻⁷²/₃₅
  −72 = y − 4
  35 x − 4
  −72x + 288 = 35y − 140
  − 72x − 35y = −428
  72x + 35y = 428
Let a bag of maize = x/= & beans y/=
1. 40x + 25y = 380000
  35x + 18y = 301500
2. 8x + 5y = 76000
  35x + 18y = 301500
  
  A bag of maize = Ksh 4500
  A bag of Beans = Ksh 8000
3. 1. Maize new price = 120 × 4500 = Sh 5400
    100
    New price of beans = 85 × 8000 = Sh 6800
    100
    (40 × 5400) + (25 × 6800)
    = Ksh 386000
  2. x students → Ksh 386000
    110x students → ?
    100
    1.1x × 386000
    x
    = Ksh 424600
H = D
h d
h + 10 = 21
h 7
L = √(15² + 10.5²)
L = 18.31cm
L = √(5² + 3.5²)
L = 6.103cm
1. 1. h + 10 = 21
    h 7
    h = 5
    H = 15cm
  2. πRL − πrl
    π(10.5 × 18.31 − 3.5 × 6.103)
    ²²/₇(192.255 − 21.3605)
    = 537.097cm²
  3. V =¹/₃πR²H − ¹/₃πr²h
    = ¹/₃ × ²²/₇(10.5² × 15 − 3.5² × 5)
    = 1668.3cm³ or 1668¹/₃cm³
2. 537.097 = 0.0537097m²
  10000
  800 × 0.0537097 = Ksh 42.96776
  42.96776 × 20 = 859.3552
  Ksh 859.3552 + (150 × 20)
  Ksh. 3859.40
1. 2. T = ⁹²/₉₆ × 60 = 57.5 minutes
2. Time for P = 57.5 + 13 = 70.5min
  Time for Q = ⁹²/₈₀ = 69min
  Q arrived at R first
1. Students No. of schools cf
  
  195 - 244 7 7
  
  245 - 294 5 12
  
  295 - 344 7 19
  
  345 - 394 7 26
  
  395 - 444 4 30
2. 294.5 + (³⁰/₂ − 12) × 50
  7
  = 316
3. Allow any other scale used
1. 1. 120000
    x
  2. 120000 or 120000 + 1000
    x−6 x
2. 120000 − 120000 + 1000
  x−6 x
  120000x − (x − 6)(120000) = 1000x²  − 6000x
  120000x  − (120000x  − 720000) = 1000x²  − 6000x
  720000 = 1000x² − 6000x
  x² − 6x  − 720 = 0
  x = 6 ±√(36 − 4 x − 720 × 1)
  2 × 1
  x = 6 ± 54 , x = 6 + 54 = 30 or x = 6 − 54 =  −24
  2 2 2
  Original number of farmers = 30
3. 120000
  (30 −6)
  =Ksh 5000
2. A''(1,3) B''(4,1) and C''(0,0)
4. 1. A'B'C' and A''B''C''
  2. A''B''C'' and A'''B'''C'''
1. B1 constructing 75° at P
  B1 constructing 90° at S
  B1 Locating S & R
  B1 completing the diagram
2. SN = 4.8cm ± 0.1cm
  Area of SPN
  ¹/₂ × 1.3 × 4.8
  = 3.12cm²
3. Radius = 3.4cm ± 0.1cm