# Mathematics Paper 1 Questions and Answers - Asumbi Girls Pre Mock Examinations 2023

INSTRUCTIONS TO THE CANDIDATES

• This paper contains two sections; Section A and Section B
• Marks may be given for correct working even if the answer is wrong.
• Show all the steps in your calculations, giving your answers at each stage

FOR EXAMINER’S USE ONLY
Section A

 Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total Marks

Section B

 Question 17 18 19 20 21 22 23 24 Total Marks

## QUESTIONS

Section A. (50mks)
Answer all the questions in this section in the spaces provided.

1. Without using a calculator evaluate. (3mks)
2. Use logarithms to evaluate (4mks)
3. Without using a calculator or mathematical tables evaluate. (3mks)
4. The base of a right pyramid is a rectangle of length 80cm and width 60cm. each slant edge of the pyramid is 150cm. Calculate the volume of the pyramid. (3mks)
5. Simply. (3mks)
6. A two digit number is such that, the sum of its digits is 13. When the digits are interchanged the original number is increased by 9. Find the original number. (4mks)
7. The size of an interior angle of a regular polygon is 3xº while its exterior is (x+20)2. Find the number of triangles that makes the polygon. (3mks)
8. In the triangle ABC below BC = 14cm < ABC= 70º and < ACB= 40º. Calculate; correct to 4 significant figures the areas of triangle ABC. (3mks)
9. A vector given by  transforms a point A(3,6) to A1. Find the distance A A1 (3mks)
10.
1. Using a rule and a pair of compasses only, construct a quadrilateral PQRS in which PQ = 6cm ,PS = 4cm QR = 5cm, <PQR= 105º and <SPQ= 90º (2mks)
2. The quadrilateral PQRS represents a plot of land drawn to a scale of 1:4000. Determine the actual length of RS in metres. (2mks)
11. Without using mathematical tables or calculator evaluate. (3mks)
12. Use matrix method to solve. (3mks)
5x+3y=35
3x-4y= -8
13. Use mathematical tables to find the reciprocal of 0.0247, hence evaluate. (4mks)
Correct to 2 decimal places.
14. A Kenyan businessman intended to buy goods worth US dollar 20,000 from South Africa. Calculate the value of the goods to the nearest south Africa (S.A) Rand given that 1 US dollar = Ksh 101.9378 and 1 S.A Rand = Ksh 7.6326. (3mks)
15. Find all integral values of x which satisfy the inequalities. (3mks)
5 - 3x ≤ x - 7 < 11 - 2x
16. Express 5184 and 2744 in terms of its prime factors hence determine the value of  (4mks)

Section B (50mks)
Answer any five questions from this section on the spaces provided.

1. Two vertices of a rectangle ABCD are A(3,6) and B( 7,12)
1. Find the equation of line AB. (3mks)
2. Find the equation of the perpendicular bisector of line AB. (4mks)
3. Given that BC is perpendicular to AB. Find the equation of BC. (3mks)
2. Three business partners Abila, Bwire and Chirchir contributed Ksh 120,000, Ksh 180,000 and Ksh 240,000 respectively to boost their business. They agreed to put 20% of the profit accrued back into the business and to use 35% of the profits for running the business. The remainder was to be shared among the business partners in the ratio of their contribution. At the end of the year, a gross profit of Ksh 225,000 was realised.
1. Calculate the amount.
1. Put back into the business. (2mks)
2. Used for official operations. (1mk)
2. Calculate the amount of profit each partner got. (4mks)
3. If the amount put back into the business was added to individual’s shares proportionately of their initial contributions, find the amount of Chirchir’s new shares. (3mks)
3. Coast bus left Nairobi at 8.00Am and travelled towards Mombasa at an average speed of 80Km/hr. At 8.300am Lamu bus left Mombasa towards Nairobi at an average speed of 120Km/hr. given that the distance between Nairobi and Mombasa is 400Km, determine.
1. The time Lamu bus arrived in Nairobi. (2mks)
2. The time the two buses met. (4mks)
3. The distance from Nairobi to the point where the buses met. (2mks)
4. How far coast bus is from Mombasa when Lamu bus arrives in Nairobi. (2mks)
4. A land is enclosed by four straight boundaries AB,BC,CD and DA. Point B is 25Km on bearing of  315º  from A, C is directly south of B on a bearing of  260º  from A and D is 30Km on a bearing of  210º from C
1. Using a scale of 1cm to represents 5Km represent the above information on a scale drawing. (3mks)
2. Using the scale drawing, determine the
1. Distance in Kilometres of D from A. (2mks)
2. Bearing of A from D. (1mk)
3. Calculate the area, correct to 1 decimal place, of the land in square kilometres. (4mks)
5.
1. Complete the table below for the functions
y= x2- 6x+7 (2mks)
 x 0 1 2 3 4 5 6 -x2 0 - - 9 16 25 - -6x - -6 -12 - - - -36 7 7 7 7 7 7 7 7 Y = x2 - 6x + 7 7 - -1 -2 - - -
2. On the grid provided below draw the graph of y= x2- 6x+7 for 0 ≤ x ≤6 and use it to estimate the roots of the equation.
x2- 6x+7=0 (4mks)
3. Use the graph above to solve the equation x2- 7x+9=0 (3mks)
4. Determine the range of values of X for which x2- 6x+7 <x-2 (1mk)
6. In the figure below PQR and S are points on the circumference of the circle centre O. TP and TR are tangents to the circle at P and R respectively. POQ is a diameter of the circle and angle PQR = 64º  (10mks)

Giving reasons on each case, find the size of
1. <ROP
2. <PSR
3. <ORP
4. <TRP
5. <RTP
7. The figure below shows two triangles ABC and BCD with a common base BC = 3,4cm. The area of triangle ABC = Area of triangle BCD and ABC =90º

1. Calculate, correct to one decimal place.
1. The area of triangle ABC. (3mks)
2. The size of <BCD (3mks)
3. The length of BD (2mks)
4. The size of <BDC (2mks)
8. The marks scored by 40 students in a mathematics test were as shown in the table below.
 Marks 48 – 52 53 – 57 58 – 62 63 – 67 68 – 72 73 – 77 Number of students 3 4 10 12 8 3
1. State the modal class. (1mk)
2. Using an assumed mean of 64, calculate the mean mark. (3mks)
3. On the grid provided, draw the cumulative frequency curve for the data. (3mks)
4. Use the graph to estimate the semi- interquartile range (3mks)

## MARKING SCHEME

1.
 No. Std Log 0.006628 6.628 x 10-3 3.8214 3      + 7.4642 193.9 1.939 x 102 2.2876 5.7518 2.2822 2.2822 x 10º 0.3584 5.3934 5        - 0.1198 1.198 x 10-1 1.0786
2. 493/2 = (√49)3 = 73 = 343
256 = (4√256)3 = 64
2401 = (4√2401)3 = 243
343 x 64/343 = 64

3. Length of diagonal
√802 + 602 = 100
1/2 diagonal  = 50
Height = 1502 - 502
= 22500 - 2500
= 20000
= √20000 = 141.4
1/3 x 80 x 60 x 141.4
= 226.2
4. 2x2 - xy - 6y2
2x2 - 4xy + 3xy - 6y2
2x (x - 2y) + 3y(x - 2y)
(2x + 3y)(x - 2y)

x2 - 4xy + 4y2
x2 - 2xy - 2xy + 4y2
x(x - 2y) - 2y(x - 2y)
(x - 2y) (x - 2y)
(2x + 3y)(x - 2y)
x - 2y ) ( x - 2y)
2x + 3y
x - 2y
5. x + y = 13
(10y + x) - (10x + y) = 9
10y + x - 10x - y = 9
9y - 9x = 9
y - x = 1
-x + y = 13
-x + y = 1
2x = 12
y = 6
x + y = 13
6 + y = 13
Y = 7
6. int + ext = 180
3x + x + 20 = 180
4x + 20 = 180
4x + 160
x = 40
Ext  = 40 + 20 = 60
Ext = 360/n = 360/n = 60
n = 360/60 = 6
Number of traingles = n - 2
= 6 - 2
= 4
7. <BAC = 180 - 110
= 70º
14    =   AC
sin 70     sin 70
AC = 14 x Sin 70
sin 70
Ac = 14
Area = 1/2 x 1414 x sin 40
= 62.99
63.0
8. Rs = 7 cm
1 : 4000
7:4 :  &.4 x 4000
= 29600
9. reciprical of
2.47 x 102
= 0.4049 x 102 = 40.49
3√3.025 = 1.446
= 40.49 x 1.446 = 58.55
10. 1US dollar = ksh 101.9378
20000 = 20000 x 101.9378
= Ksh 2038756
1 S.A = 7.6326
2038756 = 2038756
7.6326
= 267111.6
= 267112 S.A
11. 5 - 3x ≤ x - 7 < 11 - 2x
5x - 3x ≤ x - 7
5 + 7 ≤ x + 3x
12 ≤ 4x
3 ≤ x
3 ≤ x < 6
3 , 4 , 5

x - 7 < 11 - 2x
x + 2x < 11 + 7
3x < 18
x < 6
12. 5184 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3
√5184 = 2 x 2 x 2 x 3 x 3
= 72
2744 = 2 x 2 x 2 x 7 x 7 x 7
3√2744 = 2 x 7  = 14
= 72/14 = 51/7
13.
1. A(3,6) B(7,12) (x,y)
Gradient of AB = 12 - 6 = 6 = 3
7 - 3    4     2
y - 6 = 3
x - 3    2
2(y - 6) = 3(x - 3)
2y - 12 = 3x - 9
2y = 3x - 9 + 12
2y = 3x + 3
2. mid point of AB
3 + 7 , 6 + 12
2         2
(5,9)    (x,y)
perpendicular bisector gradient = -2/3
y - 9 = -2
x - 5     3
3(y - 9) = -2(x - 5)
3y - 2y = -2x + 10
3y = -2x + 10 + 27
3y = -2x + 37

AB ⊥ BC gradient of BC = -2/3, B(7,12)
(x,y) x y - 12 = -2
x - 7       3
3y = -2 + 14 + 36
3y = -2x + 50
3(y - 12) = -2(x - 7)
3y - 36 = -2x + 14
14.
1.
 A B C 120000 180000 240000 12 18 24 2 3 4

1. 20/100 x 225000 = 45000
2. 35/100 x 225000 = 78750
2. reminder = 225000 - 45000 - 78750
= 101250
2:3:4 = 9
= 101250
1. 2/9 x 101250 = 22500
2. 3/9 x 101250 = 33750
3. 4/9 x 101250 = 45000
3. 2:3:4 = 9
45000
4. 4/9 x 45000 = 20000
Chirchir new share = 240 000 + 20 000
= 260 000
15.
1. Time = D/s = 400/120 = 31/3 hrs
= 3:20 min
8:30
+3:20
11:50 am
2. Distance by coast bus for 1/2 hr
= 80 x 1/2  = 40km
R.D = 400 - 40 = 360 km
R.S = 80 + 120 = 200 km/h
R.T = 260/200 = 14/5 hr = 1hr : 48 min
Time met =
8  30
1  48
10 18 am
3. R.T = 14/5 hr
coast by distance in 14/5 hr
= 80 x 14/5 hr
= 144 km
Distance = 40 + 144
= 184
4. In 3hr 20 min coast bus covers
= 80 x 31/3
= 80 x 10/3 = 2662/3 km
Distance from mombasa = 400 - 2662/3
= 1331/3 km

16. AD = 8.7 cm = 43.5 km ± 1
Bearing = 047º
17. y = x - 6x + 7
18.
1.  <ROP = 128º
Angles on a straight line
2. <PSR = 116º
Angle on cyclic quadrilateral
3. <ORP = 26º
Base angle of isosceles triangle
4. <TRP = 64º
Angle between a chard a tangent  (angle in alternate segment)
5. <RTP = 53º
Angles on a traingle
19.
1. 10.8
2. =57.8 = BCD
3. BD = 6.37
4. BDC = 26.9
20.
 Marks x F Tx-A FT CF 48 – 52 50 3 -14 -42 3 53 – 57 55 4 -9 -36 7 58 – 62 60 10 -4 -40 17 63 – 67 65 12 1 12 29 68 – 72 70 8 6 48 37 73 – 77 75 3/40 11 33/-25 40
1. Modal class
63 - 67
2. Mean = A + ξFt / ξ F
64 + -25/40
= 633/8

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Read 3901 times Last modified on Tuesday, 16 May 2023 08:59

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