# Mathematics Paper 1 Questions and Answers - Form 4 Mid Term 2 Exams 2021

MATHEMATICS
PAPER 1
FORM 4 MID TERM 2

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 calculation, 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 I: (50 MARKS)
Answer all the question in this section in the spaces provided.

1. Without using a calculator , evaluate:(3 mks)
2. Simplify completely. (3 mks)
3. The price of an article is marked as 12,000/= Mr. Omanga sold the article at a discount of 10% and still made a profit of 8%. Calculate the cost of the article. (3 mks)
4. Three sirens wail at intervals of thirty minutes, fifty minutes and sixty minutes. If they wail together at 7.18 a.m. on Monday, what time and day will they wail together? (3 mks)
5. The table shows the frequency distribution of marks scored by students in a test.
Marks         Frequency
21 - 30          2
31 - 40          4
41 - 50          11
51 - 60          5
61 - 70          3
Determine the median mark correct to one decimal point. (3 mks)
6. A cylindrical solid whose radius and height are equal, has a total surface area of 154cm2. Calculate the diameter. (3mks).
7. The exterior angle of a regular polygon is (λ - 50)º and the interior angle is (2λ + 20)º. Find the number of sides of the polygon. (3 mks)
8. Solve the following inequalities and represent it on the number line. 6x + 2 < 3x + 11 ≤ 27x − 1 Write down the integral values that satisfy the inequality. (3mks)
9. Find the equation of a line through the point (2, 1), perpendicular to the line 1/2x + 2y = −3 (3 mks )
10. Find the value of x given that; 9x + 2 × 32x − 243 = 0 (3mks).
11. The position vectors of A and B are given as OA = 2i – 3j + 4k and OB= -2ij + 2k respectively. Find to 2 decimal places, the length of vector AB. (4 mks)
12. Use the exchange rates below to answer the question.
 Buying Selling 1 us Dollar 63.00 63.50 1 Euro 125.30 125.43
A tourist arriving in Kenya from Britain has 9600 Euros. He converts the Euros to Kenya shillings at a commission of 5%, while in Kenya he converts the money to US dollars. If he was not charged any commission from the last transaction, calculate to the nearest USA dollar what he received. (3mks).
13. Given that sin(2x − 10)o = cos 60o and x is an acute angle, find x. (3 mks)
14. The length of a rectangle is (3x+1)cm. Its width is 3cm shorter than the length. Given that area of the rectangle is 28cm2 , find its length. (3 mks)
15. The mass of two similar solid are 324g and 768g. Find
1. height of the smaller solid if the height of the bigger solid is 20cm. (2 mks)
2. the surface area of the smaller solid if the surface area of the bigger solid is 40cm2.(2 mks)
16. The cost of three pens and five books is sh. 130. Kanyoro bought 2 of the pens and 3 of the books at sh. 80. How much did he pay for each? (3mks)

SECTION II (50mks)
Answer only five questions in this section in the spaces provided.

1. A bus left Nairobi at 7a.m and travelled towards Eldoret at an average speed of 80km/h. At 7:45 a.m a car left Eldoret towards Nairobi at an average speed of 120 km/h. Given that the distance between Nairobi and Eldoret is 300km,
Calculate:
1. The time the bus arrived at Eldoret. (2 mks)
2. The time of the day, the two met. (3 mks)
3.  The distance from Nairobi to where the two met. (2 mks)
4. The distance of the bus from Eldoret when the car arrived in Nairobi. (3 mks)
2. The following measurements were recorded in a field book using XY as the base line. XY = 400m.
Y
C 60  340
300 120 D
240 160 E
220 160 F
B 100 140
A 120 80
X
1. Using a scale of 1: 4000, draw an accurate map of the farm. (4 mks)
2. Determine the actual area of the farm in hectares. (4 mks)
3. If the farm is on sale at sh.80,000 per hectare, find how much the farm costs.(2 mks)
3. Mr. Omwega is employed. His basic salary is Kshs. 21, 750 and is entitled to a house allowance of Kshs 15, 000 and a travelling allowance of Kshs 8, 000 per month. He also claims a personal monthly relief of Kshs 1, 056 per month. Other deductions are; Union dues Kshs 200 and Co-operative shares Kshs 4, 500 per month. The table below shows the tax rates for the year.
 Income (Kshs perannum) Tax rates 1 – 116, 600116, 161 – 225, 600225, 601 – 335, 040335, 041 – 444, 480Over 444, 480 10%15%20%25%30%
Calculate;
1. Mr. Omwega’s annual taxable income. (2 mks)
2. The tax paid by Mr. Omwega in the year. (6 mks)
3. Mr. Omwega’s net income per month. (2 mks)
4. A straight line L1 has a gradient -1/2 and passes through point P (-1, 3). Another line L2 passes through the points Q (1, -3) and R (4, 5). Find.
1. The equation of L1. (2 mks)
2. The gradient of L2. (1 mk)
3. The equation of L2. (2 mks)
4. The equation of a line passing through a point S (0, 5) and is perpendicular to L2 (3 mks)
5. The equation of a line through R parallel to L1. (2 mks)
5. A ship leaves port P and sails to port Q which is 80km away on a bearing of 040o. The ship then sails from Q to R on a bearing 160o where R is 150km from Q. From R, the ship returns directly to P at a speed of 25km/h.
1. Using a suitable scale show the relative positions of P, Q and R. (3 mks)
2. Find the bearing of R from P (2 mks)
3. Find the distance travelled from R and the time taken to arrive at the destination(3 mks)
4. An island S is equidistant from P, Q and R. Show its relative position. (2 mks)
6. In the figure below (not drawn to scale) AB = 8cm, AC = 6cm, AD = 7cm, CD = 2.82cm and angle CAB = 50°.

Calculate (to 2d.p.)
1. The length BC. (3 mks)
2. The size of angle ABC. (3 mks)
3. Size of angle CAD. (2 mks)
4. Calculate the area of triangle ACD. (2 mks)
7.
1. Complete the table for the function y = 1 – 2x - 3x2 in the range -3 ≤ x ≤ 3 (2 mks)
 X -3 -2 -1 0 1 2 3 -3x2 -27 -3 0 -12 -2x 4 0 -6 1 1 1 1 1 1 1 1 Y -20 -15
2. Using the table above and the graph paper provided, draw the graph of y = 1 – 2x –3x2 (4 mks)
3. Use the graph in (b) above to solve
1. 1 – 2x – 3x2 = 0 (2 mks)
2. 2 – 5x – 3x2 = 0 (2 mks)
8. The diagram below (not drawn to scale) shows the cross – section of a hexagonal solid metal prism length 20cm.

Calculate;
1. The area of the shaded region (Take hexagon to be regular). (5 mks)
2. The volume of the material used to make the metal in cm3(2 mks)
3. If the density of the metal prism is 3.5 g/ cm3, find its mass in kg. (3 mks)

MARKING SCHEME

1. Numerator
3/4 + 9/7 ÷ 3/7 x 7/3
3/4 + 9/7 x 1
3/4 + 9/7 = 21 + 27/28
48/28
12/7
Denominator
2/3(9/3/8)
=2/3(72-21/56)
2/3 x 51/56
17/28
Num/Den = 12/7 ÷ 12/28 = 12/7 x 28/17
214/17

2. Numerator
3x2-xy-xy2
3x2-3xy+2xy-2y2
3x(x-y)+2y(x-y)
(3x+2y)(x-y)
Denominator
2(9x2-4y2)=2(3x+2y)(3x-xy)
(3x+2y)(x-y)      =  x-y
2(3x+xy)(3x-xy)      6x-4y

3. Sh 12000 → 100%
S.P → 90%
SP=1200x90/100=10,800
Sh10 → 108%
B.P → 100%
B.P = 10800 x 100/108 = sh10,000

4.    LCM
 2 30 50 60 2 15 25 30 3 15 25 15 5 5 25 5 5 1 5 1
LCM= 22x3x52
=300min
Time in hrs=300/60=5hrs
7:18am
5:00
12:18pm

5.  Close limit f c.f 24.5 - 30.530.5 - 40.540.5 - 50.550.5 - 60.560.5 - 70.5 241153 26172225
Position:
25/2=12.5
Median=
40.5 + (12.5 -6/11)x10
40.5 + (6.5 x 10/11)
40.5 + 5.91
=46.4

6. r=h
2πr2 + 2πrh = 154cm2
2πr2 + 2πr(r)=154
4πr2=154
4(22/7)r2=154
r=√154x7/88 = 3.5cm
distance=2 x 3.5=7cm

7. (x-50)+(2x+20)=180º
3x-60=180º
3x=210º
x=70º
Exterior angle= 70-50=20
Number of sides= 360/20=18sides
=18sides

8. 6x+2<3x+11
3x<9
x<3
3x+11<27x-1
-24/24x <-12/24
x>1/2
1/2<x<3

9. 1/2x+2y=-3
2y=-1/2x-3
y=-1/4x-3/2
M1xM2=-1
-1/4 x M2= -1 → M2=4
Equation of perpendicular
y-1/r-2=4
y-1=4x-8
y=4x-7

10. 9x+2r32x=243
32x+2x32x=243
Let 32x=y
y+2y=243
3y=243
y=81
32x=34
2x=4
x=2

11.

Length of AB=√(-42)+22+(-2)2
√24
=4.90units

12.   Amount in Ksh= 9600 x 125.3= 1,202,880
Amount ofter commission= 1,202,880 x 95/100= 1,142,736
Amount in US\$=1,142736/63.5 = 17,996

13. (2x-10)+ 60º=90º
2x+50=90
2x=40º
x=20º

14.
(3x-2)(3x+1)=28
9x2+3x-6x-2=28
9x2-3x-30=0
3x2-x-10=0
3x2-6x+5x-10=0
3x(x-2)+(5(x-2)=0
(3x+5)(x-2)=0
Either:
3x+5=0 → x=-5/3
OR
x-2= → x=2
Length = 3(x)+1=7cm

15.
1. vsf = ratio of mass = 324/768
vsf=27/64
lsf=³√27/64
=3/4
h/20=3/4 → h=20x3/4
=15
2. a.s.f = (lsf)2
(3/4)2
=9/16
A/40 = 9/16 → h=40 x 9/16
=22.5cm2

16.
3ρ+5b=130
2ρ+3b=80

6ρ+10b=260
6ρ+ 9b=240
1b=20
3ρ+5(2x)=130
3ρ=30
ρ=10
1 pen=10KSh
1 bok=20Ksh

17.
1. Time=d/s
300km/80km/hr
=33/4
time of arrival
7:00am
3:45
10:45am

2. Distance travelled by bus for 46mins
D=sxt
80 x 3/4=60km
Remaining dist=300-60=240km
Rel Spd= 80+120=200km/hr
Time taken= 240/200= 1hr 12 mins
Time met
7:45
1:12
9:57

3. Distance travelled by cra when meeting:
D= sxt
120km/h x 11/5=144km
Dist from nai= 300-144=156km

4. Time taken to nai by car:
t= 300km/120km/hr= 2hrs 30 min
7:45
2:30
10:15am
Remaining dist= 80 x 1/2
=40km
18.
1.
2.
Area 1 = ½ x 80 x 120=4800m2
Area 2 = ½x 60 x (120+100)=6600m2
Area 3 = ½ x 200x(100+60)=1600m2
Area 4 = ½ x 60 x 60 = 1800m2
Area 5 = ½ x 100 x 120 = 6000m2
Area 6 = ½ x 60 x (160+12m2
Area 8 = ½ x 160 x 220 = 17600m2
Total area = 64400m2
Area in ha= 64400/10,000=6.44ha

3. Total cost= 6.44ha x 80,000= 515,200
19.
1. Month taxable income = 21750 + 15000 + 8000= 44750
annua taxable inc= 44750x12=537,000
2.
116600 x 10/100 = 11660
109000 x  15/100= 16350
109440 x  20/100= 21888
109440 x   25/100= 27360
92520 x 30/100 = 27756
Total tax payable = 105014
Tax due = tax payable - reliefs
105014-(1056 x 12)
105014- 12672
=92342

3. Tax p.m = 92342/12= 7695.17
Net income= gross income - all deductions
44750-12395.17= 32354.83/=
20.
1. y-3/x-(-1)= -1/2
2y-6= -x-1
2y= -x+5
y= -1/2x+ 5/2
(any form of this eqn is acceptable)

=8/3

3. Eqn of L2
y-(-3)/x-1=8/3
3y+9=8x-8
3y=8x-17
y= 8/3x - 17/3

4.
1. M1 x 8/3=-1
M=-3/8
y-5/x = -3/8
8y-40 = -3x
8y=-3x+40
y=-3/8x+5

2.  y-5/x-4 = -1/2
2y-10= -x+4
2y= -x+14
y=-1/2x+7
21.
1.
2.   Distance PR=6.5x20
130 ± 1km

3. t=d/s
120/25
=5hr 12 min

4. S is the centre of the circumference
22.
1. a2= b2+c2-2bcCosA
a2=62+82-2(6)(8)Cos50º
a2=100-61.71
a=√38.29
=6.19cm

2. b/sin Bº = a/Sin Aº
6/Sin B= 6.19/ Sin 50º
Sin Aº= 6 x Sin50º/6.19
B= Sin-10.7425
=47.95º

3. a2= 72+62-2(7)(6)CosA
2.822=85-84CosA
Cos A= 85-2.822
84
A=Cos-10.9172
=23.48º

4. Area of ΔACD= 1/2 x 6 x 7 x Sin23.48
=8.37cm2
23.
1.
 X -3 -2 -1 0 1 2 3 -3x2 -27 -12 -3 0 -3 -12 -27 -2x 6 4 2 0 -2 -4 -6 1 1 1 1 1 1 1 1 Y -20 -7 0 1 -4 -15 -32

2.
3.
1. Roots
x=-1 or x=0.2
2.
y=1-2x-3x2
0=2x-5x-3x
y=-1+3x
Roots
x=-2 or x=0.3
24.
1. Area of Shaded R= (1/2 x 14x Sin 60)-(1/2x12xSin60)
84.87-62.35
=22.52

2.  Volume= Area of cross-section length
= 135.12cm2 x 20cm
=2702.4cm2

3. Mass= density x volume
3.5g/cm2 x 2702.4cm3
9458.4g
9.458kg

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