Free Marking Schemes For Form 2 Maths Exams

MATHEMATICS

TERM 3

NAME: …………………………………………………ADM NO……………….

CLASS………………..DATE…………………

FORM TWO

MATHEMATICS

TIME: 2 ½ HOURS

Instructions

Write your name, adm no. class and date in the spaces provided above.
The paper consists of two sections: section I and section II.
Answer all the questions in section I and any five in section II
Section I has sixteen questions and section II has eight questions
All answers and work must be written on the question paper in the spaces provided below each question.
Show all the steps in your calculations, giving your answers at each stage in

the spaces below each question

KNEC Mathematical table and silent non-programmable calculators

may be used.

FOR EXAMINER’S USE ONLY

SECTION I

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

SECTION II



	GRAND TOTAL

17	18	19	20	21	22	23	24	Total

This paper consists of 14 printed pages

SECTION 1 (50 MARKS)

Answer any FIVE questions in this section in the spaces provided

1.Evaluate: + (3mks)

of ( +)

2. Express as a fraction. (2mks )

3.Simplify (3mks)

Fifteen tractors each working eight hours a day take eight days to plough a piece of land. How long would it take 24 tractors each working 10 hours a day to plough the same piece of land 3mks)

The shaded region below shows the area swept out on a flat windscreen by a wiper. Calculate the area of the shaded region. (4mks)

4cm

16cm 120^o

6. The mass of two bags of beans and three bags of salt is 410kg. If the mass of three bags of beans and two bags of salt is 390kg, find the mass of each bag. (3mks)

7. The interior angle of a regular polygon is twice the exterior angle.

Find the number of sides of the polygon. (3mks)

What is the name of the polygon? (1mks)

The angle of elevation of a church tower from point A, 50 metres away from the foot of the church is 24^o. Find the distance between A and B if the angle of elevation of the tower from B is 20^o. (4mks)

9. The figure below is a cross-section of a swimming pool 8m wide. Calculate the capacity of the pool in litres. (3mks)

30m

Three litres of water (density 1g/cm³) is added to twelve litres of alcohol (density 0.8/cm³). What is the density of the mixture? (3mks)

The volume of two similar solid spheres are 4752cm³ and 1408cm³. If the surface area of the small sphere is 352cm², find the surface area of the larger sphere. (3mks)

Solve for x in the equation = 32 (3mks)

Momanyi spent eight of his February Salary on farming, half on school fees and two-thirds of the remainder on food. Calculate his February salary and the amount he spent on school fees if he spent sh. 3200 on food. (3marks)

Form three inequalities that satisfy the unshaded region R. (3marks)

A Kenyan tourist in the US borrowed 10,000 US dollars to pay for his son’s examination.

He is expected to pay either in Kenyan shillings or through an account in the United Kingdom in

sterling pounds. If he decided to pay through the United Kingdom, how much would he save given

that

1 US dollar = 82.4 Kenyan shillings

1 Sterling pound = 1.4 US dollar

1 Sterling pound = 105 Kenyan shillings (3mks)

Solve for X in the equation. (3mks)

SECTION II (50MKS)

Answer any FIVE questions in this section in the spaces provided

The figure below shows a glass in the form of a frustum of a cone whose top and bottom diameters of 7cm and 3.5cm respectively. Its depth is 10cm. Taking,

Calculate;

a) Its total surface area. (5 marks)

b). Its capacity in litres. (5 marks)

18. Two friends Jane and Tom live 40km apart. One day Jane left her house at 9.00 a.m. and cycled towards Tom’s house at an average speed of 15km/hr. Tom left at 10:30 a.m. on the same day and cycled towards Jane’s house at an average speed of 25km/hr.

Determine;
The distance from Jane’s house, where the two friends met. (4 marks)

The time they met. (2 marks)

How far Jane was from Tom’s house when they met? (2 marks)

The two friends took 10 minutes at the meeting point and they cycled to Tom’s house at an average speed of 12km/hr. Find the time they arrived at Tom’s house. (2 marks)

Town Q is 180km on a bearing of 050^o from town P. Another town R is on a bearing 110^o from P and also on a compass bearing S 30^oE from Q. Town S is South of P and also West of R.

Using a scale of 1 cm rep. 20 km;

Draw the scale diagram to show the positions of the four towns. (6 marks)

Use your scale diagram in (a) above to find;
The distance of R from P. (1 mark)

The bearing of Q from S. (1 mark)

The distance of Q from S. (1 mark)

How far P is North of S. (1 mark)

The marks of 100 candidates for the mathematics examination were distributed as follows.

marks	No of candidates(f)	Mid-point(x)	fx	c.f
30-3435-3940-4445-4950-5455-5960-64	52426241362

(a)Calculate

(i) The mean mark (2mks)

(ii) The median (3mks)

(b) On the grid provided, draw a histogram. (3mks)

(d) Find the modal mark. (1mk)

The figure below shows two circles of radii 10.5 and 8.4cm with centres A and B respectively. The common chord PQ is 9cm.

(a) Calculate angle PAQ. (2 mks)

(b) Calculate angle PBQ. (2 mks)

Three business partners; Kamau, Tatwa and Makau contributed Ksh. 100,000, Ksh. 80,000 and Ksh. 50,000 respectively to start a business. After one year, the business realized a profit which they shared in the ratio of their contributions.
- If Makau’s share of profit was Kshs. 20,000, how much was the total amount of profit?

(3mks)

At the beginning of the second year, Makau boosted his shares by Ksh. 10,000. If the business profit increased by 20% at the end of the second year, calculate:-
- Kamau’s share of the profit. (4mks)

The difference between Kamau’s and Tatwa’s share of profit. (3mks)

(a) Show by shading the unwanted region, the region which satisfies the following inequalities (8mks)

Y > -3

4y ≤5x + 20

2y < – 5 x + 10

4y≤ -3x – 12

(b) Calculate the area of this region in square units (2mks)

Triangle ABC has the vertices A (3, 1), B (2, 2) and C (3, 4).

(a) On the grid provided draw triangle ABC and its image A¹B¹C¹ under a rotation of negative quarter turn about the point (0,0) (3 marks)

(b) (i) Draw triangle A¹¹B¹¹C¹¹ the image of A¹B¹C¹ under a reflection in the line y = -x (2 marks)

(ii) Describe fully the transformation that maps A¹¹B¹¹C¹¹ onto ABC (2 marks)

(c) (i) On the same axes draw triangle A¹¹¹B¹¹¹C¹¹¹ the image of A¹¹B¹¹C¹¹ under a translation given by translation Vector

(iii) State the coordinates of A¹¹¹B¹¹¹C¹¹¹(2 marks)

^{______________________________________________________________________________________}

MATHEMATICS FORM 2

MARKING SCHEME

1Evaluate: + + (2mks)

of ( + )

: + +

of ( + )

=25+ = 25

Let r= 0.1515

100r=15.1515

99r=15.0000

R= =

Simplify (2mks)

a(y-x)= a(y-x) = a = -a

b(y-x) – b(y-x) –b b

T D H

15 8 8

24 10

15/24 x 8/10×8= 4 days

A1 = 0/360 r²

=120/360×3.142×20²=418.933cm²

4cm = 2.68.117cm²

Area of shaded region.=418.933-268.17

16cm 120^o=150.816cm²

2b+3s=410

3b+2s=390

4b+6s=820

9b+6b=1170

5b+0=350

5b=350

5 5

2×70+35=410

140+35=410

3s=410-140

3s=270

3 3

S=90

Beans=70bags

Salt=90bag

7a). Let the exterior angle be x

X+2x=180

3x=180

X=60^o

no of sides

360/60=6

b) Hexagon

Tan 24^o= h/50

50 tan 24^o = H

Tan 20^o = H/(50 +x)

(50+x) tan 20=H

18.1999+0.364x=22.26

0.3640c=22.26-18.199

0.364x=4.061

X=4.061

0.364

=11.16m

Volume of water=Ah

A=1/2(1×3) x 30= 60m²

V=60m² x 8m=480m³

1m³=1000L

480m³=?

480m3 x 1000L

1m³

= 480,000L

Total vol = 15 litres = 15000cm³

Tota; mass = 3000g + (12000 ´ 0.8)g

= 3000g + 9600g = 12600g M1

Density = M1

= 0.84g/cm³ A1

VSF = 3.375

LSF = M1

ASF = (1.5)²

Area of the larger cylinder

= 352 x 2.25= 792cm² A1

^X ^{1- X} = 32

( ^X ^{1- X}=

13.

1. February salarySchool fees

M1 A1 B1

14.

B1B1B1

10,000 ´ 82.4 = 824000 M1

10,000 M1

824000 – 750000 =

Sh.74000 A1

L.C.M=12 24x-16-12x+6=12-10x

24x-12x+10x=12-6+16

22x=22

X=1

SECTION II

x = 9.859.853.5101.753.51.75ℓ = 9.85L = 19.69a) T.S.A = + ( RL – rL)= (r² + RL – rL)= (1.75² + 3.5² x 19.69 – 1.75 x 9.85)= x 54.18= 171.1cm² b) Vol = R²H – r²hH = 20h = 10(R²H – r²h)(3.5² x 20 – 1.75² x 10)(245 – 30.625)x 214.375cm³

B1 M1M1M1A1 B1 M1M1 M1A1

18a)	i) 10.30–9.001.30Jane travelled = x 15 = 22.5Distance before Tom starts journeyRelative speed = 15 + 25 = 40km/hrT.T.T.M == 0.4375 hrs15 x 0.4375 = 6.5625km22.5 + 6.5625= 29.0625km ii) They met after 0.4375 hrs= 0.4375 x 60= 26 minutes10.30+ 2610.56amiii) Jane had travelled 29.0625km= 40.00 – 29.0625= 10.9375km b) = 0.91146 hrs0.91146 hrs = 55 minutesAdd rest time = 10 minutes= 65 = 1 hr 5 minutes10.56+1.0512.01 pm	M1 M1 B1 A1 M1 M1 B1B1 M1 A1
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19	a) b)i) Distance R from P= 13.4cm ± 0.1But 1 cm rep 20km = 13.4 x 20 = 268km ii) Bearing of Q from S034^o± 001^o iii) Distance of Q from S12.4cm ± 0.1But 1cm rep 20km = 12.4 x 20 = 248kmiv) How far P is north of S= 4.5cmBut 1cm rep 20km = 4.5 x 20 = 90km	B1 B1 B1 B1

21	a)< PAQ = <PAM + <QAM< PAM = sinθ1 =Sin ^-1 (0.4286) = 25.38⁰< QAM = <PAM = 25.38→<LAP = 25.38×2= 50.76 b) <PBQ = < PBM + <QBM< PBM = sin∝1 =Sin^-1 (0.5357) = 32.39⁰< PBM = <QBM = 32.39⁰<PBQ = 32.39⁰x 2 = 64.78 c)i) area of segment = area of a section – area of DTaking (i)== 48.84 – 42.69 = 6.15cm²Taking (ii)== 39.89 – 31.92 = 7.97cm²= (6.15 + 7.97) cm² = 14.12cm²	M1 A1 M1 A1 M1 B1 M1M1 B1 A1

22.a) Kamau Tatwa Makau100,000 80,000 50,00010 : 8 : 55 = 20,000231 = ?20,000 x 235= 92,000 (a) (i) New Ratio5 : 4 : 3120 x 92,000100New profit = 110,400 Kamau’s share = 5 x 110,40012= 46,000 (ii) Tatwa’s share = 4 x 110,40012= 36,800Difference = 46,000 – 36,800= 9,200	B1 M1 A1 B1 B1 M1 A1 M1 M1 A1
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