221/1 Maths Paper 1 Page 1 of 10

END OF TERM II FORM FOUR EXAMINATION, 2017

Kenya Certificate of Secondary Education (K.C.S.E)

221/1

MATHEMATICS

PAPER 1

MARKING SCHEME

Solution

1. 0.216 = 1

0.216

= 1

0.216 × 10−1

= 10 x 0.4630 = 4.63

√0.5123

0.216= √0.512

3 x 4.63

= 0.8 x 4.630 = 3.704

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2. 11

5−1

1

3

1

8− (−

1

2)

2 = 6

5−

4

31

8−

1

4

= −2

15 × −8 =

16

15

16

15−

7

15 𝑜𝑓 2

16

15−

14

15

= 2

15

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3. 9 15 21

3 33 5 7

3 1 2 7

5 1 1 7

7 1 1 1

L.C.M = 315 minutes 315

60 = 5 hrs 15 minutes

11:00

5:15

5:45pm

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4 (2𝑥−3)(3𝑥−2)

(2𝑥+3)(2𝑥−3)

(2𝑥−3)(3𝑥−2)

(2𝑥+3)(2𝑥−3) =

3𝑥−2

2𝑥+3

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5 2 log10 𝑥 − 3 log10 2 + log10 32 = log10 100

log𝑥 𝑥2 − log 23 + log10 32 = log10 100

log (𝑥2

8 × 32) = log10 100

4𝑥2

4=

100

4

x2 = 25

x = 5

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221/1 Maths Paper 1 Page 2 of 10

6 2y = 4x + 5

y = 2x + 5

2

m1 = 2 m1 x m2 = 1

2m2 = 1

m2 = −1

2

Let (x, y) be a point on L 𝑦−1

𝑥−3 = −

1

2

y = 1 = −1

2 (x – 3)

y = −1

2x +

5

2

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7 a) 195250 x 12.34 = 2,409,385

b) 2,409,385 – 1,258,000 = 1,151,385

1,151,385 x 11.37

= 13,091,247.4

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8 7th = a + (7 – 1)d

a + 6d = 20

S20 = 20

2 (2a + (20 – 1)d)

= 10(2a + 19d) = 610

= 20a + 190d = 610

2a + 19d = 61

a + 6d = 20 x 2

2a + 19d = 61

2a + 12d = 40

7d = 21

d = 3, a = 2

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9 a)

OQ = OQ + PQ

PQ = OP + PQ PQ = 2

3PR

= (23

13) +

2

3(

−2−3

−13) + (

5−34

)

(23

13) +

2

3(

3−6−9

) = (2

−4−6

)

OQ = (23

13) + (

2−4−6

) = (4

−17

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221/1 Maths Paper 1 Page 3 of 10

OQ = (4

−17

)

b) |OQ| = √42 + (−1)2 + 72

= √16 + 1 + 49

= √66

= 8.12 units

10

LPM = 103o

11 AB = (𝑘 43 2

) (1 23 −4

) = (𝑘 + 12 2𝑘 − 16

9 −2)

(k + 12 x -2) – (2k – 16 x 9) = 10

-2k + 24 – 18k + 144 = 10

-2k – 18k + 24 + 144 = 10

-20k = -158

k = 158

20= 7.9

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12 In 1hr A fills 1

3 of the tank

B fills 1

5 of the tank

A and B fills 1

3+

1

5=

8

15 of the tank in 1 hr

In C empty 1

4 of the tank

8

15−

1

4=

32−15

60=

17

60

When A, B and C are opened, they fill 17

60

1hr 17

60

? 60

60

60

60 ×

60

17=

60

17= 3

9

60 hrs or 3hrs 9 minutes

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13 y = x2 – 3x + 1 (-2, 3) dy

dx = 2x – 3

At (-2, 3), gradient 2(-2)-3

= -4 – 3 = -7

Equation a normal, gradient is 1

7

y−3

x− −2 =

1

7

y – 3 = 1

7(x + 2)

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221/1 Maths Paper 1 Page 4 of 10

y – 3 = 1

7x +

2

7 or y =

1

7x +

23

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14 6x = 180

x = 30o

n = 360

30

n = 12 sides

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15 x + 8 > 4x – 6 4x – 6 ≥ 12 – 3x

x – 4x > -6 – 8 4x + 3x ≥ 12 + 6

-3x > -14 7x ≥ 18

x < 14

3 x ≥

18

7

18

7 ≤ x <

14

3

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16 Let bay A contain xkg of sugar

B contain ykg of sugar

4(x – 2)kg = y + 2

4x – 8 = y + 2

4x – y = 10 ………….(i)

2(x + 10) = y + 10

2x + 20 = y + 10

2x – y = -10 ………….(ii)

4x – y = 10

2x – y = -10 +

6x = 0

x = 0

y = 10

Bay A = No sugar

B = 10kg

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SECTION II

17 a) Modal class 40 – 44

b) i) Using the assumed mean of 37

class midpoints t = x 37 f ft

2024 22 -15 2 -30

2529 27 -10 15 -150

3034 32 -5 18 -90

3539 37 0 25 0

4044 42 5 30 150

4549 47 10 6 60

5054 52 15 3 45

5559 57 20 2 40

∑ 𝑓 = 101 ∑ 𝑓𝑡 = 25

𝑡̅ = ∑ 𝑓𝑡

∑ 𝑓 =

25

101 = 0.2475

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221/1 Maths Paper 1 Page 5 of 10

Mean = 0.2475 + 37

= 37.2475

ii) Median = 𝐿 + (𝑛

2 − 𝐶

𝑓) 𝑖

Median class 35 – 39

Height of the 51st pawpaw is 34.5 + 16

25 × 5

34.5 + 3.2 = 37.7

Difference is 37.7 – 37.2475

= 0.4525

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18 a) 1cm 60m

3 300m = 5cm

b) i) Distance between T and S

6.6cm ± 0.1

66 x 60 = 396m

ii) 222o

c) Area of A = 1

2 × 150 × 264 = 19,800m2

Area of B = 1

2 × 264 × 300 = 39,600m2

59,400m2

In ha 59,400

10,000 = 5.94 ha

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221/1 Maths Paper 1 Page 6 of 10

19 a)

b) i) P(not bitten by a rabid dog)

P(R and B) or P(R1 and B)

(186

21 ×

4

155 ) + (

3

21 ×

1

13)

8

35+

1

91=

109

455= 0.24

ii) P (He will get Rabies)

P(R and B and T1 and R) or P(R1 and B and T1 and R)

(18

21 ×

4

15 ×

1

5 ×

5

7) + (

3

21 ×

1

13 ×

1

5 ×

5

7 )

8

245+

1

637=

109

3185

= 0.034

iii) P(R and B and T1 and R1) or

P(R1 and B and T1 and R1)

(18

21 ×

4

15 ×

1

5 ×

2

7) + (

3

21 ×

1

13 ×

1

5 ×

2

7 )

16

1225+

2

3185=

218

15925

= 0.014

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20 a) V = kr2 + mr3

k + m = 54.6

4k + 8m = 226.8

4k + 4m = 218.4

4k + 8m = 226.8

-4m = -8.4

m = 2.1 and k = 52.5

V = 52.5r2 + 2.1r3

b) V = 52.5 x 42 + 2.1 x 43

= 52.5 x 16 + 2.1 x 64

= 840 + 134.4

= 974.4

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221/1 Maths Paper 1 Page 7 of 10

c) 52.5r2 = 2.1r3

(2.1r – 52.5)r3 = 0

r =25

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21 a) y = 5x 1

2x2

x 0 1 2 3 4 5 6

5x 0 5 10 15 20 25 30

1

2x2 0 -0.5 -2 -4.5 -8 -12.5 -18

y 0 4.5 8 10.5 12 12.5 12

b) ∫ (5x −1

2x2)

6

0dx

[5

2x2 −

1

6x3 + c]

0

6

(90 – 36 + c) – c

54 square units

c) i) y = 2x

x 0 1 2 3 4 5 6

2x 0 2 4 6 8 10 12

y 0 2 4 6 8 10 12

ii) Area bounded by the curve and the line y = 2x

54 square units 1

2 x 6 x 12

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221/1 Maths Paper 1 Page 8 of 10

54 – 36

= 18 square units

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22 a) Volume of hemisphere

= 1

2 ×

4

3 ×

22

7 × 5.23

Big cone V1 = 1

3 ×

22

7 × 5.22 × ℎ

Small cone V2 = 1

3 ×

22

7 × (

5.2

3) × ℎ

V1 – V2 = 1

2 ×

22

7 × 5.22 × (3 −

1

9) ℎ

= 1

2 ×

22

7 × 5.22 ×

26

9h

26

9h = 10.4

h = 10.4 ×9

26 = 3.6

Height of frustum

2h = 7.2cm

b) L = 3.62 + 5.22

3 = 3.995

L = √0.82 + 5.22 = 11.98

Area = 𝜋𝑟2 + 𝜋𝑅𝐿 − 𝜋𝑟𝐿 22

7 × 3 × 5.22 ×

11.98

7−

22

7 ×

5.2

3 × 3.998

9.429 + 195.8 – 21.76

= 183.469

= 183.5 cm

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23 a) T = 300

80= 3

3

4 ℎ𝑟𝑠

= 3hrs 45 minutes

7:00

3:45

10:45 a.m

b) Distance travelled by bus before the car started = 3

4 x 80

= 60km

Distance between the two vehicles

= 300 – 60

= 240km

Relative speed = (80 + 120) = 200km/h

Time taken to meet = 240

200= 1

1

5 hrs

= 1 hr 12 minutes

The two vehicles met at 7:45

1:12

8:57A.m

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221/1 Maths Paper 1 Page 9 of 10

c) In 1 hr 12 minutes the bus had travelled 11

5 x 80

= 6

5 × 80 = 96km + 60km

= 156km

d) 300

120=

5

2= 2 hrs 30 minutes

Distance = 5

2 x 80 = 200km

Remaining distance = 300 – (200 + 60)

= 40km

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24. a) y = 2x3 + 5x2 – x – 6

x 4 3 2 1 0 1 2

2x3 128 54 -16 -2 0 2 16

5x2 80 45 20 5 0 5 20

-x 4 3 2 1 0 -1 -2

-6 6 6 6 6 6 6 6

y 50 0 0 -2 6 0 28

b)

221/1 Maths Paper 1 Page 10 of 10

c) i) y = 2x3 + 5x2 – x – 6

0 = 2x3 + 5x2 + x – 4

y = -2x – 2

x -4 -3 -2 -1 0 1 2

-2x 8 6 4 2 0 -2 -4

-2 -2 -2 -2 -2 -2 -2 -2

y 6 4 2 0 -2 -4 -6

x = 0.5 ± 0.1 or

x = -2.3 ± 0.1

ii) y = 2x3 + 5x2 – x – 6

0 = 2x3 + 5x2 – x + 2

y = -8

Roots are x = 2.5 ± 0.1 or

x = -3.2 ± 0.1

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