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
)
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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