0713779527
MATHEMATICS PAPER 1 MARKING SCHEME
NO CALCULATIONS MARKS REMARKS
1 No log
3.196 3.5846 x 2 7.1692
0.024 2.3802 x3 5.1406 All logs
2.3098 B1 correct
204.6 2.3109 Addition and
5.9889 M1 subtraction
6 + 1.9889 = 3.9945 Splitting of –
2 M1 ve char and
= 9.874 x 10-3 A1 div. by 2
4
2 250 + 130 + 1.50 x 898 = 1724 M1 Alt.
Transport = 5/100 x 1727 105
/100 (250 +
M1 130 + 150 +
1727 + 86.35 898) M1M1
= Kshs. 1813.35 A1 = 1813.35 A1
3
3 x + 2/3 = x + 1
y y 21
x = 2
y–½ 9
9x – 2y = -1
4y – 3y2 = 0 M1 Both equations
3y = 42
y = 14 M1 Trying to
x = 2 x 14 – 1 = 3 solve equation
9 B1 X and Y correct
x=3
y 14
y = 14
x 3 A1
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4 Time taken = 30/5 = 6hrs B1 Follow thro’
2nd 3rd 4th 5th 6th
Stop Stop Stop Stop Stop Stop
Total time for rests = (10x5)mins + 30mins M1
50 + 30 = 80mins = 1hr 20mins
Time taken back = 30/20 = 1½hrs M1
Total time = 1hr 20mins + 1hr 30mins = 2hrs 50mins
Arrival time = 8:00 + 2:50 = 10:50a.m A1
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5 Let no. of girls = x
No. of boy = 3x
Boys to poor families = x
2x boys to rich families
No. of pupils from rich families
= 2x + 1/5x = 11/5x M1
% = 11/5x (¼x) x 100 M1
= 5% A1
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6 log (2 x 18) = log 36 M1 Both numerator
log(24 x 9) & denominator
log6 M1 Simplification
3log 6
2
= /3 A1
3
M1
7. 3 + T = -1
1 3
T = -1 - –3
3 1
= -4
2
A T A1
1 + -4 = -3
-1 2 1
1
A (-3, 1)
B TB1
2 + -4 = -2 Both coord. of
2 2 4 A1 A1 & B1
1
B (-2, 4)
2
8 25(x-3)
x2 3(x+4)
= 2 2
6 x M1 Expressing
5(x-3) + 3(x + 4) = 6 – x under one base
5x – 15 + 3x + 12 = 6-x
8x – 3 = 6-x M1
9x = 9
x =1
A1
9 (i) Both expressed
2 48 2 60 as products of
2 24 2 30 their prime
factors
2 12 3 15
2 6 5 5
3 3 1
1
= 24 x 3 = 22 x 3 x 5 B1
(ii) greatest value of x is the GCD Accept alt.
= 22 x 3 = 12cm2 M1 A1 method
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10 Sketch of PQR
B1 -1
q2 = p2 + r2 – 2prcosQ M1
= 22500 + 19600 + 42000 cos120o
= 42100 + 21000 M1
q = 63100
= 251.2km A1
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11
√B1
Tan 45o= gradient = 1
m1m2 = -1 m2 = -1
x 4
y= mx + c (2 , 1 ), m = -1 √B1
1 = (-1 x 2) + c c = 3
y = -x + 3 √A1
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12 (i) ( x)2 = 12 + 12 M1
x =2 A1
(ii) PS = 12 + (x2) = 1 + x M1 For PT
PT = 12 (1 + x)2 = 1 + 1 +x
=2+x
=4 =2 A1
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13 Original price = shs.x
New price = shs.(x +3)
660 – 660 = 3
x x+2
660x + 1320 = 660x =3x (x+2)
3x2 + 6x – 1320 = 0 Quadratic equn.
x = -2 4 + 1760 M1
2 M1 Accept alt.
x = -2 42 method
2
x = 20 or -22
new price = shs.22 A1
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14 2(k + 2) – 3k = 0 M1
2k + 4 – 3k = 0
k=4 A1
2
15 4(1-bx) = 3(bx +2) M1 Removal of
-4bx – 3bx = 2 denominator
-7bx = 2
x = -2/7b A1
2
16 ½ (24 – 4x) > 6(3x – 4/3)
12 – 2x > 18x – 8
20 > 20x B1 √ correct
x<1 inequality
6(3x – 4/3) -2/3 (42 + 3x)
18x – 8 -28 – 2x √ correct
20x – 20 B1 inequality
x -1 x B1 √ correct
-1 x <1 inequality
-1 0 1
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17
M1
M1
A1
M1
(a) Sin 30o = AN A1
AN = 8 sin 30o M1
AB = 2x8 sin 30o M1
= 8cm M1
A1
(b) AO2 = Sin 40o
=4
AO2
AO2 = 4 = 6.223cm
o
Sin 40
(c) Area if segement = area of sector – area of triangle
Area of two segments (unshaded area)
= 60/360 x 3.142 x 82Sin 60o + 80/360 x3.142 x 6.2332–½x6.223Sin80o
= 5.803cm2 + 23.975
= 29.778cm2
Area of shaded region = Area of the two sectors – Area of the unshaded region
= 37.860 – 29.778
= 8.082cm2
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18 (a) tan 15 = h1/100
h1 = 100tan 15 M1
= 26.8m
tan 1 = h2/100 M1
h2 = 1.75m
height of tower = 26.8 + 1.75= 28.6m A1
(b) (i) 180o – 30o=150o
(ii) a2 = 1002 + 702 – 100 x 70 2cos60 M1A1
= 14700 – 700
a = 7700 = 87.7m M1
(iii) 100 = 87.7
sin A Sin 60 A1 Follow through
sin A = 100 sin 60
87.7 M1
A = 80.9o
180 – (30 + 80.9) = 69.0o M1
Bearing of P from Q = 180 + 69.0
= 249o A1
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19 (a) (i) No. of clearer articles = 450 dozens B1
x
No. of cheaper articles = 450 dozens B1
x – 30
(ii) 450 – 450 = 9
x–30 x 12
450(12x) – 45012(x-30) = 9 x(x-30) Quad. Equ.
5400x – 5400x + 162000 = 9x2 – 270x
162000 = 9x2 – 270x
9x2 – 270x – 162000 = 0 M1
9xo
x2 – 30x – 18,000 = 0
x2 – 150 x + 120x – 18000 = 0
x(x-150) + 120(x-150) = 0 M1
(x + 120) (x-150) = 0
x + 120 = 0 x = -120 Clearer
or x – 150 = 0 x = 150 Cheaper
cost of clearer = shs. 150 per dozen
cost of cheaper = shs. 120 per dozen M1
(b) Cost of 3 dozens = 3 x 120 = shs.360 M1 s.p
Selling price = 125 x 360 = shs. 450 price per @
100 M1 M1
Price per each article = 450/36 = shs . 12.50 M1
A1
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20 (a) B2 Frequency
x Mid pt. x f fx S1 density
0–2 1 3 3 B2 Bars
2–5 3.5 6 21 B1 (atleast 5
correct bars)
5 – 12 8.5 12 102
12 – 20 16 24 384
20 – 35 27.5 30 825
35 – 60 47.5 20 950 M1
60 - 90 75 5 375 M1A1
100 2660 M1
X = 2660
100 = 26.6 A1
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21 (a) Abdul = 2/10 x 60/100 x 680,000 M1
= shs. 816,000 A1
40
(b) 5/10 x 40/100 x 6800,000 M1M1M1 /100 x 680,000
= shs. 1,360,000 A1 = 2,720,000
5
/10 x 2,720,000
(c) 3/10 x 40/100 x 6,800,000 = 816,000 M1
3 = shs.1,360,000
/10 x 6.080,000 = 1,824,000 M1
1,824,000 – 816,000 = 1,008,000 A1
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22 (a)
Area = S(S-a) (S-b) (S-c)
S = 110 M1
Area = 110 (110-50) (110 – 75) (110 – 95)
= 110 x 60 x 35 x 15 A1
= 1861.45m2
M1
(b) 502 = 952 + 752 – 2 x 95 x 75cosR
2500 = 14650 – 14250cosR M1
Cos R = 12150
14250 A1
R = 31.5o
(c) Area of PQRS = 1461.45 x 3
= 4384.35m2
Sin 31.5o = h
For height
95 A1
h = 95 sin 31.5o
h = 49.64m
let PS = xm
area = ½ (75 + x) x 49.64 M1
4384.35 = ½ (75 + x) x 49.64 A1
75 + x = 4384.35 x 2 A1
49.64 A1
x = 101.65m
let RS = y
y2 = 952 + 101.652 – 2(95 x 101.65) cos 31.5
y2 = 19357.7255 – 16467.466
y2 = 2890.2565
y = 53.76m
75 + x = 176.65
area = ½ (75 + x) x 49.64
4384.35 = ½ (75 + x) x 49.64
75 + x = 4384.35 x 2
49.64
75 + x = 176.65
x = 101.65
y = 952 + 101.652 – 2(95 x 101.65) cos 31.5
2
y2= 193577.7225 – 16467.466
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y2 = 2890.2565
y = 53.76m
perimeter = 50 + 75 + 53.76 + 101.65
= 280.41m
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23 PQ = 10 0.1cm B1
PQR = 45 1o B1 Construction of
B1 at pt. R
ST = 10.2 0.1cm B1
QTR = 24 1o B1
B1
B1
B1
B1
B1
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24 1
(a) P (-3, 1), Q1(-6, 1) R (-6, 4) S1 S
(b) P11(-3, -1) Q11(-6, -1) and R11(-6, -4) P1 cord
(c) P111(1, -3) Q111(1, -6) R111(4, -6) B1 B1 correct coord
(d) (i) P11Q11R11 maps onto PQR by reflection in the mirror line y = -x B1 B1
(ii) P111Q111R111 maps onot PQR by reflection in the mirror line y =0 B1 B1
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B1 B1
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