smaths paper 1lainaku 1 2012 marking … kcse lainaku mock maths paper 1 ms.pdfsmaths paper 1lainaku...
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Lainaku 212/1 march/april www.kcse-online.info
sMATHS PAPER 1LAINAKU 1 2012 MARKING SCHEME
NO WORKING MARKS REMARKS
1 3÷ 3-2 x ¾ - 1
6 10
3 ÷ ½ 8/1 –
1/10
= 3 – 1/10
= 2 9/10
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2 a) 1 + 1.5 ( rel. speed) = 2.5m/s
b) In 1 sec. they cover 2.5m
Thus they will cover
= 4800sec
= 1 hrs or 1hr 20mins
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3 3.
3b)
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For both
y=mx+c
For value of p
4
4 120 × 112 =13440
13440-1800 = 11640
⅔ × 11640 = 7760
7760 ÷ 114.20 = 67.95
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Lainaku 212/1 march/april www.kcse-online.info
5 Height = 4.3cm
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For 135o
For ABC
For height AX
4
6 6.
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For num
For Den
For
3
7
7.
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8 Frequency density = frequency/class interval
4.5 – 6.5 ⇒8
6.5 – 9.5 ⇒3
9.5 – 13.5 ⇒2
13.5 – 14.5 ⇒1
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formula
1st two values both correct
Last two values both correct
3
9 V= =736CM3
=
r = 5.600
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Lainaku 212/1 march/april www.kcse-online.info
SA = 4
SA = 4 ×5.6 = 394.27cm2
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10 4x – 9 < 6 + x 8 – 3x ≤ x + 4
3 x < 15 12 ≤ 4x
x < 5 3 ≤ x
3 ≤ x < 5
3 4 5
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2 in equalities
Represented
3
11 Area of original rectangle = 3× 4= 12
After change =
=24cm2
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12 12.
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13 Inverse T-1
= 1 2 -1
7 1 3
1 2 -1 3 -5 2 -2
7 1 3 -8 4 -3 1
A (2,-3) B (-2,1)
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Inverse of matrix
Both points in coordi. form
3
14
Q(11, 4) P(5, 1) M 1 2
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Lainaku 212/1 march/april www.kcse-online.info
OM = 1/3 5 +
2/3 11
1 4
= 9
3
OM = 92 + 3
2
= 9.487
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15 0.6135 + 3(0.1263) – 2(0.3521)
0.6135 + 0.3788 – 0.7042
= 0.2881
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16 16.
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17. a) Relief P.a = 1056 x 12
= 126672
Tax Pa = 18152 + 12672
= 30824
b) 4512 x 2 = 9024
4512 x 3 = 13536
2066 x 4 = 8264
£11090 30824
Taxable income = £ 11090
c) Basic salary = 11090 x 20 x 82
12 100
= 15,156.33
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18 (a) 12 = x
Sin 55 sin 50
x = 11.22km
(b) (b) Distance QP = 12sin75 = 14.15km
Sin 55
Speed = 14.15
2
= 7.075km/hr
(c) h = 11.22 sin 55 of 12 sin 50
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Sketch drawn
Calculation of perpendicular from
L to QP
→
→
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Lainaku 212/1 march/april www.kcse-online.info
= 9.191km
(d) 2R = 12
Sin 55
R = 7.32km
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Calculation of radius of the field
Radius gives the required
distance
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19 19 (a)
b)
c)
-2 +3K
d)
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Lainaku 212/1 march/april www.kcse-online.info
e)
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10
20
(a)
X -3 -2 -1 0 1 2 3
2x3
-54 -16 -2 0 2 16 54
X2
9 4 1 0 1 4 9
-5x 15 10 5 0 -5 -10 -15
+2 2 2 2 2 2 2 2
Y -28 0 6 2 0 12 50
(b
(c)
X= -2, 0.5, 1
(d)
B2
S1
P1
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B1 FOR ANY 10
CORRECT VALUES
For all three correct
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(e)
At (13,0) (2.5, 20)
=1.667
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SUBTRA.
� LINE DRAWN
� ALL THE THREE
VALUES
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21 (a) L = 8
(L + 26) 24
3L = L + 26
L = 13cm
x = ½ 242 + 18
2 = 15cm
h = 392 – 15
2 = 36cm
(b) h = 1/3 x 3t = 12
V = 1/3 (24 x18)36 –
1/3 x (8x6)12
= 5184 – 192 = 4992cm3
(c) p= m
/v
= 7488g
4992cm3 = 1.5g/cm
3
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Convert to g
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Lainaku 212/1 march/april www.kcse-online.info
22
BI
BI
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BI
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� OBJECT DRAWN
� LINE y=-x DRAWN
� PERPENDICULAR
BISECTOR
� COODINATES
� IMAGE 1 DRAWN
� IMAGE II DRAWN
� IMAGE III DRAWN
� IMAGE 1
COODINATES
� IMAGE II
COODINATES
� IMAGE III
COODINATES N
10
23
b)
SI
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� SCALE
� OA DRAWN
� AB DRAWN
� BC DRAWN
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(c)
R.S=10 km/h
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24. (a) y = (3x-2x2)(5+4x)
Y = 15x+12x2-10x
2-8x
3
Y = 15x + 2x2
- 8x3
Dy = 15+4x-24x2
Dx
(b) (i) h = -16t2+16t+32
H = 0
-16t2
+ 16t + 32 = 0
-t2
+ t + 2 = 0
t2
– t – 2 = 0
t2
- 2t + t – 2 = 0
t(t - 2) + 1(t - 2) = 0
t + 1 = 0 t - 2 = 0
t = -1 t = 2seconds
No negative sign
t = 2 seconds
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Lainaku 212/1 march/april www.kcse-online.info
(ii) dh = -32t + 16
dt
t = 2
v = -32(2) + 16
= -64 + 16
= -48m/s
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