2013 westlands mathematics p1 ms.pdf
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math mockTRANSCRIPT
WESTLANDS FORM FOUR JOINT EVALUATIONKenya Certificate of Secondary Education
MATHEMATICS Paper-121/1 ‘
July / August 2013 Marking Scheme
a- V108 x 147 x lO-4
0.21 x 4.8 V 2 x 2 x 3 x 3 x 3 x 3 x 7 x 7 x 10~4
0.21 x 4.8 2 x 3 x 3 x 7 x 10"2
0.21x4.8 2 x 3 x 3 x 7 x 1 0 1
3 x 7 x 2 x 2 x 2 x 2 x 3 10 • 1
T “ 14
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32. x + y ~ 10
l O y + • x — (lOx 4 - y ) ~ 5 4 9 y — 9 x = 5 4
9y + 9x — 9 0 1 8 y = 1 4 4 y = 8, x = 2
TTie number is 28
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10 1 1 3 3 x 174..3i x
3:752 .J 1008.5035
30 x 0.2665 ■ — 10
7.995 - 0.850357.14465
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34 1 0 - 4 6 3
= 3 - -- 7 = 1 0 = 55
m2 = — - / — 7 4 - 3 1 0 + 4 \
m ( 2 , 2 ) - ( - W y — 7 5 x - - 2 3
( y - 7 ) = - - ( x + 2 )5 1 9
y ~ —~x + -—* 3 3
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-
45 . 9 5 + 2 1 5 + 180(n - 4) = 180n - 360
150n -90 = 180n - 360 3 On = 270
n = 985 + 65 + (n - 2)30 = 360
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Accept alt two
1Z. 1cm = 2km 1cm2 = 4fem2
11.7cm2 = 11.7 x 4 = 46.8km2 Area in hectare = 46.8 x 100
= 4680 ha
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changing linear scale factor into A.S.F getting actual area.
313. Sin (3^r — 50) = cos(2x + 10)
Sin (3x - 50) - Sin (90 - (2x + 10)3x — 50 + 2x + 10 = 90
5x = 130 x = 26°
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Complementary angles droppy x near cosine
3
14. 20 B — = tan 62X
20X tan 62
f
64\ t
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Correct substitution
Attempt to solve forxX —- iU.OJ Tfl X
3
15. i?2 — r2 — 100
Area = n{R2 — r2) = 100x3.142 = 314.2cm2
1
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Method and answer (Pythagoras theorem) Substitution
316. lOOp + 800p + 200p = 22000
HOOp = 22000 200 x 20
p - 2 0 “ - 4 0 0 = 1 0
Total = 10 + 20 + 80 = 110
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Formation of equation
3
17.a)h 4 9 = 64
h = — x 9 = 6 cm 6
i 2 2 , , ^ K = -x — x 4 x 4 x 6 3 7
4VoJ. = 100-cm3
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Correct expression
Substitution in correct formula
b ) 7r x 4 x 4 x h = x 4 x 4 x 6 M1 equating
c )
/i = 2height of water — 12 — 2 = 10cm2
1FoZ. = 7 r x 4 x 4 x l Q — —7r x 6 x 6 x 9
= 1607T - 1087122 3 33 = 52 x — = 163 —cm33
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correct expressions substraction
10
- ’
- ' &
I 11 u/etctt ampiq . chrm A - MATHEMATICS - 1
18.
19a)
b)
c)
d)
520 xSin 32 ” Sin 118
520 sin 118* = Sin 32
= 866.32 m
= Cos32866.42
x — 866.32 cos 32 = 750m
Height = 750 tan 30
= 433m
—— = tan 80433
76.37tan 80
Distance = 750 — 76.37 - = 673.6m
x3 — 9x — 0 x(x2 - 9) = 0
x .= 0 or x = ±3
X -3 -2 | -1 0 i 2 3
y 0 O 00 0 -8 ,10 I 0
Area — — x 1{0 + 0 + 2(10 + 8 + 0 + 8 + 10)}2 ^
1Area = - x 2 x 36 = 36 sq units
2
o *Area = J(x3 - 9x)dx + f (x3 - 9x)dx
n 4 9I aX ~ nL4 2
X2 + c3
■»0
= 20.25 + -20.25= 40.5 sq unit
Error — 40.5 — 36 = 4.5
4.5Percentage Errors -
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Sine rule with correct substitution attempt to solve forx
substitution
substitution
substitution
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ordinates
Substitu for the coorindate
intergral expression substities in the integrated expression
<B 70n WSSTI.ANDS- FORM 4 - MATHEMATICS - I
20a)Det = 4x5 — 6x3 = 20 — 18
) = f2 “L5) J V-3 2.5 )Inverse1, 4 -3 2 '—6 5
b) (:2.55) (6
2 ) ® - ' 8 4 0 'V1040/3\(x\ _ ( 2 —1.5\ / 840 'N 4/ \yJ V—3 2.5 ) \1040'
?)©=©X =e 120,3/ = 80
Small costs = Sh, 80, Large costs 120
Large = -y-rz X 120 = 144
CoSh. 120
c)100
90Small =? x 80 = 72
« <424)
1152 4- 360 = Sh. 1,510
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^ ( - 6 1 )
matrix equation
premultiplication both sides
attempt to solve
both x cost of large and small
- ZDAB=35°Alternate segment theorem
b) ZADB=55°The radius and the tangent are perpendicular
c) Z EOD=70°Angle at the centre is twice the angle at the circumference
d) Z DCB=62°Angle in a semi circle is right angled
e) AFE=145 . ^Opposite angles of acyclic quadrilateral are suplimetary
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, B 1B 1
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Mark for the correct angle
One mark for the season
1022"
Class 10.5-20.5 20.5-25.5 25.5-40.5 40.5-50.5 50.5-55. 5i 10 5 15 10 5fd 1.2 2.6 1.4 2.2 1.2f 12 13 21 22 6
b)Modal frequency = 22
B1B1B2
B1
class width frequency densityall correct frequencies B1 - any 4 correct
modal frequency
B1 modal frequency
Class i fd X f x-A fd10.5-20.5 10 1.2 15.5 12 -17.5 -21020.5-25.5 5 2.6 23 13 -10 -130
25.5-40.5 15 1.4 33 21 0 0
40.5 - 50.5 10 2.2 45.5 22 12.5 27550.5 — 55.5 5 1.2 53 6 20 20
Total 74 55
Mean = 33 +5574
Mean = 33 4- 0.7432 = 33.7432
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mid - poin
deviations from assumed mean - (d)
fd
correct
23. 7̂ 27
(_88)
^=I(-88) = (-66)
m-i , )= (“ )D(10,-4)
^=e° 4) - l c 26)
™ - a - (-42)
=(-2)
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ND = J62 + (—2)2 = V40 ND = 6.325 units
* Time taken — %
Time — 8.40 + 2.00 = 10.30am
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10
60 80 VII80x — 3200 = 60x y[|
20x = 3200 Vfl.r — 160fcm
320 - 160 = 160/cm
160 uriTime taken = -r-r- = 2 hrs »80
Bus took = 200 = 3 hrs, 20 mins M1 W~
Saloon took = 200 = 2hrs, 30 mins M1 TRT
Repair time = 3hr 20min - 2 hrs 30 mins M1= 50 min A1
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Marking Scheme 7 0 2013 WESTLANDS - FORM 4 - MATHEMATICS -1