2013 WESTLANDS MATHEMATICS P1 MS.pdf
7
WESTLANDS FORM FOUR JOINT EVALUATION Kenya 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 V2x2x3x3x3x3x7x7x 10~ 4 0.21 x 4.8 2x3x3x7x 10" 2 0.21x4.8 2x3x3x7x 10 1 3x7x2x2x2x2x3 10 • 1 T “ 1 4 Ml Ml A1 3 2. x + y ~ 10 lOy +• x — (lOx 4- y) ~ 54 9y — 9x = 54 9y + 9x — 90 18y = 144 y = 8,x = 2 TTie number is 28 Ml Ml A1 3 10 1 1 3 3 x 174..3i x 3:752 .J 100 8.5035 30 x 0.2665 ■ — 10 7.995 - 0.85035 7.14465 Ml Ml A1 3 4 10-4 6 3 = 3 - -- 7 = 10 = 5 5 m2 = — - /—7 4- 3 10 + 4\ m( 2 , 2 )-(-W y — 7 5 x - -2 3 (y - 7) = - - (x + 2) 5 19 y ~ —~x + -— * 3 3 Ml 81 Ml A 1 - 4 5 . 9 5 + 2 1 5 + 180(n - 4) = 180n - 360 150n -90 = 180n - 360 3 On = 270 n = 9 85 + 65 + (n - 2)30 = 360 Ml Ml A1 Accept alt two
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Transcript of 2013 WESTLANDS MATHEMATICS P1 MS.pdf
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
M l
M l
A 1
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
Ml
Ml
A13
10 1 1 3 3 x 174..3i x
3:752 .J 1008.5035
30 x 0.2665 ■ — 10
7.995 - 0.850357.14465
Ml
Ml
A1
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
M l
8 1
M l
A 1
-
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
Ml
MlA1
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
M7
Ml
A
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°
MiMlA1
Complementary angles droppy x near cosine
3
14. 20 B — = tan 62X
20X tan 62
f
64\ t
Ml
MlA1
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
B1
MlA1
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
M l
A1
B 1
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
M1
A1
M1
A1
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
A1
B1
M1M1A1
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 -
Ml
Ml
A1
Ml
A1
Ml
Al
Ml
Ml
Al
Sine rule with correct substitution attempt to solve forx
substitution
substitution
substitution
10M1A1
B1
M1
A1
M1
M1
A1
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
M1
,A1
B1M1
M1
A1
M1M1M1
A1in
^ ( - 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
B1B1
B1B1
, B 1B 1
B1B1
B1B1
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
B 1
B1
B1
Ml
Al
W
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)
Ml
Al
B1
Ml
Al
B1
Ml
A1
ND = J62 + (—2)2 = V40 ND = 6.325 units
* Time taken — %
Time — 8.40 + 2.00 = 10.30am
M1
A1
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
