KCSE Past Papers Maths A 2013
4.0
4.1
4.1.1
I
2
3
4
5
6
7
8
PART TWO: THE YEAR 2013 KCSE EXAMINATION QUESTION PAPERS
MATHEMATICS ALT. A (121)
Mathematics Alt. A Paper 1 ( 121/1)
SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.
_ Z " _
Evaluate Q — (2 marks)
6 X 2 72 +' 8 X 3
The production of milk, in litres, of 14 cows on a certain day was recorded as follows:
22, 26, 15, 19, 20,16, 27,15, 19, 22, 21, 20, 22 and 28.
Detennine:
(a) the mode; (1 mark)
(b) the median. (2 marks)
Use logarithms, correct to 4 decimal places, to evaluate:
3,1794 X 0.038
1.243
Simplify the expression:
(4 marks)
1611119112
4m2  mu — 3n’ (3 111111115)
A wholesaler sold a radio to a retailer making a proﬁt of 20%. The retailer later sold the radio
for Ksh l 560 making a proﬁt of 30%. Calculate the amount of money the wholesaler had paid
for the radio. (3 marks)
A point P on the line AB shown below is such that AP = %AB. By construction locate P.
(3 marks)
c a
A B
Chelimo’s clock loses 15 seconds every hour. She sets the correct time on the clock at 0700h
on a Monday. Determine the time shown on the clock when the correct time was l900h on
Wednesday the same week. (3 marks)
Given that sin (x + 20)° = —0.7660, ﬁnd x, to the nearest degree, for 0° 3 x S 360°. (3 marks)
A number m is formed by writing all the prime numbers between 0 and 10 in an ascending
order. Another number n is formed by writing all the square numbers between 0 and 10 in a
descending order.
(a) Find m — n; (2 marks)
(b) Express (m — n) as a product of its prime factors. (1 mark)
The ﬁgure below shows a net of a solid. (Measurements are in centimetres).
3 5
3 4 5
3
3
5
Below is a part of the sketch of the solid whose net is Sh0WIl above. Complete the sketch of the
solid, sho\m'ng the hidden edges with broken lines. (3 marks)
The interior angles of an octagon are Zx, %x, (x + 40)°, 110°, 135°, 160°, (Zx + 10)” and 185°.
Find the value of x. (2 marks)
Astraight line passes through points (2, 1) and (6, 3).
Find:
(a) the equation of the line in the form y = mx + c; (2 marks)
(b) the gradient of a line perpendicular to the line in (a). (1 mark)
A triangle ABC is such that AB = 5 cm, BC = 6cm and AC = 7cm.
(a) Calculate the size of angle ACB, correct to 2 decimal places. (2 marks)
(b) A perpendicular drawn from A meets BC at N. Calculate the length AN correct to one
decimal place. (2 marks)
A cylindrical pipe 2% metres long has an intemal diameter of 2l millimetres and an extemal
diameter of 35 millimetres. The density of the material that makes the pipe is 1.25 g/cm‘.
Calculate the mass of the pipe in kilograms. (Take 1r = 3%). (4 marks)
The ﬁgure below represents a pentagonal prism of length 12cm. The crosssection is a regular
pentagon of side 5 cm.
_ _ _ _     ' ' ' ' > ' ’ _
’__._‘\
_  »  ' ' ' ' ' ' _
_ — \
_ _  — 
” \
.
\
‘\
.
\
\
___‘\ _ _ _ _ _ _ _ _ _ _ _
__
12cm
5cm
Calculate the surface area of the prism correct to 4 signiﬁcant ﬁgures. (4 marks)
Given the inequalities x — 5 5 3x — 8 < 2x — 3.
(a) Solve the inequalities; (2 marks)
(b) represent the solution on a number line. (1 mark)
SECTION ll (50 marks)
Answer only ﬁve questions in this section in the spaces provided.
A farmer had 540 bags of maize each having a mass of l I2 kg. Aﬁer drying the maize, the mass
decreased in the ratio 15: l 6.
(a) Calculate the total mass lost aﬁer the maize was dried. (3 marks)
(b) A trader bought and repacked the dried maize in 90 kg bags. He transported the maize in
a lorry which could carry a maximum of 120 bags per trip.
(i) Determine the number of trips the lorry made. (3 marks)
(ii) The buying price of a 90 kg bag of maize was Ksh 1 500. The trader paid
Ksh 2 500 per trip to transport the maize to the market. He sold the maize and
made a proﬁt of 26%. Calculate the selling price of each bag of the maize.
(4 marks)
(a) Solve the equation, L2’? = (4 marks)
X _
(b) The length of a ﬂoor of a rectangular hall is 9m more than its width. The area of the
ﬂoor is 136 ml.
(i) Calculate the perimeter of the ﬂoor. (4 marks)
(ii) A rectangular carpet is placed on the ﬂoor of the hall leaving an area of 64 mi.
If the length of the carpet is twice its width, determine the width of the carpet.
(2 marks)
A trader bought 2 cows and 9 goats for a total of Ksh 98 200. If she had bought 3 cows and
4 goats she would have spent Ksh 2 200 less.
(a) F om1 two equations to represent the above information. (2 marks)
(b) Use matrix method to determine the cost of a cow and that of a goat. (4 marks)
(c) The trader later sold the animals she had bought making a proﬁt of 30% per cow and
40% per goat.
(i) Calculate the total amount of money she received. (2 marks)
(ii) Determine, correct to 4 signiﬁcant ﬁgures, the percentage proﬁt the trader made
from the sale of the animals. (2 marks)
Two towns, A and B are 80 km apart. Juma started cycling from town A to town B at 10.00 am
at an average speed of 40 km/h. Mutuku started hisjoumey from town B to toum A at 10.30 am
and travelled by car at an average speed of 60 km/h.
(a) Calculate:
(i) the distance from town A when Juma and Mutuku met; (5 marks)
(ii) the time of the day when the two met. (2 marks)
(b) Kamau started cycling from town A to town B at l0.2lam. He met Mutuku at the same
time as Juma did. Determine Kamau's average speed. (3 marks)
s=t’—5t +3t+l0.
(a) Find s whent = 2.
(b) Determine:
(i) the velocity of the particle when t = 5 seconds;
(ii) the value of t when the particle is momentarily at rest.
(c) Find the time, when the velocity of the particle is maximum.
21 The displacement, s metres, of a moving particle from a point O, aﬁer t seconds is given by,
(2 marks)
(3 marks)
(3 marks)
(2 marks)
ln the ﬁgure below, OABC is a trapezium. AB is parallel to OC and OC = SAB. D is a point on
OC such that OD: DC = 3:2.
A q B
P
O > C
D
(a) Given that OA = p and AB = q, express in tem1s of p and q:
(i) OB;
(ii) AD;
(m) cs.
k and r are scalars. Determine the values of k and r.
(l mark)
(2 marks)
(2 marks)
(b) Lines OB and AD intersect at point X such that AX = kAD and OX = rOB, where
(5 marks)
(a) On the grid provided, draw the square whose vertices are A (6, 2), B (7, 2),
C(7,—l)andD(6,—l). (I mark)
t . _ _ _._.. +  _.,._._. . ~_»,_
k_  <_ ._ g _ v ~_<_~.. . .....
1 T if ‘ ; ‘ . .*_
tit T1 'Y‘F *f‘£'l ~ A iiT 1
T?" my vi; 51+ g.
4:_:.4T7:;fx :;;;1i1l *4"; ;;;? ;j.,i_ ._.'; 17.4;
5" .1“ 111:? y '~ A i 53,: f l‘.1*“‘...J'.":if‘f . . .
i T ‘_ 1 ‘Y ‘ ~ ' ‘ ‘ <\‘* \
V : 1i _ 4 "W ‘:1; 17;; ::.; M.
1+. LL}. 734‘ #7 _ ..v; ‘___.. ._
_t,,. t._.... . 1. >—>o4‘—~Iv _ ._, ..:.,t,T.t, _, Y7 .
*1 :; 31* "*1" ::":.1.1:::.' 111* it; :;1: i~1*t*:
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 ~*‘*** 11; .1. ,1»:
:::: »* 7:4‘Y‘1V\>'7‘%\4\7’ +
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IIIIIIIII III III \ ‘ > II
7 I II In ::II ) HIIIIIIH
I IIIII II II .J 1; _j_;_.
; I E. >1?
(b) On the same grid, draw:
(i) A’B’C'D', the image of ABCD, under an enlargement scale factor 3,
centre (9, 4); (3 marks)
"5: :5"
..
i
f"j * 111,
¢_ >0 t +4»
,:;1j;‘ ‘ 1;.
IA I
‘*1 444 t _,.‘.i
w 9—+‘
t fr" " w i
.‘_L_..7 ‘L ll
Ill 1747
'I""t" "llH' iIHQI“** ""¢'.‘;;__'f'§f.‘
.,. A
_ . _ L . .. >\
IIIII
IIII
IHIE
Hill
1' .:::
mi
It ‘=i!i
1+» i::
ii;
ﬁf} ‘
II
mii
gigs
Iii
(ii) A"B"C”D”, the image of A’B’C'D' under a reﬂection in the line x = 0; (2 marks)
(m) A'"B"’C"'D’”, the image of A"B"C"D" under a rotation of +90° about (0,0).
(2 marks)
(c) Describe a single transformation that maps A’B’C'D' onto A”'B’”C’”D"’. (2 marks)
24 The ﬁgure below represents a cone of height 12 cm and base radius of 9 cm from which a
similar smaller cone is removed, leaving a conical hole of height 4 cm.
l2cm
\
\
\
\
\\
._______;
/
A /
0 /
,
/
/
/
/
_____./,___ ____
\
\
ITI ‘L
___ /1 : "_\\ ~~____ _
— ' ' /,_.———r~——~~~__‘\ " ‘ 
c ~ > \
‘1i9CIl1—ivi
(a) Calculate:
(i)
(ii)
(b) (i)
(ii)
the base radius of the conical hole; (2 marks)
the volume, in terms of 11:, of the smaller cone that was removed. (2 marks)
Determine the slant height of the original cone. (l mark)
Calculate, in terms of 1:, the surface area of the remaining solid aﬂer the smaller
cone is removed. (5 marks)
4.1.2 Mathematics Alt. A Paper 2 (121/2)
SECTION l (S0 marks)
Answer all the questions in this section in the spaces provided.
1 The sum of n terms of the sequence; 3, 9, 15, 21, is 7500. Determine the value of n. (3 marks)
2 A quadratic curve passes through the points (—2, 0) and (l, 0). Find the equation of the curve in
the form y = ax’ + bx + c, where a, b and c are constants. (2 marks)
3 Make d the subject of the fomula,
d1
P = %mn1 — qT (3 marks)
4 Solve the equation
2 logx — log (x — 2) = 2 log 3. (3 marks)
5 (a) Using a pair of compasses and ruler only, construct an escribed circle to touch side XZ of
triangle XYZ drawn below. (3 marks)
Z
Y
X
(b) Measure the radius of the circle. (l mark)
6 The equation of a circle is given by xi + 4x + yl —— 2y — 4 = 0. Determine the centre and
radius of the circle. (3 marks)
7 (a) Expand (l — x)’. (l mark)
(b) Use the expansion in (a) up to the term in x3 to approximate the value of (0.98)5.
(2 marks)
8 The position vectors of points F, G and H are f, g and h respectively. Point H divides FG in the
ratio 4:— 1. Express h in terms off and g. (2 marks)
10
ll
12
13
14
15
6
0/W
Two machines, M and N produce 60% and 40% respectively of the total number of items
manufactured in a factory. lt is observed that 5% of the items produced by machine M are
defective while 3% of the items produced by machine N are defective.
lf an item is selected at random from the factory, ﬁnd the probability that it is defective.
(3 marks)
Two taps A and B can each ﬁll an empty tank in 3 hours and 2 hours respectively. A drainage
tap R can empty the full tank in 6 hours. Taps A and R are opened for 5 hours then closed.
(a) Determine the fraction of the tank that is still empty. (2 marks)
(b) Find how long it would take to ﬁll the remaining fraction of the tank if all the three taps
are opened. (2 marks)
4
Simplify the expression &, leaving the answer in the fonn a \/b + c where a, b and c
are integers. \/5 + \/3 (3 marks)
A point P moves inside a sector of a circle, centre O, and chord AB such that 2cm < OP 5 3 cm
and angle APB = 65°. Draw the locus of P. (4 marks)
The table below shows income tax rates in a cenain year.
Monthly income in Kenya shillings 1 Tax rate in each shilling
Up to 9 680 10%
from968l to 18800 15%
from 18 801 to 27 920 20%
from 27 921 to 37 O40 25%
over 37 040 30%
In that year, a monthly personal tax relief of Ksh 1 056 was allowed. Calculate the monthly
income tax paid by an employee who eamed a monthly salary of Ksh 32500. (4 marks)
Solve the equation 6 cos’ x + 7 sin x — 8 = 0 for 0° S x S 90°. (4 marks)
The positions of two towns are (2° S, 30° E) and (2° S, 37.4° E). Calculate, to the nearest km, the
shortest distance between the two towns. (Take the radius of the earth to be 6 370 km) (2 marks)
The vertices of a triangle T are A(l , 2), B(4, 2) and C(3, 4). The vertices of triangle T’, the
image ofT are A'(%, 1), B'(2, l) and C’(%, 2).
Determine the transformation matrix M = (3 that maps T onto T’. (3 marks)
SECTION ll (50 marks)
Answer only ﬁve questions from this section in the spaces provided.
The Hire Purchase (H.P.) price of a public address system was Ksh 276000. A deposit of
Ksh 60000 was paid followed by I8 equal monthly instalments. The cash price of the public
address system was 10% less than the H.P. price.
(=1)
(b)
(C)
Calculate:
(i) the monthly instalment; (2 marks)
(ii) the cash price. (2 marks)
A customer decided to buy the system in cash and was allowed a 5% discount on the
cash price. He took a bank loan to buy the system in cash. The bank charged compound
interest on the loan at the rate of 20% p.a. The loan was repaid in 2 years. Calculate the
amount repaid to the bank by the end of the second year. (3 marks)
Express as a percentage of the Hire Purchase price, the difference between the amount
repaid to the bank and the Hire Purchase price. (3 marks)
In the ﬁgure below, PR is a diameter of the circle centre O. Points P, Q, R and S are on the
circumference of the circle. Angle PRQ = 72°, QS = QP and line USV is a tangent to the circle
at S
.
Giving reasons, calculate the size of:
(8)
(b)
(C)
(<1)
(6)
L QPR; (2 marks)
L PQS;
L OQS;
L RTS;
L RSV.
(2 marks)
(2 marks)
(2 marks)
(2 marks)
(a) Complete the table below for y = x‘ + 4x3 — 5x — 5. (2 marks)
X 1sl—4—32l—10[2{
=x‘+4.\"5x—5l  19 l —5[ l
(b) On the grid provided, draw the graph ol'y = x‘ + 4x3 — 5.r — 5 for — 5 5x 5 2.
(3 marks)
(c) (i) Use the graph to solve the equation x’ + 4.\1 — 5x — 5 = 0. (2 marks)
(ii) By drawing a suitable straight line on the graph, solve the equation
x"‘ + 4r‘ — Sx — 5 =  4x — l. (3 marks)
The ﬁgure ABCDEF below represents a roofofa house. AB = DC = l2m, BC = AD = 6m,
AE
(8)
(b)
=BF=CF=DE=5mandEF=8m.
E
. 8 m
F
Sm
A C
l2m 6m
B
Calculate, correct to 2 decimal places, the perpendicular distance of EF from the plane
ABCD. (3 marks)
Calculate the angle between:
(i) the planes ADE and ABCD; (2 marks)
(ii) the line AE and the plane ABCD, correct to 1 decimal place; (2 marks)
(m) the planes ABF E and DCFE, correct to l decimal place. (3 marks)
(a) Complete the table below, giving the values correct to l decimal place. (2 marks)
X“ 0  40 so  120 160 I 200 240
2 sin (X + 20)“ 0.7 2.0 0.0  2.0
\/§ cosx  1.7 1.3  — 0.9 1.6
(b) On the grid provided, using the same scale and axes, draw the graphs of
y = 2 sin (x + 20)° andy = \/5 cos x for 0° Sx S 240°. (5 marks)
(c) Use the graphs drawn in (b) above to determine:
(i) the values of x for which 2 sin (x + 20) = x/5 cos x; (2 marks)
(ii) the difference in the amplitudes of y = 2 sin (x + 20) and y = \/3 cos x.
(l mark)
Three quantities R, S and T are such that R varies directly as S and inversely as the square of T.
(a) Given that R = 480 when S = 150 and T = 5, write an equation connecting R, S and T.
(4 marks)
(b) (i) Find the value of R when S = 360 and T = l,5. (2 marks)
(ii) Find the percentage change in R if S increases by 5% and T decreases by 20%.
(4 marks)
The equation of a curve is given by y = 5x — éxz.
(a) On the grid provided, draw the curve of y = 5x — fr’ for 0 5 x 5 6. (3 marks)
(b) By integration, ﬁnd the area bounded by the curve, the linex = 6 and the xaxis.
(3 marks)
(c) (i) On the same grid as in (a), draw the line y = 2x. (l mark)
(ii) Determine the area bounded by the curve and the line y = Zx. (3 marks)
105
24 The table below shows marks scored by 42 students in a test
35
69 57 S8
40
56
36 62 49
41
68 73 65
64
64 54 74
73
4] 61 S6
I
86 47 ' s1
I
(a) Starting with the mark of 25 and using equal class intervals of 10, make a frequency
distribution table. (2 marks)
(b) On the grid provided, draw the ogive for the data. (4 marks)
(c) Using the graph in (b) above, estimate:
(i) the median mark; (2 marks)
(ii) the upper quartile mark. (2 marks)
5.0
5.1
5.1.1
THE YEAR 2013 KCSE EXAMINATION MARKING SCHEMES
MATHEMATICS ALT. A (121)
Mathematicsa Iternative a Paper l (121/1
l.
36 _ 103
12 27
=34 Ml
=_7 Al
2
2.
(a) Mode
= 22 Bl
(b) Median
l5,15.16.19.19.20.20.21.22.22.22.26,27.28
20+21
medi:m=?— MI
= 20.5 Al
3
3.
N0. LO
1.794 0.2538 Ml
0.038 2.5798
2.8336 Ml
1.243 0.0945
7.7391+3 M‘ *3 ‘/
0.3799 15797 L
4
all log \/
+ and  operations V’
4.
(4m + 3n)(4m — 3n)
(4m + 3n)(m — n)
= 4m~3n
m—n
M1 faclorizing numerator v’
Ml faclorizing denominator ¢
AI
3
5.
Retailer
l30% — 1560
__. 1560 X I00 M1
l00% ——il30
= 1200
Wholesaler
120% —~ I200 00 M
1200 X 1 l
lO0% no
= 1000 Al
3
207
D
sl Bl
A P
Bl
Bl
3
construction of equal pans on AC
draw DP//CB such that AP= % AB
locating point P
From 0700 h Monday to I900 h Wednesday
=24><2+l2h
=60h
Time lost = 60 X 15 = 900 sec
= I5 min
Time shown on clock:
l900h 15 min = 184
Sh
Ml
M1
Al
3
x+20=23O°orx+20=3lO'
x = 210°
or
x = 290°
Bl
Bl
Bl
3
for 230° or 310°
(a)
2357_
941
1416
(b) l416 = 2’X3X59
Bl
Bl
Bl
3
for 2357 and 94l J
for 1416
E Bl
C
A B 3
Bl
Bl
lines AF, ED equal and parallel to
BC
lines AB, FC equal and parrallel or
lines AE and FD equal and parallel
or lines CD, EB equal and parallel.
completing the solid showing dotte
lines.
d
ZX+%.\'+.\'+4O+l0+l35+l60+2Jt+l0+185
'h= 440=x=440><2—=s0° Al
2 11 T
= I080 Ml
12 (a) Gradient of line: % =%
line equation
L3 = L
x — 6 4
_ = L _
y 3 4 (x
y = :11x + 1%
(b) Gradient of perpendicular line
—Lm' = —l
4
m’ =4
6)
Ml
Al
L.
3
13 (a) 5’=7’+6’—2><6><7cosC Ml
(b) h
cos C
C = 44.42’
= 7 sin 44.42
= 4.9 cm
= 49+3s25
24
Al
M1
i
4
Volume of pipe material
Q(1.7s=1.0s=)><2s0¢m Ml
7 Ml
I540 Cm“
mass of pipe
1540 X 1.25
1000
1.925 kg
Ml
Al
4
209
h = 2.5 tan 54° = 3.44] cm Bl
Area of pentagonal faces
=2(%><5><3.441><5) M1
= 86.025
Total area Ml
= 86.025 + 5( l2 X 5)
Al
= 386.0 4
(a) x—5S3x—8
—?_xS—3
x?.l.5 Bl
3x—8<2x—3
x<5 Bl
l.5Sx<5
(b)
0:1**5“ Bl
7. (a) Mass after decrease
112 X % Ml or equivalent
= 105 kg
Total decrease
(112105)>< 6
= l5000
Total 945000 + l5000 Ml
Selling price per bag:
960000 X L26 M1
:0
=1920 A1
l0
21 l
8. (a)
(b)
(i)
(x+3)(x—2)=24
x’+x—3O=O
(x+6)(x—5)=0 Ml
x=—6 0rx=5 Al
Ml
Ml
(x+9)x = 136
x1+9x— l36=0
(x+ l7)(x—9)= 0
x =17 or x= 8
'."x=8 A1
penmeter
=2(8+l7)=50m B1
Ml
Ml
(ii)
2xXx= l36—64
2.1:= 72
x*= 36
x= 6m Al
T
Ml
2c + 9,; = 98200 Bl
(8) Bl
3c + 4g = 96000
(b) Det.of(§ Z)=—l9
M’ =_Tl9'(43 .29) Bl
w—L(§ "§)(§ i)(Z>'=1%(_§ 'Z)(ZZ§3Z) M‘
001” _‘i9>(§)=~%<24800X1.3+9X5400Xl.4 Ml
= l32S20 Al
(ii) % proﬁt
_ 132520 — 98200
 98200 X lO0% Ml
= 34.95% Al
l0
(a) (i) Time taken by Juma = ﬁh
Time taken by Mutuku = T
Let x km be distance from A
2(5x— 160): I20
10x=440
x= 44km
(ii) Time they met
10.00 am + %h
= 10.00 +l h 6min
=ll.O6am
(b) Speed if Kamau delayed by 2 minutes
Kamau‘s time = — %)h
speed needed:
= 58% km/h Al
10
_Q:1=l
40 so 2
3x—%(80—x) __i
120 '2
Bl
80x Bl
Ml
Ml
Al
Ml
Al
M1
4 0
_l
_4h
Ml
21 (a) Displacement.s,whent=2
2"—S><2‘+3><2+lO Ml
=4 Al
(b) (i) velocity when I = 5 seconds
V=~‘1i=3r‘—l0r+3 BI
dr
whenl=5,V=3><5’—l0><5+3 M1
=23 Al
(ii) 3:’  101+ 3 = 0 M1
(3r—1)(r—3)=0 M1
= i =
1 3, r 3 A1
(c) time when velocity of particle is at its maximum
acceleration = = 6; __ 1Q = Q Ml
,=%=1%S Al
10
(a) (i) QB=I3+q
(b)
also
E(r—l)+rz1=—kE+3kQ
—k=r—l and r=3k
—k=3k—l
4/( =—1=/< =2
subxrirure r = 3 X—
(ii) Ap=—g+%><5q
(m) QB=—5Ql +g+q
A_X=/<(A.D)
=/<<2+32>
=—kE+ 3kg
A_X=—@+'(QB)
=—n+r(11+q)
=E(r——l)+
Bl
Ml
=—Lv+3Q Al
Ml
_ “Hp Al
Bl
Bl
'21
Ml
Ml
I
._
zi Al
¥>~
A
.
O
or equivalent
' /
i I
6/1 ll \ _
\\
I l B
.r>” A"'l"+ , .
“°l
<»¢,¢,~ val l1s“4$"e‘?"’ 9*
‘Er
ii 4 s
C/H 811/ 3
. 9 "
2 ’ 4
\/
(a) ABCD J drawn Bl
(b) (i) Centre identiﬁed and used J Bl
Bl AA‘, BB’, CC‘ and DD‘ drawn J
Bl completion of square A'B‘C'D'
and labelled
(ii) A"B"C"D" B2 A"B"C"D" drawn J
(m) A...B."c...D,.. B2 A...B...C...D...drawn
(c) Reﬂection on line y = —x Bl reﬂection
Bl line y =—x
l0
v
H) (i)
L=A_
9 u
9x4
r=T=3c‘m Al
(ii) volume of material drilled out
=%nx?x4 Ml
=l27I
(b) Slant height of cone
=ﬂUTF=wm» m
(c) Surface area of solid after conical has been drilled
rr><9><15+1r><(9’3’)+rr><3x5 Ml
Ml
Al
for7ZX9><3><5
Ml summing up
Al
TF'
5.1.2 Mathematics Alternative A Paper 2 (121/2)
1. 1“term,a
_l
7500 2
3n’ =75O0
n = \/2500 =50
= 3; common difference,d=6 B1
{2><3+(n—1)><6} Ml
Al
?
2. y=(x+2)(x1)
y=x’+x2
Ml
L
2
3. qdz
_i Li
PZmn n
qd’_1
i___ 1_p
n 2mn
dz = %mn"‘ — nP
q
dz /%mn’— nP
q
Ml
Ml
AI
T
4.
Log = log 3’
(0%)
9
_I’_=
x—2
X2—9X+18=0
(x6)(x3)=0
x=6orx=3
Ml
Al
Ml ‘
?
5 (=1) I
W
\
\
\
\
\
B1 extending YX and YZ
X
\ Z Bl bisecting 4; vxz and
XZW
Bl escribed circle drawn
/ Y
Y
z’
}?
z
V Bl
(b) radius = 3.1
4
allow 1 0.]
Completing square on L.H.S.
xZ+4x+4+y22y+l=4+4+1 B1
(x+2)2+(y—l)2=9 Bl
centre of circle : (2, 1)
B1
radius of circle: 3 units
3
(a) (1  X)5 = l + 5(x) + 10(X)’ + l0(x)’ + 5(x)‘ + (x)5
=l5x+l0x210x3+5x“x’ B1
(b) (0.98)’ = (1  0.02)‘ = x = 0.02
(O.98)5 = 1 — 5(0.02) + lO(O.O2)2 » lO(O.O2)3 Ml
=1 0.1+ 0.004  0.00008
= 0.90392 Al
?
220
By: _“@2
0:'23 150 115
_ 1 4 Ml
I3‘ 4+(1)f+ 4+(1)?
__1 A
'3f+3$ A1
?
P(dcfective) : M  0.6 >< 0.05 = 0.03
N  0.4 x 0.03 =0.012
P(defective) 0.03 + 0.02 = 0.042 Ml
?
0.03
0‘ 0;
A1 04 Q91
4!
.g3
Ml For 0.6 X 0.05 or 0.4 X
good
defective
good
defective
(a) Fraction ﬁlled if A and R are open for 5h
i_i=i
5X<3 6) 6
Fraction of tank still empty=1—%=% B1
Bl
(b) Fraction ﬁlled ifA,BandRare0pen forlh
L L_i=l
3+2 6 3
 1.2 I 3
T1metakentoﬁllthetank=€¢€=gx3 M1
= %h or 15 min Al
4
»/E = 4~/§(~/31/5) M1
/§+/5 (1/5+./'3')(\/§—~/5)
=4/§(/5E) M1
53
=2./§(~/3/5)
:2‘/E_6 A1
T
221
N
A A
AAOB = 130°
arc AB  solid curve
arc A'B'  broken curve
region shown
Locus of P
I
1
C
130' q
Bl
B1
B1
B1
T
9680 X 0.1 = 968
9120 X 0.15; 9120 X 0
=1368 =1824
Net tax
= (968 + 1368 + 1824
= 4249
Ml
.2; 4580 X 0.25 Ml
= 1145
+1145)1056 Ml
A1
4
6(1sin2x)+7sinx8=0 Ml
66sin2x+7sinx8
6sin2x7sinx+2=0
(3sinx2)(2sinx1)
sinx=% 0rsinx=%
x=41.81° orx=30°
=0
=0 M1
Ml
Al
4
Distance between town
= 21: X 6370 cos 2° ><
= 822.212128l
= 822 km
s K and S
L M1
360
Al
<“ "><1 4 3>~<‘
c d 2 2 4 _
a+2b=§
4a+2b=2
3a =%=>a=%
~_)
%+2b= =b=0
c+2d=l
4c+2d=1
3c =0=>c=0
0+2d =1==d=%
M=(’5 Z)
N
[Q
gum.
T
Ml
Ml
L
3
J f0l'lTl8llOl'l and solution
of simultaneous equations
J formation and solution
of simultaneous equations
(a) 0) 2760001; 600
=l2000
(ii) 276000 >< 0.9
= 248400
(b) 248400 X 0.95
= 235980
235980 X 1.22
= 3398ll.2
(c) 3398112  276000
63811.2
276000 X100
=23.12%
00 Ml
A1
M1
A1
Ml
M1
Al
Ml
Ml
Al
T
223
0'/'23 450 425
(b) 4PQS = I80”  2(72) = 36° Bl
(c) LOQS =36°  l8°= 18° Bl
(d) ARTS = 180 (36+ 18) = l26° B1
(e) ARSV = 90°  36° = 54° Bl
a) AQPR=90°72°=l8° B1
LPQR = 90°  angle subtended by diameter Bl
APSQ = 72°  angle subtended at the circumference by
chord PQ equal and base is of isosceles AQPS = 72° B1
base angles of isosceles A OPQ = 18° Bl
extension angle RTS equal to sum of opposite interior B1
angles TSP and TPS
4RSV = 4RPS  angle in altemate segment. Bl
10
or equivalent
19. (a)
x 54321012
y 5 15 13 3 5 9
=x»‘+4x15x5
(b)
(c) (i) X = 4.8, 0.7, 1.5 B2
B2
A
\ ze
I Sl
/ to. Pl
/ >\ c1
I5! 4» 3 2 ii?’5k 2 3)
* ‘k
10* .
J
‘J= 41:
(ii) y=4x1 Pl
Solutions Ll
x=4,l, l. Bl
10
allow Bl for 4 correct
Suitable scale
All correctly plotted
10.1 allow Bl for 2
values \/ plotting for line
224
(a) = distance of EF from place ABCD
slant height from F to BC
= J5’ 3’ M1
E 9§w\ F
=4 ‘ ‘\“\\5/t1
I
= distance of EF from planeABCD ) It i _ '
1,_ _ —  * ‘
=J4’ 2’ M1 ’
‘L _ _
l /
1/ ' ,,
mm 15
fl
=/ﬁ=3.46m Al
(b) (i) angle between planes
ADE and ABCD
_ tan_, /§ Ml or equivalent
‘ 2
=60° Al
(ii) angle between line AE
and plane ABCD
= sin‘ ‘Z? Ml or equivalent
= 43 .9° A 1
(m) angle between planes
ABFE and DCFE
= 2¢tan' ' im M1 tan' ' ;,— 0r equivalent
/E ¢ 12
= s1.s° M1 doubling
Al
10
21. (a)
x 0 40
80 120 160 ZOO 240
y= 1.7 1.3 1.3
2sinx+20 B1
y = 0.3 1.6 0.9
/gcos x
Bl
(b)
“73€é§§f5?‘ L s5sy¢y?1
1° +'k f. .7 1 ;__ A A ., . ,__._ ._.__,
.» ff‘: ' 1;: '. ,..:;,l*"ﬁ
1
.
1.5
~— ~ s »
.=.,,,¢ .. .,.. .. ..i,~
IIi_Z'IlVl§._‘ “If I_, __ . .f‘ I“ _ .1 II: II _' ..
1.0
’.T.:i’:":_‘.f¢_‘:’:;: :1." :.; .1.:":t _:;' _:_I1
0.5 ,
51 11;
§_§‘::::‘..;:;;,=;__;:_;: .1: I .,:‘:;"_iQ ' :: 1,1:
0.0 ?
0.5 V
‘vb
_,,__
o
511201130? 1_ 1.01»
\...,_
1.0
¢;;:1?:;:j5§ 59% my ;. Q YE
1; ;, ";.:_ gs, J
V1
k .
1.5
_._. . .1 : .,.,, _..,._. . %qk.,,.,. . ,. . w __ _ _ __
.4.:‘>__,.l . ,..,._._ 4 . ., _. _ ,
~ l 4   é— ‘5%°
=5 k
(b) (i)
= s0><360 
R <15)’
=l2800
(ii)
R1
R1
(i)><100%= >< 100%
R 80;,
= 80S
R T
S kS Bl
G? = R=?
Ml
Z 150/<
25
480 >< 25
= i= A1
150 8°
l Bl
M1
_ 80 >< 360
_ 2.25 A1
s2 = l.05s,T2 = 0.8T B1
= 80 ><1.05s
(0.81?
Z s0>< 1.05 X i
(0.s)’ 7“
= L
131.25 T2
M1
i_Q§
13125 Ml
_ X 100
%, so
= 64.0625
Al
= 64.06 %
23. (a)
x O I 2 3 4 5 6
y 0 45 8 105 12 125 12 _
=5X%X1 B1 table may be lmplled
(b) 6
[0
A
I9

.1
5' c1
\i14:¢Y
__ 5X6’_i 3 _ _
_[T2 6x6] [0 0]
= [90—36][0]=54
(c) (i) Drawing liney=2x
(ii) AreaofA: %><6>< 12
Bounded area = 54  36 = l8
Sx
/t< +~>¢~
:;1_ 1 .=
[zx 2><3" [="36" m1 ml al ll a1 l i0 \ j integral substitution marks frequency 2534 3544 4554 8 5564 l2 6574 9 7584 3 8594 .l'piiif !!"' (a .5 ) cfs v> 
,_ _ __
*1
%
i V l._. 1,... __ Z __ _ _ _
—+~4 ~»ﬂi A 4 »__~ _.k_A~.i .%.__€__ ___..____
""*—:—_~:*:'f_—">%' ff_~—~f~* '> <'r—— —~ *—< — —i—v————~~/ii
J § _ , _
'T‘T‘_’§.i. ‘M * ‘ _ _ ‘ _ _ ,,_ A
_:‘.f::tTttL;“t;'t1; ..'. _ ._ . 1 , 3 _ '1 _~;;: ‘H: :,
,...1_._._... Ha. _ a _
» ~ ~:'~l—~_;~ ~
£4"?
(c) (1) Identiﬁcation of median
= 575 1 O 5
(ll) Identiﬁcation of upper quartile mark
= 665 1: 0 5
Bl \/ marks class column
Bl J frequency column
___ ___ _._ __ __ _ ‘___ ‘_ _
_ ._ ._¢ _, 4.
‘ >_._t_.a.,__a 
czrgs t" *,:::E1: 2;; 3:::l:*—;“