Download (i) k = 2

2003 E Maths Paper 1
16
4
SR. A is a point on PS such
7
that PS = 3AS. B is a point on PC such that AB is parallel to SC.
PQRS is a parallelogram. C is a point on SR such that SC =
P
Q
A
B
S
C
R
Given that the area of PSC = 36 cm2, calculate the area of
(a) PAB,
(b) QCR,
(c) parallelogram PQRS.
18
19
The intensity of illumination, I lumens/m2, at a point on a screen varies inversely as the
square of the distance, d m, of the light source from the point.
(i)
What happens to I when d is doubled?
(ii)
Given that d = 2.5 m when I = 0.8 lumens/m2, form an equation connecting I and d.
(iii) Find the value of d when I = 0.2 lumens/m².
Tom and Harry are playing a game with cards labelled 1, 2, 3, 4. They each take one card
without looking.
(a) List all the possible outcomes. The first three have been done for you.
Tom
1
2
1
Harry
2
1
3
(b) What is the probability that
(i) neither card is odd?
(ii) the number on Harry’s card is greater than the number on Tom’s card?
20
 4
In the diagram, OZYX is a parallelogram with position vectors of Z and X given as   and
 3
 3
  respectively. Find
 4
(a)
OX ,
(b)
(c)
the coordinates of Y.
P is a point on ZY produced such that ZP  3ZY .
(i) Find the coordinates of P.
area of OZP
(ii) Find the ratio
, given that OKP is a straight line.
area of parallelogram OZYX
y
P
Y
K
X
Z
O
Solution:
area of PAB
2
4
 ( )2 
16. a)
area of PSC
3
9
area of QCR 3
b)


area of PSC 4
x
4
Area of PAB   36  16cm 2
9
3
Area of QCR   36  27cm 2
4

c)area of PQRS  (36  27)  2  126cm2
1
18(i) I is of the orig amt Or I decreases by 75%
4
k
4 5 5
5
(ii) I  2  k     5  I  2
(iii)
d
5 2 2
d
19(a)Tom 1 2 1 3 4 1 2 3 2 4 3 4
Harry2 1 3 1 1 4 3 2 4 2 4 3
d
5
 25  5
0.2
(i)
2 1

12 6
(ii)
6 1

12 2
20a) OX  32  42  5 units
 4  3  7
b) OY  OZ  ZY  OZ  OX         
 3  4  7
Coordinates of Y = (7,7)
ci)
ZP  3ZY
 7   4  13 
OP  OZ  3(OY  OZ )  OP  3OY  2OZ  3    2     
 7   3  15 
Coordinates of P = (13,15)
2003 E Maths Paper 2
1
A straight line crosses the x-axis at point A(8,0) and the y-axis at the point B(0,6).
(i)
(ii)
(iii)
(iv)
(v)
Find the gradient of the line AB.
[1]
Find the equation of the line AB.
[1]
Calculate the area of AOB .
[1]
Calculate the perpendicular distance from the origin, O, to the line AB.
[2]
Find the equation of the line which is the reflection of the line AB in the y-axis.
[2]
(vi) Given that (4, h) lies in the line AB produced, what is the value of h?
[1]
___________________________________________________________________________
2
(a)
Factorise completely, b(b  1)  c(c  1) .
(b)
Solve the simultaneous equations
5 x  2 y  29,
x  4 y  3.
(c)
Solve the equation
3
5
1


 4.
2  x 4  2x x  2
[2]
[3]
[3]
___________________________________________________________________________


ABCD is a rectangle in which AD = 48 m, DE = 100 m, ECD = 90o and EDC = 32o.
3
E
100 m
o
32
D
C
48 m
A
B
Calculate, giving your answers to 3 significant figures,
(a)
AB,
[1]
(b)
EC,
[1]
(c)
AE,
[2]

(d)
6
EAC .
[4]
In the diagram, OR represents a vertical cliff. From a boat, which is at a point P due south of
R, the angle of elevation of O is 35o.
O
o
North
74
R
60 m
o
35
Q
P
45 m
S
(a) The boat now sails 60 m due north to a point Q from which the angle of
elevation is 74o. Calculate, in metres the height of the cliff.
(b)
[4]
From P, a swimmer swims to a point S, 45 m due west of Q. Calculate
(i)
the bearing of P from S,
[2]
(ii)
the shortest distance between the swimmer and Q.
[2]
If the swimmer continues in the direction of PS, reached a point X, 90 m from P. Find
the distance QX and the bearing of X from Q.
[4]
8
Diagram I shows 2 circular arcs BCD and BPD. Diagram II shows the major arc BCD has
centre O and radius 5 cm and the minor arc BPD has centre A and radius 3 cm. A is a point
on the circumference BDC.
Taking  to be 3.142.
A
B
D
3 cm
D
5 cm
P
O
B
P
O
C
C
Diagram I
Diagram II

(a)
Show that BOD = 69.8o.
[2]
(b)
Find the length of the arc BPD.
[3]
(c)
In Diagram II, find
(i)
(d)
Solution:
the area of sector BOD,
(ii) the area of minor segment BAD.
[3]
In Diagram I, find the area BPDC enclosed by the 2 circular arcs BCD and BPD.[3]
60
3

0  (8) 4
3
1
x6
(iii)  8  6  24unit 2
4
2
1
3
h0 3
3
 or h  (4)  6  9
(iv)  h 10  24  h  4.8 (v) y   x  6 (vi)
2
4
48 4
4
2a)
b(b  1)  c(c  1)  b2  b  c 2  c  b2  c 2  b  c  (b  c)(b  c)  b  c  (b  c)(b  c  1)
5 x  2 y  29
10 x  4 y  58
2b) 

 11x  55  x = 5,y = -2
 x  4 y  3
 x  4 y  3
1(i) mAB 
(ii) c  6 , y 
3
5
1


4
2  x 4  2x x  2
3
3
5
1



 4  3(2)  5  2  4(2)(2  x)  16  8x  13  x 
8
2  x 2(2  x) 2  x
3a)
DC  100cos32  84.805 84.8m  AB  84.8m(3sf )
b) EC  100sin 32  52.992 53.0m (3sf)
2c)
c) AE  AB 2  BE 2  84.8052  (48  52.992) 2  131.876 132m
100.992
48
 EAB  49.979 , tan CAB 
d) tan EAB 
 CAB  29.510
84.805
84.805
EAC  49.979  29.510  20.5
6a)
POQ  74  35  39
OQ
60
60sin 35

 OQ 
 54.6853
sin 35 sin 39
sin 39
OR
sin 74 
 OR  52.5669 52.6m
54.6853
60
 PSQ  53.130 
bi) tan PSQ 
Bearing of P from S = 143.1
45
bii)
shortest distance
4
sin 53.130 
 shortest distance  45   36m
45
5
4
QX 2  902  602  2  90  60   QX  55.3m
5
45
3060  602  902
sin PQX
 PQX  102.529
 75  PQX  102.529 or cos PQX 
2  55.3173  60
90
3060
Bearing of X from Q  180  102.529  282.5
8a)
52  52  32
32  52  52  2(5)(5) cos AOD  AOD  cos 1
 34.9152
2(5)(5)
BOD  2(34.9152 )  69.8304
69.8
b)
360  69.8304
145.085
 145.0848  Arc Length BPD 
 2 (3)  7.5976
2
360
69.8304
   52 =15.2366=15.2 sq cm
ci) Area of sector BOD=
360
cii)
1
Area of minor segment BAD  15.2366  (5)2 sin 69.8304 3.50cm 2
2
DAO 
7.60cm
d) Area of circle = (5)(5)=25 
Area of minor segment BAD = 3.50315
145.0848
1
  (3) 2   (3) 2  sin145.085  8.819326
Area of minor segment BPD 
360
2
Required area = 25 -3.50315-8.819326 = 66.2 sq cm
2003 E Maths Common Test
2
(a)
(b)
5
(a)
A ship sails from port A on a bearing of 040 for 20 km to a port B. At B, it alters
course and sails on a bearing of 300 for 30 km to a port C. Find
(i)
the distance of port C from port A,
[2]
(ii)
the bearing of port C from port A.
[2]
The ship sails back straight from port C to port A. What is the shortest distance of
the ship from port B during its journey back?
[2]
The diagram below shows the origin O and five points O, A, B, P and Q. Given that
OA = 2a + 3b and OB = 4a  b.
Q
A
B
P
O
(i)
(ii)
(b)
(c)
Express OP in terms of a and b.
Express OQ in terms of a and b.
[2]
 8 
 m
It is given that p =   and q =   .
6
n
 10 
(i)
If 2p + q = 
 , find the values of m and n.
 20 
(ii)
If instead, p = kq where k is a negative constant and |q| = 8 units, find the
values of k, m and n.
[5]
In the diagram OAB is a triangle and C lies on AB produced and MN produced.
The point M is the mid-point of OA and N is a point on OB such that ON:OB = 3:4.
Given that the area of NBC is 48cm2.
O
Calculate
(i)
the area of ANB,
[1]
(ii)
the area of OMN.
[2]
M
N
C
B
A
10
(a)
On a mini-market shelf there are 15 boxes containing white table-tennis balls and n
boxes containing orange table-tennis balls. The probability of picking a box of
5
white table-tennis balls is . Find the value of n.
[2]
7
Calculate the probability of
(b)
(i)
picking two boxes containing white table-tennis balls,
[1]
(ii)
picking two boxes which are different.
[2]
Six cards with numbers 0, 1, 2, 3, 4 and 5 written on them are placed in a box and
are well-mixed. Two cards are to be drawn without looking into the box. The sum
of the two numbers on the cards drawn is to be obtained.
+
0
0
1
4
2
1
2
1
2
3
1
3
4
5
4
5
4
3
5
3
7
8
5
6
7
5
6
8
9
(i) Copy and complete the possibility diagram shown above, giving all the possible sums of the
two numbers on the cards drawn. The shaded boxes indicate that it is not possible to obtain 2
cards with the same number.
[2]
(ii) What is the probability that the sum of the two numbers will be a prime number? [1]
(iii) What is the probability that the sum of the two numbers will be a factor of 48?
[1]
(iv) Which sum is more likely to occur, the sum of 7 or the sum of 8? Explain your answer.[2]
Solution:
2(ai)
AC 2  202  302  2  20  30  cos80  AC  33.0km(6sf )
sin CAB sin 80

 CAB  63.4064 (6sf )
30
33.0397
2(aii)
Bearing of port C from port A
= 360 – (63.4064 – 40) = 336.6o (1dp)
2(b)
5(ai)
sin 63.4064 
OP = 6a – 5b
d
 d  17.9km(3sf )
20
5(aii)
5(bi)
5(bii)
1
OQ = 2a + 6 b
2
m=6
n=8
Since, p = kq  p  k q 10 = k( 8 ) k = 1.25
Since k < 0, k = -1.25
m=-
5(ci)
4
2
x ( -8 ) = 6
5
5
n=-
4
4
x6=-4
5
5
Area of ANB = 2 (48) = 96 cm2
Area of OAN = 3 (96)
5(cii)
Since
areaOMN 1
3  96
  Area of OMN =
= 144 cm2
areaAMN 1
2
10(a)
15
5
  n  15  21  n  6
n  15 7
10(ai)
15 14 1


21 20 2
10(aii)
2
15 6 3


21 20 7
+
10(bi)
0
0
1
2
3
4
5
1
2
3
4
5
3
4
5
6
5
6
7
7
8
1
1
2
2
3
3
3
4
5
4
4
5
6
7
5
5
6
7
8
9
9
10(bii)
10(biii)
10(biv)
16 8

30 15
18 9 3
 
30 15 5
Sum of 7 is more likely to occur as it occurs 4 times out of 30 while the sum of
8 occurs 2 times out of 30.
2002 E Maths Paper 1
21 (a) Select from the five given graphs (A) to (E), one which illustrates each of the following
statements:
(i) The area, y, of a circle is proportional to the square of the radius, x.
(ii)
The density, y, of a material is inversely proportional to the volume, x.
y
y
y
x
(A)
(b)
y
x
(B)
y
x
x
(C)
x
(D)
(E)
The diagram shows the graph of the curve y  x( x  4)  2 and line y  h where h is a
real number. The dashed line is the line of symmetry of the curve. The point A lies on
the y-axis, on the curve and on the line y  h . The point B lies on the curve and the line
y  h . State
y
(i) the value of h,
(ii)
y  x( x  4)  2
the coordinates of B,
(iii) the equation of the line of
symmetry of the curve,
A
(iv) the least value of the curve.
0
Answer:
(a)
(i)
C (ii) E (b)(i) 2 (ii) (4, 2), (iii) x = 2 (iv) –2
y=h
B
x
2002 E Maths Paper 2
2
Given that A is (k, 3) and B is (5, 7), where k is a real number less than 5. The length of
straight line AB is 5 units.
(i) Show that k = 2.
[3]
(ii)
Find the mid-point of AB,
[1]
(iii) Find the equation of the straight line that passes through A and B,
[2]
(iv) Find the point of intersection of the line AB with another straight line with gradient –3
and passes through the point (0, 4).
3
The diagram shows part of a hill
represented by rectangle PQRS
where the top of the hill PQ is
parallel to the ground. The rectangle
MNRS is on level ground. The
angle of elevation of the top of the
hill from the foot is 28 and the
highest point of the hill from the
ground is 65 m.
(a) Jerry walked straight up to the
[2]
Q
P
65 m
N
M
28
S
R
top of the hill along RQ, calculate, to the nearest metre,
(i)
the distance he travelled,
[2]
(ii)
the horizontal distance he had covered.
[2]
(b) T is the top of a vertical flagpole TP (not shown in the diagram). Given that the angle of
elevation of T from S is 33 and the angle of elevation of T from R is 30, find, to one
decimal place,
(i) the height of the vertical flagpole,
[2]
(ii)
6
the distance SR.
[3]
In the diagram P, Q and R represent three islands. P is 25 km from Q on a bearing of 200. R
is 30 km from Q on a bearing of 118.
North
(a) Calculate the distance of R from P.
[3]
118
(b)
Calculate the
Q
bearing of R from P.
[2]
200
(c) A ship leaves P at 12 55 and sails directly to Q at
a steady speed of 20 km/h.
30
(i)Calculate the distance between the ship and island R when it is due west of R.
25
(ii)
Find the time, to the nearest
minutes, at which the ship is
closest to R.
[3]
D
P
[2]
R
9
(b)
A special pack of playing cards contains the two, three, four, five, six and seven of
each of the four suits, namely clubs, diamonds, hearts and spades.
If a single card is turned up from the shuffled pack, find the probability that it is
(i)
a six,
[1]
(ii)
a club.
[1]
Three cards are turned up in succession. Find the probability that they are
(iii) all showing the number six,
[2]
(iv) all of the same suit.
[2]
Solution
2
(i)
(ii)
3
(5  k )2  9  5  k  3 or 5  k  3  k = 2
(3.5, 5)
y 3 4
4
1
  y  x
(iii)
x2 3
3
3
(iv)
equation y = 3x + 4 solve with equation in (iii)
11
19
x=
& y=
13
13
65
65
(a)(i) RQ 
= 138 m (ii) RN 
= 122 m
sin 28
tan 28
(b)(i)
TP  122.24tan33  65 = 14.4 m
(ii)
MR =
their (bi )  65
tan 30
SR = [(137.49)2  (122.24)2]= 62.9 m
6
(a)
RP  252  302  2(25)(30) cos82 = 36.3 km
(b)
 QPR  sin 1
30sin 82
= 31.6 m (ii) 25  30 cos 82 = 20.82
sin 70
25  30 cos82
= 1357
20
110  2
 20 = 24 m/s (iii)
(a)(i) 17 m/s (ii)
 4.8 m/s2
5
4 3 2
6 5 4
(b) (i) 1/6 (ii)1/4 (iii)  
= 1/506 (iv) 4(   ) = 10/253
24 23 22
24 23 22
(c)(i)
9
30sin 82
= 54.97 = (200 –180 + 55) 075.0
their (a)
RD 
2002 E Maths Common Test
8.
An unbiased six-sided Green die is numbered 1, 9, 11, 12, 15 and 18.
An unbiased six-sided Yellow die is numbered 2, 3, 4, 16, 19 and 21.
Tom and Jerry decide to play a game where each one will choose a different die and the
one who throws a smaller number on the die will win.
Tom lets Jerry choose the die he wants. Jerry chooses the yellow die, thinking that "the
Yellow die has more smaller numbers compared with the Green die".
The possibility diagram when the two dice are thrown is shown below. The letter Y
indicates that the number (4) thrown on the Yellow die is smaller than the number (11) on
the Green die.
GREEN
1
9
11
12
15
18
2
3
YELLOW
4
Y
16
19
(a)
(b)
(c)
(d)
21
Copy the possibility diagram and complete it by putting a letter Y in each box
where the number thrown on the Yellow die is smaller than the number thrown on
the Green die.
[2]
Justify whether Jerry has made the correct choice, showing your method clearly.[1]
The two dice are each thrown twice. Calculate the probability that the number
thrown on the Yellow die is smaller than the number thrown on the Green die
(i)
both times, [1]
(ii) just once.
[2]
An unbiased Red die is numbered 5, 6, 8, 10, 14 and 17.
Showing your method clearly, find out whether the Green or Yellow die is more
likely to show a number smaller than the number shown on the Red die. [3]
Answer
1
2
3
4
16
19
21
9
Y
Y
Y
8(a)
11
Y
Y
Y
12
Y
Y
Y
15
Y
Y
Y
18
Y
Y
Y
Y
4
5
and P(green win) = .
9
9
Since the probability that yellow will win is higher
than probability that green will win, Jerry has
made the wrong choice.
16
40
8(c) i)
ii)
81
81
19
14
8(d) P(yellow < red) =
, P(green < red) =
36
36
therefore yellow die is more likely to show a
smaller number than the number on the red die.
8(b) P(yellow win) =
2001 E Maths Paper 2 (Solution not available
yet)
7.
A, B, C and D are points on level ground,
with A due North of B.
Angle BAD = 38o, angle ADB = 56o,
BD = 6 km and CD = 10 km.
Calculate
(a)
(b)
BC,
angle DBC,
[2]
[2]
(c)
the bearing of B from C,
[2]
(d)
the shortest distance from
D to BC,
[2]
North
A
38
T
56
D
TD represent a vertical tower. The angle
of elevation of T from B is 3.5o.
Find
o
o
B
6 km
10 km
[4]
(e)
the height of the tower in metres,
(f)
the maximum angle of elevation
of the top of the tower, T, when
observed along BC.
C




 4    8 
1
10. (a) Given that OA =   , OB =   and OC =   where O is the origin, calculate
k 
1
 3


(i)
| AB | ,
[2]
(ii)
the value of k given that A, B and C are collinear.
[2]
(b)
B
M
b
a
O
A
N


In the triangle OAB, M is the midpoint of AB and N is the midpoint of OA. Given that OA


= a and OB = b, express in terms of a and b,








(i)
AB ,[1] (ii) AM , [1] (iii) OM , [1] (iv) BN . [1]
2
P lies on OM such that OP = OM.
3


(v)
Express BP in terms of a and b.
[2]
(vi)
Express BP in terms of BN . Explain the geometrical significance of this
relationship.
[2]




2001 E Maths Paper 1
13
Which of the following figure below could be the graph of
1
(a) y  3  x , (b) y  2 , (c) y  2 x .
x
23
8 girls and 12 boys, including John himself was invited to take part in a lucky draw. The
organiser will draw three winners from the group.
(i)
What is the probability of getting all winners with the same gender if each winner is
only entitled to one prize?
(ii)
Find the probability of John winning the first prize only.
(iii) What is the probability of John winning all the three prizes?
Answers:
13(a) 6
23
23(i)
95
(b) 2
(ii)
1
20
(c) 5
(iii) 0
or
(ii)
361
8000
(iii)
1
8000