June 2009
Paper 4
1) Tania went to Italy. She changed
£325 into euros. The exchange rate
was £1 = 1.68 euros
a)
Change £325 into euros
325 x 1.68 = 546
1)
When she came home she changed
117 euros into pounds.
The new exchange rate was £1 =
1.5 euros.
b) Change 117 euros into pounds
117 ÷ 1.5 = 78
2)a)
2)b)
Reflection
in the y axis
(1 mark)
(1 mark)
3)
The nth term of a number sequence
is n2 + 1. Write down the first three
terms of the sequence.
n=1
12 + 1 = 2
n=2
22 + 1 = 5
n=3
32 + 1 = 10
4)a)
4)b)
Positive correlation
or the taller the sheep, the longer it is
4)c)
108cm
5)
Julie buys 19 identical calculators.
The total cost is £143.64
Work out the total cost of 31 of
these calculators
£143.64 ÷ 19 = £7.56
£7.56 x 31 = £234.36
6) F = 1.8C + 32
a) Work out the value of F when C = −8
F = 1.8 x -8 + 32
F = -14.4 + 32
F = 17.6
(1 mark)
(1 mark)
6) F = 1.8C + 32
b) Work out the value of C when F = 68
68 = 1.8C + 32
36 = 1.8C
C = 36 ÷ 1.8
C = 20
(1 mark)
(1 mark)
7)
R
310°
60°
The bearing of a boat R from boat P is
060°.
The bearing of boat R from boat Q is
310°.
8)a) There are some sweets in a bag.
18 of the sweets are toffees.
12 of the sweets are mints.
Write down the ratio of the number of
toffees to the number of mints.
Give your ratio in its simplest form.
18 : 12
3:2
8)b) There are some oranges and
apples in a box. The total number of
oranges and apples is 54. The ratio of
the number of oranges to the number of
apples is 1 : 5
Work out the number of apples in the
box.
1+5=6
54 ÷ 6 = 9
5 x 9 = 45
45 apples in the box
9) The equation x3 + 20x = 71 has a
solution between 2 and 3. Use a trial
and improvement method to find this
solution. Give your answer correct to 1
decimal place.
x
x3 + 20x
2.5
65.625
too low (1 mark)
2.6
69.576
too low
(1 mark)
2.7
73.683
too high
2.65 71.609625 too high (1 mark)
Answer = 2.6
(1 mark)
10)
Leave in your construction lines!
11) Tarish says, ‘The sum of two prime
numbers is always an even number’. He
is wrong. Explain why.
Give a counter example
e.g. 2 + 3 = 5
12) Sethina recorded the times, in minutes,
taken to repair 80 car tyres. Information
about these times is shown in the table.
Time
Frequency
0<t<6
6<t<12
12<t<18
18<t<24
24<t<30
15
25
20
12
8
Calculate an estimate for the mean time
taken to repair each car tyre.
12) Sethina recorded the times, in minutes,
taken to repair 80 car tyres.
Time
Frequency
Midpoint
0<t<6
6<t<12
12<t<18
18<t<24
24<t<30
15
25
20
12
8
80
3
9
15
21
27
Estimated Mean =
=
45
225
300
252
216
1038
1038 ÷ 80
12.975
13)
8cm
Perimeter
=
=
=
(π x 8 ÷ 2) + 8
12.568 + 8
20.57cm (2dp)
14)
a) Simplify a × a × a
a3
b) Expand 5(3x – 2)
15x – 10
c) Expand 3y(y + 4)
3y2 + 12y
d) Expand and simplify 2(x – 4) + 3(x + 2)
2x – 8 + 3x + 6
5x – 2
e) Expand and simplify (x + 4)(x – 3)
x2 + 4x – 3x – 12
x2 + x – 12
15) Work out
4.6 + 3.85
3.22 – 6.51
Write down all the numbers on your
calculator display.
8.45
3.73
=
2.26541555
16)a) Simplify t6 × t2
t8
b) Simplify
m8
3
m
m5
c) Simplify (2x)3
8x3
d) Simplify 3a2h × 4a5h4 12a7h5
17)
AB2 = 92 – 62
AB2 = 81 – 36
AB2 = 45
AB = 45
AB = 6.71(3sf)
18)a)
The box plot gives information about the
distribution of the weights of bags on a
plane. Jean says the heaviest bag
weighs 23 kg. She is wrong. Explain
why.
The heaviest bag weighs 29kg
18)
b) Write down the median weight. 17
c) Work out the interquartile range of
the weights.
23 – 10 = 13
18)
There are 240 bags on the plane.
d) Work out the number of bags with a
weight of 10 kg or less.
25% x 240 = 60
19)a) Toby invested £4500 for 2 years in
a savings account. He was paid 4% per
annum compound interest. How much
did Toby have in his savings account
after 2 years?
4% of 4500 = 180
After 1 year = 4500 + 180 = 4680
4% of 4680 = 187.20
After 2 years = 4680 + 187.20 = £4867.20
19)b) Jaspir invested £2400 for n years
in a savings account. He was paid 7.5%
per annum compound interest. At the
end of the n years he had £3445.51 in
the savings account. Work out the
value of n.
2400 x (1.075)n = 3445.51
n=5
20)a)
cos x = 5
8
x = cos-1(5/8)
x = 51.3178...
x = 51.3°
20)b)
tan 40 = y
12.5
y = tan 40 x 12.5
= 10.4887...
= 10.5cm (1dp)
21)a)
Language studied
German French Spanish
Male
45
52
26
Female
25
48
62
A sample, stratified by the language studied
and by gender, of 50 of the 258 students
is taken. Work out the number of male
students studying Spanish in the sample.
26 x 50 = 5.03...
258
Therefore 5 students in the sample
21)b)
Language studied
German French Spanish
Male
45
52
26
Female
25
48
62
A sample, stratified by the language studied
and by gender, of 50 of the 258 students
is taken. Work out the number of female
students in the sample.
25+48+62 x 50 = 26.16...
258
26 (or 27) students in the sample
22)Prove that (3n + 1)2 – (3n –1)2 is a
multiple of 4, for all positive integer
values of n.
(3n + 1)2 – (3n –1)2
= (9n2 + 3n + 3n + 1) – (9n2 – 3n – 3n + 1)
= 9n2 + 6n + 1 – 9n2 + 6n – 1
= 12n
12n is a multiple of 4
or 12n = 4n x 3
23)a)
AB = -a + b
23)b)
P is the point on AB such that
AP:PB = 3:2. Show that OP = 1/5(2a + 3b)
OP =
=
=
=
=
OA + AP
a + 3/5(-a+b)
a – 3/5a + 3/5b
2/ a + 3 / b
5
5
1/ (2a + 3b)
24)
Area of triangle ABC = ½ x 6 x 6 x sin 60
= 15.588...
Area of sector APQ = 60/360 x π x 32
= 4.712...
Shaded Area = 15.588... – 4.712...
= 10.9cm2 (3sf)
25)
x2 – 8x + 15
2x2 – 7x – 15
=
(x – 3)(x – 5)
(2x + 3)(x – 5)
=
(x – 3)
(2x + 3)
26) Phil has 20 sweets in a bag. 5 of
the sweets are orange. 7 of the sweets
are red. 8 of the sweets are yellow. Phil
takes at random two sweets from the
bag. Work out the probability that the
sweets will not be the same colour.
P(not same colour) = 1 – P(same colour)
P(same colour) = (5/20 x 4/19) + (7/20 x
6/19) + (8/20 x 7/19) = 0.311
P(not same colour) = 1 – 0.3105 = 0.689