Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
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Question 1
a) 6012
b) 6789 is closer to 6800 than it is to 6700
6800
c) 80 or eighty
d) to find a number exactly halfway, add the two numbers and then divide by 2
(742 + 864) ÷ 2 = 1606 ÷ 2 = 803
Question 2
a) we can see that each number is 9 more than the previous number (eg 45 = 36 + 9)
45 + 9 = 54
54 + 9 = 63
The next two numbers in the sequence are 54 and 63
b) if you look at the difference between each term, we are adding 9 to the previous term
or you can see that this is just the 9 times table.
c) the 20th term will be 20 x 9 = 180
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 3
a) i) twenty to ten
9:40 pm
ii) 9 + 12 = 21
21:40
b) -2⁰C
c)
Question 4
a) France is 75 million
b) USA was 46 million
c)
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 5
a)
i)
ii) the triangle has two equal angles and one line of symmetry so is isosceles
isosceles
b)
c) i) an 8 sided polygon is called an octagon
ii) in order to be a regular polygon then all the sides must be the same length and all the angles must
be the same size as well
the shaded polygon does not have equal angles so is not regular
d) i) flag C has no lines of symmetry
ii) Flag C has rotational symmetry of order 2 (it will look exactly the same in two positions when
turned)
e) i) there are five stripes and 2 of them are white
ii) = = 0.4
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 6
a)i) 22 and 24 have a sum of 46 (that means that the two numbers add together to give 46)
ii) a multiple of 7 is something that is in the 7 times table
28 is a multiple of 7
iii) a square number is a number such as 1, 4, 9, 16, 25 etc (5 x 5 = 5 squared = 25 so 25 is a square
number)
25 is a square number
iv) a prime number is a number that has no factors (nothing goes into it exactly) apart from 1 and
itself
23 and 29 are both prime numbers
b)i) there are 9 numbers and only 1 of them is 22
ii) there are 9 numbers and 5 of them are even (22, 24, 26, 28 and 30)
Question 7
a) i) √7 = 2.645751311
ii) 2.65 (to 2 decimal places)
b) i) 0.292 = 0.0841
ii) 0.08 (1 significant figure)
c) 3.375 + 0.4 = 3.775
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 8
a) to find the median we must first rewrite all the numbers in order from smallest to largest
1, 4, 4, 5, 6, 10, 10, 10, 10, 10
The median will be the middle number
We have 10 numbers so the median is between the 5th and the 6th ((10 +1) ÷ 2 = 5.5)
The 5th number is 6 and the 6th number is 10 so
The median is 8
b) to find the range we take the difference between the largest number and the smallest number
range = 10 – 1 = 9
Question 9
a) 4q
b) 5np
c) divide both sides by 6
x = 42 ÷ 6 = 7
d) add 1 to both sides
8y = 6
divide both sides by 8
y==
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 10
a) in order to be able to compare the fractions we need them all to have the same denominator
(number on the bottom). We need to find a number that is divisible by 3, 10, 20 and 8. This could
be 120.
=
(we multiplied the denominator by 40 to get 120 so we multiply the numerator by 40 to get
80)
=
(we multiplied the denominator by 12 to get 120 so we multiply the numerator by 12 to get
84)
=
=
(we multiplied the denominator by 6 to get 120 so we multiply the numerator by 6 to get 78)
(we multiplied the denominator by 15 to get 120 so we multiply the numerator by 15 to get
75)
Now we can compare the fractions:
In order
, , ,
converting back, we have
, , ,
b) we need a common denominator. We must find a number that both 4 and 12 divide exactly into.
This could be 12.
= (we multiplied the denominator by 3 to get 12 so we multiply the numerator by 3 to get 9)
- =
=
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 11
a) triangle ABC is isosceles since AB = AC. That means that angle B will equal angle C.
Angles in a triangle add up to 180⁰.
180 – 48 = 132⁰
132 ÷ 2 = 66⁰
b) angles on a straight line add up to 180⁰ so the angle inside the quadrilateral ACDE at C will be 114⁰
180 – 66 = 114⁰
angles in a quadrilateral add up to 360⁰
69 + 106 + 114 = 289⁰
360 – 289 = 71⁰
y = 71⁰
Question 12
a)
Mathematics Biology
5
2
80
32 note 1
Total
7
Note 1: whatever we did to 5 to get to 80, we do to 2
80 ÷ 5 = 16
2 x 16 = 32
32 candidates for IGCSE biology
b)
Foundation
1
20 note 2
Higher
3
Total
4
80
Note 2: whatever we did to 4 to get to 80, we do to 1
80 ÷ 4 = 20
1 x 20 = 20
20 candidates for Foundation
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 13
speed = distance ÷ time
speed = 40, time = 13.25 (note that 15 minutes is 0.25 not 0.15)
40 = distance ÷ 13.25
multiply both sides by 13.25
40 x 13.25 = distance
distance = 530 km
Question 14
We look at each point from (0,0). The point (4, 4) was 4 along and 4 up (from (0,0)) so it will now be
10 along and 10 up. The point (6,4) was 6 along and 4 up so will now be 15 along and 10 up. The
point (6,8) was 6 along and 8 up so will now be 15 along and 20 up. The dashed tram lines help us to
be sure that we have enlarged the triangle correctly.
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 15
a) A = 2((12 x 7) + (5 x 7) + (5 x 12))
A = 2 x (84 + 35 + 60) = 2 x 179 = 358
b) 70 = 2(4L + (4 x 2) + 2L)
70 = 2(6L + 8)
divide both sides by 2
35 = 6L + 8
rewrite
6L + 8 = 35
subtract 8 from both sides
6L = 27
divide both sides by 6
L = 4.5
b) A = 2LW + 2HW + 2HL
subtract 2WH from both sides
A – 2HW = 2LW + 2HL
rewrite
2LW + 2HL = A – 2HW
factorise
2L(W + H) = A – 2HW
divide both sides by 2 and by (W + H)
L = Question 16
a) x 850 = $119
b) x 100 = 35%
c) $204 represents 30%
204 ÷ 30 = $6.80 represents 1%
6.80 x 100 = $680 represents 100%
Weekly pay is $680
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 17
If we work out 0.35 of 10 we get 0.35 x 10 = 3.5. As this is not a whole number we know that this
probability must be wrong because we can’t have 3.5 red beads (or 3.5 beads of any colour).
Question 18
a) p(p + 7)
b) add 5x to both sides
4 = 5x + 2
rewrite
5x + 2 = 4
subtract 2 from both sides
5x = 2
divide both sides by 5
x=
c) t3+6 = t9
d)3(4y + 5) – 5(2y + 3)
12y + 15 – 10y – 15
group terms
2y
Question 19
Distance (d km)
Frequency
Midpoint
0 d 20
20 d 40
40 d 60
60 d 80
80 d 100
Total
8
24
5
2
1
40
10
30
50
70
90
Midpoint x
frequency
80
720
250
140
90
1280
Estimate of mean = 1280 ÷ 40 = 32 km
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 20
Total surface area will be the area of the five faces added together. We have two end triangles, one
rectangular base and two side rectangles
End triangles
Area of a triangle is ½ x base x height
½ x 10 x 12 = 60 cm2
We have two triangles so 2 x 60 = 120 cm2
Rectangular base
15 x 10 = 150 cm2
Side rectangles
15 x 13 = 195 cm2
We have two of these so 2 x 195 = 390 cm2
total surface area = 120 + 150 + 390 = 660 cm2
Question 21
A = 1, 3, 9, 27
B = 1, 3, 9
C = 2, 4, 6, 8
a) A ! B is everything that is in A together with everything that is in B
1, 3, 9, 27
b)A " C is everything that is in both of A and B
there is nothing that is in both of A and B so this set is empty
the statement is true A " C = #
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Mathematics IGCSE Higher Tier, May 2009
4400/1F (Paper 1F)
Question 22
label all the sides (opp, adj, hyp) from the point of view of the angle
hyp
3.6 cm
7.9 cm
opp
x⁰
adj
From SOHCAHTOA, we don’t have adj so we are left with SOHCAHTOA
$%% .
sin x⁰ =
= = 0.455696…
&'%
.
take the inverse sin (sin-1) of both sides
x = sin-1(0.455696…) = 27.1⁰
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