Core Sample Worked Solutions
Core Sample Worked Solutions
Core Sample Worked Solutions
These are worked solutions of some selected questions from the Core Student Book.
Revision exercise 1A
7. Volume of a cylinder = πr2 h
Substituting: 2000 = π × 82 × h
2000
Hence h =
π × 82
= 9.947…
= 9.95 cm (3 s.f.)
π × 102 = 100π
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CORE Worked Solutions 2
Revision exercise 2A
3. (a) 3x − 1 = 20
⇒ 3x = 21
⇒x=7
(b) 4x + 3 = 4
⇒ 4x = 1
1
⇒x=
4
(c) 5x − 7 = −3
⇒ 5x = 4
4
⇒x=
5
8. y
0 2 4 6 x
–1
y = 3x and x + y = 6 intersect at (1.5, 4.5), giving the height of the triangle as 4.5 − 2 = 2.5 units.
2
x + y = 6 and y = 2 intersect at (4, 2) and y = 3x and y = 2 intersect at , 2 , giving the base of the triangle
3
2 1
as 4 − = 3 units.
3 3
1
Area of triangle = bh
2
1 1
= × 3 × 2.5
2 3
1
= 4 square units
6
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CORE Worked Solutions 3
Revision exercise 3A
9. (a) Cost for 4 hour disco (from 8 p.m. until 12 midnight):
250 + (4 × 50) + (200 × 0.5) + (200 × 0.6) + 50
= $720
720
(b) = 144 tickets
5
(c) Total revenue:
(400 × 5) + (200 × 0.75) + (200 × 1)
= $2350
Total profit:
2350 − 720
= $1630
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CORE Worked Solutions 4
Revision exercise 4A
3. (a) ( i) 10 × 56 = 560 kg
(560 + 67)
(ii) = 57 kg
11
(10 × 56) + (20 × 47)
(b) = 50 kg
30
45
5. × 200 = 25
(a)
360
135
(b) × 200 = 75
360
36
(c) × 200 = 20
360
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CORE Worked Solutions 5
Revision exercise 5A
1. Using Pythagoras’ theorem: 32 + 52 = 34
⇒ hypotenuse = 34
The minimum side length of the card needed = 20 + (2 × 14.14) = 48.28 cm (2 d.p.)
Note: 14.14 is a rounded value from the previous line. You should work with all of the numbers on
your calculator display to prevent errors caused by rounding.
Note: both 48.28 and 14.14 are rounded values from the previous part of the question.
You should work with all of the numbers on your calculator display, unless told otherwise, to prevent
errors caused by rounding. Only round your final answer to the required degree of accuracy.
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CORE Worked Solutions 6
Revision exercise 6A
4. (a) Adding 3t to both sides gives:
s = rt + 3t
6. (a) 7c + 3d = 29 (1)
5c − 4d = 33 (2)
(b) 2x − 3y = 7 (1)
2y − 3x = −8 (2)
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CORE Worked Solutions 7
Revision exercise 7A
1. (a) $1.80 × 1.1 = $1.98
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CORE Worked Solutions 8
Revision exercise 8A
4. Possible outcomes with at least three ‘heads’ are:
HHHT
HHTH
HTHH
THHH
HHHH
5
There are 16 total possible outcomes, so the probability of at least three heads is .
16
1
6. × 300 = 50 times
6
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CORE Worked Solutions 9
Revision exercise 9A
6.
42°
32°
y z
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CORE Worked Solutions 10
3 1 3 2
14. (a) + (b) ×
5 3 8 3
9 5 3×2
= + =
15 15 8×3
9+5 6
= =
15 24
14 1
= =
15 4
1 1 2 1
(c) − (d) ÷
5 10 3 4
2 1 4 2
= − = ×
10 10 1 3
1 8 2
= = 2
10 3 3
1 2 1 3
(e) 1 − (f) 2 ×
2 5 4 4
3 2 9 3
= − = ×
2 5 4 4
15 4 27 11
= − = =1
10 10 16 16
11 1
= = 1
10 10
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