Answers to all pages - Hodder Plus Home

Model answers 

Integers (page 3) 

1 (a) 14 (b) 12 or 18 (c) 11, 13, 17 or 19 

2 264 132 66 33 11 

2 2 2 3 

264 = 2 × 2 × 2 × 3 × 11 

3 12 = 2 × 2 × 3 

16 = 2 × 2 × 2 × 2 

HCF () = 2 × 2 = 4 

LCM () = 2 × 2 × 2 × 2 × 3 = 48 

4 10 = 2 × 5 

12 = 2 × 2 × 3 

20 = 2 × 2 × 5 

HCF () = 2 

LCM () = 2 × 2 × 3 × 5 = 60 

1 

5 4 × – 

4 = 1 

6 25 + – 12 = 13 

Powers and roots (page 5) 

1 

1 (a) –– 

81 (b) 1 (c) 3 

2 (a) (i) 1 (ii) 3 (b) 24 × 10 3 = 2·4 × 10 4 

1 1 

3 (a) × 4 = (b) p 6 q –2 = p 3 q –1 p 3 

–8 – or –– 

2 

q 

8 

4 (a) 8pq 3 (b) ––-– 

343 

5 (a) 18 × 10 7 = 1·8 × 10 8 (b) 1·2 × 10 – 4 

6 8·75 × 10 – 3 

7 80 × 0.85 6 = 30·2 litres 

Fractions and decimals (page 7) 

3 3 6 3 9 1 1 2 1 4 3 1 

1 (a) 2 –4 + – = 2 + –8 + – = 2 + –8 = 2 + 1 – = 3 – (b) 3 – – 2 – = 1 + – – –6 = 1 –6 

8 

8 

8 8 

3 2 6 

3 2 8 20 8 4 32 5 1 3 5 12 17 

2 (a) 1 –5 × 2 –9 = – × ––9 = – × – = ––9 = 3 – (b) 1 – + 2 – = 3 + –– + –– = 3 –– 

5 1 9 

9 

4 5 20 20 20 

1 4 4 

3 (a) Reduction of – so sale price is – of previous price. Sale price = –5 

5 

5 

× $27·50 = 4 × $5·50 = $22 

4 4 5 

(b) Previous price = sale price ÷ –5 = $72 ÷ –5 = $72 × – 

4 = 18 × 5 = $90 

3 1 3 4 8 3 4 7 1 4 9 9 9 5 1 5 5 1 

4 (a) 2 – – 1 – = 1 + – – –8 = –8 + – – –8 = –8 

(b) 2 – ÷ 1 –5 = –4 ÷ –5 = –4 × –9 = –4 × –1 = –4 = 1 –4 

8 2 8 

8 

4 

2 2 8 

5 2xy = 2 × –5 × –7 = –– 

35 

2 2 14 10 24 

x + y = –5 + – = –– + –– = –– 

7 35 35 35 

8 24 8 35 1 

h = –– ÷ –– = –– × –– = –3 

35 35 35 24 

6 (a) Let x = 0·141 414 14 .... so 100x = 14.141 414 ... 

Subtract 99x = 14 

14 

x = –– 

99 

(b) (i) Recurring (ii) Terminating (iii) Recurring (iv) Recurring 

The one that terminates has a denominator with just two prime factors 2 and 5. The other denominators 

have other prime factors. 

3 

7 – 

4 of the total voters = 13 845 

3 4 

so total voters = 13 845 ÷ –4 = 13 845 × – 

3 = 18 460 

IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational 

1

Percentages and ratios (page 9) 

Model answers 

1 (a) Falls by 30% each year so after one year = 70% = 0·7 = 0·7 × 899 = $629·30 

(b) After four years it will be $899 × (0·7) 4 = $215·8499 

One quarter of the original is 899 ÷ 4 = $224·75. So it is less than one quarter of the original. 

2 A reduction of 15% so sale price = 85% = 0·85 of pre-sale price 

The pre-sale price = sale price ÷ 0·85 = $68 ÷ 0·85 = $80. 

3 Increase of 8% so multiply by 108% = 1·08 each year. 

After six years = 84 000 × (1·08) 6 = $133 297·44 = $133 300 to nearest $100. 

4 Price including VAT = 1·175 × basic price 

(a) Price including VAT = 1·175 × 650 = $763·75. 

(b) If store pays VAT then their price is $650 rather than $763·75. This is a discount of $113·75. 

113·75 

As a percentage = –––––––– × 100 = 14·8936% = 14·9% 

763·75 

5 Total parts = 20 

8 

20 

(a) Amount of water = –– × 250 = 100 ml (b) Amount of Eau de Parfum = ––3 

20 

× 45 = 300 litres 

6 Scale 5 cm to 2 km 

5 

(a) So 7 km is drawn as 7 × – 

2 = 17·5 cm (b) 5cm:2km = 5 cm : 200 000 cm = 1 : 40 000 

2800 

7 (a) Percentage in favour = ––––– × 100 = 70% 

4000 

7 

(b) Total parts = 15 Number of men = –– 

15 × 300 = 140 and number of women = 15 × 300 =160 

8 –– 

Mental methods (page 11) 

1 750 

2 (a) 2 11 (b) 2·05 × 10 3 

3 2·4 × 10 10 

1 –2 

1 – 

2 

1 –2 

4 (50 + 80 ) × (10 + 25 ) = (7 + 9) × (3 + 5) = 16 × 8 = 128 

5 35 × 1·496 × 10 11 = 52·36 × 10 11 = 5·236 × 10 12 

Written methods (page 14) 

1 450 × 1·06 3 = 535·9572 = $535·96 

2 160·48 × 1·025 3 = 172·819... = $172·82 

3 T ∝ L 

⇒ T = k L 

⇒ 1·6 = k 64 

⇒ 1·6 = k × 8 

⇒ k = 1·6 ÷ 8 = 0·2 

⇒ T = 0·2 L 

4 (a) S ∝ L 3 

⇒ S = kL 3 

⇒ 2·4 = k × 4 3 

⇒ 2·4 = k × 64 

⇒ k = 2·4 ÷ 64 = 0·0375 

⇒ S = 0·0375L 3 

(b) S = 0·0375 × 6 3 = 8·1 cm 

(c) 0·3 = 0·0375 × L 3 

⇒ L 3 = 0·3 ÷ 0·0375 = 8 

⇒ L = 2 m 

5 

1 

P ∝ ––– 

Q 

⇒ 

k 

P = ––– 

Q 

k 

⇒ 12 = –––– 

49 

⇒ k = 12 × 49 = 12 × 7 = 84 

⇒ 

84 

P = ––– 

Q 

1 – 

2 

2 

IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational

Model answers 

Calculator methods (page 17) 

1 (a) 3 ÷ 5 × 200 = 120 g 

19 

(b) 2 ab/c 3 ab/c 4 – 1 ab/c 4 ab/c 5 = –– 

20 or 0·95 

(c) 4 ÷ 7 × 2 6 · 6 = $15·20 

1 

2 1 ÷ ( 3 x 2 ) = – 

9 

3 (a) ( 1 8 · 6 – 2 · 7 5 ) ÷ ( 3 · 5 + 1 · 0 4 3 ) = 3·49 

(b) 1 ÷ ( 4 · 5 + 6 · 8 ) = 0·088 

(c) 3 · 2 × ( 5 · 2 – ( 1 ÷ 1 · 6 ) ) = 14·64 

4 (a) ( 4 · 2 x y 5 – 1 · 7 x y 4 ) ÷ 1 · 2 5 x 2 = 831·08 

(b) √ ( 3 x 2 + 2 · 7 x y 3 ) = 5·36 

5 (a) 6 · 3 EXP 9 + 5 · 8 EXP 1 0 = 6·43 × 10 10 

(b) ( 9 · 5 2 EXP 1 4 ) x 2 ÷ 8 EXP +/– 3 = 1·13 × 10 32 

6 (a) 6 × ( cos 2 7 + sin 2 7 ) = 8·07 

(b) 3 – ( 4 × sin 2 8 · 7 ) = 1·08 

7 6 4 x y ( 4 ab/c 3 ) = 256 

Solving problems (page 18) 

1 (a) $58·80 ÷ 12 × 5 = $24·50 (b) $5·49 ÷ 3 × 4 = $7·32 

2 (a) $4000 × 1·07 3 = $4900·17 

(b) 8 years. Found from repeated multiplication by 1·07. 

90 

3 –– = 1·5 h = 1 h 30 min 

60 

Total time = 1 h 30 min + 50 min + 1 h 40 min = 4 h 

180 

––– = 45 km/h 

4 

4 (a) Total = 4·47 × 10 7 or 4·5 × 10 7 (b) 5·14 × 10 6 ÷43000 = 120 

5 $900 = 120% of 1998 bonus = 1·2 × 1998 bonus 

1998 bonus = $900 ÷ 1·2 = $750 

165 × 40 

6 –––––––– = 44·9 cm to 1 d.p. 

147 

Symbols, indices, factors and expansions (page 21) 

1 3pq(4p – 5q) 

2 (a) 2(3x + 1) – 4(x – 3) = 6x + 2 – 4x + 12 = 2x + 14 (b) (3 – x) 2 =(3–x)(3 – x) =9–3x –3x + x 2 =9–6x + x 2 

3 (a) 4s 2 – 2s – 5 (b) 2e 2 + 6ef – ef – 3f 2 = 2e 2 + 5ef – 3f 2 (c) 8pq 3 

4 Let a small radiator cost $x. A large radiator will cost $(x + 45) 

Therefore 3x + 2(x + 45) = 415 

3x + 2x + 90 = 415 

5x = 415 – 90 = 325 

x = 65 

Small radiators cost $65 each 

5 (a) 6a (b) 2a(2a – 1) 

6 ( a + b)( a – b) = a + a b – a b – b = a – b 

7 (a) 12a 3 b 4 (b) 

7a 

––– 

2 

b 

8 (a) (3x – 2)(x + 4) = 3x 2 + 12x – 2x – 8 = 3x 2 + 10x – 8 (b) (2y – 1)(y – 3) = 2y 2 – 6y – y + 3 = 2y 2 – 7y + 3 

9 (a) 15a – 9 – 10ab + 6b (b) 3a – 6 + 10b – 5ab 

= (15a – 10ab) – (9 – 6b) = (3a – 6) + (10b – 5ab) 

= 5a(3 – 2b) – 3(3 – 2b) = 3(a – 2) + 5b(2 – a) 

= (5a – 3)(3 – 2b) = 3(a – 2) – 5b(a – 2) 

= (a – 2)(3 – 5b) 


3

Model answers 

10 (a) 50a 2 – 18b 2 = 2(25a 2 – 9b 2 ) (b) x 4 – y 4 = (x 2 – y 2 )(x 2 + y 2 ) 

= 2(5a – 3b)(5a + 3b) = (x – y)(x + y)(x 2 + y 2 ) 

11 (a) 2 –3 1 

× 16 = × 4 = (b) p 4 q –3 p 

× = 6 p 3 

– 1 1 q 

2 – – –– –– = ––– 

8 2 

q 

2 

p –2 

q 

12 (a) ab(b – 3a) (b) 9 –1 2 – 1 

× 27 = × 3 2 1 

3 – = –9 

9 × 9 = 1 

Linear equations (page 23) 

1 3x = x + 1 

2x = 1 

1 

x = –2 

2 3p – 4 = p + 8 

2p = 12 

p = 6 

3m 

3 ––– 4 = 9 

3m = 36 

m = 12 

4 2(y + 3) = 5y 

2y + 6 = 5y 

6 = 3y 

y = 2 

5 4(x – 1) = 2x + 3 

4x – 4 = 2x + 3 

2x = 7 

x = 3·5 

6 4(x + 2) + 2(3x – 2)= 14 

4x + 8 + 6x – 4 = 14 

10x = 10 

x = 1 

7 (a) 2(x + (x + 2)) = 36 

2(2x + 2) = 36 

4x + 4 = 36 

(b) 4x + 4 = 36 

4x = 32 

x = 8 cm 

(c) Area = x(x + 2) 

= 8(8 + 2) 

= 8 × 10 

= 80 cm 2 

Formulae (page 25) 

1 P = 120 + 4n 

⇒ P – 120 = 4n 

P – 120 

⇒ –––––––– = n 

4 

P – 120 

⇒ n = –––––––– 

4 

1 

2 s = ut + – 

2 at 2 

1 1 – –4 

2 

= 432 – 288 

= 144 

3 T = 20 + 30W 

1 

4 V = –3 πr 2 h 

⇒ 3V = πr 2 h 

3V 

⇒ –– = r 2 

πh 

= 9 × 48 + × ( – ) × 48 2 

⇒ r = 

3V 

–– πh 

4 


Model answers 

5 Sequence goes up in 3s so U n 

= 3n + k. 

Testing e.g. n = 2, 3 × 2 + k = 4 

so k = – 2 

so U n 

= 3n – 2 

6 (a) V = 25 × 3·1 × (7·3 2 – 5·9 2 ) 

= 25 × 3·1 × 18·48 = 1432·2 

(b) x = 4y 2 – 3 

⇒ x + 3 = 4y 2 

⇒ 

x + 3 

–––– = y 

4 

2 

⇒ 

x 

–––– 

+ 3 

= y 

4 

⇒ y = 

x 

–––– 

+ 3 

4 

7 (a) e = 5d + 3 

⇒ e – 3 = 5d 

⇒ 

e 

–––– 

– 3 

= d 

5 

⇒ d = 

e 

–––– 

– 3 

5 

(b) –––––– 

3d – 7 

= e 

4 + 5d 

⇒ 3d – 7 = e(4 + 5d) 

⇒ 3d – 7 = 4e + 5ed 

⇒ 3d = 4e + 5ed + 7 

⇒ 3d – 5ed = 4e + 7 

⇒ d(3 – 5e) = 4e + 7 

⇒ d = 

4e 

––––– 

+ 7 

3 – 5e 

14·9 × ( – 10·2) – 151·98 

8 (a) f = –––––––––––– = ––––––– 

14·9 + ( – 10·2) 4.7 

= – 32·336... 

= – 32·3 to 3 s.f. 

(b) f = –––– 

uv 

u + v 

⇒ f(u + v) = uv 

⇒ fu + fv = uv 

⇒ fu = uv – fv 

⇒ fu = v (u – f) 

⇒ –––– 

fu 

= v 

u – f 

⇒ v = –––– 

fu 

u – f 

9 t = 1, n = 2 × 3 

t = 2, n = 2 × 3 2 

t = 3, n = 2 × 3 3 and so on 

n = 2 × 3 t 

Direct and inverse proportion 

(page 27) 

1 y ∝ x 2 

so y = kx 2 

so 75 = k × 5 2 

so k = 3 

so y = 3 × 10 2 

= 300 

1 

2 N ∝ –– 

d 2 

k 

so N = –– 

d 2 

k 

so 8000 = –– 2 

2 

so k = 4 × 8000 = 32 000 

32 000 

so N = –––––– 

4 2 

= 2000 

3 D ∝ V 2 

so D = kV 2 

so 12·5 = k × 50 2 

12·5 

so k = –––– = 0·005 

2500 

D = 0·005 × 120 2 

= 72 m 

4 R ∝ V 2 

r = kV 2 

100 = k × 300 2 

100 

k = –––––– = 0·001 11 

90 000 

(a) R = 0·001 11 × 600 2 

= 400 N 

(b) 200 = 0·001 11 × V 2 

V 2 = 180 000 

V = 424 m/s 

Simultaneous equations and 

linear inequalities (page 29) 

1 10x + 8y = 26 

3x + 8y = 5 

Subtract 

7x = 21 

x = 3 

Substitute 

3 × 3 + 8y = 5 

8y = 5 – 9 

y = – 0·5 

Solution is x = 3, y = – 0·5 

2 4x + 3y = 5 

6x + 3y = 3 

Subtract 

– 2x = 2 

x = – 1 

Substitute 

2 × – 1 + y = 1 

y = 1 + 2 

y = 3 

Solution is x = – 1, y = 3 

3 (a) Replace x with 3 and y with 2. 

(b) 9p + 11q = 5 

(c) 3p + 2q = 5 

9p + 11q = 5 

9p + 6q = 15 

9p + 11q = 5 


5

Subtract 

– 5q = 10 

q = – 2 

Substitute 

3p + 2 × – 2= 5 

3p = 5 + 4 

p = 3 

Solution is p = 3, q = – 2 

4 8x 25 – 5 

8x 20 

x 20 ÷ 8 

x 2·5 

5 2x – 4x 6 – 17 

– 2x – 11 

x – 11 ÷ – 2 

x 5·5 

6 x 6 or x – 6 or – 6 x 6 

Quadratic equations (page 31) 

1 (a) (x – 2)(x – 4) = 0 

so either x – 2 = 0 or x – 4 = 0 

so x = 2 or 4 

(b) (2x – 3)(x + 3) = 0 

so either 2x – 3 = 0 or x + 3 = 0 

1 

so x = 1 – 

2 or – 3 

2 (a) 5(x 2 – 4) = 5(x + 2)(x – 2) 

(b) (i) (x – 8)(x – 1) 

(ii) either x – 8 = 0 or x – 1 = 0 

so x = 8 or 1 

3 (a) 6x 2 – 12x + 21x – 42 = 6x 2 + 9x – 42 

(b) (i) x(x + 6) 

(ii) either x = 0 or x + 6 = 0 

so x = 0 or – 6 

4 Using the quadratic formula, 

a = 2, b = – 38, c = 45 

38 ± 38 2 – 4 × 2 × 45 

x = –––––––––––––––––––––––– = 17·73 or 1·27 to 2 d.p. 

4 

5 (a) 3x 2 –12x + 2 

= 3(x 2 – 4x) + 2 

= 3[(x – 2) 2 – 4] + 2 

= 3(x – 2) 2 – 10 

(b) 3(x – 2) 2 –10 = 0 

3(x – 2) 2 = 10 

(x – 2) 2 10 

= ––3 

x – 2 = ± 

x = 2 ± 

10 ––3 

10 ––3 

= 3·83 or 0·17 to 2 d.p. 

6 (a) y(15 – y) = 55 

15y – y 2 = 55 

y 2 – 15y + 55 = 0 

15 ± 15 2 – 4 × 1 × 55 

(b) y = –––––––––––––––––––––––– , 

2 

8·62 and 6·38 cm 

7 (a) Length in m = 22 – 2x 

area in m 2 = x(22 – 2x) = 60 

so 22x – 2x 2 = 60 

so 11x – x 2 = 30 

so x 2 – 11x + 30 = 0 as required 

(b) (x – 6)(x – 5) = 0 

so x – 6 or x – 5 = 0 

so x = 6 or 5 

(c) The pen is 6 m by 10 m or 5 m by 12 m 

Model answers 

Algebraic fractions (page 32) 

1 (a) 2x – 3 = ––––––– 

3(x + 1) 

x + 4 

(2x – 3)(x + 4) = 3(x + 1) 

2x 2 + 5x – 12 = 3x + 3 

2x 2 + 2x – 15 = 0 

– 2 ± 4 + 120 

(b) x = –––––––––––––– 

4 

= 2·28 or – 3·28 

x(x + 3) 

2 –––––––– 

x 2 + 3x 

= ––––––––––– 

x 2 + x – 6 (x – 2)(x + 3) 

= –––– 

x 

x – 2 

3 (a) –––––––– 

x 2 – 9 (x 

= ––––––––––– 

– 3)(x + 3) 

x 2 – x – 6 (x – 3)(x + 2) 

x 

= –––– 

+ 3 

x + 2 

12 5 

(b) ––––– – –––– = 1 

3x + 1 x + 1 

12(x + 1) – 5(3x + 1) = (3x + 1)(x + 1) 

12x + 12 – 15x – 5 = 3x 2 + 4x + 1 

3x 2 + 7x – 6 = 0 

(3x – 2)(x + 3) = 0 

2 

x = – or – 3 3 

Sequences (page 34) 

1 Position Term Difference 

1 3 

5 

2 8 

5 

3 13 

So rule is 5n – 2. 

If n = 50 then 50th term is 5 × 50 – 2 = 248 

2 Position Term Difference 

1 1 

5 

2 6 

5 

3 11 


3 (a) Position Term Difference 

1 2 

7 

2 9 

7 

3 16 


(b) If 7n – 5 = 300 then 7n = 305 and n = 305 ÷ 7. 

Therefore n = 43·57 i.e. n is not a whole number 

therefore 300 is not in the sequence. 

6 


Model answers 

4 (a) n 2 

(b) The clue is in the word ‘hence’ – by inspection each number is 1 more than 3 times n 2 . 

Therefore the rule is 3n 2 + 1. 

5 (a) Position Term Difference 

1 3 

2 

2 5 

2 

3 7 

So the rule is 2n + 1. 

(b) By inspection: n × (2n + 1) + 1 = 2n 2 + n + 1. 

6 (a) 1, 8, 27 should be recognised as n 3 . 

(b) 2, 16, 54 should now be recognised as 2n 3 . 

(c) From 2, 16, 54 

to – 1, 10, 45 

––––––––––– 

subtract 3, 6, 9 i.e. 3n. 

Therefore the rule = 2n 3 – 3n. 

Graphs of linear functions (page 36) 

1 (a) Gradient = – 5 

= – 2·5 

(b) c = 5, m = – 2.5 so y = – 2·5x + 5 

2 (a) Compare with y = mx + c 

m = – 2 c = 4 

So the gradient is – 2 and the y-intercept is 4. 

(b) m = – 2 c = – 1 

y = – 2x – 1 

3 y = mx + c 

– 16 

m = –– = – 8 2 

y = – 2x + c 

( – 3, 12) means x = – 3 and y = 12 

12 = – 2 × – 3 + c 

12 = 6 + c 

c = 6 

y = – 2x + 6 

––2 

x 0 1 2 

Graphical solution of simultaneous equations and 

linear inequalities (page 38) 

1 (a) y = 3 – x 

y 3 2 1 

4 

y 

y = 3x – 2 

x 0 1 2 

2 

y 

– 2 1 4 O 

– 2 

1 2 

3 

x 

1 

x = 1 , y = 1 3 – – 

4 4 


7

(b) y = 3 – x 

x 0 1 2 

y 

Model answers 

y 3 2 1 

7 

x 2 + y 2 = 25 

x 0 1 2 

6 

x = 4·7, y = – 1·7 

x = – 1·7, y = 4·7 

y 5 4·9 4·6 

5 

4 

3 

2 

1 

– 2 

– 1 

1 

2 3 4 5 6 

x 

– 1 

– 2 

2 (a) y = 3x + 4 x 

y 

0 

4 

1 

7 

2 

10 

x + y = 2 x 

y 

0 

2 

2 

0 

1 

1 

x = – 0·5, y = 2·5 

y 

10 

8 

6 

4 

2 

– 1 

O 

1 

2 

x 

3 x + y = 5 

x 0 5 1 

y 5 0 4 

y 

6 

4 

y = 2x – 1 

x 0 1 2 

2 

y – 1 1 3 

y = 0 is the horizontal line through (0,0) 

O 

– 2 

2 

4 

6 

x 

Choose (0,1) to check line 

0 + 1 5 true, shade opposite side 

1 0 – 1 false, shade same side 

1 0 true, shade opposite side 

4 Choose (0,1) to check lines. 

y = x 1 0 so y x 

x + y = 3 1 + 0 3 so x + y 3 

x = – 2 0 – 2 so x – 2 

8 


Model answers 

Linear programming (page 41) 

1 (a) Type A takes 1 hour of machine time and type B takes 2 hours. There are 28 hours of 

machine time in a day. 

x + 2y 28 

Type A takes 3 hours of workers’ time and type B takes 1 hour. There are 24 hours of workers’ 

time in a day. 

3x + y 24 

(b) y 

30 

24 

20 

14 

10 

O 

3x + y = 24 x + 2y = 28 

8 28 x 

10 20 

30 

(c) (i) The greatest profit will result from either the greatest number of type A, the greatest number of type 

B or the greatest total number of items so test the points (8, 0), (0, 14) and (4, 12). 

Profit at (8, 0) = $160. 

Profit at (0, 14) = $140. 

Profit at (4, 12) = $80 + $120 = $200. 

So the greatest profit that can be made is $200. 

(ii) 4 type A and 12 type B. 

2 (a) (i) There is $1000 available; paperback books cost $10 and hardback books cost $25. 

So 10x + 25y 1000. 

Dividing by 5 gives 2x + 5y 200. 

(ii) y > x and x + y 50 

(b) y 

50 

y = x 

40 

20 

O 

(c) 33 hardback books 

50 

x + y = 50 

x 

100 

2x + 5y = 200 


9

Model answers 

3 (a) (i) x small vehicles will carry 5x people and y large vehicles will carrry 8y people. 60 people must 

fit into these vehicles so 5x + 8y 60. 

(ii) 600x + 300y 4500 or 2x + y 15 

x 8 

y 7 

(b) y 

15 

x = 8 

10 

7.5 

7 

5 

y = 7 

O 

2x + y = 15 

5x + 8y = 60 

7 8 12 

5 10 15 

x 

(c) (i) The points (4, 7), (5, 7), (6, 7), (7, 7), (8, 7), (5, 6), (6, 6), (7, 6), (8, 6), (5, 5), (6, 5), (7, 5), 

(8, 5), (6, 4), (7, 4), (8, 4), and (8, 3) marked. 

(ii) 10 

(iii) 5 of each type or 6 small and 4 large vehicles. 

4 (a) x > y 

10x + 30y 300 or x + 3y 30 

(b) y 

20 y = x 

10 

(9, 7) 

x + 3y = 30 

O 

28 x 

10 20 

30 

(c) (i) The shortest amount of time spent must result from either the highest or lowest integer 

point on the line x + y = 30 so test the points (9, 7) and (30, 0). 

Time at (9, 7) is 9 + 14 = 23 hours. 

Time at (30, 0) is 30 hours. 

So the shortest time is 23 hours. 

(ii) 9 individual photos and 7 sets. 

10 


Model answers 

Interpreting graphs (page 43) 

1 

1 (a) 6 km/h means that in 10 minutes ( –6 th of an hour) Steve walks 1 km. 

32 km/h for 8 km. 

1 

Time = 8 ÷ 32 = – 

4 hour = 15 mins. 

10 

9 

8 

7 

Distance in km 

6 

5 

4 

3 

2 

1 

0800 0810 0820 

0830 

Time 

2 

(b) 3 – 

3 km or 3·7 km 

2 e.g. A car accelerates and then travels at a steady speed and then stops suddenly (crashes, etc.). 

3 (a) d 

(b) d 

(c) 

d 

t 

t 

t 


11

Quadratic and other functions (page 47) 

Model answers 

1 (a) – 5 0 [3] 4 3 [0] – 5 

y 

4 

3 

2 

1 

– 2 

– O 1 

1 2 3 4 

5 

x 

– 1 

– 2 

– 3 

– 4 

– 5 

(b) (i) 2 (ii) 0·6 and 3·4 

2 (a) T (b) Q (c) U 

3 (a) 

x 1 2 3 4 5 6 

V 7 32 81 160 275 432 

v 

500 

450 

400 

350 

300 

250 

200 

150 

100 

50 

0 

0 0 . 5 1 1 . 5 2 2 . 5 3 3 . 5 4 4 . 5 5 5 . 5 6 

(b) 4·4 cm, 4·4 cm, 10·4 cm 

x 

12 


Model answers 

4 (a) 

y 

5 

4 

3 

2 

1 

– 5 

– 4 

– 3 – O 

2 – 1 1 2 3 4 5 

x 

– 1 

– 2 

– 3 

– 4 

– 5 

(b) 2·5 

5 (a) ( – 2, – 1) in table 

y 

10 

– – 2 

– 

3 1 

O 

1 2 3 

– 10 

x 

(b) – 2 to – 1·7, – 0·4 to – 0·2, 2·0 to 2·2 

(c) y = 2x + 2 drawn, 

– 2·3 to – 2·0, – 0·6 to – 0·3, 2·5 to 2·8 

6 Manipulate x 3 – 2x – 1 = 0 to give x 3 = 2x + 1. 

The line to draw is y = 2x + 1. 

y 

8 

4 

y = 2x + 1 

– 2 – 1 O 

1 2 

– 4 

x 

y = x 3 

– 8 

The solution is where the line cuts the curve. 

x = – 1, – 0·6 or 1·6 


13

7 Manipulating 2x 3 + 7x 2 – 6 = 0 gives 

6 – 2x 3 = 7x 2 . 

Dividing by 3x 2 gives 

2 2x 7 

–– – –– = – . 

x 

2 3 3 

Manipulating this gives 

2 

– 1 1 

–– – x = –– x + 2 – . 

x 

2 3 3 

Model answers 

2 

y = – x 

x 2 

y 

4 

3 

2 

1 

y = – 1 

x + 2 1 3 3 

– 4 – 3 – 2 

– 1 O 1 2 3 4 

– 1 

2 

y = – x 

x 2 

x 

The solution is x = – 3·2, – 1·1 or 0·8. 

Graphs – gradient and area (pages 51–52) 

4 

1 (a) ––– = 1.6 ms 

2·5 

–2 

1 

(b) Area under graph = 4 × 3.5 + – × 2.5 × 4 = 14 + 5 = 19 m. 

2 

2 (a) 

Time (t hours) 0 1 2 3 4 5 

Number of bacteria (n) 20 60 180 540 1620 4860 

(b) Each hour the number of bacteria multiplies by 3 so n = 20 × 3 t . 

(c) n 

5000 

4000 

3000 

2000 

1000 

0 

1 2 3 4 5 

t 

2600 – 450 

(d) (i) Gradient = –––––––––– = 1075 

5 – 3 

(ii) The rate of increase in bacteria per hour. 

3 t = 0, d = 1 2 48 

+ –– – 20 = 29; p = 29 

1 

t = 5, d = 6 2 48 

+ –– – 20 = 24; q = 24 

6 

t = 7, d = 8 2 48 

+ ––8 – 20 = 50; r = 50 

14 


Model answers 

(b) 

d (m) 

50 

40 

30 

20 

10 

0 

F 

1 2 3 4 5 

6 7 

t (s) 

(c) F on graph (d must be increasing because the fish 

is swimming away from Dimitra.) 

t = 3·6 seconds 

(d) 3·35 – 0·8 = 2.55 s 

(e) Tangent at t = 2·5 drawn on graph. 

18·5 – 0 

Gradient = speed = ––––––– = 3·7 m/s. 

6 – 1 

4 (a) Tangent at t = 30. 

33·5 – 12 

Gradient = acceleration = –––––––– = 1·08 m/s 

35 – 15 

2 

(b) Taking strips of width = 10 s 

1 – × 5 × 10 

2 

= 25 

1 – (5 + 14) × 10 

2 

= 95 

1 – (14 + 28) × 10 

2 

= 210 

1 – (28 + 32) × 10 

2 

= 300 

10 × 32 = 320 

10 × 32 = 320 

1 – (32 + 25) × 10 

2 

= 285 

1 – × 25 × 10 

2 

= 125 

Total = 1680 

Approximate distance = 1680 m 

1 

5 (a) Using strips of width –2 hour 

1 1 – × –2 × 2 = 0·5 

2 

1 1 – (2 + 8) × –2 

2 

= 2.5 

Approximate distance = 19.85 km or 20 km 

(b) The acceleration is greatest when the graph is 

steepest. This is approximately at 9.45 a.m. 

Functions (page 55) 

1 (a) f(x) = 2x – 1 

y = 2x – 1 

2x = y + 1 

y + 1 

x = ––––– 

2 

(b) 

f –1 x + 1 

(x) = ––––– 

2 

gf(x) = g[f(x)] 

= g[2x – 1] 

= (2x – 1) 2 – 1 

= 4x 2 – 4x 

2 (a) (i) g(4) = 2 × 4 2 – 5 

= 27 

1 – 

3 

(ii) fg(4) = 27 

= 3 

(b) (i) gf(x)= g[f(x)] 

1 – 

3 

= g[x ] 

1 –3 1 – 

3 

= 2 × x × x – 5 

1 – 

3 

= 2x – 5 

(ii) x = y 

x = y 3 

f –1 (x)= x 3 

3 (a) f( – 1) = 3 × – 1 – 5 

= – 8 

(b) 3x – 5 = y 

y + 5 

x = ––––– 

3 

f –1 x + 5 

(x)=––––– 

3 

(c) fg(x)= f[x + 1] 

= 3(x + 1) – 5 

= 3x – 2 

(d) 3f(x) = 5g(x) 

3(3x – 5) = 5(x + 1) 

9x – 15 = 5x + 5 

4x = 20 

x = 5 

4 (a) f( – 1 

4) = – (2 × – 4 + 5) 

3 

2 – 

3 

1 1 – (8 + 9.1) × –2 

2 

= 4.275 

1 1 – (9.1 + 8.7) × –2 

2 

= 4.45 

1 1 – (8.7 + 8.2) × –2 

2 

= 4.225 

1 1 – (8.2 + 7.4) × – 

2 

2 

= 3.9 

Total = 19.85 

(b) 

= – 1 

1 – (2x + 5) = y 

3 

2x + 5 = 3y 

2x = 3y – 5 

3y – 5 

x= –––––– 

2 

f –1 (x) = –––––– 

3x – 5 

2 


15

5 (a) 3x + 1 = y 

3x = y – 1 

y – 1 

x = ––––– 

3 

f –1 x – 1 

(x) = ––––– 

3 

(b) fg(2) = f[g(2)] 

= f[2 × 2 2 ] 

= 3 × 8 + 1 

= 25 

(c) gf(x) = g[3x + 1] 

= 2(3x + 1) 2 

= 2(9x 2 + 6x + 1) 

= 18x 2 + 12x + 2 

(d) ff(x) = f[3x + 1] 

= 3(3x + 1) + 1 

= 9x + 4 

6 (a) f( – 1) = 2 × – 1 + 1 

= – 1 

(b) 2x + 1 = y 

2x = y – 1 

y – 1 

x = ––––– 

2 

f –1 x – 1 

(x) = ––––– 

2 

x 

(c) – – 5 = y 

3 

x 

– = y + 5 

3 

x = 3y + 15 

g –1 (x) = 3x + 15 

0 

– 

3 

(d) fg(0) = f[ 5] 

– 

= f[–5] 

= 2 × – 5 + 1 

= – 9 

(e) fg(x) = f[ – 5] 

= 2[ – 5] + 1 

x –3 

x –3 

2x 

= –– – 9 3 

Sets and Venn diagrams (page 58) 

1 

A 

2 (a) {6. 12} 

(b) {1, 5, 7, 11} 

B 

C 

3 

4 

Total = 5 + 6 + 9 + 10 = 30 

J 

S 

(b) S' ∪ T 

(c) Let y students study both. 

Model answers 

n() = 32 

12 – y + 18 + 9 = 32 

39 – y = 32 

y = 7 

12 – y = 5 

so 5 students study chemistry but not French 

Matrices (page 61) 

1 (a) AB = ( 

3×5 + 4×1 

2×5 + 4×1 ) 

= ( 

19 

14 ) 

(b) |A| = (3 × 4) – (2 × 4) 

= 4 

( 

4 – 4 

) 

1 

– – –4 

1 

A –1 4 

– 1 3 

= 

– 

= ( –2 – 

2 3 –4 – 

) 

4 

4 

2 (a) The elements in the corresponding positions in A 

and M add up to zero. 

M = ( – 3 2 

) 

– 1 

– 2 

1 0 

0 1 

(b) ( ) is the identity matrix so N is the inverse of A. 

|A| = (3 × 2) – (1 × – 2) 

= 8 

A –1 = N 

( 

5 6 9 

C 

) 

1 1 –4 –4 

– 1 3 –8 –8 

L 

K 

W 

10 

12 – y y 18 – y 

9 

 

F 

 

 

16 


Model answers 

3 (a) ( 

1 0 

) + ( 

3 

– 2 

) = ( – 2 

x 0 

) 

– 2 3 

– 6 3 

So x + – 2 = – 6 

x = – 4 

(b) |R| = 4 × 3 – ( – 6 × – 2) 

= 0 

The determinant of R is zero. 

(b) |Q| = 3 × 3 – (–2 × –2) 

= 5 

Q –1 = 

( 

) 

4 (a) XY = Z 

So 3p – 8 = q 1 

and – 4p + 4 = 16 2 

From 2 

– 4p = 12 

p = – 3 

Substituting in 1 

3 × – 3 – 8 = q 

q = – 17 

(b) |X| = 3 × – 1 – ( – 4 × 2) 

= 5 

) 

5 (a) The number of columns in A is not the same as the number of rows in B. 

X –1 = 

(– – 

(b) (i) B 2 = ( 

3×3 + – 4× – 2 3× – 4 + – 4×5 

) 

2×3 + 5× – 2 – 2× – 4 + 5×5 

= ( ) 

16 33 

(ii) |B| = 3 × 5 – ( – 2 × – 4) 

= 7 

B –1 = 

( 

3 –5 

4 

3 –5 

2 –5 

2 –5 

1 2 – –5 

5 

4 3 –5 – 

5 

) 

5 –7 

17 – 32 

3 – 

7 

2 –7 

4 –7 

Properties of triangles and other shapes (page 64) 

1 (a) Five equal sides so centre will split in to five equal angles. 

360° 

x = –––– 

5 = 72° 

(b) As the pentagon is regular, all the lines from the vertices to the centre are equal. 

So triangle ABC is isosceles. 

2 x = 53° (corresponding angles are equal) 

y = 43° + 53° = 96° (exterior angle = sum of opposite interior angles) 

360º 

3 Hexagon: exterior angle = ––– 6 = 60° (sum of exterior angles = 360°) 

interior angle = 180° – 60° = 120° (exterior + interior = 180°) 

n-sided polygon: interior angle = 120° + 48° = 168° (hexagon + 48°) 

exterior angle = 180° – 168° = 12º (exterior + interior = 180°) 

360 

number of sides, n = ––– 

12 = 30 (sum of exterior angles = 360°) 

4 Interior angles of triangle are 64°, 64° and 52°. (isosceles triangle) 

So x = 180° – 64° = 116°, y = 180° – 52° = 128°. 

5 x = 96° – 27° = 69° (exterior angle = sum of interior angles) 

Both base angles are y (isosceles triangle) 

2y = 27° (exterior angle = sum of interior angles) 

y = 13·5° 


17

6 In triangle ABC and triangle ECD 

BC = CD (given) 

angle ABC = angle EDC (alternate angles) 

angle ACB = angle DCE (vertically opposite) 

So triangle ABC is congruent to triangle EDC 

7 (a) Angle ADE = angle ABC (corresponding angles, DE parallel to BC) 

Angle AED = angle ACB (corresponding angles, DE parallel to BC) 

Angle A is common 

Triangles are similar as corresponding angles are equal. 

(b) 

AD 1 

––– = – , BC = 4 × DE = 12 cm 

AB 4 

Model answers 

Pythagoras and trigonometry (page 67) 

1 (a) BC 2 = BD 2 + DC 2 – 2 × BD × BD × cos160° 

= 10 2 + 8 2 – 2 × 10 × 8 × cos160° 

= 314·35 

BC = 17·7 m 

(b) PA 2 = PD 2 + DA 2 

= 4 2 + 8 2 

= 16 + 64 

= 80 

PA = 8·94 m 

2 (a) AB 2 = AC 2 + BC 2 – 2AC × BC × cos15° 

= 8 2 + 3 2 – 2 × 8 × 3 × cos15° 

= 64 + 9 – 48 cos15° 

= 26·64 

AB = 5·16 cm or 5.2 cm 

1 

(b) Area = –2 × BC × AC × sinC 

1 

= –2 × 8 × 3 × sin15° 

= 3·11 cm 2 or 3.1 cm 2 

3 (a) BC 2 = 15 2 + 23 2 – 2 × 15 × 23 × cos64° 

= 451.52 

BC = 21.2 km 

1 

(b) Area = – 

2 × 15 × 23 × sin64° 

= 155 km 2 

Properties of circles (page 70) 

1 x = 36° (angles in the same segment) 

y = 2 × 36° = 72° (angle at centre = twice angle at circumference) 

z = 90° – 28° = 62° (radius at right angles to tangent) 

2 (a) Angle ACO = 34° (isosceles triangle) ⇒ angle OCB = 90° – 34° = 56° (angle in a semi-circle) 

(b) Angle CBO = 56° (isosceles triangle) ⇒ angle CBT = 180° – 56° = 124° 

(c) Angle BCT = 90° – 56° = 34° (tangent and radius) ⇒ angle CTA = 180° – 34° – 124° = 22° 

or angle COB = 2 × 34° = 68° (angle at centre = twice angle at circumference) ⇒ angle 

CTA = 180° – 90° – 68° = 22° 

3 a = 90° – 63° = 27° 

b = 63° (isosceles triangle) 

c = 180° – 63° – 63° = 54° 

d = 180° – 90° – 54° = 36° 

4 (a) Angle ADE = x° (angles in the same segment) 

(b) Angle DAE = 180° – 90° – x° = 90°– x° (angle in a semi-circle) 

(c) Angle EAB = 90° – (90° – x°) = 90° – 90° + x° = x° (tangent and radius) 

(d) Angle AOE = 2 × x° = 2x° (angle at centre = twice angle at circumference) 

5 (a) Angle ACB = 90° (angle in a semi-circle) 

(b) Angle AOB = 2x° (angle at centre = twice angle at circumference) 

Angle ATB = 360° – 90° – 90° – 2x° (since angle TAO = angle TBO = 90°) 

= 180° – 2x° 

18 


Model answers 

3D shapes (page 73) 

1 

1 Area of end = 0·8 × 0·6 + –2 × π × 0·4 2 = 0·73 m 2 

Volume = 0·73 × 1·2 = 0·88 m 3 

1 

2 Area of end = –2 × 4 × 3 = 6 cm 2 

V = A × L 

L = 42 ÷ 6 = 7 cm 

1 –2 

Surface area = 2 ( × 4 × 3) + (5 × 7) + (3 × 7) + (4 × 7) = 96 cm 2 

1 – 

3 

3 × π × 3·5 2 × 11 + ( × π × 3·5 3 ) = 230·9 cm 3 

4 Base diagonal = (4 2 + 4 2 ) = 5·66 cm 

–3 Volume = × 4 × 4 × 5·29 = 28·2 cm 3 

Perpendicular height = 

1 

(6 2 1 

– ( –2 × 5·66) 2 ) = 5·29 cm 

5 (a) 40 ÷ 1·25 = 32 cm (b) 32 000 × (1·25) 3 = 62 500 cm 3 

6 (a) 2cm (b) 

π –3 (4 2 × 10 – 2 2 140 

× 5) = ––– π = 147 cm 3 

3 

Transformations and coordinates (pages 80–81) 

1 (a) ( – 0·5, 2) 

(b) BC 2 = 3 2 + 5 2 = 34 

BC = 34 = 5·83 units to 2 d.p. 

2 (a) Enlargement centre (0, 0), scale factor 0·5. 

(b) Rotation through 180° about (4, 1). 

3 (a) and (d) 

4 Volume scale factor = 0·5 

Length scale factor = 3 0·5 

Height = 24 × 3 0·5 = 19·0 cm 

5 Length scale factor = 24 ÷ 16 = 1·5 

(a) Armour plate area = 9 × 1·5 2 = 20·3 cm 2 

6 (a) ( 

a b 

)( – 4 – 8 – 6 

) = ( 

4 8 6 

c d 

2 3 4 ) 

– 2 3 4 

– 4a + 2b =4 

– 8a + 3b = 8 so by inspection a = – 1 and b = 0 

– 6a + 4b = 6 

also 

– 4c + 2d = 2 

– 8c + 3d = 3 so by inspection c = 0 and d = 1 

– 6c + 4d = 4 

so ( 

a b 

) = ( – 1 0 

c d 0 1 ) 

(b) ( 

a b 

)( 

4 8 6 

) = ( – 2 3 4 

c d 2 3 4 

) 

– 4 – 8 – 6 

4a + 2b = 2 

8a + 3b = 3 so by inspection a = 0 and b = 1 

6a + 4b = 4 

also 

4c + 2d = – 4 

8c + 3d = – 8 so by inspection c = – 1 and d = 0 

6c + 4d = – 6 

so ( 

a b 

) = ( 

0 1 

c d 

) 

– 1 0 

(b) Weight = 270 × 1·5 3 = 911 g 

(c) From the diagram, the transformation represented by matrix P followed by the transformation represented by 

matrix Q gives a reflection in the line y = x. 

0 1 

– 1 0 0 1 

– 1 0 0 1 1 0 

0 1 

– 1 0 0 1 

– 1 0 0 1 1 0 

or ( )( ) = ( ) , which is the matrix that produces a reflection in the line y = x. 

(d) QP = ( )( ) = ( ) , reflection in line y = x 

– 1 0 0 1 

0 1 

– 1 0 

1 4 –2 – 

3 

– 1 

0 

– 1 0 

PQ = ( )( ) = ( ) , reflection in line y = – x. 


19

PQ applied to T 3 

give ( )( ) = ( ) and this is a reflection in the line y = – x. 

The inverse of a reflection in the line y = x is itself a reflection in the line y = x. 

7 (a) Reflection in the line x + y = 0, i.e. y = – x 

(Hint: if in doubt, join corresponding points on each triangle and bisect these lines.) 

(b) A = ( 

3 4 3 

) B = ( – 2 2 5 

2 2 5 

) 

– 3 – 4 – 3 

a b 3 4 3 

( )( ) = ( – 2 2 5 

c d 

) 

2 2 5 – 3 – 4 – 3 

3a + 2b 4a + 2b 3a + 5b 

( ) = ( – 2 2 5 

) 

3c + 2d 4c + 2d 3c + 5d – 3 – 4 – 3 

Comparing elements: 

3a + 2b = 2 

4a + 2b = 2 

so a = 0 and b = 1 

Also 3c + 2d = – 3 

4c + 2d = – 4 

so c = – 1 and d = 0 

Hence the transformation matrix = ( 

0 1 

) 

– 1 0 

Model answers 

(c) By comparing corresponding lengths, the scale factor is – 2. By joining corresponding points with straight lines 

and finding the intersection of these lines, the centre is at (1, 5). 

2 1 3 4 3 

(d) (i) ( 0 3)( 2 2 5) = ( 6 6 15) 

1 1 

(ii) Area = (10 – 8) × – (15 – 6) = – 

2 

2 × 2 × 9 = 9 square units 

(iii) ( 

a b 

)( 

8 10 11 

) = ( 

3 4 3 

c d 6 6 15 2 2 5 ) 

8a + 6b 10a + 6b 11a + 15b 

( ) = 3 4 3 

( ) 

8c + 6d 10c + 6d 11c + 15d 2 2 5 

Comparing elements: 

8a + 6b = 3 

10a + 6b = 4 

1 

so 2a = 1, thus a = – 

2 

3 – 4 

and b = = 

– 1 

–––––– ––6 

6 

Also 8c + 6d = 2 

10c + 6d = 2 

1 

so c = 0 = d = – 

3 

8 (a) (i) A shear along the x-axis direction, shear factor = – 2 (or a shear with the x-axis as the invariant line) 

(ii) A shear along the y-axis direction, shear factor = 2 (or a shear with the y-axis as the invariant line) 

(b) ( 

a b 

)( 

0 3 5 2 

) = ( 

0 3 3 0 

c d 0 0 1 1 0 0 2 2 ) 

3a = 3 thus a = 1 

5a + b = 3 thus b = – 2 

Also 

5c + d = 2 thus c = 0 

2c + d = 2 thus d = 2 

so the matrix is ( 

1 

– 2 

0 2 ) 

9 (a) (i) A parallelogram 

0 

– 1 

– 2 3 4 

– 1 0 

– 4 – 8 – 6 

1 –2 

8 10 11 

1 1 1 3 3 1 

( )( ) = 2 4 7 5 

( ) 

0 1 1 1 4 4 1 1 4 4 

(ii) A shear with the x-axis as the invariant line and shear factor 1 

(b) (i) The original rectangle 

1 

( – 1 2 4 7 5 

)( ) = 1 3 3 1 

0 1 

( ) 

1 1 4 4 1 1 4 4 

– 4 8 6 

– 2 – 3 – 4 

(ii) A shear with the x-axis as the invariant line and shear factor = – 1 

20 


Model answers 

Vectors (page 84) 

– 2 

– 8 

1 (a) (4, 6) (1, 3·5) (0, – 0·5) (b) $ BC = ( ), 

2 (a) (i) 2b (ii) a – b 

1 

(iii) b + – 

2a 

(b) 

1 – 

3 (a + 2b) (c) Collinear, and OM = 1·5 × OH. 

3 (a) 

y 

4 

$ 

– 1 

– 4 

MN = ( ) (c) MN is parallel to BC and BC = 2 × MN. 

2 

B 

C 

– 8 – 6 – 4 – 2 

A 

O 2 4 6 8 

– 2 

x 

– 4 

D 

(b) (i) 

$ 14 

AC = ( ) (ii) DB $ = ( – 1 

2 

5) 

(b) | AC| $ = (14 2 + 2 2 ) 

= 200 

= 14.1 

4 (a) 

y 

Y 

4 

2 

Z 

– 8 – 6 – 4 – 2 

O 

O 2 4 6 8 

– 2 

X 

– 4 

x 

(b) $ 10 

YX = ( ) 

– 6 

(c) OZ $ 4 

= ( 2) 

Measures (page 86) 

1 1·4 m = 1400 mm, 0·9 m = 900 mm 

(a) Volume = 3·5 × 1400 × 900 = 4 410 000 mm 3 

(b) Volume = 4 410 000 ÷ 1 000 000 000 m 3 = 0·004 41 m 3 

2 Mass = volume × density = 600 × 15 g = 9000 g = 9 kg 

v – u 

3 a = ––––– 

t 

To find maximum value of a, use maximum v, minimum u and minimum t. 

Maximum v = 30·35, minimum u = 17·35, minimum t = 2·55. 

30·35 – 17·35 13 

Maximum a = –––––––––––––– = ––––– = 5·098 039 2 = 5·10. 

2·55 2·55 

4 Population density = population ÷ area 

Lower bound = minimum population ÷ maximum area = 2·55 × 10 7 ÷ (5·85 × 10 5 ) = 43·589 = 43·6 people/km 2 

Upper bound = maximum population ÷ minimum area = 2·65 × 10 7 ÷ (5·75 × 10 5 ) = 46·086 = 46·1 people/km 2 

5 1·2 billion litres = 1·2 × 10 9 litres, 390 m 3 = 390 000 litres = 3·9 × 10 5 litres 

Greatest possible number of swimming pools = maximum volume of drink ÷ minimum volume of a pool 

1·25 × 10 9 

= ––––––––––– = 0·324 67 × 10 4 = 3246·7 

3·85 × 10 5 

So 3246 pools. 


21

Constructions (page 88) 

1 AX = 4·8 cm 

2 MZ = 2·9 cm 

Model answers 

2D shapes (page 90) 

1 

1 A = ( – 

2 × 5 × 4·6) + (0·7 × 6) = 15·7 cm 2 

2 C = π × d 

40 200 

d = ––––––– = 12 796 km 

π 

r = 6398 km 

1 –2 

3 (24 × 28) – ( × π × 12 2 ) = 445·8 m 2 

4 (39 × 30) – 8( × 16 × 12) = 402 m 2 

144 

5 × π × 14·5 2 1 

––– 

– – 

360 

2 × 14·5 × 14·5 × sin144° = 202·4 m 2 

Loci (page 92) 

All diagrams are shown full size. 

1 

1 –2 

A 

B 

Scale 1 cm to 10 km 

2 

A 

Fence 

C 

Hedge 

House 

B 

Fence 

D 

Scale 1 cm to 2 m 

22 


Model answers 

3 

Gate 

Toilets 

The pavilion 

can be built 

on this section 

of the line. 

Gate 

Scale 1 cm to 50 m 

Processing and representing data (pages 96–97) 

1 (a) (i) 

Weight (w kg) w 5 w 10 w 15 w 20 w 25 w 30 w 35 

Cumulative frequency 9 22 43 60 70 78 80 

(ii) 

80 

Cummulative frequency 

60 

40 

20 

Upper quartile 

Median 

Lower quartile 

0 

(c) 

5 10 

15 20 25 30 

35 

Weight (kg) 

2 

(b) (i) 14 to 14·8 kg (ii) 20 – 9 = 11 (10·5 to 11 is acceptable) 

(c) Reading off from 8 gives 17. 80 – 17 = 63 (63 to 65 is acceptable) 

Weight (w grams) Number of apples Mid-interval value w × f 

50 w 60 23 55 1265 

60 w 70 42 65 2730 

70 w 80 50 75 3750 

80 w 90 20 85 1700 

90 w 100 15 95 1425 

Total 150 10 870 

Mean = 10 870 ÷ 150 = 72·5 kg to 3 s.f. 


23

Model answers 

3 (a) (i) 4 t 5 

(ii) 

100 

80 

Frequency 

60 

40 

20 

0 

1 2 

3 

4 

5 6 7 

Time (minutes) 

1·5 × 5 + 2·5 × 25 + 3·5 × 45 + 4·5 × 82 + 5·5 × 33 + 6·5 × 10 843 

(b) ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– = ––– = 4·215 minutes (or 4·22) 

5 + 25 + 45 + 82 + 33 + 10 

200 

(c) (i) 

Time (t minutes) t 1 t 2 t 3 t 4 t 5 t 6 t 7 

Number of tracks 0 5 30 75 157 190 200 

(ii) 

200 

Cumulative frequency 

150 

100 

Median 

50 

0 

1 2 

3 4 

5 

6 

Time (minutes) 

7 

24 


Model answers 

(d) (i) 

143 

4·3 minutes (4·2 to 4·5 is acceptable) (ii) 200 – 57 = 143, –––– = 0·715 (0·69 to 0·75 is acceptable) 

200 

4 

Age (x) in years 0 x 5 5 x 15 15 x 25 25 x 45 45 x 75 

Number of patients 14 41 59 70 16 

Frequency density 2·8 4·1 5·9 3·5 0·53 

6 

5 

Frequency density 

4 

3 

2 

1 

0 

10 20 30 40 

50 

60 70 80 

Age (years) 

5 

Time in minutes (t) 0 t 1 1 t 3 3 t 5 5 t 10 10 t 20 

Number of calls 12 32 19 20 15 

Frequency density 12 16 9·5 4 1·5 

20 

Frequency per one minute interval 

15 

10 

5 

0 

0 

5 10 15 20 

Time (t minutes) 

6 Since area = frequency, missing frequencies are 20 × 1·4 = 28, 15 × 8·8 = 132, 15 × 7·2 = 108 and 30 × 2 = 60. 

Time (t minutes) 0–20 20–30 30–45 45–60 60–90 

Frequency 28 60 132 108 60 


25

Probability (page 101) 

1 (a) 

Ellen 

Schweta 

Model answers 

0 . 8 

Score 

0 . 7 

Score 

0 . 2 

Miss 

0 . 8 

Score 

0 . 3 

Miss 

0 . 2 

Miss 

(b) (i) 0·7 × 0·8 = 0·56 

(ii) 0·7 × 0·2 + 0·3 × 0·8 = 0·38 

6 4 

2 1 1 3 

2 (a) –– and –– on first set of branches, – , – , – , – on second 

10 10 

3 3 4 4 

set with ‘goes’, ‘not go’ or other suitable labels. 

6 2 2 

(b) (i) –– × –3 = –5 or equivalent 

10 

4 3 3 

(ii) –– × – = –– or equivalent 

10 4 10 

6 5 1 

3 (a) –– × –– = – or equivalent 

15 14 7 

4 3 5 4 31 

(b) (a) + –– × –– + –– × –– = ––– 

15 14 15 14 105 

74 

(c) 1 – (b) = ––– 

105 

1 1 

4 (a) –6 × –6 = 36 

5 –6 

1 –– 

5 – 

6 

(b) (i) × × ... n times = ( ) n 

(ii) P(at least 1 six) = 1 – P(no six) 

5 – 

6 

5 – 

6 

= 1 – ( ) n 

5 (a) 

Box A 

Box B 

50% 

Red 

Red 

15% 

Blue 

20% 

35% 

Black 

50% 

Blue 

50% 

15% 

Red 

Blue 

35% 

Black 

30% 

50% 

Red 

Black 

15% 

Blue 

35% 

Black 

(b) 0·2 × 0·5 = 0·1 

= 10% 

(c) 0·2 × 0·5 + 0·5 × 0·15 + 0·3 × 0·35 

= 0·1 + 0·075 + 0·105 

= 0·28 

= 28% 

26

Answers to all pages - Hodder Plus Home

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