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Cambridge Essentials Mathematics Core 8 A1.1 Homework Answers

Original material © Cambridge University Press 2009 1

A1.1 Homework Answers

1 a Pattern

Position 1 2 3 4 5

Term 1 5 9 13 17

b Pattern

Position 1 2 3 4 5

Term 4 6 8 10 12

c Pattern

Position 1 2 3 4 5

Term 17 14 11 8 5

2 a add 4 or + 4 b add 2 or + 2 c subtract 3 or − 3



3 a i + 3 ii 18, 21, 24

b i + 4 ii 22, 26, 30

c i − 2 ii 1, −1, −3

d i − 4 ii −16, −20, −24

e i + 14

ii 74

, 2, 94

f i – 1.5 ii −2.5, −4, −5.5

4 a i − 2 ii 13

b i − 3 ii 5

5 a i + 6 ii −1

b i + 9 ii 7

c i − 23

ii 0

d i − 0.05 ii 2.30

6 a i + 3 ii 4 7 10 13

b i + 5 ii –2 3 8 13 18

c i − 4 ii 7 3 −1 −5 −9 −13

d i + 0.2 ii 5 5.2 5.4 5.6 5.8 6

e i + 9 ii 6 15 24 33

f i + 6 ii 0 6 12 18 24 30

g i + 3 ii –3 0 3 6 9 12 15 18 21 24

h i + 7 ii 9 16 23 30 37 44

i i + 1 12 ii 1

2 2 3 1

2 5

j i + 13

ii –1 32

− 31

− 0 31 2

3




Position 1 2 3 4 5 1 a

Term 4 5 6 7 8

Position 1 2 3 4 5 b

Term 3 6 9 12 15

Position 1 2 3 4 5 c

Term −3 −1 1 3 5

Position 1 2 3 4 5 d

Term −1 −5 −9 −13 −17

Position 1 2 3 4 5 e

Term 37

38 3

39=

310

311

f Position 1 2 3 4 5

Term

37

38 3

39=

310

311

g Position 1 2 3 4 5

Term 1.5 4 6.5 9 11.5

2 The term values are the same. This is because × 13

is the same as ÷ 3.



3 a Position × 8 Term

b Position × 2 + 3 Term

c Position × 3 − 3 Term

d Position × −3 + 1 Term

e Position × 4 − 11 Term

f Position × 0.5 + 3.5 Term

4 a n + 5 b 4n

c 3n + 2 d −2n + 5 or 5 − 2n

e 3n + 4 f

5n − 3

5 a 4, 8, 12, 16 b 3, 5, 7, 9

c −3, −6, −9, −12 d −1, −3, −5, −7

e 14

, 12

, 34

, 1 f 132

, 4, 142

, 5

g 3, 2, 1, 0 h 243

, 143

, 4, 233

6 a i + 6 ii 6n

b i + 4 ii 4n + 1

c i + 6 ii 6n − 5

d i + 3 ii 3n − 5

e i + 15

ii 15

n − 25

7 Jacqui is correct. The term-to-term rule becomes the multiplier part of the nth term rule,

so it must be 5n (then + or − anything is optional).

8 a 4n + 3 b 10n − 1

c 3n − 7 d −5n + 17 or 17 − 5n

Cambridge Essentials Mathematics Core 8 A2.1 Homework 1 Answers


A2.1 Homework 1 Answers

1 a 4 b 4 c 1

d 13 e 7 f 21

2 a Multiply m by 3 then add n.

b Square p and subtract from 12.

c Divide x by y and add to 2.

d Add d to e and divide the answer into 12.

3 c a – b ÷ c

4 a Tom is not right because in m + n ÷ 2, only the n is divided by 2,

not the sum of m + n.

b Tom should have written 2

nm + or (m + n) ÷ 2

5 b 2

48 −

6 5

7 a 9m b 7q + 7p c 3x + 5y d 7 + 6d

8 a 6x + 2y b 2x + 2y – 4 c 4a + b




1 a m4 b t2 w3 c p5 q2 d ab3 c4

2 a x × x × x b b × b × c × c × c c d × d × d × d × e

d m × n × n × n × p × p

3 a t × t × t b t × t × t × t c t × t × t × t × t × t × t d t7

4 a 12t + 8 b 6 – 12m c 10x + 5y

d px + qx e 4e + 3ef f 2h + h2

5 a 11x + 10 b am + 5an c 4pq + 7pr d ab + 2ac

6 a 21p + 3 b 8m – 16

7

8 a 28 – 2x b 4(7 – x) + 2x c 14 + 2(7 – x)

d From b, area = 4(7 – x) + 2x

= 28 – 4x + 2x

= 28 – 2x

From c, area = 14 + 2(7 – x)

= 14 + 14 – 2x

= 28 – 2x

All answers for the area in parts a, b and c are equivalent to 28 – 2x.




1 a p = 2 b x = 9 c b = 4

d k = 6 e q = −2 f t = 4

2 a x = 2 b t = 3 c m = 5 d g = 7

3 a 3x + 10 + x + 20 + x = 180°

5x + 30 = 180° Angles are 100°, 50° and 30°

x = 30°

b 4a – 12° + 3a – 3° + 90° = 180°

7a + 75° = 180° Angles are 48°, 42° and 90°

a = 15°

c p + 20° + 3p – 15° + p + 55° + 2p + 20° = 360°

7p + 80° = 360° Angles are 60°, 105°, 95° and 100°

p = 40°

4 a (x + 6) cm + (x + 6) cm + (x – 3) cm = 30 cm

b x = 7 cm c AB = 13 cm, BC = 4 cm

5 a 2(4K – 3) cm + 2K cm = 24 cm

b k = 3 cm c 9 cm and 3 cm

6 a (y + 2) cm + (2y + 3) cm + 2y cm = 40 cm

b y = 7 cm c 9 cm, 7 cm, 17 cm, 7 cm




1 a i 5p + 4 = 39 ii p = 7

b i 3(m + 6) = 27 ii m = 3

c i 2(2x – 7) = 22 ii x = 9

2 a 3(k + k + 4) = 72

b Jack is 10 years old.

3 a 9m + 325 = 2800

b The mass of one packet of biscuits is 275 g.

4 a = 2

5 y = 5

6 a n + 2, n + 3

b n + n + 1 + n + 2 + n + 3 = 4n + 6 = 42

n = 9

7 i m + 6 = 14 m = 8 side lengths are: 14 cm, 14 cm, 20 cm

ii 3m – 4 = 14 m = 6 side lengths are: 14 cm, 12 cm, 14 cm

iii m + 6 = 3m – 4 m = 5 side lengths are: 14 cm, 11 cm, 11 cm




1 a h = 11 b h = 12 c h = 11

d h = 1 e h = 30 f h = 5

2 a Card S, (4 + 1)2 = 25

b Card T, 4 – 4 = 0

c Card Q and Card T.

3 a n = 95 b n = 215 c n = 1015

4 a 190 minutes or 3 hours 10 minutes

b 290 minutes or 4 hours 50 minutes

5 £275

6 a A = p2 – q2

b i 300 cm2 ii 1600 cm2




1 a y = x + 6 b y = 4x c y = 2x + 3

d y = 5x e y =

4x – 3 f y = 3x + 4

g y = 3x – 7 h y =

5x × 3 = 3

5x i y = 2x – 5

j y = (x – 5) × 2 = 2(x – 5)

2 a i The rule is × 4 ii Input → × 4 → Output

b i The rule is × 3 then + 2 ii Input → × 3 → + 2 → Output

c i The rule is × 5 then – 5 ii Input → × 5 → – 5 → Output

3 a y = 4x b y = 3x + 2 c y = 5x – 5

4 a i The rule is + 4 ii Input → + 4 → Output

iii y = x + 4 iv P = 6 Q = 14

b i The rule is × 3 then + 4 ii Input → × 3 → + 4 → Output

iii y = 3x + 4 iv R = 16 S = 7

5 a Rule is ‘Multiply by 3 and then add 1’. x → 3x + 1

b Rule is ‘Multiply by 45

’ or ‘Multiply by 4 then divide by 5’.

x → 45x

c Rule is ‘Multiply by 0.5 then add 2’ or ‘Divide by 2 then add 2’.

x → 0.5x + 2

6 a y = x × 2 × 5 y = 10x b y = x + 5 – 2 y = x + 3

c y = x ÷ 4 × 8 y = 2x d y = x ÷ 5 × 5 y = x




1 a i Input → + 2 → Output

ii Input Output 1 3 2 4 3 5 4 6 5 7

b i Input → – 4 → Output

ii Input Output 1 –3 2 –2 3 –1 4 0 5 1

c i Input → × 3 → –1 → Output

ii Input Output 1 2 2 5 3 8 4 11 5 14

iii

2 a Parallel means the lines never meet; they remain the same distance apart.

b The lines are parallel because the output value is either made by just adding or by just subtracting a constant. In part c, multiplication is involved.

iii

iii



3 a i Input → × –1 → + 1 → Output

b i Input → × –3 → Output

c i Input → × –2 → + 5 → Output

ii Input Output 1 0 2 –1 3 –2 4 –3 5 –4

ii Input Output 1 –3 2 –6 3 –9 4 –12 5 –15

ii Input Output 1 3 2 1 3 –1 4 –3 5 –5

iii

iii

iii



d i Input → × –3 → + 4 → Output

4 Dwayne is wrong – the outputs are the same for an input of 1 but not for other values.

One way to show this would be to draw input/output tables:

Input → × 2 → + 3 → Output y = 3x + 2

Input Output Input Output

1 5 1 5

2 7 2 8

3 9 3 11

4 11

4 14

Instead of drawing input/output tables, you could say that × 2 means the gap between

terms is + 2 but × 3 means the gap between terms is + 3, so the two expressions cannot be

the same.

ii Input Output 1 1 2 –2 3 –5 4 –8 5 –11

iii




11 aa x y

–3 –1 –1 1 0 2 1 3 2 4

2 a i The rule for the y-coordinate is: x-coordinate + 2.

ii y = x + 2

b i The rule for the y-coordinate is: x-coordinate × 3 then –2.

ii y = 3x –2

c i The rule for the y-coordinate is: x-coordinate × 2.

ii y = 2x

d i The rule for the y-coordinate is: x-coordinate × –1 then ÷ 2, or × −21 .

ii y = −21 x

e i The rule for the y-coordinate is: x-coordinate × –2 then + 3.

ii y = –2x + 3

f i The rule for the y-coordinate is: x-coordinate × 31 then – 2.

ii y = 31 x – 2

b x y –1 –5 0 –2 1 1 2 4 3 7

c x y –2 –4 –1 –2 0 0 1 2 2 4

d x y –4 2 –2 1 0 0 2 –1 4 –2

f x y –3 –3 0 –2 3 –1 6 0 9 1

e x y –1 5 0 3 1 1 2 –1 3 –3



3 a x y –2 2 –1 3 0 4 1 5 2 6

4 a b

b x y –2 –10 –1 –5 0 0 1 5 2 10

d x y –2 –9 –1 –7 0 –5 1 –3 2 –1

c x y –2 –5 –1 –4 0 –3 1 –2 2 –1

e x y –2 2

–1 221

0 3

1 321

2 4

f x y –2 6 –1 5 0 4 1 3 2 2



c d

e f




1 a i x = 1

ii y = 2

iii x = –3

iv y = –3

v x = 4

b x-coordinate of point on line i is always 1.

c y-coordinate of point on line ii is always 2.

d x-coordinate of point on y axis is always 0.

2 (1, 4) is also on line y = x + 3.

(0, 3) is on line y = x + 3.

(1, –1) is on lines y = x –2 and y = –x.

(2, 5) is on lines y = x + 3 and y = 3x –1.

(2, 0) is on line y = x –2.

3 a Line v (y = 5x) is steepest; the value of m (gradient) is the biggest.

b Line i (y = 3x + 4), line iii (y = 3x – 5) and line vi (y = 3x) all have the same steepness.

c i The word parallel describes lines which have the same steepness.

ii The word ‘gradient’ is used to mean ‘steepness’ of a line.

d Line iv (y = 2x + 3) crosses the y-axis at point (0, 3).

e Line iii (y = 3x – 5) crosses the y-axis at (0, –5).

f Lines v (y = 5x) and vi (y = 3x) cross the y-axis at (0,0).

The value of c in y = mx + c is 0 for both these lines.




1 a

b

2 a Length of larger rectangle is 4 + n. Width of smaller rectangle is 3 – m.

b 4m + 3n

3 a i 20 cm2 ii 5(k – 1) cm2 or 5k – 5 cm2 iii k(k – 1) cm2 or k2 – k cm2

b k2 + 8k + 15 cm2

4 a Width is 3d – 3 and length is c + 7.

b 8c + 10d + 10

5 a They are both correct. Paul has doubled each side and then added.

Mena has added the sides together and then doubled.

Paul : 2(2x + 4) + 2(6 – x) Mena 2(2x + 4 + 6 – x)

= 4x + 8 + 12 – 2x = 4x + 8 + 12 – 2x

= 2x + 20 = 2x + 20

b i Length is 4x + 8 cm and width is 3 – 12

x cm.

ii 7x + 22

c i Length is12 – 2x cm and width is x + 2 cm.

ii 28 – 2x




1 a 1539 to 1542 b 30 km c 60 km/hour

d 15 km e 30 km/hour f S = 60t

g J = 30t + 20

h 60t = 30t + 20

t = 23

so they pass 23

of an hour after 1500 which is 1540.

2 a

b 58 g c w = 25t + 40

3 Alternative suggestions for the missing labels may also be correct.

A Jane is filling the bath B Jane turns off the taps

C Jane gets in the bath D Jane is sitting in the bath

E Jane runs in extra hot water F Jane pulls out the plug and the bath

empties

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