Cambridge Essentials Mathematics Extension 7 A1.1 …essentials.cambridge.org/media/CEMKS3_E7_A1_1_WS_ANS1.pdf · Cambridge Essentials Mathematics Extension 7 A1.1 Answers Original

Cambridge Essentials Mathematics Extension 7 A1.1 Answers

Original Material © Cambridge University Press 2008 1

A1.1 Answers

1 a 2 b 10 c 4 d 12

e 7 f 11 g 12 h 30

i 8 j 7 k 5 l 20

2 a 20 b 12 c 3 d 6

e 2 f 6 g 15 h 4

3 a 9 b 14 c 3 d 10

e 11 f 5 g 15 h 1

i 12 j 2 k 35 l 49

m 25 n 1 o 1 p 18

4 a 12 b 12 c 24 d 24

5

6 a 24 b 40 c 17 d 10

e 11 f 24 g 100 h 40

7 a 214 b 1 c

212− d

41

e 2130 f

214 g

212 h 1

Part a will always be equal to part f and part b will always be equal to part h.

8 2 and 8



9

10 a e = 5 b f = 8 c g = 33

d k = 27 e i = 25 f j = 20

g h = 80 h m = 25 i t = 11

11 a 15 b 2 c 48

d 49 e 18 f 3

g 21 h 96 i 96

12



13 x + y

14 x – 3

15 a 20, 22; increasing b 15, 12; decreasing c 412, 312; decreasing

d 88, 104; increasing e 398, 418; increasing f 35, 24; decreasing

g 20,24; increasing h 64, 128; increasing i 10 000, 100 000; increasing

j 16, 8; decreasing k 3, 1; decreasing l 8, 13; increasing

16 a 1, 6, 11, 16, 21, 26 b 4, 7, 10, 13, 16, 19 c 2, 5, 8, 11, 14, 17, 20

d 8, 13, 18, 23, 28, 33 e 40, 31, 22, 13, 4 f 52, 48, 44, 40, 36, 32

g 2.5, 3, 3.5, 4, 4.5, 5 h 10, 9.5, 9, 8.5, 8, 7.5, 7 i 16, 17.5, 19, 20.5, 22, 23.5

17

18

19 a

b 1, 4, 9, 16, 25, 36, 49, ...

c 144 is a term in the new sequence because it is a square number.



20 The 4th term is 3 × 4 = 12 The 10th term is 3 × 10 = 30

The 50th term is 3 × 50 = 150 The nth term is 3 × n

21 a 6, 7, 8, 9 b 11, 12, 13, 14 c 101, 102, 103, 104

d 0, 1, 2, 3 e 2, 4, 6, 8 f 5, 10, 15, 20

g 3, 13, 23, 33 h 11, 22, 33, 44 i 3.5, 6.5, 9.5, 12.5

j 4.5, 6.5, 8.5, 10.5 k 2.5, 5.5, 8.5, 11.5 l 10.5, 11.5, 12.5, 13.5

22 a i nth term = 3 × n − 1 ii 30th term = 3 × 30 − 1 = 89

b i nth term = 2 × n − 1 ii 30th term = 2 × 30 − 1 = 59

c i nth term = 2 × n ii 30th term = 2 × 30 = 60

d i nth term = 5 × n + 5 ii 30th term = 5 × 30 + 5 = 155

e i nth term = 4 × n − 3 ii 30th term = 4 × 30 − 3 = 117

f i nth term = 10 × n ii 30th term = 10 × 30 = 300

g i nth term = n + 6 ii 30th term = 30 + 6 = 36

h i nth term = 6 × n + 14 ii 30th term = 6 × 30 + 14 = 194

23 a 36 b 12, 24, 36, 48, 60, ... c 120

d nth term = 12 × n e 600

24 a 16 b 8, 16, 24, 32, … c 8 × n

d 160 e 50th border which is green.



A1.2 Answers

1 a i 15 ii 55 iii 35 iv 5 × ♠

b i 4 ii 6 iii 7 iv 4 ×

2 a

b

c

d

3 a

b

c

d

4 a

b

d

d

5 a x → x +3 b x → 4x c x → x − 6

d x → 11x e x → x + 100

6 x → x +12 x → 4x 2xx →



7 a 0.25

b 2xx → 25.0−→ xx 2xx →

8 There is more than one input with corresponding outputs and the rule must work in each case.

9

10

11

12 a

b

13 a 10 → 2 b 27 → 12 c 5 → 5 d 0 → 7

12 → 3 36 → 16 8 → 14 2 → 23

18 → 6 99 → 44 12 → 26 10 → 87

32−→

xx 2

4xx → 103 −→ xx 78 +→ xx

14 a

10xx →



14 b

c

15 a

b

c

d

16 a − 5 b × 9 c + 7.2 d ÷ 99

e − 4 f × 4 g ÷ 3

17 a

b

18 a

b

19 a

b

20 a y = x + 5 b y = 6x

21 a

b 35 °C c 68 °F



A2.1 Answers

1 a −8 b 1 c −2

d 14 e −12 f 40

g 39 h 10 i −1.5

2 a 8 b 15 c −7

d 5 e −6 f 16

3 a 4 b 5 c 9

d 5 e 2 f 3

4 w = 10, x = 4, y = 8 and z = 4

5 a 34 b 18 c 75

d 3 e −16 f 47

6 a When y is 4, 5y is 20.

b When y is 3, 10y2 is 90.

c When y is 3, 10y – 1 is 89 (there are other possible answers).

7 Any three expressions with value 20.

8 a x + 5 b x + 7 c x + 12 d 2x

e 3x f x g 3x h x

i 5x j −2x k 7x l 3x

9 a 39 + 27 + 1 = 39 + 1 + 27 = 67

b 28 + 75 + 25 + 11 = 28 + 100 + 11 = 139

c 327 – 98 – 2 + 30 = 327 – 100 + 30 = 257

d 579 + 86 – 79 = 579 – 79 + 86 = 586

e 49 + 78 + 51 + 65 + 22 = 49 + 51 + 78 + 22 + 65 = 265



10 The additions and subtractions can be done in any order. This allows us to choose the order that makes the calculation as simple as possible.

For instance, in 9c, –98 − 2 is the same as − 100. So we can write 327 − 100 + 30.

11 a 288 b 167 c 30

12 a 2x + 3 b 2x + 10 c 3x + 2

d x e 4x + 4 f 5x

13 a x + 2y b 2x c 2x + 2y

d x + y + 8 e 2x + 3 f x + 5y − 3

g 2xy h 5xy i 2xyz

14 a Like terms are terms with the same combination of symbols.

b 3x2 + 4y2 + 2x + 5y + 7

c The number 7 has no like terms. It remains unaltered.

15 a − 4m b −7k − 3m + 2 c −5k + 6 + 7m

d 11 − 9p + 8q e 5q − 6 + 4p f p − 3q − 17

g −4k − 8 + n h 5k + 2 + 7n i 11k − 5 + 5n

j 2x2 + 3x + 6 k 4y2 − y l 5a2 + 11a + 1

16 a P = 4x b P = 2x + 10 c P = 4y + 6

17 a P = 3y + 6 b P = 6g + 3 c P = 6t + 10

d P = 7m − 1 e P = 6u + 8v f P = 12h + 1

18

This is one possible answer – there are others,

19 a B and F b B and E c A and E d B and D

e D, E and F f B, D and F g 4x + 10

20 a s = 4 b s = 7 c s = 6 d d = 220



21 a F = 30 b F = 48 c F = 6.5 d m = 6

22 a m = 11 b m = 4.5 c m = 4

d The formula finds the number that is halfway between x and y.

23 a d = 45 b d = 80 c d = 180

d t = 7.7 seconds.

24 a Find 2v and 20 × f, then divide the two answers.

Your calculator will do this if you type 10 2x ÷ ( 20×0.8 ) = .

b 11.616

c 18.75

25

26 a

c The formula may be written as D + R − A = 2 or any equivalent form.



A2.2 Answers

1 a x + 2 b 3x c 5w d p − 9

2 a 6x + 5 b 4x − 3 c 3x − 7 d

8x + 1

e 5(m+ 4) f 10

)5( +w

3 a

b

c

d

e f

g

h

i

j

k

l

4 a x = 39 b x = 2

5 a Subtract 6 b Divide by 5 c Add 1 d Multiply by 2

e − 5 f ÷ 6 g × 3 h + 9

i ÷ 10 j − 3.4 k + 9.8 l × 7.2

6 a

b x = 44

7 a i

ii

iii x = 2

b i

ii

iii n = 12

c i ii iii y = 10



7 d i ii iii p = 6

e i

ii

iii t = 5.5

f i

ii

iii m = 9.5

g i

ii

iii y = 16

h i

ii

iii r = 150

i i

ii

iii a = 42

8 a Yes b m = 7.5

c p = 8 (Divide both sides by 5 and then subtract 12 from each side.)

9 a t = 7 b n = 20 c y = 5

d y = 2 e p = 2.5 f x = 10

g w = 2.5 h m = 1 i m = −1

10 a Rashid has subtracted 7 from the right hand side instead of adding 7.

b 6y – 7 = 11. Add 7 to both sides to give 6y = 18 then divide both sides by 6 to give the answer y = 3

11 a Katherine has divided part of the equation by 2 but she should have subtracted 7 from each side first.

b 2x + 7 = 16 (subtract 7 from each side) 2x = 9 (divide each side by 2) x = 4.5



A3.1 Answers

1 a 511−x b 2(x + 7) c

453 +x

d 3(10 − x) e 3x + 5

2 a 100 – n b 2n – 1 c 107+

n

d 2(n + 3) e 2

2n

3 a 6x + 5y b 20 − (6x + 5y) or 20 − 6x − 5y

c 7

56 yx + d 100w e

1003wx +

4 a Subtract 5 and multiply by 2

b Multiply by 5 and subtract from 21

c Divide by 3 and add 4

d Add 8 and divide the answer by 4

e Add 1 then multiply by 4 and subtract the answer from 35

f Subtract 3. Divide 18 by the answer.

5

6 a p + 10 = 2p + 6 b 2p + 1 = p +5 c p + 12 = 5p – 4

7 a 2x b 4x c x

d x e x f x + 3



8 a x + 8 − 8 b x − 9 + 9 c x − 6 + 6

d 4

4x e 4

4x + 5 − 5 f 7

9−x × 7 + 9

9 a y − 4 + 4 b y + 12 − 12 c y – 3 + 3

d 22t e

5y × 5 f

66y

10 a

b

c

11 a x = 170 b x = 16 c x = 48

d x = 15 e m = 26 f n = 3

12 a Step 1: add 4 to both sides.

Step 2: divide both sides by 2.

b i 6.5 ii 9 iii 28 iv 13

13 a 3x − 17 = 16 b 2(y + 12) = 68 c 5x + 112 = 126

3x = 33 y + 12 = 34 5x = 14

x = 11 y = 22 x = 70

d 3

5x = 15 e 11 = 619−x f 75 = 5(p – 32)

5x = 45 66 = x − 19 15 = p – 32

x = 9 x = 85 p = 47



14 a 80 b 4 c 6

d 280 e 19 f 6632

15 a 21 b 2

32 c

53

d 43 e

54 f

32

16 a 6x + 34 = 76 b x = 7 c 21 cm

17 a 3x + 60° = 180° b x = 40° c 40°, 60° and 80°

18 a 4x + 40 = 180° b x = 35° c 35°, 65° and 70°

19 a n – 7 b 2n – 7 = 29 c 18

d First tray has 18 chocolates, second tray has 11 chocolates.

20 a x = 3.5 b k = 6 c n = 11 d m = 17

e p = 1 f y = 7 g u = 6.5 h x = 12

i y = −4 j x = 6 k t = 8 l y = 2

21 a 3xy b 6xz c 4xy

d 2pq e 7pr − 2 f 2pq + pr

22 a i 60 ii 60 iii 60

b (xy)z = x(yz)

23 a Answers should all be the same but will depend on the values chosen by the pupil.

b When you multiply three numbers together it does not matter what order you do the multiplication. This means that xyz is the same as zyx or yzx or any other permutation.

24 a 5pqr b 7pqr c 9pqr

25 a 8p b 15q c 28t

d 24r e 36n f 30k

26 a 22g b 26w c 11v

d x e 19y f 8a



27 a 12x b 38m c 39h

d 29p e 30n f 3pq

28 Total area = area A + area B

5(x + 3) = 5x + 15

29 Coloured area = total area – white area

3(y − 4) = 3y − 12

30 24 × 19 = 24(20 − 1)

= 24 × 20 − 24 × 1

= 480 − 24

= 456

31 a 32 × 29 = 32 × (30 − 1) = 960 − 32 = 928

b 14 × 49 = 14 × (50 − 1) = 700 − 14 = 686

c 18 × 9.9 = 18 × (10 − 0.1) = 180 − 1.8 = 178.2

32 a 4x + 20 b 6n − 24 c 15 + 3t

d 120 − 10h e 56 + 8p f 99 − 9b

g 3x + xy h 5r − rt i kn + 2k

33 a When you expand brackets, the number outside the brackets should multiply each term inside the brackets. Jim has only multiplied the first term by this number when he really should multiply all the terms inside the brackets.

b ii 8y + 4 iii 21m – 14 iv 40n + 32 v 3 – 6k

34 a 6x + 3 b 12n − 8 c 20 − 10k

d 60 + 18j e 18 − 10e f 21d + 28

g 2ab + 5a h 9g − 3gt i 10z + 6yz

35 4(x + y + 5) = 4x + 4y + 20

36 a 5p + 5q + 15 b 6a + 3b − 18 c 48 − 4m + 8n

d 20h + 50 − 10k e 30 − 18c − 24d + 12e f 6np + 9nq − 9n



37 a 5x + 15 b 7n − 12 c 14 − 10t

d 8a + 10 e 9w + 7q f 7h + 15

38 a 23t + 11 b 7m + 20n + 18

c 19 + 12k + 6h d 11 + x + 9y

39 a x = 4 b x = 8 c x = 4 d x = 3

e x = 9 f x = 7 g x = 1 h x = 4

i x = 3 j x = 5



A3.2 Answers

1 a

b i y = x + 5 ii x → x +5

2 a

b i 13 ii 18 iii 35

c i y = 2x – 3 ii x → x – 3

3 a i y = 2x + 3 ii x → 2x + 3

b i 3 ii 5 iii 7

iv 23 v −1

c i 6 ii 30 iii 8.5

iv 23.5 v −3

d

4 a In the rule y = x + 3 × 2, only the 3 is doubled.



It should be the value of x + 3 that is doubled.

b y = 2(x + 3)

c y = −2, 0, 2, 4, 6, 8, 10 and 12.

d

5 a

b i

ii

iii

6 a i y = 10 − x ii y = x + 3 iii y = 2x − 5 iv y = 20 − 2x

b i

ii



6 b iii

iv

7 a i y = x – 1; x → x – 1

ii y = 2x – 10; x → 2x – 10

iii y = 6x + 5; x → 6x + 5

b i none. ii 10 iii −1

8 a x → x, x → 2(x − 2) or x → 8 − x.

b i x → 8 − x ii x → x iii x → 2(x − 2)

9 a 4 b 6



A4.1 Answers

1 a A(2, 4), B(2, 3), C(2, 1.5), D(2, −1.5)

b The x-coordinate is always 2.

c x = 2

2 a 1: x = −2 2: x = 1 3: x = 2.5

b x = 0 lies on top of the y-axis.

c x = −½ or x = −0.5

d x = 1.75

3 a A(−2, 4), B(1, 4), C(2.5, 4), D(4.5, 4)

b y = 4

c 1: y = 2 2: y = −1.5

d y = 0 is the line that lies on top of the x-axis.

e y = ¼ or y = 0.25

4 a, b

c The coordinates of the corners are (−3, 0), (−3, 3), (4, 3) and (4, 0).

The coordinates of the centre of the rectangle are (0.5, 1.5).



5

6 a b

c, d

e The lines are parallel. y = x passes through (0, 0) and y = x – 8 passes through (0,1).

6 f

All the lines are parallel to the line y = x but cut the y-axis at 2, 3 and −3 respectively.

g The line y = x + 10 is parallel to the line y = x but cuts the y-axis at 10. The line y = x − 8 is also parallel to y = x but cuts the y-axis at −8.



7 a

b

The line y = − x + 2 is parallel to y = − x but cuts the y-axis at 2. The line y = − x + 4 is also parallel to y = − x but cuts the y-axis at 4.

c Yes

d y = 10 − x is parallel to the line y = −x and cuts the y-axis at 10.

8 a y = 4 b x = −2 c y = x

d y = x + 1 e y = 2 − x

9 a y = x b y = 5 − x c y = 4

d y = x + 5 e y = −x

10 a, b

c (0, 0), (3, 3), (3, 6), (−3, 6), (−3, 3)



11 a

The line cuts the x-axis at (−7, 0) and cuts the y-axis at (0, 7).

b

The line cuts the x-axis at (−9, 0) and cuts the y-axis at (0, 9).

c

The line cuts the x-axis at (6, 0) and cuts the y-axis at (0, 6).

12 b

c, d

e

(2, 3)



13 a i (−2, 5) ii (0, 3) iii (1, 2) iv (5, −2)

d

13 b, c, e

f (−1, 4)

14 b

c

d The origin (0, 0).

e i 6 ii 1

f

g Graphs like y = 3x, y = 2x and y = ½ x all pass through the origin. The bigger the number in front of the x, the steeper the line.



15 a i (−2, 1) ii (0, 2) iii (2, 3) iv (6, 5)

b, d

c

e (4, 4)

16 b

c

d, e

f The coordinates of the corners are (−1, 3), (1, 5) and (2, 3). The area is 3 units2.



A4.2 Answers

1 a 23 m b 5 m c 59 m d 45 m

2 a 52% b i 70% ii 35% iii 44%

c 40 d 8 e 36

3 a 68 °F b 30 °C c 32 °F

d 23 °F e 37 °C

4 a If 6¼ miles is approximately 10 km then 12.5 miles is approximately 20 km.

b

c i 42 km ii 32 miles iii 69 km

d Convert 70 km to miles (43.73 or about 44 on the graph) and double the answer.

So 140 km ≈ 87.5 (or) 88 miles.

5 a x = 4

b i x = 2 ii x = 1.5 iii x = 0 iv x = 1

6 a i x = −2 ii x = −8 iii x = −10

iv x = −5 v x = 0 vi x = −12

b Equation i: 4521

=+x



6 c 5.192121

=+x

5.3521

=+x

x = −3

7 a

b

c i t = 3 ii t = 0 iii t = −3

d 672

3=+

+t

12

3−=

+t

t = −5

8 a i 1 ii 4 iii 9 iv 16

b x = −3 and x = 3, there are two numbers on the x-axis that are mapped to 9.

c i x = −2 and x = 2 ii x = −2.7 and x = 2.7

iii x = −1 and x = 1 iv x = −1.6 and x = 1.6

d x = −2.4 and x = 2.4

e i x = −10 and x = 10 ii x = −13 and x = 13 iii x = −5 and x = 5

f Might say that x2 is always positive. Therefore, x2 + 4 will always be more than 4 so it cannot equal zero. Or similar argument.

g Any equation equivalent to the form x2 = a, where a is bigger or equal to zero.



A5.1 Answers

1 a m + 3 b m − 5 c 2m

d m21 or

2m e 7 + m or m + 7 f 11 − m

2 a mn b m + n or m + n c n + m or m + n

d m − n e n − m f mn

3 a n + 2 b n + 1 c n − 1 d n − 2

4 a i

ii

iii

b Sum = 9n. The numbers all cancel each other out to leave the 9n.

c 180.

5 a n − 1, n + 2, n + 5 b n + 2 c 3n + 6

d n + 2 e 6

6 a n + 1 b n c 4n

d 2

5+n e 2

9+n f 2

nm +

7 a i 2t + 3 ii 2t + 2 iii 8t + 12 iv 2t + 3

b The median and the mean have the same value.

8 a A has the same y-coordinate as P and it has x-coordinate b less than P so the coordinates are (5 – b, 5).

b G

c B(5 – b, 5 + b) C(5, 5 + b) D(5 + b, 5 + b) E(5 + b, 5) F(5 + b, 5 – b) H( 5 – b, 5 – b)

d P′(5 + 2b, 5)



9 a i P(1, 2) P′(2, 1) ii Q(4, 3) Q′(3, 4) iii R(−1, −3) R′(−3, −1)

b (y, x)

10 a y = 3x + 1 b y =7(x − 4)

11 3

52 +=

nT

12 a A = b(h − a)

b P = 2(h − a + b)

13 S = 3m

14

15 a 1, 3, 5, 7, 9 b nth term = 2n − 1

16 a 2n , 60 b 10n , 300 c n + 6 , 36

d 3n + 2 , 92 e 5n + 5 , 155 f 4n − 3 , 117

17 a 60, 50; −10n + 110 b 29, 27; −2n + 39

18 a Number of triangles (t) 1 2 3 4 5

Number of matches (m) 3 5 7 9 11

b m = 2n + 1

c 31

d Pattern with 150 triangles.

19 a The two faces in the base and two faces where the cubes join are hidden.

b There are seven hidden faces.



19 c

d f = 3n − 2

e 34

20 a

b g = 2r + 6

21 a 50

b 14, 26, 38, 50…

c nth term = 12n +2. 30th border contains 362 tiles.



A5.2 Answers

1 a 513=r b

321=h c

762=d

d 1194=x e

912=s f

412=k

2 a g = 3.5 b w = 2.25 c x = 0.6

d b = 8.9 e s = 1.5 f y = 4.5

3 a 5x + 3 = 33; x = 6 b 4x + 6 = 34; x = 7

4 a x + 3 b 3x c 8x + 3

d 8x + 3 = 99 e x = 12 f 15 years

5 a x + 10°, x + 20°, x + 30°

b 360°

c x + ( x + 10° ) + ( x + 20°) + ( x + 30°) = 4x + 60° = 360°

d x = 75° ; 75°, 85°, 95° and 105°

6 a, b

c 4x + 4 = 100

d x = 24

e

7 a n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 2010

b n = 400; numbers are 400, 401, 402, 403 and 404



8 a

b 2y + 2 + (2y + 6) = 4y + 8 = 68

c y = 15; numbers are 15, 17 and 19

9 a x + 1

b x + x + (x + 1) + (x + 1) = 12 or 4x + 2 = 12

c x = 2.5; length = 3.5 cm and width = 2.5 cm

10 a

b 4x + 40

c 4x + 40 = 164 x = 31

d 3x + 30 = 66 x = 12

11 a n, 3n − 6 , 3n

b n + (3n − 6) + 3n = 50 or 7n − 6 = 50; n = 8

c 8, 18 and 24

12 a (x + 6) m b 4x + 12 m c 4x + 12 = 70

d x = 14.5 e length 20.5 m, width 14.5 m



13 a The top circle and the bottom-left circle must add to 20, so if the number in the top circle is n, the number in the bottom-left circle should be n less than 20.

b 23 − n

c (20 − n ) + ( 23 − n ) = 27

d 43 − 2n = 27; n = 8

e

14 a (14 − n) + (22 − n) = 28; n = 4;

b (37 − n) + (41 − n) = 16; n = 31

c (15 − n) + (9 − n) = 4; n = 10

15 a x = 5 b x = 15 c x = 9 d x = 2

16 a h = 5 b r = 7 c d =10

d g = 6 e p = 11 f v = 5

17 a x = 17 b u = 55 c t = 19

d y = 40 e a = 37 f e = 20



18 4(x − 2) = 14; x = 5.5

19 a 4.5517

=+x ; x = 10

b 20 − 6x = 11.5 ; x = 51

20 a y = 5 b n = 8 c p = 9 d p = 5

21 a n + 2; n − 4

b n + 30

c The sum of the ages of the three children is equal to the sum of the ages of the father and the middle child.

d 3n − 2 = 30 + n; n = 16

e Paul is 46 and the children are 12, 16 and 18.

22 a First shape has perimeter 10n and the second shape has perimeter 8n.

b 10n = 8n + 18 cm

c n = 9 cm so the large square has sides of 18 cm.

23 a First shape has perimeter 8x and the second shape has perimeter 6x.

b 8x = 6x + 5 cm

c x = 2.5 cm



A5.3 Answers

1 a 2.30 p.m. b 30 minutes c 9 km

d 1 hour, 30 minutes e 5.30 p.m. f 4 hours

2 a Mike b 25 m c 6 m

d Alex e Alex swam faster over the last 10 m.

3 a 60 m b About 1.5 seconds c 12.5 m

d 4.3 times its natural length.

4 a The speed is constant at 5 m/s.

b The highest point at the top of the first slope.

c BC shows a rapid increase in speed from 5 m/s to 20 m/s over 5 seconds.

d CD shows a reduction in speed from 20 m/s to 12.5 m/s over 5 seconds.

e 27.5 m/s

5 a The rocket is launched from rest and reaches a speed of 70 m/s in 3 seconds.

b 70 m/s

c The rocket slows down to a stop.

d B. The greatest height is reached when the rocket has temporarily stopped at the top of its flight path.

e 315 m

6 a B b D

c A d C

7 A Both cold tap and hot tap on full B Hot tap on full, cold tap off

C Both taps off D Sally gets in the bath

E Sally stays in bath F Sally gets out of the bath

G Level of water after sally has got out of the bath

H Sally pulls out the plug.



8 a 2.5 m

b 16.7 m

c 12 boxes

d Place the first box just over 2 m from the ramp.

e 12 boxes, start 5.5 m from the ramp, 6 boxes side by side and another 6 on top.

9

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Cambridge Essentials Mathematics Extension 7 A1.1 …essentials.cambridge.org/media/CEMKS3_E7_A1_1_WS_ANS1.pdf · Cambridge Essentials Mathematics Extension 7 A1.1 Answers Original