IB MYP 4 and 5 Mathematics Extended (2nd Edition) Answers [2 ed.] 1382010915, 9781382010917

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ANSWERS

Unit 1 Answers

 4 a  2 6

E1.1

e  215

b  3 6

 1 a  x  6

b 28

c 8x  6

 2 a  3 5 1  3 a  4

b 4 1 b  36

c  6 5 1 c  8

3 e  2

f 

( 1 + m1 ) = a( mmn+ n )

b 3x

c 

e  1

−1 f  x 6

−1 g  x 4

−1  3 a  x 2

1 b  x 3

−1 c  x 4

 4 a  x = 8

b  y = 100

1

1 2x 2

b 81

e  25

f 49

4

 3 a  x 3

12

g  4 2

h  5 5

10 h  3

 3 a x = 3

b  y = 3 3

c  z = 3

1 125 125 e  216  2 a  5 e  81

3

b  x 4 13

b  x 3

 1 a  5 32 5 3

d 3

h 3x

1 e  6 7 7

3 2

j  7

b  32

c  1000

e  8

f  1000

1 1 0.2 5  2 32 = (32) 5 = (2 ) 5 = 2

 3 a 2.4

c 125 3

c  x 2

 5 a 3

d 16

b  1

c  4.6

5

d  x 2

7

c  x 6

1 8 8 f  27 b 

b  2 f  3

c 

64 27

d 32 = 9

d  256

c  2401

d  4

7

−

5

 1 a  2 2

b  2 3

c  2 2

d  2 3

8  2 a  715

3 b  3 2

1 c  2 4

1 d  2 4

1 1  3 a  2 2 × 3 2 1 1 e  2 4 × 52

1 1 b  2 3 × 7 3 1 1 2 2 × 33

1 c  3 × 2 4

1 1 d  2 2 × 5 2

f  

b  3

 2 a  7x3

b 

 3 a 1000

b  100

c 

1 9

d 

1 27

 4 a 4

b 

27 8

c 

25 16

d 

25 27

f 2 1 4x

d 3

5

 5 a  2 2

b  3 4

 6 a 25

b 

7

c  2

7

d  2 3 × 3 3

1 2

c  7

 7 a  27

b  2 6

c  2

d  3 3

e  2

f  8

1 g  12 12

1 h  12 3

3

c 9 d 2

1 c  5 5 g  2 3 c  2x

1  1 a  3 3 e  4

5

Practice 4 7

b  2

Mixed practice

f 16

 3 Student’s own answers 1  4 a  b 4 9

7

i 

9 2

Practice 6

10 d  2 x

Practice 3  1 a 

j 

6 3 6 b  12 37 × 2 c  24 × 37 d  25

f 55

4

e  x 15

i  26 5

3

4

h  10 37

 4 x = 1024     

 1 a 4 2

e  7

g  15 25

d 5

Practice 2

 2 a  x 3

6

1 d  6 11

6

 2 a  25

 2 a  2 x 2

c  6 5

b  3

 1 a 2

Practice 1 1 f  3

( 1 − m1 ) = a( mmn− n )

h  a n

Practice 5 d 2

f 35

1 c  2 5 g  2

d  2 6 3 6

f      2 n

g  a n

25 4

b 2

1 1

3

c  710

( 1 _ m1 ) × 5 ( 1n _ m1 )= 2( mmn− n ) × 5( mmn− n )  

4

You should already know how to:

1  1 a  2 2 e  3

7

5

 8 Student’s own answers 3

1 2

 9 a  27

b  12

c 

1 10 a  x = 3

b  x = 2

c  x = 27

b  5

c  9

5

11 a  59

d 

12 5 4 5

2

d  1

e  26 × 33 = 1728 12 Mathematics is as old as man.

Answers 201092_answers.indd 1

1 05/03/21 9:32 AM

E1.2

Mixed practice

You should already know how to:

 1 a 35 ≤ x < 45 b  65 ≤ x < 75 c  115 ≤ x < 125

 1 a  5200     b 23.961     c 18.24  2 a  –4

–2

2

4

b  –2



0

–1

0

1

2

3

c   –5

–4

–3

–2

–1

0

1

 3 a  424.3   b 0.001125   c 98.231   d 5.236

Practice 1  1 1.55 m ≤ length of wood < 1.65 m  2 375 kg ≤ mass of crate < 425 kg  3 59.5 m ≤ length of rope < 60.5 m  4 55 km ≤ length of island of Mauritius < 65 km  5 It is acceptable since ‘within 3 mm of 7 cm’ means that 67 mm < h < 73 mm.  6 The card may not fit in the envelope. 9.45 cm ≤ length of card < 9.55 cm and 9.5 cm ≤ length of envelope < 10.5 cm. The card will not fit if its length is greater than the length of the envelope, which happens when 9.5 cm ≤ length of envelope < length of card < 9.55 cm  7 21.5 kg ≤ mass < 22.5 kg 12.55 cm ≤ length < 12.65 cm 2.035 cm3 ≤ volume < 2.045 cm3 0.65 tonnes ≤ mass < 0.75 tonnes 132.5 cm ≤ height < 133.5 cm 51 500 m2 ≤ area < 52 500 m2 58.65° ≤ temperature < 58.75°

 2 a 1.5 ≤ x < 2.5 b  16.5 ≤ x < 17.5 c  −85.5 ≤ x < −84.5  3 a  12.45 cm ≤ x < 12.55 cm b  21.65 cm ≤ x < 21.75 cm c  34.75 cm ≤ x < 35.85 cm d  52.05 cm ≤ x < 52.15 cm e  80.35 cm ≤ x < 80.45 cm  4 a  458 m ≤ perimeter < 462 m b  12 210.25 m2 ≤ area < 12 440.25 m2  5 3.76 cm ≤ side length ≤ 4.26 cm  6 12.16 km/liter  7 0.027  8 a  14.6 c  52.4 e  14.8 g  53.9

b 1.8 d 1.28 f   2 h 1.31

Review in context  1 a  6850 km ≤ radius of satellite’s orbit < 6900 km b  43 332 km c  314 km  2 22 fuel tanks are needed, since it needs a minimum of 21.66 fuel tanks (21 is therefore not enough).  3 a 229 days (the actual time is 228.4 days, but it would not have reached Mars after only 228 days). b  252.9 days rounded to 253 days.

Practice 2  1 34 boxes  2 34 cm ≤ perimeter of photoframe < 34.4 cm  3 59.9 m ≤ perimeter of garden < 62.1m, and 193.725 m2 ≤ area of garden < 205.325 m2  4 25 minutes  5 She can fill between 10 and 12 full glasses.

2 201092_answers.indd 2

Answers 05/03/21 9:32 AM

Unit 2 Answers

Practice 3

E2.1 You should already know how to:  1 a 6 < 8, valid c  6 < 12, valid

b 2 < 4, valid d  −3 < −6, not valid

 2 a  x = 2, addition and multiplication principles. b  x = 12, addition and multiplication principles.  3 a  x > 3 c  x≥3

b  x < 2 d  −2 < x < 2

 4 a 

y 5 4 3 1 −4 −3 −2 −1 0 −1

2

1

−2

3

6 5

−4

3

4 3

x + 2y = 6

2

2

1

1 −2 −1 0 −1

1 2 3 4 5 6 7 8 9 10 x

7

4 x

y = 3x − 2

−3 y 4

−4 −3 −2 −1 0 −1 −2 −3 −4 −5 −6 −7 y b  8

2

b 

 1 The unshaded area is the solution set. a  y 7 6 5 4 3 2 1

1

2

3

4

5

6

7 x

0

Practice 1

17

0

Practice 2

−4

 1 Solutions the same as in Practice 1

−5

 2 y ≥ 6 − x and y > 4

−6

 3 3x + 2 < 9x + 6 −6x < 4

−7



which is an infinite set of solutions.

4

5

6

7

x

8

1

2

3

4

5

6

7

8 x

−2

−2 3

3

−1

 7 a > 4   8  b ≤ 1.25  9 k ≤ 6.4

x>

2

c  y

 1 x < 4   2  x > 54  3 m < 1  4  x ≥ 8 7  5 x ≥ 2   6  x < −



1

−3

−8 d 

y 4 3 2 1

−4 −3 −2 −1 0 −1

1

2

3

4

x

−2 −3 −4

Answers 201092_answers.indd 3

3 05/03/21 9:32 AM

e 

 2 x < 6; y ≤ x; y ≥ 6 − 2x

y 4

 3 The unshaded area is the solution set. a  y 6

3 2

5

1

4

−4 −3 −2 −1 0 −1

1

2

3

4

x

3 2

−2

1

−3

0 −6 −5 −4 −3 −2 −1 −1

−4

1

2

3

4

x

1

2

3

4

5

y 6

b 

5

−2

4

−3

3

−4

2

y 4

1

3

−6 −5 −4 −3 −2 −1 0 −1

2

−2

1

−3 −3

−4 −3 −2 −1 0 −1

1

2

3

4

6

x

−4 −4

x

−5 −5

−2

−6 − 6

−3

 4 The unshaded area is the solution set. a  y

−4

8

y

7 6

1.4

5

1.2

4

1

3

0.8

2

0.6

1

0.4

−4 −3 −2 −1 0 −1

0.2 0.1 0.2 0.3

−0.4

201092_answers.indd 4

6 x

−6

−4 −3 −2 −1 0 −1

4

5

−5

1

−0.3−0.2−0.10

4

−4

2

h 

3

−3

3

g 

2

−2

y 4

f 

1

x

1

2

3

4

5

6

7

8

x

−2 −3 −4

Answers 05/03/21 9:32 AM

b 

d  y 20

y 30 28

18

26

16

24

14

22

12

20

10

18 16

8

14

6

12

4

10

2

8

−2 0 −2

6 4

c 

4

6

8

10

−6 2

4

6

8 10 12

x

−8

−4

−10

−6

−12 −14

y 16

−16

14

−18

12

−20

10

−22

8

−24

6

−26

4

Practice 4

2

 1 Maximum = 780 at (15, 1.5)

−6 −4 −2 0 −2 −4

x

−4

2 0 −6 −4 −2 −2

2

2

4

6

8

10 12 14 16

x

 2 Maximum = 19.6 at (1, 6)  3 Maximum = 707 at (99, 181)

−6

 4 The manufacturer should make 105 mid-top shoes and 45 high-top shoes, which would make a profit of $2085 per day.

−8

 5 6 small and 6 large minibuses

−10

 6 100 algebraic solvers and 170 graphing programs

−12

 7 $640 (from 40 downhill skis and 30 cross-country skis)

−14

 8 $75 000 in municipal bonds and $25 000 in bank mutual fund

−16

Answers 201092_answers.indd 5

5 05/03/21 9:32 AM

Mixed practice

c  y 6

 1 a  x ≥ 1 b  x < 18 c  x > −9 d  x