Sec 2 Maths Direct and inverse proportion worksheet with answers

A free Sec 2 Maths worksheet on direct and inverse proportion: 6 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 2 Maths worksheet: Direct and inverse proportion

Total: 30 marksSuggested time: 45 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The graph of

y

against

x

is a straight line through the points

(1, 4)

and

(5, 20)

. Is

y

directly proportional to

x

Answer[2]

A map is drawn to a scale of

1 : 80\,000

. Two places are

1.2

km apart. Find the distance, in centimetres, between the two places on the map.

Answercm[2]

(a)

y

is inversely proportional to the square of

x

. Given that

y = 2

when

x = 3

, express

y

in terms of

x

Answer (a)[3]

(b)

y

is directly proportional to the square root of

x

. When

x

is divided by

4

y

is multiplied by a number. Find this number.

Answer (b)[2]

Section B

(a)On a map, an area of

1

cm² represents an actual area of

4

km². Express the scale of the map in the form

1 : n

Answer (a)[2]

(b)The extension of a spring is directly proportional to the mass hung on it. A mass of 200 g stretches the spring by 5 cm. Find the extension, in millimetres, when a mass of 1.2 kg is hung on it.

Answer (b)mm[3]

(a)9 workers take 17 days to paint a block of flats. Find the number of days 3 workers would take, working at the same rate.

Answer (a)days[2]

(b)(i)

y

is directly proportional to

x

. Complete the table of values.

Answer (b)(i)[3]

(b)(ii)The volume of water flowing from a tap is directly proportional to the time it is on. In 5 minutes, 15 litres of water flow out. Find the volume, in millilitres, that flows out in 15 seconds.

Answer (b)(ii)ml[3]

(a)A car travelling at an average speed of 64 km/h takes 3 hours for a journey. Find the average speed needed to make the same journey in 2 hours.

Answer (a)km/h[2]

(b)(i)

y

is directly proportional to the square root of

x

. Given that

y = 42

when

x = 49

, find the value of

y

when

x = 81

Answer (b)(i)[3]

(b)(ii)

y

is directly proportional to the square of

x

. When

x

increases from

1

6

y

increases by

70

. Find the value of

y

when

x = 10

Answer (b)(ii)[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer: YES[2]

(1, 4)

\dfrac{y}{x} = \dfrac{4}{1}

\dfrac{y}{x} = 4

.At

(5, 20)

\dfrac{y}{x} = \dfrac{20}{5}

\dfrac{y}{x} = 4

\dfrac{y}{x}

is the same, so the line passes through the origin:

y

is directly proportional to

x

Question 2

Answer:

1.5\text{ cm} : 1.2\text{ km}

[2]

\mathstrut 1 : 80\,000

\mathstrut 1\text{ cm} : 80\,000\text{ cm}

1\text{ cm} : 0.8\text{ km}

\mathstrut \dfrac{1.2}{0.8}\text{ cm} : 1.2\text{ km}

1.5\text{ cm} : 1.2\text{ km}

Question 3

(a)Answer:

y = \dfrac{18}{x^2}

[3]

y = \dfrac{k}{x^2}

2 = \dfrac{k}{(3)^2}

2 = \dfrac{k}{9}

k = 2 \times 9

= 18

y = \dfrac{18}{x^2}

(b)Answer:

y

is multiplied by

\dfrac{1}{2}

.[2]

y = k\sqrt{x}

\text{New } y = k\sqrt{\dfrac{1}{4}x}

= \dfrac{1}{2}k\sqrt{x}

y

is multiplied by

\dfrac{1}{2}

Question 4

(a)Answer: So the scale is

1 : 200\,000

.[2]

\mathstrut 1\text{ cm}^2 : 4\text{ km}^2

\mathstrut 1\text{ cm} : \sqrt{4}\text{ km}

1\text{ cm} : 2\text{ km}

1\text{ cm} : 200\,000\text{ cm}

So the scale is

1 : 200\,000

(b)Answer: The extension is

300

mm.[3]

1.2

= 1.2 \times 1000

= 1200

\text{Extension} = \dfrac{5}{200} \times 1200

= 30

30 \times 10 = 300

The extension is

300

mm.

Question 5

(a)Answer:

= 51

days[2]

\text{1 worker} \to 9 \times 17

= 153

days

\text{Number of days} = \dfrac{153}{3}

= 51

days

(b)(i)Answer:

= 24

[3]

k = \dfrac{9}{12}

= \dfrac{3}{4}

When

x = 4

y = \dfrac{3}{4} \times 4

= 3

When

x = 16

y = \dfrac{3}{4} \times 16

= 12

When

x = 32

y = \dfrac{3}{4} \times 32

= 24

(b)(ii)Answer: The volume is

750

ml.[3]

15

= \dfrac{15}{60}

= \dfrac{1}{4}

min

\text{Volume} = \dfrac{15}{5} \times \dfrac{1}{4}

= 0.75

litres

0.75 \times 1000 = 750

The volume is

750

ml.

Question 6

(a)Answer:

= 96

km/h[2]

\text{Distance} = 64 \times 3

= 192

\text{Speed} = 192 \div 2

= 96

km/h

(b)(i)Answer:

= 54

[3]

y = k\sqrt{x}

42 = k\sqrt{49}

42 = 7k

k = \dfrac{42}{7}

= 6

y = 6\sqrt{x}

y = 6\sqrt{81}

= 54

(b)(ii)Answer:

= 200

[3]

y = kx^2

When

x = 1

y = k(1)^2

y = 1k

.When

x = 6

y = k(6)^2

y = 36k

36k - 1k = 70

35k = 70

k = \dfrac{70}{35}

= 2

y = 2x^2

y = 2(10)^2

= 200

Practice questions by skill

Sec 2 Maths Direct and inverse proportion worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Direct and inverse proportion notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)On a map, an area of

1

cm² represents an actual area of

4

km². Express the scale of the map in the form

1 : n

Answer (a)[2]

(b)The extension of a spring is directly proportional to the mass hung on it. A mass of 200 g stretches the spring by 5 cm. Find the extension, in millimetres, when a mass of 1.2 kg is hung on it.

Answer (b)mm[3]

Answers and worked solutions

Question 1

Answer: YES[2]

(1, 4)

\dfrac{y}{x} = \dfrac{4}{1}

\dfrac{y}{x} = 4

.At

(5, 20)

\dfrac{y}{x} = \dfrac{20}{5}

\dfrac{y}{x} = 4

\dfrac{y}{x}

is the same, so the line passes through the origin:

y

is directly proportional to

x

Question 2

Answer:

1.5\text{ cm} : 1.2\text{ km}

[2]

\mathstrut 1 : 80\,000

\mathstrut 1\text{ cm} : 80\,000\text{ cm}

1\text{ cm} : 0.8\text{ km}

\mathstrut \dfrac{1.2}{0.8}\text{ cm} : 1.2\text{ km}

1.5\text{ cm} : 1.2\text{ km}

Question 3

(a)Answer:

y = \dfrac{18}{x^2}

[3]

y = \dfrac{k}{x^2}

2 = \dfrac{k}{(3)^2}

2 = \dfrac{k}{9}

k = 2 \times 9

= 18

y = \dfrac{18}{x^2}

(b)Answer:

y

is multiplied by

\dfrac{1}{2}

.[2]

y = k\sqrt{x}

\text{New } y = k\sqrt{\dfrac{1}{4}x}

= \dfrac{1}{2}k\sqrt{x}

y

is multiplied by

\dfrac{1}{2}

Question 4

(a)Answer: So the scale is

1 : 200\,000

.[2]

\mathstrut 1\text{ cm}^2 : 4\text{ km}^2

\mathstrut 1\text{ cm} : \sqrt{4}\text{ km}

1\text{ cm} : 2\text{ km}

1\text{ cm} : 200\,000\text{ cm}

So the scale is

1 : 200\,000

(b)Answer: The extension is

300

mm.[3]

1.2

= 1.2 \times 1000

= 1200

\text{Extension} = \dfrac{5}{200} \times 1200

= 30

30 \times 10 = 300

The extension is

300

mm.

Question 5

(a)Answer:

= 51

days[2]

\text{1 worker} \to 9 \times 17

= 153

days

\text{Number of days} = \dfrac{153}{3}

= 51

days

(b)(i)Answer:

= 24

[3]

k = \dfrac{9}{12}

= \dfrac{3}{4}

When

x = 4

y = \dfrac{3}{4} \times 4

= 3

When

x = 16

y = \dfrac{3}{4} \times 16

= 12

When

x = 32

y = \dfrac{3}{4} \times 32

= 24

(b)(ii)Answer: The volume is

750

ml.[3]

15

= \dfrac{15}{60}

= \dfrac{1}{4}

min

\text{Volume} = \dfrac{15}{5} \times \dfrac{1}{4}

= 0.75

litres

0.75 \times 1000 = 750

The volume is

750

ml.

Question 6

(a)Answer:

= 96

km/h[2]

\text{Distance} = 64 \times 3

= 192

\text{Speed} = 192 \div 2

= 96

km/h

(b)(i)Answer:

= 54

[3]

y = k\sqrt{x}

42 = k\sqrt{49}

42 = 7k

k = \dfrac{42}{7}

= 6

y = 6\sqrt{x}

y = 6\sqrt{81}

= 54

(b)(ii)Answer:

= 200

[3]

y = kx^2

When

x = 1

y = k(1)^2

y = 1k

.When

x = 6

y = k(6)^2

y = 36k

36k - 1k = 70

35k = 70

k = \dfrac{70}{35}

= 2

y = 2x^2

y = 2(10)^2

= 200