20178

Question

{{{question}}}

Worked Solution

{{{workedSolution}}}

Variant 0

DifficultyLevel

635

Question

Manoj is making a number of curries.

He has 8 cups of coconut milk.

Manoj uses 23\dfrac{2}{3} of a cup of coconut milk for every curry.

What is the maximum number of curries Manoj can make?

Worked Solution
Maximum curries = 8÷238 \div \dfrac{2}{3}
= 8×328 \times \dfrac{3}{2}
= 12

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Manoj is making a number of curries. He has 8 cups of coconut milk. Manoj uses $\dfrac{2}{3}$ of a cup of coconut milk for every curry. What is the maximum number of curries Manoj can make?
workedSolution
| | | | --------------------- | ---------------------------------------| | Maximum curries | = $8 \div \dfrac{2}{3}$ | | | = $8 \times \dfrac{3}{2}$ | | | = {{correctAnswer}} |
correctAnswer
12

Answers

Is Correct?Answer
x

8238\dfrac{2}{3}

x

9139\dfrac{1}{3}

✓

12

x

16

Variant 1

DifficultyLevel

640

Question

Benjamin is making a number of apple pies.

He has 12 kilograms of apples.

Benjamin uses 25\dfrac{2}{5} of a kilogram of apples for every apple pie.

What is the maximum number of apple pies Benjamin can make?

Worked Solution
Maximum apple pies = 12÷2512 \div \dfrac{2}{5}
= 12×5212 \times \dfrac{5}{2}
= 30

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Benjamin is making a number of apple pies. He has 12 kilograms of apples. Benjamin uses $\dfrac{2}{5}$ of a kilogram of apples for every apple pie. What is the maximum number of apple pies Benjamin can make?
workedSolution
| | | | ---------------------: | ---------------------------------------| | Maximum apple pies| = $12 \div \dfrac{2}{5}$ | | | = $12 \times \dfrac{5}{2}$ | | | = {{correctAnswer}} |
correctAnswer
30

Answers

Is Correct?Answer
x

4454\dfrac{4}{5}

x

122512\dfrac{2}{5}

x

131213\dfrac{1}{2}

✓

30

Variant 2

DifficultyLevel

642

Question

Bailey is making a number of bread boards.

He has 15 metres of timber.

Bailey uses 37\dfrac{3}{7} of a metre of timber for every bread board.

What is the maximum number of bread boards Bailey can make?

Worked Solution
Maximum bread boards = 15÷3715 \div \dfrac{3}{7}
= 15×7315 \times \dfrac{7}{3}
= 35

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Bailey is making a number of bread boards. He has 15 metres of timber. Bailey uses $\dfrac{3}{7}$ of a metre of timber for every bread board. What is the maximum number of bread boards Bailey can make?
workedSolution
| | | | --------------------- | ---------------------------------------| | Maximum bread boards| = $15 \div \dfrac{3}{7}$ | | | = $15 \times \dfrac{7}{3}$ | | | = {{correctAnswer}} |
correctAnswer
35

Answers

Is Correct?Answer
✓

35

x

153715\dfrac{3}{7}

x

144714\dfrac{4}{7}

x

6376\dfrac{3}{7}

Variant 3

DifficultyLevel

678

Question

Sandra has 2 kilograms of flour to make cupcakes.

Sandra uses 25\dfrac{2}{5} of a kilogram of flour for every batch of 24 cupcakes she makes.

What is the maximum number of cupcakes Sandra can make?

Worked Solution
Maximum batches of cupcakes = 2÷252 \div \dfrac{2}{5}
= 2×522 \times \dfrac{5}{2}
= 5 batches
Maximum cupcakes = 24×524 \times 5
= 120

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Sandra has 2 kilograms of flour to make cupcakes. Sandra uses $\dfrac{2}{5}$ of a kilogram of flour for every batch of 24 cupcakes she makes. What is the maximum number of cupcakes Sandra can make?
workedSolution
| | | | --------------------- | ---------------------------------------| | Maximum batches of cupcakes| = $2 \div \dfrac{2}{5}$ | | | = $2 \times \dfrac{5}{2}$ | | | = 5 batches | | | |
| | | | --------------------- | ---------------------------------------| | Maximum cupcakes| = $24 \times 5$ | | | = {{correctAnswer}} |
correctAnswer
120

Answers

Is Correct?Answer
x

60

x

90

✓

120

x

240

Variant 4

DifficultyLevel

703

Question

Xanto is laying a number of driveways.

His cement truck holds 7.2 cubic metres of concrete.

Xanto uses 2252\dfrac{2}{5} cubic metres of concrete for every driveway he lays.

What is the maximum number of driveways Xanto can lay?

Worked Solution
Maximum driveways = 7.2÷2257.2 \div 2\dfrac{2}{5}
= 715÷125\dfrac{1}{5}\div \dfrac{12}{5}
= 365×512\dfrac{36}{5} \times \dfrac{5}{12}
= 3

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Xanto is laying a number of driveways. His cement truck holds 7.2 cubic metres of concrete. Xanto uses $2\dfrac{2}{5}$ cubic metres of concrete for every driveway he lays. What is the maximum number of driveways Xanto can lay?
workedSolution
| | | | --------------------- | ---------------------------------------| | Maximum driveways| = $7.2 \div 2\dfrac{2}{5}$ | | | = 7$\dfrac{1}{5}\div \dfrac{12}{5}$ | | | = $\dfrac{36}{5} \times \dfrac{5}{12}$ | | | = {{correctAnswer}} |
correctAnswer
3

Answers

Is Correct?Answer
x

2.6

✓

3

x

3123 \dfrac{1}{2}

x

1772517\dfrac{7}{25}

Variant 5

DifficultyLevel

680

Question

A cement company is laying driveways in a housing development.

One cement truck holds 7.2 cubic metres of concrete.

Each driveway uses 3353\dfrac{3}{5} cubic metres of concrete.

What is the maximum number of driveways that can be laid if 3 trucks are used?

Worked Solution

Maximum driveways (one truck)
= 7.2÷3357.2 \div 3\dfrac{3}{5}
=715÷1857\dfrac{1}{5}\div \dfrac{18}{5}
= 365×518\dfrac{36}{5} \times \dfrac{5}{18}
= 2

∴\therefore Maximum driveways (three trucks) = 6

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
A cement company is laying driveways in a housing development. One cement truck holds 7.2 cubic metres of concrete. Each driveway uses $3\dfrac{3}{5}$ cubic metres of concrete. What is the maximum number of driveways that can be laid if 3 trucks are used?
workedSolution
sm_nogap Maximum driveways (one truck)
>| | | | --------------------- | ---------------------------------------| | | = $7.2 \div 3\dfrac{3}{5}$ | | =$7\dfrac{1}{5}\div \dfrac{18}{5}$ | | = $\dfrac{36}{5} \times \dfrac{5}{18}$ | | | = 2 |

$\therefore$ Maximum driveways (three trucks) = {{{correctAnswer}}}
correctAnswer
6

Answers

Is Correct?Answer
x

2

✓

6

x

8

x

12

Tags