20187

Question

A {{event}} has {{number1}} {{pens}} that can each hold {{number2}} {{animal1}}.

If each yard is {{frac1}} full, how many {{animal1}}, in total, are for sale at the auction?

Worked Solution

{{animal2}} in each yard
= {{frac2}} ×\times {{number2}}
= {{total}}

∴\therefore Total {{animal1}} for sale
= {{total}} ×\times {{number1}}
= {{{correctAnswer}}}

Variant 0

DifficultyLevel

564

Question

A cattle auction has 90 holding pens that can each hold 24 cattle.

If each yard is one-third full, how many cattle, in total, are for sale at the auction?

Worked Solution

Cattle in each yard
= 13\dfrac{1}{3} ×\times 24
= 8

∴\therefore Total cattle for sale
= 8 ×\times 90
= 720

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
event
cattle auction
number1
90
pens
holding pens
number2
24
animal1
cattle
frac1
one-third
animal2
Cattle
frac2
$\dfrac{1}{3}$
total
8
correctAnswer
720

Answers

Is Correct?Answer
x

360

x

440

✓

720

x

1440

Variant 1

DifficultyLevel

564

Question

A sheep auction has 140 holding pens that can each hold 12 sheep.

If each yard is one-quarter full, how many sheep, in total, are for sale at the auction?

Worked Solution

Sheep in each yard
= 14\dfrac{1}{4} ×\times 12
= 3

∴\therefore Total sheep for sale
= 3 ×\times 140
= 420

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
event
sheep auction
number1
140
pens
holding pens
number2
12
animal1
sheep
frac1
one-quarter
animal2
Sheep
frac2
$\dfrac{1}{4}$
total
3
correctAnswer
420

Answers

Is Correct?Answer
✓

420

x

640

x

860

x

1260

Variant 2

DifficultyLevel

564

Question

A cattle auction has 110 holding pens that can each hold 36 cattle.

If each yard is one-quarter full, how many cattle, in total, are for sale at the auction?

Worked Solution

Cattle in each yard
= 14\dfrac{1}{4} ×\times 36
= 9

∴\therefore Total cattle for sale
= 9 ×\times 110
= 990

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
event
cattle auction
number1
110
pens
holding pens
number2
36
animal1
cattle
frac1
one-quarter
animal2
Cattle
frac2
$\dfrac{1}{4}$
total
9
correctAnswer
990

Answers

Is Correct?Answer
x

840

✓

990

x

1240

x

2120

Variant 3

DifficultyLevel

564

Question

A sheep auction has 70 holding pens that can each hold 12 sheep.

If each yard is two-thirds full, how many sheep, in total, are for sale at the auction?

Worked Solution

Sheep in each yard
= 23\dfrac{2}{3} ×\times 12
= 8

∴\therefore Total sheep for sale
= 8 ×\times 70
= 560

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
event
sheep auction
number1
70
pens
holding pens
number2
12
animal1
sheep
frac1
two-thirds
animal2
Sheep
frac2
$\dfrac{2}{3}$
total
8
correctAnswer
560

Answers

Is Correct?Answer
x

280

x

420

✓

560

x

720

Variant 4

DifficultyLevel

564

Question

A cattle auction has 160 holding pens that can each hold 20 cattle.

If each yard is one-quarter full, how many cattle, in total, are for sale at the auction?

Worked Solution

Cattle in each yard
= 14\dfrac{1}{4} ×\times 20
= 5

∴\therefore Total cattle for sale
= 5 ×\times 160
= 800

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
event
cattle auction
number1
160
pens
holding pens
number2
20
animal1
cattle
frac1
one-quarter
animal2
Cattle
frac2
$\dfrac{1}{4}$
total
5
correctAnswer
800

Answers

Is Correct?Answer
x

320

✓

800

x

1050

x

3200