Number, NAPX-G3-CA32

Question

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Worked Solution

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Variant 0

DifficultyLevel

685

Question

Dawn had a bag of different coloured marbles.

14\dfrac{1}{4} of the marbles were clear and 13\dfrac{1}{3} of the marbles were black.

Her bag contained 5 more black marbles than clear marbles.

How many marbles were in Dawn's bag?

Worked Solution

Strategy 1

Test each option:

14 ×60\dfrac{1}{4}\ \times 60 = 15

13 ×60\dfrac{1}{3}\ \times 60 = 20

Difference = 20 −- 15 = 5

∴\therefore 60 marbles are in the bag.


Strategy 2 (advanced)

Let  n\ \large n = number of marbles in bag
13n\dfrac{1}{3}\large n −- 14n\dfrac{1}{4}\large n = 5
412n\dfrac{4}{12}\large n −- 312n\dfrac{3}{12}\large n = 5
112n\dfrac{1}{12}\large n = 5
∴n\therefore \large n = 60

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Dawn had a bag of different coloured marbles. $\dfrac{1}{4}$ of the marbles were clear and $\dfrac{1}{3}$ of the marbles were black. Her bag contained 5 more black marbles than clear marbles. How many marbles were in Dawn's bag?
workedSolution
Strategy 1 Test each option: $\dfrac{1}{4}\ \times 60$ = 15 $\dfrac{1}{3}\ \times 60$ = 20 Difference = 20 $-$ 15 = 5 $\therefore$ 60 marbles are in the bag.
Strategy 2 (advanced) Let $\ \large n$ = number of marbles in bag
| | | | ---------------------: | -------------- | | $\dfrac{1}{3}\large n$ $-$ $\dfrac{1}{4}\large n$ | \= 5 | | $\dfrac{4}{12}\large n$ $-$ $\dfrac{3}{12}\large n$ | \= 5 | | $\dfrac{1}{12}\large n$| = 5| |$\therefore \large n$| \= {{{correctAnswer}}}|
correctAnswer
60

Answers

Is Correct?Answer
x

36

x

48

✓

60

x

72

Variant 1

DifficultyLevel

683

Question

Mica had a barrel of different coloured discs.

15\dfrac{1}{5} of the discs were yellow and 14\dfrac{1}{4} of the discs were green.

His barrel contained 7 more green discs than yellow.

How many discs were in Mica's barrel?

Worked Solution

Strategy 1

Test each option:

15 ×140\dfrac{1}{5}\ \times 140 = 28

14 ×140\dfrac{1}{4}\ \times 140 = 35

Difference = 35 −- 28 = 7

∴\therefore 140 discs are in the barrel.


Strategy 2 (advanced)

Let  n\ \large n = number of discs in barrel
14n\dfrac{1}{4}\large n −- 15n\dfrac{1}{5}\large n = 7
520n\dfrac{5}{20}\large n −- 420n\dfrac{4}{20}\large n = 7
120n\dfrac{1}{20}\large n = 7
∴n\therefore \large n = 140

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Mica had a barrel of different coloured discs. $\dfrac{1}{5}$ of the discs were yellow and $\dfrac{1}{4}$ of the discs were green. His barrel contained 7 more green discs than yellow. How many discs were in Mica's barrel?
workedSolution
Strategy 1 Test each option: $\dfrac{1}{5}\ \times 140$ = 28 $\dfrac{1}{4}\ \times 140$ = 35 Difference = 35 $-$ 28 = 7 $\therefore$ 140 discs are in the barrel.
Strategy 2 (advanced) Let $\ \large n$ = number of discs in barrel
| | | | ---------------------: | -------------- | | $\dfrac{1}{4}\large n$ $-$ $\dfrac{1}{5}\large n$ | \= 7 | | $\dfrac{5}{20}\large n$ $-$ $\dfrac{4}{20}\large n$ | \= 7 | | $\dfrac{1}{20}\large n$| = 7| |$\therefore \large n$| \= {{{correctAnswer}}}|
correctAnswer
140

Answers

Is Correct?Answer
x

120

✓

140

x

160

x

180

Variant 2

DifficultyLevel

680

Question

Ed had a hobby farm growing two different varieties of cabbages.

13\dfrac{1}{3} of the cabbages were purple and the rest were green.

Last year Ed planted 120 more green cabbages than purple.

How many cabbages did Ed plant last year?

Worked Solution

Strategy 1

13\dfrac{1}{3} of cabbages are purple

23\dfrac{2}{3} of cabbages are green

Test each option:

13 ×360\dfrac{1}{3}\ \times 360 = 120

23 ×360\dfrac{2}{3}\ \times 360 = 240

Difference = 240 −- 120 = 120

∴\therefore 360 cabbages were planted last year.


Strategy 2 (advanced)

Let  n\ \large n = number of cabbages planted
23n\dfrac{2}{3}\large n −- 13n\dfrac{1}{3}\large n = 120
13n\dfrac{1}{3}\large n = 120
∴n\therefore \large n = 360

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Ed had a hobby farm growing two different varieties of cabbages. $\dfrac{1}{3}$ of the cabbages were purple and the rest were green. Last year Ed planted 120 more green cabbages than purple. How many cabbages did Ed plant last year?
workedSolution
Strategy 1 $\dfrac{1}{3}$ of cabbages are purple $\dfrac{2}{3}$ of cabbages are green Test each option: $\dfrac{1}{3}\ \times 360$ = 120 $\dfrac{2}{3}\ \times 360$ = 240 Difference = 240 $-$ 120 = 120 $\therefore$ 360 cabbages were planted last year.
Strategy 2 (advanced) Let $\ \large n$ = number of cabbages planted
| | | | ---------------------: | -------------- | | $\dfrac{2}{3}\large n$ $-$ $\dfrac{1}{3}\large n$ | \= 120 | | $\dfrac{1}{3}\large n$| = 120| |$\therefore \large n$| \= {{{correctAnswer}}}|
correctAnswer
360

Answers

Is Correct?Answer
x

240

x

300

x

330

✓

360

Variant 3

DifficultyLevel

681

Question

The local animal shelter has a number of different animals for adoption.

16\dfrac{1}{6} of the animals are cats and 13\dfrac{1}{3} of the animals are dogs.

There are currently 25 more dogs than cats in the shelter.

How many animals are currently available for adoption in the shelter?

Worked Solution

Strategy 1

Test each option:

16 ×150\dfrac{1}{6}\ \times 150 = 25

13 ×150\dfrac{1}{3}\ \times 150 = 50

Difference = 50 −- 25 = 25

∴\therefore 150 animals were in the shelter.


Strategy 2 (advanced)

Let  n\ \large n = number of animals in the shelter
13n\dfrac{1}{3}\large n −- 16n\dfrac{1}{6}\large n = 25
26n\dfrac{2}{6}\large n −- 16n\dfrac{1}{6}\large n = 25
16n\dfrac{1}{6}\large n = 25
∴n\therefore \large n = 150

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
The local animal shelter has a number of different animals for adoption. $\dfrac{1}{6}$ of the animals are cats and $\dfrac{1}{3}$ of the animals are dogs. There are currently 25 more dogs than cats in the shelter. How many animals are currently available for adoption in the shelter?
workedSolution
Strategy 1 Test each option: $\dfrac{1}{6}\ \times 150$ = 25 $\dfrac{1}{3}\ \times 150$ = 50 Difference = 50 $-$ 25 = 25 $\therefore$ 150 animals were in the shelter.
Strategy 2 (advanced) Let $\ \large n$ = number of animals in the shelter
| | | | ---------------------: | -------------- | | $\dfrac{1}{3}\large n$ $-$ $\dfrac{1}{6}\large n$ | \= 25 | | $\dfrac{2}{6}\large n$ $-$ $\dfrac{1}{6}\large n$ | \= 25 | | $\dfrac{1}{6}\large n$| = 25| |$\therefore \large n$| \= {{{correctAnswer}}}|
correctAnswer
150

Answers

Is Correct?Answer
x

25

x

50

x

100

✓

150

Variant 4

DifficultyLevel

677

Question

At a swimming carnival there were two grandstands.

Students were seated in the grandstands based on whether they were competitors or non-competitors.

58\dfrac{5}{8} of the students were seated in the non-competitors' grandstand and the rest were seated in the competitors' grandstand.

There were 60 more non-competitors than competitors at the carnival.

How many students, in total, were at the swimming carnival?

Worked Solution

Strategy 1

58\dfrac{5}{8} are non-competitors

38\dfrac{3}{8} are competitors

Test each option:

58 ×240\dfrac{5}{8}\ \times 240 = 150

38 ×240\dfrac{3}{8}\ \times 240 = 90

Difference = 150 −- 90 = 60

∴\therefore 240 students were at the carnival.


Strategy 2 (advanced)

Let  n\ \large n = number of students at the carnival
58n\dfrac{5}{8}\large n −- 38n\dfrac{3}{8}\large n = 60
28n\dfrac{2}{8}\large n = 60
14n\dfrac{1}{4}\large n = 60
∴n\therefore \large n = 240

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
At a swimming carnival there were two grandstands. Students were seated in the grandstands based on whether they were competitors or non-competitors. $\dfrac{5}{8}$ of the students were seated in the non-competitors' grandstand and the rest were seated in the competitors' grandstand. There were 60 more non-competitors than competitors at the carnival. How many students, in total, were at the swimming carnival?
workedSolution
Strategy 1 $\dfrac{5}{8}$ are non-competitors $\dfrac{3}{8}$ are competitors Test each option: $\dfrac{5}{8}\ \times 240$ = 150 $\dfrac{3}{8}\ \times 240$ = 90 Difference = 150 $-$ 90 = 60 $\therefore$ 240 students were at the carnival.
Strategy 2 (advanced) Let $\ \large n$ = number of students at the carnival
| | | | ---------------------: | -------------- | | $\dfrac{5}{8}\large n$ $-$ $\dfrac{3}{8}\large n$ | \= 60 | | $\dfrac{2}{8}\large n$| = 60| | $\dfrac{1}{4}\large n$| = 60| |$\therefore \large n$| \= {{{correctAnswer}}}|
correctAnswer
240

Answers

Is Correct?Answer
x

180

✓

240

x

360

x

400

Tags