Algebra, NAPX-I4-CA32 SA

Question

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Worked Solution

{{{workedSolution}}}

Variant 0

DifficultyLevel

784

Question

A miniature engine contains two types of ball bearings.

There are two times as many small ball bearings as large ball bearings.

The mass of a small ball bearing is 190 milligrams and the mass of a large ball bearing is 300 milligrams.

The total mass of all the ball bearings in the engine is 238 grams.

How many ball bearings in total are used in the engine?

Worked Solution
Let  2x\ 2\large x = number of small bearings
x\large x = number of large bearings

2x2\large x(0.190) + x\large x(0.300) = 238
0.38x0.38\large x + 0.3x\large x = 238
0.68x0.68\large x = 238
∴ x\therefore \ \large x = 2380.68\dfrac{238}{0.68}
= 350

∴\therefore Total ball bearings = (2 ×\times 350) + 350
= 1050

Question Type

Answer Box

Variables

Variable nameVariable value
question
A miniature engine contains two types of ball bearings. There are two times as many small ball bearings as large ball bearings. The mass of a small ball bearing is 190 milligrams and the mass of a large ball bearing is 300 milligrams. The total mass of all the ball bearings in the engine is 238 grams. How many ball bearings in total are used in the engine?
workedSolution
| | | | ------------: | ---------- | | Let $\ 2\large x$ | \= number of small bearings | | $\large x$ | \= number of large bearings |

| | | | ------------: | ---------- | | $2\large x$(0.190) + $\large x$(0.300) | \= 238 | | $0.38\large x$ + 0.3$\large x$ | \= 238 | | $0.68\large x$ | \= 238 | | $\therefore \ \large x$ | \= $\dfrac{238}{0.68}$ | | | \= 350 |

| | | | ------------: | ---------- | | $\therefore$ Total ball bearings | \= (2 $\times$ 350) + 350 | | | \= {{{correctAnswer0}}} |
correctAnswer0
1050
prefix0
suffix0
ball bearings

Answers

Specify one or more 'ANSWER' block(s) as exampled below.
Note: correctAnswer is required, the rest are optional. ("correctAnswer" is what the student would need to type in to the box to get the answer correct.)

For example:
correctAnswer: 123.40

And optionally, specify the following, but only if you need something different to the defaults: 'width' defaults to 5 if not present, and valid values are 3 to 10; 'prefix' and 'suffix' default to nothing if not present.
prefix: $
suffix: mm$^2$
width: 5

correctAnswerNcorrectAnswerValueAnswer
correctAnswer01050

ball bearings

Variant 1

DifficultyLevel

790

Question

A recipe contains two types of dry ingredients.

There is two times as much almond meal as plain flour.

The mass of a cup of almond meal is 84 grams and the mass of a cup of plain flour is 120 grams.

The total mass of all the dry ingredients in the recipe is 2.304 kilograms.

How many cups of dry ingredients in total are used in the recipe?

Worked Solution
Let  2x\ 2\large x = amount of almond meal
x\large x = amount of plain flour

2x2\large x(0.084) + x\large x(0.120) = 2.304
0.168x0.168\large x + 0.120x\large x = 2.304
0.288x0.288\large x = 2.304
∴ x\therefore \ \large x = 2.3040.288\dfrac{2.304}{0.288}
= 8

∴\therefore Total cups of dry ingredients = (2 ×\times 8) + 8
= 24

Question Type

Answer Box

Variables

Variable nameVariable value
question
A recipe contains two types of dry ingredients. There is two times as much almond meal as plain flour. The mass of a cup of almond meal is 84 grams and the mass of a cup of plain flour is 120 grams. The total mass of all the dry ingredients in the recipe is 2.304 kilograms. How many cups of dry ingredients in total are used in the recipe?
workedSolution
| | | | ------------: | ---------- | | Let $\ 2\large x$ | \= amount of almond meal | | $\large x$ | \= amount of plain flour |

| | | | ------------: | ---------- | | $2\large x$(0.084) + $\large x$(0.120) | \= 2.304 | | $0.168\large x$ + 0.120$\large x$ | \= 2.304 | | $0.288\large x$ | \= 2.304 | | $\therefore \ \large x$ | \= $\dfrac{2.304}{0.288}$ | | | \= 8 |

| | | | ------------: | ---------- | | $\therefore$ Total cups of dry ingredients | \= (2 $\times$ 8) + 8 | | | \= {{{correctAnswer0}}} |
correctAnswer0
24
prefix0
suffix0
cups

Answers

Specify one or more 'ANSWER' block(s) as exampled below.
Note: correctAnswer is required, the rest are optional. ("correctAnswer" is what the student would need to type in to the box to get the answer correct.)

For example:
correctAnswer: 123.40

And optionally, specify the following, but only if you need something different to the defaults: 'width' defaults to 5 if not present, and valid values are 3 to 10; 'prefix' and 'suffix' default to nothing if not present.
prefix: $
suffix: mm$^2$
width: 5

correctAnswerNcorrectAnswerValueAnswer
correctAnswer024

cups

Variant 2

DifficultyLevel

786

Question

A BMX race track is made up of straight sections and hill climb sections.

There are three times as many straight sections as hill climb sections.

The length of each straight section is 2380 metres and the length of each hill climb section is 860 metres.

The total length of all sections is 24 kilometres.

How many racing sections in total make up the BMX track?

Worked Solution
Let  3x\ 3\large x = number of straight sections
x\large x = number of hill climb sections

3x3\large x(2.38) + x\large x(0.86) = 24
7.14x7.14\large x + 0.86x\large x = 24
8x8\large x = 24
∴ x\therefore \ \large x = 248\dfrac{24}{8}
= 3

∴\therefore Total racing sections = (3 ×\times 3) + 3
= 12

Question Type

Answer Box

Variables

Variable nameVariable value
question
A BMX race track is made up of straight sections and hill climb sections. There are three times as many straight sections as hill climb sections. The length of each straight section is 2380 metres and the length of each hill climb section is 860 metres. The total length of all sections is 24 kilometres. How many racing sections in total make up the BMX track?
workedSolution
| | | | ------------: | ---------- | | Let $\ 3\large x$ | \= number of straight sections | | $\large x$ | \= number of hill climb sections |

| | | | ------------: | ---------- | | $3\large x$(2.38) + $\large x$(0.86) | \= 24 | | $7.14\large x$ + 0.86$\large x$ | \= 24 | | $8\large x$ | \= 24 | | $\therefore \ \large x$ | \= $\dfrac{24}{8}$ | | | \= 3 |

| | | | ------------: | ---------- | | $\therefore$ Total racing sections | \= (3 $\times$ 3) + 3 | | | \= {{{correctAnswer0}}} |
correctAnswer0
12
prefix0
suffix0
racing sections

Answers

Specify one or more 'ANSWER' block(s) as exampled below.
Note: correctAnswer is required, the rest are optional. ("correctAnswer" is what the student would need to type in to the box to get the answer correct.)

For example:
correctAnswer: 123.40

And optionally, specify the following, but only if you need something different to the defaults: 'width' defaults to 5 if not present, and valid values are 3 to 10; 'prefix' and 'suffix' default to nothing if not present.
prefix: $
suffix: mm$^2$
width: 5

correctAnswerNcorrectAnswerValueAnswer
correctAnswer012

racing sections

Variant 3

DifficultyLevel

778

Question

A herb garden contains two types of herbs.

There are four times as many parsley plants as oregano plants.

The parsley plant requires 120 millilitres of liquid fertiliser per week and an oregano plant requires 150 millilitres of liquid fertiliser per week.

The total number of litres of liquid fertiliser used in the herb garden each week is 50.4 litres.

How many herb plants in total are in the garden?

Worked Solution
Let  4x\ 4\large x = number of parsley plants
x\large x = number of oregano plants

4x4\large x(0.120) + x\large x(0.150) = 50.4
0.48x0.48\large x + 0.15x\large x = 50.4
0.63x0.63\large x = 50.4
∴ x\therefore \ \large x = 50.40.63\dfrac{50.4}{0.63}
= 80

∴\therefore Total herb plants = (4 ×\times 80) + 80
= 400

Question Type

Answer Box

Variables

Variable nameVariable value
question
A herb garden contains two types of herbs. There are four times as many parsley plants as oregano plants. The parsley plant requires 120 millilitres of liquid fertiliser per week and an oregano plant requires 150 millilitres of liquid fertiliser per week. The total number of litres of liquid fertiliser used in the herb garden each week is 50.4 litres. How many herb plants in total are in the garden?
workedSolution
| | | | ------------: | ---------- | | Let $\ 4\large x$ | \= number of parsley plants| | $\large x$ | \= number of oregano plants |

| | | | ------------: | ---------- | | $4\large x$(0.120) + $\large x$(0.150) | \= 50.4 | | $0.48\large x$ + 0.15$\large x$ | \= 50.4 | | $0.63\large x$ | \= 50.4 | | $\therefore \ \large x$ | \= $\dfrac{50.4}{0.63}$ | | | \= 80 |

| | | | ------------: | ---------- | | $\therefore$ Total herb plants | \= (4 $\times$ 80) + 80 | | | \= {{{correctAnswer0}}} |
correctAnswer0
400
prefix0
suffix0
herb plants

Answers

Specify one or more 'ANSWER' block(s) as exampled below.
Note: correctAnswer is required, the rest are optional. ("correctAnswer" is what the student would need to type in to the box to get the answer correct.)

For example:
correctAnswer: 123.40

And optionally, specify the following, but only if you need something different to the defaults: 'width' defaults to 5 if not present, and valid values are 3 to 10; 'prefix' and 'suffix' default to nothing if not present.
prefix: $
suffix: mm$^2$
width: 5

correctAnswerNcorrectAnswerValueAnswer
correctAnswer0400

herb plants

Variant 4

DifficultyLevel

782

Question

A diamond necklace contains two types of diamonds.

There are three times as many square cut diamonds as round cut diamonds.

The size of each square cut diamond is 0.9 carats and the size of each round cut diamond is 0.75 carats.

The total mass, in carats, of all the diamonds in the necklace is 51.75.

How many diamonds in total are used in the necklace?

Worked Solution
Let  3x\ 3\large x = number of square cut diamonds
x\large x = number of round cut diamonds

3x3\large x(0.90) + x\large x(0.75) = 51.75
2.7x2.7\large x + 0.75x\large x = 51.75
3.45x3.45\large x = 51.75
∴ x\therefore \ \large x = 51.753.45\dfrac{51.75}{3.45}
= 15

∴\therefore Total diamonds = (3 ×\times 15) + 15
= 60

Question Type

Answer Box

Variables

Variable nameVariable value
question
A diamond necklace contains two types of diamonds. There are three times as many square cut diamonds as round cut diamonds. The size of each square cut diamond is 0.9 carats and the size of each round cut diamond is 0.75 carats. The total mass, in carats, of all the diamonds in the necklace is 51.75. How many diamonds in total are used in the necklace?
workedSolution
| | | | ------------: | ---------- | | Let $\ 3\large x$ | \= number of square cut diamonds | | $\large x$ | \= number of round cut diamonds |

| | | | ------------: | ---------- | | $3\large x$(0.90) + $\large x$(0.75) | \= 51.75 | | $2.7\large x$ + 0.75$\large x$ | \= 51.75 | | $3.45\large x$ | \= 51.75 | | $\therefore \ \large x$ | \= $\dfrac{51.75}{3.45}$ | | | \= 15 |

| | | | ------------: | ---------- | | $\therefore$ Total diamonds | \= (3 $\times$ 15) + 15 | | | \= {{{correctAnswer0}}} |
correctAnswer0
60
prefix0
suffix0
diamonds

Answers

Specify one or more 'ANSWER' block(s) as exampled below.
Note: correctAnswer is required, the rest are optional. ("correctAnswer" is what the student would need to type in to the box to get the answer correct.)

For example:
correctAnswer: 123.40

And optionally, specify the following, but only if you need something different to the defaults: 'width' defaults to 5 if not present, and valid values are 3 to 10; 'prefix' and 'suffix' default to nothing if not present.
prefix: $
suffix: mm$^2$
width: 5

correctAnswerNcorrectAnswerValueAnswer
correctAnswer060

diamonds

Variant 5

DifficultyLevel

780

Question

A coffee shop sells two types of muffins.

There are two times as many blueberry muffins as choc chip muffins.

The energy, in kilojoules, of a blueberry muffin is 2140 kilojoules and the energy of a choc chip muffin is 2470 kilojoules.

The total energy of all the muffins in the coffee shop is 101 250101\ 250 kilojoules.

How many muffins in total are for sale in the coffee shop?

Worked Solution
Let  2x\ 2\large x = number of blueberry muffins
x\large x = number of choc chip muffins

2x2\large x(2140) + x\large x(2470) = 101 250
4280x4280\large x + 2470x\large x = 101 250
6750x6750\large x = 101 250101\ 250
∴ x\therefore \ \large x = 101 2506750\dfrac{101\ 250}{6750}
= 15

∴\therefore Total muffins = (2 ×\times 15) + 15
= 45

Question Type

Answer Box

Variables

Variable nameVariable value
question
A coffee shop sells two types of muffins. There are two times as many blueberry muffins as choc chip muffins. The energy, in kilojoules, of a blueberry muffin is 2140 kilojoules and the energy of a choc chip muffin is 2470 kilojoules. The total energy of all the muffins in the coffee shop is $101\ 250$ kilojoules. How many muffins in total are for sale in the coffee shop?
workedSolution
| | | | ------------: | ---------- | | Let $\ 2\large x$ | \= number of blueberry muffins | | $\large x$ | \= number of choc chip muffins |

| | | | ------------: | ---------- | | $2\large x$(2140) + $\large x$(2470) | \= 101 250 | | $4280\large x$ + 2470$\large x$ | \= 101 250 | | $6750\large x$ | \= $101\ 250$ | | $\therefore \ \large x$ | \= $\dfrac{101\ 250}{6750}$ | | | \= 15 |

| | | | ------------: | ---------- | | $\therefore$ Total muffins | \= (2 $\times$ 15) + 15 | | | \= {{{correctAnswer0}}} |
correctAnswer0
45
prefix0
suffix0
muffins

Answers

Specify one or more 'ANSWER' block(s) as exampled below.
Note: correctAnswer is required, the rest are optional. ("correctAnswer" is what the student would need to type in to the box to get the answer correct.)

For example:
correctAnswer: 123.40

And optionally, specify the following, but only if you need something different to the defaults: 'width' defaults to 5 if not present, and valid values are 3 to 10; 'prefix' and 'suffix' default to nothing if not present.
prefix: $
suffix: mm$^2$
width: 5

correctAnswerNcorrectAnswerValueAnswer
correctAnswer045

muffins

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