randRange( 2, 10 ) randRangeExclude( 2, 15, [ N1 ] ) ( N1 * randRange( 80, 199 ) / 100 ).toFixed( 2 ) (function() { var solutions = [ "\\dfrac{" + N1 + "}{\\\$" + C + "} = \\dfrac{" + N2 + "}{x}", "\\dfrac{" + N2 + "}{x} = \\dfrac{" + N1 + "}{\\\$" + C + "}", "\\dfrac{" + N1 + "}{" + N2 + "} = \\dfrac{\\\$" + C + "}{x}", "\\dfrac{x}{" + N2 + "} = \\dfrac{\\\$" + C + "}{" + N1 + "}" ] return solutions; })() randRange( 0, SOLUTIONS.length - 1 ) SOLUTIONS[ IDX ]

plural( N1, deskItem( 0 ) ) cost \$C.

Which equation would help determine the cost of plural( N2, deskItem( 0 ) )?

`SOLUTION`

• `\dfrac{N2}{\\$C} = \dfrac{x}{N1}`
• `\dfrac{N2}{N1} = \dfrac{\\$C}{x}`
• `\dfrac{N1}{N2} = \dfrac{x}{\\$C}`
• `\dfrac{x}{N2} = \dfrac{N1}{\\$C}`
• `\dfrac{N2}{x} = \dfrac{\\$C}{N1}`

There are several equations that could help determine the cost, each with a slightly different approach.

We can write the fact that plural( N1, deskItem( 0 ) ) cost \$C as a proportion:

`\dfrac{N1}{\\$C}`

Let `x` represent the unknown cost of plural( N2, deskItem( 0 ) ). Since plural( N2, deskItem( 0 ) ) cost `x`, we have the following proportion:

`\dfrac{N2}{x}`

The cost changes along with the number of deskItem( 0 )s purchased, and so the two proportions are equivalent.

Let `x` represent the unknown cost of plural( N2, deskItem( 0 ) ). Since plural( N2, deskItem( 0 ) ) cost `x`, we have the following proportion:

`\dfrac{N2}{x}`

We can write the fact that plural( N1, deskItem( 0 ) ) cost \$C as a proportion:

`\dfrac{N1}{\\$C}`

The cost changes along with the number of deskItem( 0 )s purchased, and so the two proportions are equivalent.

We know the cost of plural( N1, deskItem( 0 ) ), and we want to know the cost of plural( N2, deskItem( 0 ) ). We can write the numbers of deskItem( 0 )s as a proportion:

`\dfrac{N1}{N2}`

We know plural( N1, deskItem( 0 ) ) costs \$C, and we can let `x` represent the unknown cost of plural( N2, deskItem( 0 ) ). The proportion of these costs can be expressed as:

`\dfrac{\\$C}{x}`

The cost changes along with the number of deskItem( 0 )s purchased, and so the two proportions are equivalent.

If we let `x` represent the cost of plural( N2, deskItem( 0 ) ), we have the following proportion:

`\dfrac{x}{N2}`

We have to pay \$C for plural( N1, deskItem( 0 ) ), and that can be written as a proportion:

`\dfrac{\\$C}{N1}`

Since the price per folder stays the same, these two proportions are equivalent.

`SOLUTIONS[ IDX ]`