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 ]