# The number of calculators Mrs. Hopkins can buy for the classroom varies inversely as the cost of...

## Question:

The number of calculators Mrs. Hopkins can buy for the classroom varies inversely as the cost of each calculator. She can buy 24 calculators that cost $60 each. How many calculators can she buy if they cost$80 each?

## Inverse Variation:

In mathematics, if y varies inversely as x, we have that {eq}y=\frac{k}{x} {/eq}. In this inverse variation equation, k is a constant called the constant of variation. We can solve inverse variation problems by finding the constant of variation, plugging the constant of variation into the inverse variation equation, and then using the inverse variation equation to answer the problem.

Mrs. Hopkins can buy 18 calculators if they cost $80 each. We are given that the number of calculators that Mrs. Hopkins can buy for the classroom varies inversely as the cost of each calculator. If we let y be the number of calculators she can buy, and we let x be the cost of each calculator, then we have that y varies inversely as x, so {eq}y=\frac{k}{x} {/eq}. The problem states that Mrs. Hopkins can buy 24 calculators that cost$60 each. Therefore, when y = 24, x = 60. Plugging these into our inverse variation equation gives {eq}24=\frac{k}{60} {/eq}. We can solve this for k to find our constant of variation.

• {eq}24=\frac{k}{60} {/eq}

Multiply each side of the equation by 60.

• {eq}1440=k {/eq}

We get that k = 1440, so our inverse variation equation for this problem is {eq}y=\frac{1440}{x} {/eq}. We can use this equation to solve the problem.

We want to know how many calculators Mrs. Hopkins can buy if they cost $80 each, so we want to know what y is when x = 80. Thus, we plug x = 80 into our inverse variation equation, and solve for y. • {eq}y=\frac{1440}{x} {/eq} Plug in x = 80. • {eq}y=\frac{1440}{80} {/eq} Simplify. • {eq}y=18 {/eq} We get that y = 18 when x = 80, so Mrs. Hopkins can buy 18 calculators for the classroom when the cost of each calculator is$80. 