1997 AJHSME Problems/Problem 18

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Problem

At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for $$5$. This week they are on sale at 5 boxes for $$4$. The percent decrease in the price per box during the sale was closest to

$\text{(A)}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 65\%$

Solution

Last week, each box was $\frac{5}{4} = 1.25$

This week, each box is $\frac{4}{5} = 0.80$

Percent decrease is given by $\frac{X_{old} - X_{new}}{X_{old}} \cdot 100\%$

This, the percent decrease is $\frac{1.25 - 0.8}{1.25}\cdot 100\% = \frac{45}{125} \cdot 100\% = 36\%$, which is closest to $\boxed{B}$

See also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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