Difference between revisions of "1997 AJHSME Problems/Problem 18"
(→Solution) |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for <math>5. This week they are on sale at 5 boxes for </math> | + | At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for <math>\$5</math>. This week they are on sale at 5 boxes for <math>\$4</math>. The percent decrease in the price per box during the sale was closest to |
<math>\text{(A)}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 65\%</math> | <math>\text{(A)}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 65\%</math> | ||
Line 11: | Line 11: | ||
This week, each box is <math>\frac{4}{5} = 0.80</math> | This week, each box is <math>\frac{4}{5} = 0.80</math> | ||
− | Percent decrease is given by <math>\frac{X_{old} - X_}{X_}} \cdot 100\%</math> | + | Percent decrease is given by <math>\frac{X_{old} - X_{new}}{X_{old}} \cdot 100\%</math> |
This, the percent decrease is <math>\frac{1.25 - 0.8}{1.25}\cdot 100\% = \frac{45}{125} \cdot 100\% = 36\%</math>, which is closest to <math>\boxed{B}</math> | This, the percent decrease is <math>\frac{1.25 - 0.8}{1.25}\cdot 100\% = \frac{45}{125} \cdot 100\% = 36\%</math>, which is closest to <math>\boxed{B}</math> |
Latest revision as of 18:49, 31 October 2016
Problem
At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for . This week they are on sale at 5 boxes for . The percent decrease in the price per box during the sale was closest to
Solution
Last week, each box was
This week, each box is
Percent decrease is given by
This, the percent decrease is , which is closest to
See also
1997 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.