Difference between revisions of "1997 AJHSME Problems/Problem 18"
Talkinaway (talk | contribs) (Created page with "==Problem== At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for <dollar/>5. This week they are on sale at 5 boxes for <dollar/>4. The perce...") |
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==Solution== | ==Solution== | ||
| − | Last week, each box was <math>\frac{5}{4} = 1. | + | Last week, each box was <math>\frac{5}{4} = 1.25</math> |
This week, each box is <math>\frac{4}{5} = 0.80</math> | This week, each box is <math>\frac{4}{5} = 0.80</math> | ||
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Percent decrease is given by <math>\frac{X_{old} - X_{new}}{X_{old}} \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. | + | 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> |
== See also == | == See also == | ||
Revision as of 03:00, 23 December 2012
Problem
At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for <dollar/>5. This week they are on sale at 5 boxes for <dollar/>4. 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 | ||
