Difference between revisions of "1997 AJHSME Problems/Problem 18"
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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_ | + | Percent decrease is given by <math>\frac{X_{old} - X_}{X_}} \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> | ||
Revision as of 20:05, 26 October 2016
Problem
At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for 
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 $\frac{X_{old} - X_}{X_}} \cdot 100\%$ (Error compiling LaTeX. Unknown error_msg)
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. 
