Difference between revisions of "1997 AJHSME Problems/Problem 18"

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(Solution)
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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_{new}}{X_{old}} \cdot 100\%</math>
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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 21:05, 26 October 2016

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 $\frac{X_{old} - X_}{X_}} \cdot 100\%$ (Error compiling LaTeX. ! Missing { inserted.)

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
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

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