Difference between revisions of "1992 AJHSME Problems/Problem 8"

(Created page with "1500*0.1=150, so the store owner is <math>150 below profit. Therefore he needs to sell 150+100=</math>250 worth of pencils. Selling them at $0.25 each gives 250/0.25=1000 which i...")
 
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1500*0.1=150, so the store owner is <math>150 below profit. Therefore he needs to sell 150+100=</math>250 worth of pencils. Selling them at $0.25 each gives 250/0.25=1000 which is choice C.
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== Problem ==
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A store owner bought <math>1500</math> pencils at <dollar/>0.10 each.  If he sells them for <dollar/>0.25 each, how many of them must he sell to make a profit of exactly <dollar/>100.00?
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<math>\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900</math>
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== Solution ==
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<math> 1500\times 0.1=150 </math>, so the store owner is <dollar/>150 below profit. Therefore he needs to sell 150+100=<dollar/>250 worth of pencils. Selling them at <dollar/>0.25 each gives 250/0.25=1000 which is choice C.

Revision as of 15:29, 29 July 2011

Problem

A store owner bought $1500$ pencils at <dollar/>0.10 each. If he sells them for <dollar/>0.25 each, how many of them must he sell to make a profit of exactly <dollar/>100.00?

$\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900$

Solution

$1500\times 0.1=150$, so the store owner is <dollar/>150 below profit. Therefore he needs to sell 150+100=<dollar/>250 worth of pencils. Selling them at <dollar/>0.25 each gives 250/0.25=1000 which is choice C.