1992 AJHSME Problems/Problem 8: Difference between revisions
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== Solution == | == Solution == | ||
<math> 1500\times 0.1=150 </math>, so the store owner is <dollar/>150 below profit. Therefore he needs to sell 150+100=< | <math> 1500\times 0.1=150 </math>, so the store owner is <dollar/>150 below profit. Therefore he needs to sell <math>150+100= 250</math> dollars worth of pencils. Selling them at <dollar/>0.25 each gives <math>250/0.25= \boxed{\textbf{(C)}\ 1000}</math>. | ||
==See Also== | |||
{{AJHSME box|year=1992|num-b=7|num-a=9}} | |||
Revision as of 20:30, 22 December 2012
Problem
A store owner bought 
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?
Solution
, so the store owner is <dollar/>150 below profit. Therefore he needs to sell 
dollars worth of pencils. Selling them at <dollar/>0.25 each gives 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
