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 | + | == Problem == |
+ | |||
+ | A store owner bought <math>1500</math> pencils at <math>\$ 0.10</math> each. If he sells them for <math>\$ 0.25</math> each, how many of them must he sell to make a profit of exactly <math>\$ 100.00</math>? | ||
+ | |||
+ | <math>\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900</math> | ||
+ | |||
+ | == Solution == | ||
+ | <math> 1500\times 0.1=150 </math>, so the store owner is <math>\$150</math> below profit. Therefore he needs to sell <math>150+100= 250</math> dollars worth of pencils. Selling them at <math>\$0.25</math> each gives <math>250/0.25= \boxed{\textbf{(C)}\ 1000}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1992|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:12, 3 July 2017
Problem
A store owner bought pencils at
each. If he sells them for
each, how many of them must he sell to make a profit of exactly
?
Solution
, so the store owner is
below profit. Therefore he needs to sell
dollars worth of pencils. Selling them at
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.