Difference between revisions of "1969 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | Let <math>c</math> be the cost. If selling the item for <math>x</math> dollars equates to losing <math>15\%</math> of <math>c</math>, then <math>x=.85c</math>. If selling the item for <math>y</math> dollars equates to profiting by <math>15\%</math> of <math>c</math>, then <math>y=1.15c</math>. Therefore <math>\frac{y}{x}=\frac{1.15c}{.85c}=23 | + | Let <math>c</math> be the cost. If selling the item for <math>x</math> dollars equates to losing <math>15\%</math> of <math>c</math>, then <math>x=.85c</math>. If selling the item for <math>y</math> dollars equates to profiting by <math>15\%</math> of <math>c</math>, then <math>y=1.15c</math>. Therefore <math>\frac{y}{x}=\frac{1.15c}{.85c}=\frac{23}{17}</math>. The answer is <math>\fbox{A}</math>. |
== See also == | == See also == | ||
Revision as of 17:15, 10 July 2015
Problem
If an item is sold for 
dollars, there is a loss of 
based on the cost. If, however, the same item is sold for 
dollars, there is a profit of based on the cost. The ratio of 
is:
Solution
Let 
be the cost. If selling the item for dollars equates to losing of , then 
. If selling the item for dollars equates to profiting by of , then 
. Therefore 
. The answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
