Difference between revisions of "1953 AHSME Problems/Problem 2"
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The first discount takes off <math>20\%</math> of the price, so the cost of the refrigerator is <math>0.8\cdot250</math>. The next discount takes off <math>15\%</math>, so the cost of the refrigerator is now <math>0.8\cdot0.85\cdot250=0.68\cdot250</math>. Thus, the sale price of the refrigerator is <math>\boxed{\textbf{(D)} \text{68\% of 250.00}}</math>. | The first discount takes off <math>20\%</math> of the price, so the cost of the refrigerator is <math>0.8\cdot250</math>. The next discount takes off <math>15\%</math>, so the cost of the refrigerator is now <math>0.8\cdot0.85\cdot250=0.68\cdot250</math>. Thus, the sale price of the refrigerator is <math>\boxed{\textbf{(D)} \text{68\% of 250.00}}</math>. | ||
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| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 19:15, 1 April 2017
Problem
A refrigerator is offered at sale at $250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:
Solution
The first discount takes off 
of the price, so the cost of the refrigerator is 
. The next discount takes off 
, so the cost of the refrigerator is now 
. Thus, the sale price of the refrigerator is 
.
See Also
| 1953 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
