Difference between revisions of "1950 AHSME Problems/Problem 22"
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− | Let the object cost <math>x</math> dollars. After the <math>10%</math> discount, it's worth <math>(1-10%)x=0.9x</math> dollars. After a <math>20%</math> discount on that, it's worth <math>(1-20%)(0.9x)=0.72x</math>. Say the single discount is of <math>k</math>. Then <math>(1-k)x=0.72x</math>. So <math>k=0.28</math>, or <math>k=28%</math>. So select <math>\boxed{D}</math>. | + | Let the object cost <math>x</math> dollars. After the <math>10\%</math> discount, it's worth <math>(1-10\%)x=0.9x</math> dollars. After a <math>20\%</math> discount on that, it's worth <math>(1-20\%)(0.9x)=0.72x</math> dollars. Say the single discount is of <math>k</math>. Then <math>(1-k)x=0.72x</math>. So <math>k=0.28</math>, or <math>k=28\%</math>. So select <math>\boxed{D}</math>. |
~hastapasta | ~hastapasta | ||
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==See Also== | ==See Also== | ||
Latest revision as of 11:16, 31 March 2022
Problem
Successive discounts of and are equivalent to a single discount of:
Solution 1 (Kind of Lame)
Without loss of generality, assume something costs dollars. Then with each successive discount, it would cost dollars, then dollars. This amounts to a total of dollars off, so the single discount would be
Solution 2 (Technical)
Let the object cost dollars. After the discount, it's worth dollars. After a discount on that, it's worth dollars. Say the single discount is of . Then . So , or . So select .
~hastapasta
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AHSME Problems and Solutions |
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