Difference between revisions of "1950 AHSME Problems/Problem 22"
Hastapasta (talk | contribs) (→Solution 2 (Technical)) |
Hastapasta (talk | contribs) (→Solution 2 (Technical)) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 10: | Line 10: | ||
==Solution 2 (Technical)== | ==Solution 2 (Technical)== | ||
| − | Let the object cost <math>x</math> dollars. After the 10% discount, it's worth <math>(1- | + | 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 | ||
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 | |
| 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. 
