Difference between revisions of "1950 AHSME Problems/Problem 22"
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so the single discount would be <math>\boxed{\mathrm{(D)}\ 28\%.}</math> | so the single discount would be <math>\boxed{\mathrm{(D)}\ 28\%.}</math> | ||
− | ==Solution | + | ==Solution 3 (Technical)== |
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>. | 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>. |
Revision as of 10:42, 21 September 2025
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
Total discount if the Product is dollars is 10/100 of 100 + 20/100 of (100 - 10/100 of 100)
= 10/100 * 100 + 20/100 * (100 - 10/100 * 100) = 10 + 2/10 * 90 = 10 + 9*2 = 10+18 = 28
So discount is dollars
so the single discount would be
Solution 3 (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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.