Difference between revisions of "2002 AMC 8 Problems/Problem 14"
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<math> \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math> | <math> \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let's assume that each item is <math>100</math> dollars. First we take off <math>30\%</math> off of <math>100</math> dollars. <math>100\cdot0.7=70</math> | Let's assume that each item is <math>100</math> dollars. First we take off <math>30\%</math> off of <math>100</math> dollars. <math>100\cdot0.7=70</math> | ||
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So the final price of an item is \$56. We have to do <math>100-56</math> because <math>56</math> was the final price and we wanted the discount. So the final answer is <math>44\%</math>, which is answer choice <math>\boxed{(B) 44 \%}</math>. | So the final price of an item is \$56. We have to do <math>100-56</math> because <math>56</math> was the final price and we wanted the discount. So the final answer is <math>44\%</math>, which is answer choice <math>\boxed{(B) 44 \%}</math>. | ||
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| + | ==Solution 2== | ||
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| + | Let <math>x</math> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 00:04, 9 July 2024
Problem
A merchant offers a large group of items at 
off. Later, the merchant takes 
off these sale prices. The total discount is
Solution 1
Let's assume that each item is 
dollars. First we take off off of dollars. 
Next, we take off the extra as asked by the problem. 
So the final price of an item is $56. We have to do 
because 
was the final price and we wanted the discount. So the final answer is 
, which is answer choice 
.
Solution 2
Let 
Video Solution
https://youtu.be/DUqszaQ01lM Soo, DRMS, NM
https://www.youtube.com/watch?v=UR2aLmJHoIs ~David
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
