Difference between revisions of "2002 AMC 8 Problems/Problem 14"
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+ | == Problem == | ||
− | + | A merchant offers a large group of items at <math>30\%</math> off. Later, the merchant takes <math>20\%</math> off these sale prices. The total discount is | |
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− | A merchant offers a large group of items at <math>30\%</math> off. Later, the merchant takes <math>20\%</math> off these sale prices | ||
<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== |
− | + | 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> | ||
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− | <math> | + | Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math> |
− | <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}</math>. |
− | ==Solution | + | ==Video Solution== |
+ | https://youtu.be/DUqszaQ01lM Soo, DRMS, NM | ||
− | + | https://www.youtube.com/watch?v=UR2aLmJHoIs | |
− | + | ==See Also== | |
+ | {{AMC8 box|year=2002|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:33, 4 December 2022
Contents
Problem
A merchant offers a large group of items at off. Later, the merchant takes off these sale prices. The total discount is
Solution
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 .
Video Solution
https://youtu.be/DUqszaQ01lM Soo, DRMS, NM
https://www.youtube.com/watch?v=UR2aLmJHoIs
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.