Difference between revisions of "2002 AMC 8 Problems/Problem 14"
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| − | + | == Problem == | |
| − | == 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 and claims that the final price of these items is <math>50\%</math> off the original price. The total discount is | A merchant offers a large group of items at <math>30\%</math> off. Later, the merchant takes <math>20\%</math> off these sale prices and claims that the final price of these items is <math>50\%</math> off the original price. The total discount is | ||
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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> | ||
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| + | Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</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. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2002|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Revision as of 21:56, 29 June 2021
Contents
Problem
A merchant offers a large group of items at 
off. Later, the merchant takes 
off these sale prices and claims that the final price of these items is 
off the original price. The total discount is
Solution
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.
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. 
