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== |
− | + | ===Solution=== | |
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− | <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> |
− | = | + | Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</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>B</math>. | |
− | + | ==See Also== | |
+ | {{AMC8 box|year=2002|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:19, 22 December 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
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 .
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 |
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