Difference between revisions of "2002 AMC 8 Problems/Problem 14"

(Solution #1)
(Solution #1)
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==Solution #1==
 
==Solution #1==
  
Let's assume that each item is <math>100</math> dollars. First we take off <math>30\%</math> off of &#036;100. <math>&#036;100\cdot0.70=</math> &#036;70
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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>&#036;100\cdot0.70=</math> &#036;70
  
 
Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math>
 
Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math>

Revision as of 07:56, 14 August 2012

Problem 14

A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices and claims that the final price of these items is $50\%$ off the original price. The total discount is


$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$

Solution #1

Let's assume that each item is $100$ dollars. First we take off $30\%$ off of $100$ dollars. $&#036;100\cdot0.70=$ $70

Next, we take off the extra $20\%$ as asked by the problem. $70\cdot0.80=56$

So the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount.

$100-56=44$ so the final discount was $44\%$

$\text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$

Solution #2

Assume the price was $100. We can just do $100\cdot0.7\cdot0.8=56$ and then do $100-56=44$ That is the discount percentage wise.

$\text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$

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