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

(Solution #1)
(Solution #2)
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==Solution #2==
 
==Solution #2==
  
Assume the price was <math>&#036;100</math>. We can just do <math>100\cdot0.7\cdot0.8=56</math> and then do <math>100-56=44</math> That is the discount percentage wise.
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Assume the price was &#036;100. We can just do <math>100\cdot0.7\cdot0.8=56</math> and then do <math>100-56=44</math> That is the discount percentage wise.
  
 
<math> \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math>
 
<math> \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math>

Revision as of 22:07, 29 July 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 $&#036;100$. First we take off $30\%$ off of $&#036;100$ $&#036;100\cdot0.70=&#036;70$ (Error compiling LaTeX. Unknown error_msg)

Next, we take off the extra $20\%$ as asked by the problem. $&#036;70\cdot0.80=&#036;56$ (Error compiling LaTeX. Unknown error_msg)

So the final price of an item is $&#036;56$. 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.

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