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

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<math>100-56=44</math> so the final discount was <math>44\%</math> | <math>100-56=44</math> so the final discount was <math>44\%</math> | ||

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+ | <math> \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math> | ||

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+ | ==Solution #2== | ||

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+ | Assume the price was <math>$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. | ||

<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 19:42, 28 July 2012

## Problem 14

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 #1

Let's assume that each item is . First we take off off of $$100\cdot0.70=$70$ (Error compiling LaTeX. ! Missing $ inserted.)

Next, we take off the extra as asked by the problem. $$70\cdot0.80=$56$ (Error compiling LaTeX. ! Missing $ inserted.)

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

so the final discount was

## Solution #2

Assume the price was . We can just do and then do That is the discount percentage wise.