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

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==Solution #1== | ==Solution #1== | ||

− | Let's assume that each item is | + | Let's assume that each item is $100. First we take off <math>30\%</math> off of $100. $100\cdot0.70=<math> $70 |

− | Next, we take off the extra <math>20\%< | + | Next, we take off the extra </math>20\%<math> as asked by the problem. $70\cdot0.80=$56 |

− | So the final price of an item is | + | 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. |

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

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

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

## Revision as of 22:09, 29 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 $100. First we take off off of $100. $100\cdot0.70=$$70

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

So the final price of an item is $56. We have to do$ (Error compiling LaTeX. )100-5656100-56=4444\%$$ (Error compiling LaTeX. ) \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 and then do That is the discount percentage wise.