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 21:09, 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=$$70
Next, we take off the extra$ (Error compiling LaTeX. Unknown error_msg)20\%$as asked by the problem. $70\cdot0.80=$56
So the final price of an item is $56. We have to do$ (Error compiling LaTeX. Unknown error_msg)100-56
56
100-56=44
44\%$$ (Error compiling LaTeX. Unknown error_msg) \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.
