Difference between revisions of "2002 AMC 8 Problems/Problem 14"
Sinhaarunabh (talk | contribs) (→Problem 14) |
Sinhaarunabh (talk | contribs) (→Solution #1) |
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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 $$100\cdot0.70=$70$ (Error compiling LaTeX. )
Next, we take off the extra as asked by the problem. $$70\cdot0.80=$56$ (Error compiling LaTeX. )
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.
