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

m (Solution 2)
(Solution)
 
Line 14: Line 14:
  
 
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.
 
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> so the final discount was <math>\boxed{\text{(B)}\ 44\%}</math>.
 
 
===Solution 2===
 
 
Assume the price was \$100. We can just do <math>100\cdot0.7\cdot0.8=56</math> and then do <math>100-56=\boxed{\text{(B)}\ 44\%}</math>. That is the discount percentage wise.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=13|num-a=15}}
 
{{AMC8 box|year=2002|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:56, 29 June 2021

Problem

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

Solution 1

Let's assume that each item is $100$ dollars. First we take off $30\%$ off of $100$ dollars. $100\cdot0.7=70$

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

So the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS