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

(Video Solution)
(Solution)
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Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math>
 
Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math>
  
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 answer is <math>44\%</math>, which is answer choice <math>\boxed{B}</math>.
+
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 answer is <math>44\%</math>, which is answer choice <math>\boxed{(B) 44 \%}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 19:32, 11 January 2024

Problem

A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices. The total discount is


$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$

Solution

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. So the final answer is $44\%$, which is answer choice $\boxed{(B) 44 \%}$.

Video Solution

https://youtu.be/DUqszaQ01lM Soo, DRMS, NM

https://www.youtube.com/watch?v=UR2aLmJHoIs ~David

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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