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

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== Problem ==
  
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== Problem 14 ==
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A merchant offers a large group of items at <math>30\%</math> off. Later, the merchant takes <math>20\%</math> off these sale prices. The total discount is
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A merchant offers a large group of items at <math>30\%</math> off. Later, the merchant takes <math>20\%</math> off these sale prices and claims that the final price of these items is <math>50\%</math> off the original price. The total discount is
 
  
  
 
<math> \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math>
 
<math> \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math>
  
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==Solution #1==
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==Solution 1==
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Let's assume that each item is <math>100</math> dollars. First we take off <math>30\%</math> off of <math>100</math> dollars. <math>100\cdot0.7=70</math>
  
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Let's assume that each item is <math>100</math> dollars. First we take off <math>30\%</math> off of &#036;100. <math>&#036;100\cdot0.70=</math> &#036;70
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Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math>
  
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Next, we take off the extra <math>20\%</math> as asked by the problem. <math>70\cdot0.80=56</math>
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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>.
  
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So the final price of an item is &#036;56. We have to do <math>100-56</math> because <math>56</math> was the final price and we wanted the discount.
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==Solution 2==
  
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<math>100-56=44</math> so the final discount was <math>44\%</math>
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Let <math>x</math> be the price of an item on sale. When the item is <math>30\%</math> off, its new price is
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<cmath>(1-0.3)x=0.7x.</cmath>
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When <math>20\%</math> is taken off of that price, the item's final sales price is
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<cmath>(1-0.2)\cdot0.7x=0.8\cdot0.7x=0.56x=(1-0.44)x.</cmath>
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Therefore, the item was <math>44\%</math> off, so the answer is <math>\boxed{(B)}</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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==Video Solution==
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https://youtu.be/DUqszaQ01lM Soo, DRMS, NM
  
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==Solution #2==
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https://www.youtube.com/watch?v=UR2aLmJHoIs  ~David
  
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Assume the price was &#036;100. 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.
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==Video Solution by WhyMath==
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https://youtu.be/zLXShtJzFM4
  
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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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==See Also==
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{{AMC8 box|year=2002|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 13:32, 29 October 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 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. So the final answer is $44\%$, which is answer choice $\boxed{(B) 44 \%}$.

Solution 2

Let $x$ be the price of an item on sale. When the item is $30\%$ off, its new price is \[(1-0.3)x=0.7x.\] When $20\%$ is taken off of that price, the item's final sales price is \[(1-0.2)\cdot0.7x=0.8\cdot0.7x=0.56x=(1-0.44)x.\] Therefore, the item was $44\%$ off, so the answer is $\boxed{(B)}$.

Video Solution

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

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

Video Solution by WhyMath

https://youtu.be/zLXShtJzFM4

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
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

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