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

(Solution #1)
Line 1: Line 1:
 
+
== Problem ==
== Problem 14 ==
 
  
 
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
 
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
Line 7: Line 6:
 
<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>
  
==Solution #1==
+
==Solution==
 +
===Solution 1===
  
 
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>
 
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>
Line 15: Line 15:
 
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.
 
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.
  
<math>100-56=44</math> so the final discount was <math>44\%</math>
+
<math>100-56=44</math> so the final discount was <math>\boxed{\text{(B)}\ 44\%}</math>.
 
 
<math> \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math>
 
  
==Solution #2==
+
===Solution 2===
  
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.
+
Assume the price was &#036;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.
  
<math> \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% </math>
+
==See Also==
 +
{{AMC8 box|year=2002|num-b=13|num-a=15}}

Revision as of 18:45, 23 December 2012

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.

$100-56=44$ so the final discount was $\boxed{\text{(B)}\ 44\%}$.

Solution 2

Assume the price was $100. We can just do $100\cdot0.7\cdot0.8=56$ and then do $100-56=\boxed{\text{(B)}\ 44}$. That is the discount percentage wise.

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
Invalid username
Login to AoPS