Difference between revisions of "2012 AMC 8 Problems/Problem 8"
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<math> \textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70 </math> | <math> \textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70 </math> | ||
| − | + | ==Solution 1: With Algebra== | |
| − | |||
Let the original price of an item be <math>x</math>. | Let the original price of an item be <math>x</math>. | ||
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The price of the item after all discounts have been applied is <math>0.4x = 40\% \cdot x</math>. However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract <math>40\% x</math> from <math>100\% x</math> (the original price of the item), to find the answer, <math> \boxed{\textbf{(D)}\ 60} </math>. | The price of the item after all discounts have been applied is <math>0.4x = 40\% \cdot x</math>. However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract <math>40\% x</math> from <math>100\% x</math> (the original price of the item), to find the answer, <math> \boxed{\textbf{(D)}\ 60} </math>. | ||
| − | + | ==Solution 2: Fakesolving== | |
Since the problem implies that the percentage off the original price will be the same for every item in the store, [[fakesolving]] is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of <math>\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{\textbf{(D)}\ 60} \%</math>. | Since the problem implies that the percentage off the original price will be the same for every item in the store, [[fakesolving]] is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of <math>\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{\textbf{(D)}\ 60} \%</math>. | ||
Revision as of 17:16, 15 February 2021
Problem
A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
Solution 1: With Algebra
Let the original price of an item be 
.
First, everything is half-off, so the price is now 
.
Next, the extra coupon applies 20% off on the sale price, so the price after this discount will be 
of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)
The price of the item after all discounts have been applied is 
. However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract 
from 
(the original price of the item), to find the answer, 
.
Solution 2: Fakesolving
Since the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of 
.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
