2012 AMC 8 Problems/Problem 8
Contents
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 .
Solution 3:
The percent off would be 50% of 20% because it has an additional 20% off of half the price. The answer to that would be 10%. You could do 10% + 50% to find the answer. This answer is (D)60
~ SmartGrowth
Solution 4:
The price is off, which means the price would be of what it was before. Then, you get off. Sidenote: A common mistake is finding of the price, but forgetting to subtract it. off the price is Now, we can multiply the two discounts as decimals. Our final answer is
~ ENDERGAMER25/Endy
Video Solution
https://youtu.be/SYou4OZcq1E ~savannahsolver
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 |
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