2012 AMC 8 Problems/Problem 8

Revision as of 18:36, 20 December 2021 by Smartgrowth (talk | contribs) (Solution 2: Fakesolving)

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?

$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$

Solution 1: With Algebra

Let the original price of an item be $x$.

First, everything is half-off, so the price is now $\frac{x}{2} = 0.5x$.

Next, the extra coupon applies 20% off on the sale price, so the price after this discount will be $100\% - 20\% = 80\%$ 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.)

$80\% \cdot 0.5 x = \frac{4}{5} \cdot 0.5 x = 0.4x$

The price of the item after all discounts have been applied is $0.4x = 40\% \cdot x$. However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract $40\% x$ from $100\% x$ (the original price of the item), to find the answer, $\boxed{\textbf{(D)}\ 60}$.

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 $\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{\textbf{(D)}\ 60} \%$.


Solution 3:

The percent off would be 50% of 20% because it has and 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 60%.

Solution 3~ SmartGrowth

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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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