During the first quarter of the year, a stock increased by x percent. During the second quarter, the stock decreased by y percent. If the stock began the year with a price of s, which of the following represents the price per share at the end of the second quarter, in terms of s, x, and y?
(A) s(1+x)(1-y)
(B) s(x/100)(y/100)
(C) s(1+[x/100])(1-[y/100])
(D) 100s(x+y)
(E) xy(100+[s/100])
Percent change algebra problem
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- TheGmatTutor
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The answer is C. A simple percent increase and percent decrease question. 1 ± percent/100 is all you need.
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Let y=100, implying that the stock price decreases by 100% in the second quarter.TheGmatTutor wrote:During the first quarter of the year, a stock increased by x percent. During the second quarter, the stock decreased by y percent. If the stock began the year with a price of s, which of the following represents the price per share at the end of the second quarter, in terms of s, x, and y?
(A) s(1+x)(1-y)
(B) s(x/100)(y/100)
(C) s(1+[x/100])(1-[y/100])
(D) 100s(x+y)
(E) xy(100+[s/100])
Since the stock loses 100% of its value, the resulting price = $0.
Implication:
When y=100 is plugged into the correct answer choice, a price of $0 must be yielded.
Only C includes a factor that will be equal to 0 when y=100:
(1-[y/100]) = (1- [100/100]) = (1 - 1) = 0.
The correct answer is C.
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We can also use the INPUT-OUTPUT approach.TheGmatTutor wrote:During the first quarter of the year, a stock increased by x percent. During the second quarter, the stock decreased by y percent. If the stock began the year with a price of s, which of the following represents the price per share at the end of the second quarter, in terms of s, x, and y?
(A) s(1+x)(1-y)
(B) s(x/100)(y/100)
(C) s(1+[x/100])(1-[y/100])
(D) 100s(x+y)
(E) xy(100+[s/100])
I might take a risk and try some EXTREME VALUES to see if I can eliminate some answer choices.
If x and y both equal zero, then the value of the stock is UNCHANGED. That is, it still equals s at the end of the second quarter.
Now, plug x = 0 and y = 0 into answer choices and see which one yields s as the OUTPUT.
(A) s(1+x)(1-y) = s works - keep
(B) s(x/100)(y/100) = 0 ELIMINATE
(C) s(1+[x/100])(1-[y/100]) = s works - keep
(D) 100s(x+y) = 0 ELIMINATE
(E) xy(100+[s/100]) = 0 ELIMINATE
Great we're QUICKLY down to 2 options:
(A) s(1+x)(1-y)
(C) s(1+[x/100])(1-[y/100])
Let's try another set of extreme values.
If x and y both equal 100, we have a situation where the stock increases 100% (i.e. DOUBLES) in first quarter and decreases by 100% in second quarter. A decrease of 100% means the stock loses ALL OF ITS VALUE.
So, the stock is is worth 0 at the end of the second quarter.
Now, plug x = 100 and y = 100 into answer choices and see which one yields 0 as the OUTPUT.
(A) s(1+x)(1-y) = some negative value - ELIMINATE
(C) s(1+[x/100])(1-[y/100]) = 0 perfect
Answer: C
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Brent
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