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From 1992 to 1993, the price of a home increased by x %. Fr
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13 Jul 2020, 07:45
13 Jul 2020, 07:45
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71% (00:57) correct
28% (00:18) wrong based on 57 sessions
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From 1992 to 1993, the price of a home increased by x %. From 1993 to 1994, the price of the home then decreased by x %.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
The price of the home at the beginning of 1992 |
The price of the home at the beginning of 1994 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
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GMAT 1: 700 Q51 V31
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Re: From 1992 to 1993, the price of a home increased by x %. Fr
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16 Jul 2020, 10:11
16 Jul 2020, 10:11
1
For simplifying the solution lets assume the price of the home in 1992 is 100
From 1992 to 1993, the price of a home increased by x %
=> Price in 1993 = Price in 1992 + x% of Price of 1992 = 100 + x% of 100 = 100 + \(\frac{x}{100}\) * 100 = 100 + x
From 1993 to 1994, the price of the home then decreased by x %
=> price in 1994 = Price in 1993 - x% of price in 1993 = Price in 1993 * (1 - x%) = (100 + x) * (1 - \(\frac{x}{100}\))
=> Price in 1994 = (100 + x) * (\(\frac{100 - x}{100}\))
[ Using (a+b) * (a-b) = \(a^2\) - \(b^2\), we have]
=> Price in 1994 = \(\frac{(100^2 - x^2)}{ 100}\) = 100 - \(\frac{x^2}{100}\)
Now, we know that x is a positive number => \(\frac{x^2}{100}\) > 0
=> Price in 1994 < 100
=> Price in 1992 (which is 100) > Price in 1994
=> Quantity A > Quantity B
So, answer will be A
Hope it helps!
From 1992 to 1993, the price of a home increased by x %
=> Price in 1993 = Price in 1992 + x% of Price of 1992 = 100 + x% of 100 = 100 + \(\frac{x}{100}\) * 100 = 100 + x
From 1993 to 1994, the price of the home then decreased by x %
=> price in 1994 = Price in 1993 - x% of price in 1993 = Price in 1993 * (1 - x%) = (100 + x) * (1 - \(\frac{x}{100}\))
=> Price in 1994 = (100 + x) * (\(\frac{100 - x}{100}\))
[ Using (a+b) * (a-b) = \(a^2\) - \(b^2\), we have]
=> Price in 1994 = \(\frac{(100^2 - x^2)}{ 100}\) = 100 - \(\frac{x^2}{100}\)
Now, we know that x is a positive number => \(\frac{x^2}{100}\) > 0
=> Price in 1994 < 100
=> Price in 1992 (which is 100) > Price in 1994
=> Quantity A > Quantity B
So, answer will be A
Hope it helps!
Re: From 1992 to 1993, the price of a home increased by x %. Fr
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21 Jul 2020, 07:15
21 Jul 2020, 07:15
Expert Reply
Sir, please provide more information about the issue ??
What does not work ??
Many thanks
What does not work ??
Many thanks
