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The value of an investment increases by x% during January and decrease
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31 Mar 2021, 06:00
31 Mar 2021, 06:00
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The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
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Re: The value of an investment increases by x% during January and decrease
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31 Mar 2021, 06:11
31 Mar 2021, 06:11
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GeminiHeat wrote:
The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
Let the investment at the beginning of January be \(I\)
Value at the end of January \(= (1 + \frac{x}{100})I\)
Value at the end of February \(= (1 - \frac{y}{100)}(1 + \frac{x}{100})I\)
Now, \(I = (1 - \frac{y}{100)}(1 + \frac{x}{100})I\)
\(1 = \frac{(100 - y)}{100}\frac{(100 + x)}{100}\)
\(10000 = (100 - y))(100 + x)\)
\(\frac{10000}{(100 + x)} = 100 - y\)
\(y = 100 - \frac{10000}{(100 + x)}\)
Hence, option E
gmatclubot
Re: The value of an investment increases by x% during January and decrease [#permalink]
31 Mar 2021, 06:11
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