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Bunuel
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The value of an investment increases by p% each month. If the investme 15 Jun 2017, 05:05
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E
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77% (02:14) correct 23% (02:32) wrong based on 164 sessions

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The value of an investment increases by p% each month. If the investment was worth x dollars initially and y dollars after one month, then in terms of x and y, how much was it worth after two months?

A. y^2−x

B. x^2−y

C. xy/100

D. x^2/y

E. y^2/x
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E
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The value of an investment increases by p% each month. If the investme 15 Jun 2017, 05:26
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Bunuel
The value of an investment increases by p% each month. If the investment was worth x dollars initially and y dollars after one month, then in terms of x and y, how much was it worth after two months?

A. y^2−x

B. x^2−y

C. xy/100

D. x^2/y

E. y^2/x

Initial worth = $x
Value is y dollars after one month
Investment increases by p% each month.

y = p% of x = \(\frac{xp}{100}\)
p = \(\frac{100y}{x}\)

Value after two months = p% of y
\(\frac{p}{100} * y\) => (Can also write as \(p * \frac{y}{100}\))

Substituting value of p in above equation. we get;

\(\frac{100y}{x} * \frac{y}{100} = \frac{y^2}{x}.\) Answer E...
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The value of an investment increases by p% each month. If the investme 15 Jun 2017, 06:27
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Let x=100 and p=20. So after one month value of investment = 120, which is y. So y=120.

But next month again there is an increase of 20% over 120. So final value = 120% of 120 = 144.

So we are looking for an option where if we put x=100 and y=120, we should get '144'.

That is only option E.

y^2/x = (120^2)/100 = 144. Hence E answer
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The value of an investment increases by p% each month. If the investme 17 Jun 2017, 05:09
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Bunuel
The value of an investment increases by p% each month. If the investment was worth x dollars initially and y dollars after one month, then in terms of x and y, how much was it worth after two months?

A. y^2−x

B. x^2−y

C. xy/100

D. x^2/y

E. y^2/x

We can create the following equation:

y = x(1 + p/100)

The above is the equation for the price at the end of the first month. So, after two months, the price should be x(1 + p/100)(1 + p/100) or x(1 + p/100)^2.

Since y = x(1 + p/100), we see that 1 + p/100 = y/x. Thus, the price after two months is x(1 + p/100)^2 = x(y/x)^2 = x(y^2/x^2) = y^2/x.

Answer: E
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The value of an investment increases by p% each month. If the investme 16 Nov 2020, 09:22
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sashiim20
Bunuel
The value of an investment increases by p% each month. If the investment was worth x dollars initially and y dollars after one month, then in terms of x and y, how much was it worth after two months?

A. y^2−x

B. x^2−y

C. xy/100

D. x^2/y

E. y^2/x

Initial worth = $x
Value is y dollars after one month
Investment increases by p% each month.

y = p% of x = \(\frac{xp}{100}\)
p = \(\frac{100y}{x}\)

Value after two months = p% of y
\(\frac{p}{100} * y\) => (Can also write as \(p * \frac{y}{100}\))

Substituting value of p in above equation. we get;

\(\frac{100y}{x} * \frac{y}{100} = \frac{y^2}{x}.\) Answer E...


y = Intial value + Interest earned
= x + p% of x
=x + \(\frac{xp}{100}\)
=x(1 + \(\frac{p}{100}\))
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The value of an investment increases by p% each month. If the investme 12 Feb 2024, 04:52
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Hello from the GMAT Club BumpBot!

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