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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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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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From the question we get
For first month:
y = x + x*p/100
y = x (1 + p/100)
Therefore p/100 = y/x - 1
For second month
= y + y * p/100
Substituting above eqn
= y + y(y/x -1)
= y + y^2/x -y
= y^2/x
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Assume easy values
Let p=100% and initial amount =100
After 1 month, the value = 100*2 =200 =y
After 2 months, the value = 200*2=400

Here we need to put the value of x & y in option to get 400. Only E works.

Remember when the increase is 100 percent you multiply by 2; when increase is 200 percent you multiply by 3 and this goes on.
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