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XimeSol

Given: Hans invested 10,000 at an annual interest rate of x percent, compounded annually.

Asked: If the annual interest rate of x percent had been compounded semiannually, how much more interest, in dollars, would he have earned of his 10,000 investment for the first year in terms of x?

Compounded annually: -
Principal P = 10000
Rate of interest = r = x/100
Period of compounding n = 1
A = P (1 + r)^n = 10000 (1 + x/100)^1
Interest = A - P = 10000 (1 + x/100 - 1) = 100x

Compounded semiannually: -
Principal P = 10000
Rate of interest = r = x/200
Period of compounding n = 2
A = P (1 + r)^n = 10000 (1 + x/200)^2
Interest = A - P = 10000 (1 + x/200)^2 - 10000 = 10000 (1 + x^2/40000 + x/100 - 1) = 10000 (x/1000 + x^2/40000)

More interest = 10000 (x/100 + x^2/40000) - 100x = x^2/4

IMO E
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10,000(1 + x/200)^2 - 10,000(1 + x/100) =
10,000((1 + x/200)^2 - (1 + x/100)) =
10,000(1 + x/100 + (x^2)/(200^2) -1 - x/100) =
10,000(x^2/(4*100^2) =
x^2 / 4
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PRO TIP: If you are short on time, you can easily reject the first 3 choices because the difference in compounded and simple interest for 1 year cannot be this high.
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XimeSol
Hans invested 10,000 at an annual interest rate of x percent, compounded annually. If the annual interest rate of x percent had been compounded semiannually, how much more interest, in dollars, would he have earned of his 10,000 investment for the first year in terms of x?

(A) \(50x\)

(B) \(100x\)

(C) \(100x + \frac{x^2}{4}\)

(D) \(\frac{x^2}{2}\)

(E) \(\frac{x^2}{4}\)

In case of semi annual compounding, the only extra component is interest on interest earned in the first 6 months.
Interest earned in first 6 months will be x/2% of 10,000. Interest earned on this interest in the next 6 months will be x/2% of x/2% of 10,000 which is

\(\frac{x}{200} * \frac{x}{200} * 10,000 = \frac{x^2}{4}\)

Answer (E)

Here is a discussion on difference between simple and compound interest and how to handle the differential efficiently: https://youtu.be/2h_luc7buWE
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