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John deposited $10,000 to open a new savings account that earned 4 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in John's account 6 months after the account was opened?

Re: John deposited $10,000 to open a new savings account that [#permalink]

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16 Jul 2012, 10:16

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john amount is compounded quarterly so the formula is A=P*(1+R/100)^N her p=principle R=rate of interest N=no of terms rate of interest is 4% for a yr so for quarter R=4/4=1% john receive amt after 6 mnts so there are two quarter so N=2 Amount A=P(1+R/100)^2 =10000(1+1/100)^2 =10000(1+0.01)^2 =10201

John deposited $10,000 to open a new savings account that earned 4 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in John's account 6 months after the account was opened?

For the first 3 moths interest was 1% of $10,000, so $100; For the next 3 moths interest was 1% of $10,000, plus 1% earned on previous interest of $100, so $100+$1=$101;

Total interest for 6 months was $100+$101=$201, hence balance after 6 months was $10,000+ $201=$10,201.

Answer: D.

Approach #2: If the interest were compounded every 6 moths instead of every 3 months (quarterly) then in 6 months the interest would be 4%/2=2% of $10,000, so $200. Now, since the interest is compounded quarterly then there would be interest earned on interest (very small amount) thus the actual interest should be a little bit more than $200, only answer choice D fits.

Although this is the method they are looking for, the question is poorly worded. By explicitly saying that 4% was earned, this implies that 4% was the effective rate, and not nominal. Would have been better to say "the account earned a nominal annual interest rate of 4%". No more confusion.

John deposited $10,000 to open a new savings account that earned 4 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in John's account 6 months after the account was opened?

Since the account compounds quarterly, John earns 0.4/4 = 0.1 or 1 percent interest each quarter.

After Q1, he earns 10,000 x 0.01 = 100 dollars interest, and thus he has a total of 10,000 + 100 = 10,100 dollars in the account after the first quarter or the first 3 months.

After Q2, John earns another 10,100 x 0.01 = 101 dollars interest.

So, after 6 months, the total amount of money in John’s account is 10,100 + 101 = 10,201 dollars.

Alternate Solution:

We can use the compound interest formula A = P[(1 + (r/n)]^nt, with P = 10,000, r = 0.04, n = 4, and t = 0.5. Thus, we have P = 10,000[1+(0.04/4)]^2 = 10,000(1.01)^2 = 10,201.

Answer: D
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

We're told that a $10,000 investment earns 4% annual interest, compounded QUARTERLY (which means that we calculate interest 4 times per year). We're asked for the total value of the investment after 6 MONTHS. This question requires that you understand the concept behind the Compound Interest Formula (and this concept is sometimes referred to as "interest on top of interest").

If we were using SIMPLE Interest, then we'd calculate just once (and the total interest would be 4% of $10,000 = $400). However, when using Compound Interest, you calculate interest more than once per year, and you have to adjust the 'math' a bit. Based on the number of calculations that you do each year, you then have to divide the interest rate by that number of terms.

Here, we calculate 4 times per year, so each interest calculation is 4%/4 = 1%.

First 3 months = $10,000(1.01) = $10,100 Second 3 months = $10,100(1.01) = $10,201

Since the prompt asks for the total after 6 months, we don't have to do any additional work.

We can use the compound interest formula A = P[(1 + (r/n)]^nt, with P = 10,000, r = 0.04, n = 4, and t = 0.5. Thus, we have P = 10,000[1+(0.04/4)]^2 = 10,000(1.01)^2 = 10,201.

For simplification of 10,000 * \((1.01)^2\) is my below approach valid:

10,000 * \((1.01)^2\)

= 101 * 101 or \((101)^2\) or\((100+1)^2\)

Now using \((a+b)^2\) we get \(100^2 + 2 * 100 * 1 + 1^2\) ie 10,201.
_________________

We can use the compound interest formula A = P[(1 + (r/n)]^nt, with P = 10,000, r = 0.04, n = 4, and t = 0.5. Thus, we have P = 10,000[1+(0.04/4)]^2 = 10,000(1.01)^2 = 10,201.

For simplification of 10,000 * \((1.01)^2\) is my below approach valid:

10,000 * \((1.01)^2\)

= 101 * 101 or \((101)^2\) or\((100+1)^2\)

Now using \((a+b)^2\) we get \(100^2 + 2 * 100 * 1 + 1^2\) ie 10,201.