Compound Interest

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Compound Interest

by [email protected] » Sun Feb 02, 2014 7:42 pm
At the beginning of January 2003, Lisa invested $10,000 in an account that collected interest, compounding monthly. Assume the annual percentage rate of interest remained constant. She withdrew the money, with interest, at the end of December, 2008. What was the annual percentage of interest paid on this account?

Statement #1: Lisa earned $61.83 in interest in November, 2003.

Statement #2: At the end of December, 2008, Lisa had gained a total of $4176.25 in interest
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by Patrick_GMATFix » Sun Feb 02, 2014 8:23 pm
Compound interest implies that each month the account balance grows by a fixed percentage (a constant multiplier). For instance if the interest is 10%, each month the balance will be multiplied by 1.10. Let's call this multiplier g. The following table shows the account balance at the start of each month

Jan 2003: $10,000
Feb 2003: 10,000g
Mar 2003: 10,000g^2
...

So the balance at any point can be expressed as 10,000 * g^k where k is the amount of time (in months) that has expired.

Since knowing the monthly percentage interest is enough to find the annual percentage interest, we can rephrase the question to ask: "What is g?"

Statement 1:
Interest accrued from start of Nov to the start of Dec 2003 was $61.83

The interest earned in a month is just the difference between balances at the beginning of the month and the end of the month. At the start of Nov 2003, the balance is 10,000 * g^10. At the end of that month, it is 10,000 * g^10 * g. We can think of this logically. As g gets bigger, the difference between 10,000g^10 and 10,000g^11 always gets bigger (we know g is positive since the account accrues interest). This means that there must be a unique value of g for which the difference between the two balances is exactly 61.83. since we can demonstrate that there is a unique solution (even if we don't know how to find it), we can be confident that there is sufficiency.

Algebraically, the relevant equation is 10,000g^11 - 10,000g^10 = 61.83

Statement 1 is SUFFICIENT.

Statement 2:
Again, a logical approach is simplest. Given that we know the total duration of the investment, and since every incremental increase in g will result in an always greater total interest, there must be a unique value of g for which the total interest accumulated is $4,176.25. This means that logically, we have enough data to find g because we've demonstrated that there is a unique value of g that will fit the statement.

Algebraically, it is easy to build an equation. There are a total of 66 months, so the total interest (difference between final balance and initial investment) can be written as: 10,000g^66 - 10,000 = 4,176.25

Statement 2 is SUFFICIENT.
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The answer is D[/spoiler]
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by [email protected] » Mon Feb 03, 2014 12:27 am
Hi shibsriz,

Since Patrick has already provided an explanation, I won't rehash that here. What I will do is help you to learn how the interest rate formula "works", especially in situations in which the interest is compounded more than once per year.

The standard formula for Compound Interest is (Principle) x (1 + R)^T; R is the yearly interest rate and T is the number of years.

So, if you invest $10,000 at 12% interest for 2 years, then the total value of the account + interest =

10,000 x (1.12)^2 = 10,000 x (1.44) = $14,400

Now, if you calculate interest every 6 months, you have to adjust the math. Over the course of 2 years, you would have 4 periods of interest; since you're doubling the value of T, the math rule dictates that you cut the interest rate in half. Under these conditions, your calculation would be:

10,000 x (1.06)^4

The GMAT won't ask you to calculate that result though.

If you're calculating once every MONTH, then you multiply T by 12 and divide the interest rate by 12:

10,000 x (1.01)^24

While these are rare situations on the GMAT, the basic rules aren't hard to learn and remember.

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Rich
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