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Compound Percentage Increase

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by devpath » Mon Oct 19, 2009 4:12 am
In 1989 the price of a new model S car was x dollars. If the price of the model S car increased each year by 10 percent of the previous year's price, what was the price of the car, in dollars, in 1991?

(A) 1.10x
(B) 1.20x
(C) 1.21x
(D) 1.25x
(E) 1.33x

Help :shock:
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by BuckeyeT » Mon Oct 19, 2009 5:21 am
Price of S in 1989 = x
Price of S in 1990 = 110% * x or 1.1x
Price of S in 1991 = 110% * 1.1x or 1.1 * 1.1x = 1.21x

C.
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Re: Compound Percentage Increase

by JeffB » Mon Oct 19, 2009 8:25 am
devpath wrote:In 1989 the price of a new model S car was x dollars. If the price of the model S car increased each year by 10 percent of the previous year's price, what was the price of the car, in dollars, in 1991?

(A) 1.10x
(B) 1.20x
(C) 1.21x
(D) 1.25x
(E) 1.33x

Help :shock:
Try to pick a number for X if you can't understand the algebra.

Say X = 100

10% of 100 = 10, so 110 in 1990
10% of 110 = 11, so 121 in 1991
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by Stuart@KaplanGMAT » Mon Oct 19, 2009 12:19 pm
Great solutions (especially JeffB - picking 100 on percent questions is often the quickest way to solve)!

While compound interest isn't frequently tested on the GMAT, it's still not a bad idea to know the formula:

Total value of the investment = P(1+r)^t

P = original amount of the investment (principal)
r = interest rate per compound period (as a decimal)
t = number of compound periods

In the question you posted:

P = x
r = .1
t = 2

So we'd get:

x(1.1)^2 = 1.21x
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