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TheMechanic
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On January 1, 1994, Jill invested P dollars in an account that pays in 17 Oct 2017, 06:33
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On January 1, 1994, Jill invested P dollars in an account that pays interest at a rate of 8 percent per year, compounded annually on December 31. If there were no other deposits or withdrawals in the account, how many dollars were in the account on January 1, 1998, in terms of P?

(A) \(0.32P\)

(B) \(4.32P\)

(C) \((0.08)^4P\)

(D) \((1.08)^4P\)

(E) \((1.08P)^4\)
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D
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On January 1, 1994, Jill invested P dollars in an account that pays in 17 Oct 2017, 11:51
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TheMechanic
On January 1, 1994, Jill invested P dollars in an account that pays interest at a rate of 8 percent per year, compounded annually on December 31. If there were no other deposits or withdrawals in the account, how many dollars were in the account on January 1, 1998, in terms of P?
A) \(0.32P\)
B) \(4.32P\)
C) \((0.08)^4P\)
D) \((1.08)^4P\)
E) \((1.08P)^4\)
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Between 1st Jan 1994 & 1st Jan 1998, time elapsed = 4 years

Use the Compound interest formula to calculate the amount

\(A=P(1+\frac{r}{100})^n\) \(=> P(1+\frac{8}{100})^4\)

or \(A=P(1.08)^4\)

Option D
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On January 1, 1994, Jill invested P dollars in an account that pays in 17 Oct 2017, 12:20
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TheMechanic
On January 1, 1994, Jill invested P dollars in an account that pays interest at a rate of 8 percent per year, compounded annually on December 31. If there were no other deposits or withdrawals in the account, how many dollars were in the account on January 1, 1998, in terms of P?
A) \(0.32P\)
B) \(4.32P\)
C) \((0.08)^4P\)
D) \((1.08)^4P\)
E) \((1.08P)^4\)
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Knowing the formula for compound interest helps a lot here.

Compound interest is given by

\(A = P(1 +\frac{r}{n})^{nt}\)

A = final amount
P = principal invested
r = interest rate in decimal form
n = number of compounding periods per year
t = time

Here interest compounds annually; it earns "interest on interest," and pays one time per year, on December 31.

So r = .08
Because \(n = 1\), \((\frac{.08}{1}) = .08\)

Then \(1 + .08 = 1.08\)

Time t, = 4 years: she gets paid December 31 of 1994, 1995, 1996, and 1997

Where n = 1 and t = 4, thus: \((1.08)^{1*4} = (1.08)^4\)

Finally, multiply by the principal: \(P(1.08)^4\)

Answer D
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On January 1, 1994, Jill invested P dollars in an account that pays in 19 Oct 2017, 10:16
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TheMechanic
On January 1, 1994, Jill invested P dollars in an account that pays interest at a rate of 8 percent per year, compounded annually on December 31. If there were no other deposits or withdrawals in the account, how many dollars were in the account on January 1, 1998, in terms of P?
A) \(0.32P\)
B) \(4.32P\)
C) \((0.08)^4P\)
D) \((1.08)^4P\)
E) \((1.08P)^4\)
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Since the investment of P dollars was compounded annually for 4 years, the new value of the original investment is P(1.08)^4.

Answer: D
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On January 1, 1994, Jill invested P dollars in an account that pays in Updated on: 13 Feb 2019, 17:23
Originally posted by BrentGMATPrepNow on 13 Feb 2019, 16:55.
Last edited by BrentGMATPrepNow on 13 Feb 2019, 17:23, edited 1 time in total.
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TheMechanic
On January 1, 1994, Jill invested P dollars in an account that pays interest at a rate of 8 percent per year, compounded annually on December 31. If there were no other deposits or withdrawals in the account, how many dollars were in the account on January 1, 1998, in terms of P?
A) \(0.32P\)
B) \(4.32P\)
C) \((0.08)^4P\)
D) \((1.08)^4P\)
E) \((1.08P)^4\)
Show more

The easiest/fastest solution involves plugging the given values into the compound interest formula (as others have done).

However, if you forgot that formula, you can always create a table to arrive at the correct answer.

# of year elapsed | value of investment
0 | P
1 | P(1.08)
2 | P(1.08)(1.08) = P(1.08)²
3 | P(1.08)(1.08)(1.08) = P(1.08)³
4 | P(1.08)(1.08)(1.08)(1.08) = P(1.08)⁴

Answer: D

Cheers,
Brent
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On January 1, 1994, Jill invested P dollars in an account that pays in 01 Jan 2023, 00:49
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