GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
On January 1, 1994, Jill invested P dollars in an account that pays in
[#permalink]
25 Jan 2023, 00:03
25 Jan 2023, 00:03
Expert Reply
1
Bookmarks
00:00
Question Stats:
75% (01:17) correct
25% (04:15) wrong based on 4 sessions
Hide Show timer Statistics
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) \((1.08P)^4\)
(A) \(0.32P\)
(B) \(4.32P\)
(C) \((0.08)^4P\)
(D) \((1.08)^4P\)
(E) \((1.08P)^4\)
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: On January 1, 1994, Jill invested P dollars in an account that pays in
[#permalink]
27 Jan 2023, 11:18
27 Jan 2023, 11:18
1
Expert Reply
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
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
gmatclubot
Re: On January 1, 1994, Jill invested P dollars in an account that pays in [#permalink]
27 Jan 2023, 11:18
Moderators:
