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Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the duratio

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Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the duratio

by M7MBA » Wed Jun 24, 2020 6:24 am

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Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the duration of investment. If she earned a total interest of $4000 in 6 years, after how many years of investment did she earn a total interest of $52000?

A.18

B.28

C.30

D.56

E. Cannot be determined

[spoiler]OA=A[/spoiler]

Source: e-GMAT
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Re: Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the dur

by deloitte247 » Sat Jun 27, 2020 3:04 pm
Given that: p = 2000
Total interest in 6 years = 4000
Money in the account after 6 years = 2000 + 4000 = 6000
Hence, money in the account is going to be; principal * 3 every 6 years


Target question => After how many years of investment did she earn a total interest of $52000?


Let the number of years to earn $52000 = x
Total money in the account after x years =>
$52000 + 2000 = 54000


Knowing that investment triples every 6 years
$$For\ the\ first\ 6000:\ \frac{6000}{2000}=3=3^1$$
$$Number\ of\ years\ =\ 1\cdot6=6\ years$$
$$For\ the\ new\ 54000\ =\ \frac{54000}{2000}=27=3^3$$
$$Number\ of\ years\ =3\cdot6=18\ years$$
$$Answer\ =\ A$$
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Re: Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the dur

by swerve » Sun Jun 28, 2020 5:11 am
M7MBA wrote: ↑
Wed Jun 24, 2020 6:24 am
 Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the duration of investment. If she earned a total interest of $4000 in 6 years, after how many years of investment did she earn a total interest of $52000?

A.18

B.28

C.30

D.56

E. Cannot be determined

[spoiler]OA=A[/spoiler]

Source: e-GMAT
\(2000(1+R/100)^6=6000\)
\((1+R/100)^6 = 3\)
\((1+R/100) =3^{1/6}\) (don't solve further)

To find \(n\)

\(2000 (1+R/100)^n = 54000\)
\(3^{n/6}=27\)
\(3^{n/6}=3^3\)
\(n/6=3\)
\(n=18\)

Therefore, A
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Re: Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the dur

by Scott@TargetTestPrep » Thu Jul 02, 2020 3:36 pm
M7MBA wrote: ↑
Wed Jun 24, 2020 6:24 am
 Roxie invested $2000 in an account that yielded a fixed rate of annual return compounded annually throughout the duration of investment. If she earned a total interest of $4000 in 6 years, after how many years of investment did she earn a total interest of $52000?

A.18

B.28

C.30

D.56

E. Cannot be determined

[spoiler]OA=A[/spoiler]

Solution:


We can see that in 6 years, the amount of money- principal plus interest - in the account will be 2,000 + 4,000 = $6,000. This means the amount of money in the account triples every 6 years. Therefore, in 12 years, the amount of money in the account will be 6,000 x 3 = $18,000, and in 18 years, the amount of money in the account will be 18,000 x 3 = $54,000, which is $2,000 in principal and $52,000 in interest.

Answer: A

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