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Maggie bought a 1-year certificate of deposit, and annual interest ear
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04 Jun 2025, 10:57
04 Jun 2025, 10:57
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Maggie bought a 1-year certificate of deposit, and annual interest earned on the deposit is 6 percent compounded semi-annually. What was the amount invested, if interest earned on the certificate at maturity is $\(\$ 60.9\)$ ?
(A) $\(\$ 1100\)$
(B) $\(\$ 1010\)$
(C) $\(\$ 1000\)$
(D) $\(\$ 110\)$
(E) $\(\$ 100\)$
(A) $\(\$ 1100\)$
(B) $\(\$ 1010\)$
(C) $\(\$ 1000\)$
(D) $\(\$ 110\)$
(E) $\(\$ 100\)$
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Re: Maggie bought a 1-year certificate of deposit, and annual interest ear
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06 Jun 2025, 04:00
06 Jun 2025, 04:00
Expert Reply
Let $P$ be the amount invested (principal).
The annual interest rate is $\(6 \%\)$, compounded semi-annually.
This means the interest rate per compounding period is $\(6 \% / 2=3 \%\)$.
The certificate is for 1 year, and it's compounded semi-annually, so there are $\(1 \times 2=2\)$ compounding periods.
The formula for the future value (amount at maturity) with compound interest is:
$$
\(A=P(1+r)^n\)
$$
where:
$A=$ amount at maturity
$P=$ principal amount (amount invested)
$r=$ interest rate per compounding period (as a decimal)
$n=$ number of compounding periods
In this case, we know the interest earned, not the total amount at maturity.
Interest earned $=A-P$
We are given that the interest earned is $\(\$ 60.9\)$.
So, $\(A-P=60.9\)$.
This means $\(A=P+60.9\)$.
Now substitute the values into the compound interest formula:
$$
\(\begin{aligned}
& P+60.9=P(1+0.03)^2 \\
& P+60.9=P(1.03)^2 \\
& P+60.9=P(1.0609)
\end{aligned}\)
$$
Now, we need to solve for $P$ :
$$
\(\begin{aligned}
& 60.9=1.0609 P-P \\
& 60.9=(1.0609-1) P \\
& 60.9=0.0609 P
\end{aligned}\)
$$
$$
\(\begin{aligned}
P & =\frac{60.9}{0.0609} \\
P & =1000
\end{aligned}\)
$$
So, the amount invested was $\(\$ 1000\)$.
The final answer is C .
The annual interest rate is $\(6 \%\)$, compounded semi-annually.
This means the interest rate per compounding period is $\(6 \% / 2=3 \%\)$.
The certificate is for 1 year, and it's compounded semi-annually, so there are $\(1 \times 2=2\)$ compounding periods.
The formula for the future value (amount at maturity) with compound interest is:
$$
\(A=P(1+r)^n\)
$$
where:
$A=$ amount at maturity
$P=$ principal amount (amount invested)
$r=$ interest rate per compounding period (as a decimal)
$n=$ number of compounding periods
In this case, we know the interest earned, not the total amount at maturity.
Interest earned $=A-P$
We are given that the interest earned is $\(\$ 60.9\)$.
So, $\(A-P=60.9\)$.
This means $\(A=P+60.9\)$.
Now substitute the values into the compound interest formula:
$$
\(\begin{aligned}
& P+60.9=P(1+0.03)^2 \\
& P+60.9=P(1.03)^2 \\
& P+60.9=P(1.0609)
\end{aligned}\)
$$
Now, we need to solve for $P$ :
$$
\(\begin{aligned}
& 60.9=1.0609 P-P \\
& 60.9=(1.0609-1) P \\
& 60.9=0.0609 P
\end{aligned}\)
$$
$$
\(\begin{aligned}
P & =\frac{60.9}{0.0609} \\
P & =1000
\end{aligned}\)
$$
So, the amount invested was $\(\$ 1000\)$.
The final answer is C .
gmatclubot
Re: Maggie bought a 1-year certificate of deposit, and annual interest ear [#permalink]
06 Jun 2025, 04:00
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