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18 Apr 2022, 02:24
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
71% (02:27) correct
29%
(02:45)
wrong
based on 435
sessions
History
Date
Time
Result
Not Attempted Yet
How much more interest will Patty receive if she invests $2000 for one year at \(p\)% annual interest rate compounded semi-annually, than if invests $2000 for one year at \(p\)% annual interest rate compounded annually?
(A) \(10p\)
(B) \(20p\)
(C) \(20p + \frac{p^2}{10}\)
(D) \(\frac{p^2}{20}\)
(E) \(\frac{p^2}{10}\)
(A) \(10p\)
(B) \(20p\)
(C) \(20p + \frac{p^2}{10}\)
(D) \(\frac{p^2}{20}\)
(E) \(\frac{p^2}{10}\)
18 Apr 2022, 03:13
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from given info we have
A= P * ( 1+r/n) ^ nt
2000*(1+p/200) ^2 - 2000 * (1+p/100 )
solve we get
p^2/20
A= P * ( 1+r/n) ^ nt
2000*(1+p/200) ^2 - 2000 * (1+p/100 )
solve we get
p^2/20
GMATNinja
07 Mar 2026, 07:07
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Deconstructing the Question
Patty invests \(2000\) dollars for one year at \(p\%\) interest.
We compare
annual compounding
and
semiannual compounding.
The question asks how much extra interest she earns with semiannual compounding.
Step-by-step
Convert the rate
\(p\%=\frac{p}{100}\)
Interest with annual compounding
\(2000\times\frac{p}{100}=20p\)
Now consider semiannual compounding.
Each half-year rate is
\(\frac{p}{2}\%=\frac{p}{200}\)
Amount after one year
\(2000\left(1+\frac{p}{200}\right)^2\)
Expand
\((1+x)^2=1+2x+x^2\)
with
\(x=\frac{p}{200}\)
\(1+\frac{2p}{200}+\frac{p^2}{40000}\)
\(1+\frac{p}{100}+\frac{p^2}{40000}\)
Multiply by \(2000\)
Interest earned
\(2000\left(\frac{p}{100}+\frac{p^2}{40000}\right)\)
\(=20p+\frac{p^2}{20}\)
Difference from annual compounding
\((20p+\frac{p^2}{20})-20p\)
\(\frac{p^2}{20}\)
Answer D
Patty invests \(2000\) dollars for one year at \(p\%\) interest.
We compare
annual compounding
and
semiannual compounding.
The question asks how much extra interest she earns with semiannual compounding.
Step-by-step
Convert the rate
\(p\%=\frac{p}{100}\)
Interest with annual compounding
\(2000\times\frac{p}{100}=20p\)
Now consider semiannual compounding.
Each half-year rate is
\(\frac{p}{2}\%=\frac{p}{200}\)
Amount after one year
\(2000\left(1+\frac{p}{200}\right)^2\)
Expand
\((1+x)^2=1+2x+x^2\)
with
\(x=\frac{p}{200}\)
\(1+\frac{2p}{200}+\frac{p^2}{40000}\)
\(1+\frac{p}{100}+\frac{p^2}{40000}\)
Multiply by \(2000\)
Interest earned
\(2000\left(\frac{p}{100}+\frac{p^2}{40000}\right)\)
\(=20p+\frac{p^2}{20}\)
Difference from annual compounding
\((20p+\frac{p^2}{20})-20p\)
\(\frac{p^2}{20}\)
Answer D
