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30 Jul 2018, 00:51
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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:
45%
(medium)
Question Stats:
71% (02:18) correct
29%
(02:33)
wrong
based on 742
sessions
History
Date
Time
Result
Not Attempted Yet
Amy deposited $1,000 into an account that earns 8% annual interest compounded every 6 months.
Bob deposited $1,000 into an account that earns 8% annual interest compounded quarterly.
If neither Amy nor Bob makes any additional deposits or withdrawals, in 6 months how much more money will Bob have in his account than will Amy have in hers?
(A) $40
(B) $8
(C) $4
(D) $0.40
(E) $0.04
Bob deposited $1,000 into an account that earns 8% annual interest compounded quarterly.
If neither Amy nor Bob makes any additional deposits or withdrawals, in 6 months how much more money will Bob have in his account than will Amy have in hers?
(A) $40
(B) $8
(C) $4
(D) $0.40
(E) $0.04
30 Jul 2018, 14:07
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Bunuel
Amy - interest only
Every 6 months, Amy's interest is paid at the annual 8% rate divided by 2 (rate/# of compounding periods).
Every six months she earns \(\frac{.08}{2}=.04\) on her account balance (principal + any accrued interest)
At the 6-month mark, Amy gets her first interest payment of 4% (on $1,000).
Amy's INTEREST EARNED at 6 months: \((.04 * $1,000)= $40\)
Bob - interest only
Every 3 months, Bob's interest is paid at the annual 8% rate divided by 4 (rate/# of compounding periods).
Every 3 months, he earns \(\frac{.08}{4}=.02\) interest on his accumulated balance
1) After 3 months he gets paid \((.02 * $1,000) = $20\)
Now Bob has \($1,020\)
2) At the 6-month mark, he gets paid \((.02 * $1,020) = $20.40\)
Bob's INTEREST EARNED after six months: \(($20 + $20.40) = $40.40\)
Difference between Amy and Bob?
\($40.40 - $40.00 = $0.40\)
Answer D
Compound interest formula: \(A= P(1+\frac{r}{n})^{nt}\)
A = total amount, P = principal, r = annual interest rate in decimal form, n = number of interest payments in a year, and t = time in years
Amy - 8% annual compounded every 6 months
In half a year: \(A= $1,000(1+\frac{.08}{2})^{(2*\frac{1}{2})}\)
\(A=$1,000(1.04)^1=($1,000*1.04) = $1,040\)
After six months, Amy's total is \($1,040\)
Bob - 8% annual compounded quarterly
In half a year, i.e., two quarters \(A=$1,000(1+\frac{.08}{4})^{(4*\frac{1}{2})}\)
\(A=$1,000(1.02)^2\)
\(1.02*1.02=1.0404\)
After 6 months Bob has \((1.0404 * $1,000)=$1,040.40\)
Difference? \($1,040.40 - $1,040.00 = $0.40\)
Answer D
General Discussion
30 Jul 2018, 03:08
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Bunuel
When compounded semiannually, time gets doubled and rate halted
If compounded quarterly, time gets four times and rate 1/4th
Since we are looking for 6 months..
The one compounded every six months..
t is 1 and r=8/2=4
Amount = 1000(1+4/100)=1000*104/100=1040
The one compounded quarterly..
t=2 and r=8/4=2
Amount =1000(1+2/100}^2=1000*1.02*1.02=1040.4
So excess = 1040.4-1040=$0.40
D
