12 multiple-choice questions, progressively harder.
You invest 1,0001{,}0001,000 dollars at 6%6\%6% compounded continuously for 111 year. Using e0.06=1.061837e^{0.06} = 1.061837e0.06=1.061837, what is the balance?
Solution
Correct answer: A
Continuous compounding uses A=PertA = Pe^{rt}A=Pert with rt=0.06×1=0.06rt = 0.06 \times 1 = 0.06rt=0.06×1=0.06.
A=1000 e0.06=1000×1.061837=1061.84A = 1000\,e^{0.06} = 1000 \times 1.061837 = 1061.84A=1000e0.06=1000×1.061837=1061.84
The balance is 1,061.841{,}061.841,061.84 dollars, the ceiling that annual (1,0601{,}0601,060) and quarterly (1,061.361{,}061.361,061.36) fall just below.
On 1,0001{,}0001,000 dollars at 5%5\%5% for 222 years, how much MORE does annual compounding earn than simple interest? Use 1.052=1.10251.05^2 = 1.10251.052=1.1025.
Correct answer: C
Compound gives 1000×1.1025=1102.501000 \times 1.1025 = 1102.501000×1.1025=1102.50; simple gives 1000(1+0.05×2)=11001000(1 + 0.05 \times 2) = 11001000(1+0.05×2)=1100.
1102.50−1100=2.501102.50 - 1100 = 2.501102.50−1100=2.50
The extra 2.502.502.50 dollars is interest earned on the first year's interest.
You invest 4,0004{,}0004,000 dollars at 10%10\%10% compounded annually for 333 years. Using 1.103=1.3311.10^3 = 1.3311.103=1.331, what is the balance?
Correct answer: B
Multiply the principal by the three-year growth factor.
A=4000×1.331=5324A = 4000 \times 1.331 = 5324A=4000×1.331=5324
The balance is exactly 5,3245{,}3245,324 dollars.
Which deposit now grows to exactly 2,0002{,}0002,000 dollars in 222 years at 5%5\%5% compounded annually? Use 1.052=1.10251.05^2 = 1.10251.052=1.1025.
Correct answer: D
Divide the goal by the two-year growth factor to recover the principal.
P=20001.1025=1814.06P = \frac{2000}{1.1025} = 1814.06P=1.10252000=1814.06
A deposit of about 1,814.061{,}814.061,814.06 dollars reaches 2,0002{,}0002,000 dollars in two years.
You invest 1,0001{,}0001,000 dollars at 10%10\%10% compounded continuously for 111 year. Using e0.10=1.105171e^{0.10} = 1.105171e0.10=1.105171, what is the balance?
Use A=PertA = Pe^{rt}A=Pert with rt=0.10rt = 0.10rt=0.10.
A=1000 e0.10=1000×1.105171=1105.17A = 1000\,e^{0.10} = 1000 \times 1.105171 = 1105.17A=1000e0.10=1000×1.105171=1105.17
The balance is 1,105.171{,}105.171,105.17 dollars, more than the 1,1001{,}1001,100 that annual compounding gives.
You invest 5,0005{,}0005,000 dollars at 4%4\%4% compounded quarterly for 111 year. Using 1.014=1.0406041.01^4 = 1.0406041.014=1.040604, what is the balance?
Each quarter pays 0.044=0.01\frac{0.04}{4} = 0.0140.04=0.01 over nt=4nt = 4nt=4 periods.
A=5000 (1.01)4=5000×1.040604=5203.02A = 5000\,(1.01)^4 = 5000 \times 1.040604 = 5203.02A=5000(1.01)4=5000×1.040604=5203.02
The balance is 5,203.025{,}203.025,203.02 dollars.
A principal of 1,0001{,}0001,000 dollars at 5%5\%5% reaches 1,157.631{,}157.631,157.63 after 333 years compounded annually. Compounded continuously at 5%5\%5% for 333 years, using e0.15=1.161834e^{0.15} = 1.161834e0.15=1.161834, it reaches:
Use A=PertA = Pe^{rt}A=Pert with rt=0.05×3=0.15rt = 0.05 \times 3 = 0.15rt=0.05×3=0.15.
A=1000 e0.15=1000×1.161834=1161.83A = 1000\,e^{0.15} = 1000 \times 1.161834 = 1161.83A=1000e0.15=1000×1.161834=1161.83
Continuous compounding earns 1,161.831{,}161.831,161.83 dollars, a few dollars above the annual result.
On 2,0002{,}0002,000 dollars at 8%8\%8% for 111 year, how much more does quarterly compounding earn than annual? Use 1.024=1.082432161.02^4 = 1.082432161.024=1.08243216.
Annual gives 2000×1.08=21602000 \times 1.08 = 21602000×1.08=2160; quarterly gives 2000×1.08243216=2164.862000 \times 1.08243216 = 2164.862000×1.08243216=2164.86.
2164.86−2160=4.862164.86 - 2160 = 4.862164.86−2160=4.86
Splitting the same 8%8\%8% into quarters earns an extra 4.864.864.86 dollars.
Compounding PPP at rate rrr for ttt years nnn times a year, then letting nnn grow without bound, the balance approaches:
The factor (1+rn)n\left(1 + \frac{r}{n}\right)^{n}(1+nr)n approaches ere^{r}er as nnn grows, so the balance approaches PPP times that power.
A⟶P(er)t=PertA \longrightarrow P\left(e^{r}\right)^{t} = Pe^{rt}A⟶P(er)t=Pert
The limit is finite, the continuous-compounding value, not infinity.
You invest 3,0003{,}0003,000 dollars at 4%4\%4% compounded annually for 333 years. Using 1.043=1.1248641.04^3 = 1.1248641.043=1.124864, what is the balance?
A=3000×1.124864=3374.59A = 3000 \times 1.124864 = 3374.59A=3000×1.124864=3374.59
The balance is 3,374.593{,}374.593,374.59 dollars.
You invest 1,0001{,}0001,000 dollars at 20%20\%20% compounded annually for 222 years. How much more does this earn than simple interest? Use 1.202=1.441.20^2 = 1.441.202=1.44.
Compound gives 1000×1.44=14401000 \times 1.44 = 14401000×1.44=1440; simple gives 1000(1+0.20×2)=14001000(1 + 0.20 \times 2) = 14001000(1+0.20×2)=1400.
1440−1400=401440 - 1400 = 401440−1400=40
Compounding earns an extra 404040 dollars over the two years.
Compounded annually at 5%5\%5%, in what year does 1,0001{,}0001,000 dollars first exceed 1,2001{,}2001,200 dollars? Use 1.053=1.1576251.05^3 = 1.1576251.053=1.157625 and 1.054=1.2155061.05^4 = 1.2155061.054=1.215506.
Run the formula forward and watch for the balance to pass 1,2001{,}2001,200. After year 3 it is 1000×1.157625=1157.631000 \times 1.157625 = 1157.631000×1.157625=1157.63, still under.
1000 (1.05)4=1000×1.215506=1215.51>12001000\,(1.05)^4 = 1000 \times 1.215506 = 1215.51 > 12001000(1.05)4=1000×1.215506=1215.51>1200
The balance first exceeds 1,2001{,}2001,200 dollars in year 4.
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