12 multiple-choice questions, progressively harder.
How much must you deposit now to have 5,0005{,}0005,000 dollars in 222 years at 6%6\%6% compounded annually? Use 1.062=1.12361.06^2 = 1.12361.062=1.1236.
Solution
Correct answer: B
Divide the goal by the two-year growth factor to recover the principal.
P=50001.1236=4449.98P = \frac{5000}{1.1236} = 4449.98P=1.12365000=4449.98
You need to deposit about 4,449.984{,}449.984,449.98 dollars today.
You invest 1,0001{,}0001,000 dollars at 9%9\%9% compounded continuously for 111 year. Using e0.09=1.094174e^{0.09} = 1.094174e0.09=1.094174, what is the balance?
Use A=PertA = Pe^{rt}A=Pert with rt=0.09rt = 0.09rt=0.09.
A=1000 e0.09=1000×1.094174=1094.17A = 1000\,e^{0.09} = 1000 \times 1.094174 = 1094.17A=1000e0.09=1000×1.094174=1094.17
The balance is 1,094.171{,}094.171,094.17 dollars.
You invest 5,0005{,}0005,000 dollars at 6%6\%6% compounded annually for 333 years. Using 1.063=1.1910161.06^3 = 1.1910161.063=1.191016, what is the balance?
Correct answer: C
Multiply the principal by the three-year growth factor.
A=5000×1.191016=5955.08A = 5000 \times 1.191016 = 5955.08A=5000×1.191016=5955.08
The balance is 5,955.085{,}955.085,955.08 dollars.
Compounded daily (n=365n = 365n=365) versus continuously, on the same principal, rate, and time, the daily balance is:
Daily compounding is very frequent but still finite, so it sits just under the continuous ceiling.
P(1+r365)365t<PertP\left(1 + \frac{r}{365}\right)^{365t} < Pe^{rt}P(1+365r)365t<Pert
The two are extremely close, with the continuous value always a hair higher.
You invest 2,5002{,}5002,500 dollars at 4%4\%4% compounded quarterly for 222 years. Using 1.018=1.0828571.01^8 = 1.0828571.018=1.082857, what is the balance?
Correct answer: D
Each quarter pays 0.044=0.01\frac{0.04}{4} = 0.0140.04=0.01 over nt=4×2=8nt = 4 \times 2 = 8nt=4×2=8 periods.
A=2500 (1.01)8=2500×1.082857=2707.14A = 2500\,(1.01)^8 = 2500 \times 1.082857 = 2707.14A=2500(1.01)8=2500×1.082857=2707.14
The balance is 2,707.142{,}707.142,707.14 dollars.
The step (1+rn)nt=[(1+rn)n]t\left(1 + \frac{r}{n}\right)^{nt} = \left[\left(1 + \frac{r}{n}\right)^{n}\right]^{t}(1+nr)nt=[(1+nr)n]t relies on which exponent law?
Correct answer: A
Splitting the exponent ntntnt into nnn times ttt regroups the factors using the power-of-a-power rule.
(1+rn)nt=[(1+rn)n]t\left(1 + \frac{r}{n}\right)^{nt} = \left[\left(1 + \frac{r}{n}\right)^{n}\right]^{t}(1+nr)nt=[(1+nr)n]t
This is exactly bnt=(bn)tb^{nt} = (b^n)^tbnt=(bn)t, the step that leads to erte^{rt}ert.
On 1,0001{,}0001,000 dollars at 10%10\%10% for 333 years, how much more does annual compounding earn than simple interest? Use 1.103=1.3311.10^3 = 1.3311.103=1.331.
Compound gives 1000×1.331=13311000 \times 1.331 = 13311000×1.331=1331; simple gives 1000(1+0.10×3)=13001000(1 + 0.10 \times 3) = 13001000(1+0.10×3)=1300.
1331−1300=311331 - 1300 = 311331−1300=31
Compounding earns an extra 313131 dollars over the three years.
You want 10,00010{,}00010,000 dollars in 333 years at 5%5\%5% compounded annually. How much must you deposit now? Use 1.053=1.1576251.05^3 = 1.1576251.053=1.157625.
Divide the goal by the three-year growth factor.
P=100001.157625=8638.38P = \frac{10000}{1.157625} = 8638.38P=1.15762510000=8638.38
A deposit of about 8,638.388{,}638.388,638.38 dollars grows to 10,00010{,}00010,000 dollars in three years.
You invest 4,0004{,}0004,000 dollars at 5%5\%5% compounded continuously for 222 years. 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=0.05×2=0.10rt = 0.05 \times 2 = 0.10rt=0.05×2=0.10.
A=4000 e0.10=4000×1.105171=4420.68A = 4000\,e^{0.10} = 4000 \times 1.105171 = 4420.68A=4000e0.10=4000×1.105171=4420.68
The balance is 4,420.684{,}420.684,420.68 dollars.
You invest 6,0006{,}0006,000 dollars at 8%8\%8% compounded semiannually for 111 year. Using 1.042=1.08161.04^2 = 1.08161.042=1.0816, what is the balance?
Semiannual gives n=2n = 2n=2, each period paying 0.082=0.04\frac{0.08}{2} = 0.0420.08=0.04 over nt=2nt = 2nt=2 periods.
A=6000 (1.04)2=6000×1.0816=6489.60A = 6000\,(1.04)^2 = 6000 \times 1.0816 = 6489.60A=6000(1.04)2=6000×1.0816=6489.60
The balance is 6,489.606{,}489.606,489.60 dollars.
Compounded annually at 6%6\%6%, in what year does 1,0001{,}0001,000 dollars first pass 1,5001{,}5001,500 dollars? Use 1.066=1.4185191.06^6 = 1.4185191.066=1.418519 and 1.067=1.5036301.06^7 = 1.5036301.067=1.503630.
Run the formula forward. After year 6 the balance is 1000×1.418519=1418.521000 \times 1.418519 = 1418.521000×1.418519=1418.52, still under 1,5001{,}5001,500.
1000 (1.06)7=1000×1.503630=1503.63>15001000\,(1.06)^7 = 1000 \times 1.503630 = 1503.63 > 15001000(1.06)7=1000×1.503630=1503.63>1500
The balance first passes 1,5001{,}5001,500 dollars in year 7.
You invest 1,0001{,}0001,000 dollars at 5%5\%5% for 101010 years. Which earns more, compounded annually or simple interest, and by roughly how much? Use 1.0510=1.6288951.05^{10} = 1.6288951.0510=1.628895.
Compound gives 1000×1.628895=1628.901000 \times 1.628895 = 1628.901000×1.628895=1628.90; simple gives 1000(1+0.05×10)=15001000(1 + 0.05 \times 10) = 15001000(1+0.05×10)=1500.
1628.90−1500=128.90≈1291628.90 - 1500 = 128.90 \approx 1291628.90−1500=128.90≈129
Compound interest wins by about 129129129 dollars over ten years.
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