Compound Interest - Finite Mathematics

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Question

Money is deposited in a bank account at 7% interest per year, compounded bimonthly. If no money is deposited into or withdrawn from the account, then how long will it take for the money to double?

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Answer

If is deposited in a bank account with interest rate ( converted to a decimal) compounded times a year for years, then the final balance of the account is

$P = $P_{0}$ \left (1+ $\frac{r}{n}$ \right ) ^{nt}$

Dividing both sides of the formula by , this is

$\left (1+ $\frac{r}{n}$ \right ) ^{nt} = $$\frac{P}{P_{0}$$ }$

The money doubles, so substitute 2 for . Also, set and (bimonthly), and calculate as follows:

$\left (1+ $\frac{0.07}{6}$ \right ) ^{6t} = 2$

$\ln \left [ \left (1+ $\frac{0.07}{6}$ \right ) ^{6t} \right ]= \ln 2$

$6t \ln \left (1+ $\frac{0.07}{6}$ \right ) = \ln 2$

$t = $\frac{\ln 2}{ 6 \ln\left(1+ \frac{0.07}${6} \right ) }$

$\approx $\frac{\ln 2}{ 6 \ln\left(1.011666667 \right) }$$

$\approx $\frac{0.693147181 }{ 6 (0.011599136) }$$

$\approx 9.960$

$0.960 \times 12 \approx 11.52$ , so this is 9 years and 11.52 months; rounding up to the next bimonth, this is 10 years even.

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Question

Money is deposited in a bank account at 7% interest per year, compounded bimonthly. If no money is deposited into or withdrawn from the account, then how long will it take for the money to double?

Tap to reveal answer

Answer

If is deposited in a bank account with interest rate ( converted to a decimal) compounded times a year for years, then the final balance of the account is

$P = $P_{0}$ \left (1+ $\frac{r}{n}$ \right ) ^{nt}$

Dividing both sides of the formula by , this is

$\left (1+ $\frac{r}{n}$ \right ) ^{nt} = $$\frac{P}{P_{0}$$ }$

The money doubles, so substitute 2 for . Also, set and (bimonthly), and calculate as follows:

$\left (1+ $\frac{0.07}{6}$ \right ) ^{6t} = 2$

$\ln \left [ \left (1+ $\frac{0.07}{6}$ \right ) ^{6t} \right ]= \ln 2$

$6t \ln \left (1+ $\frac{0.07}{6}$ \right ) = \ln 2$

$t = $\frac{\ln 2}{ 6 \ln\left(1+ \frac{0.07}${6} \right ) }$

$\approx $\frac{\ln 2}{ 6 \ln\left(1.011666667 \right) }$$

$\approx $\frac{0.693147181 }{ 6 (0.011599136) }$$

$\approx 9.960$

$0.960 \times 12 \approx 11.52$ , so this is 9 years and 11.52 months; rounding up to the next bimonth, this is 10 years even.

← Didn't Know|Knew It →

Question

Money is deposited in a bank account at 7% interest per year, compounded bimonthly. If no money is deposited into or withdrawn from the account, then how long will it take for the money to double?

Tap to reveal answer

Answer

If is deposited in a bank account with interest rate ( converted to a decimal) compounded times a year for years, then the final balance of the account is

$P = $P_{0}$ \left (1+ $\frac{r}{n}$ \right ) ^{nt}$

Dividing both sides of the formula by , this is

$\left (1+ $\frac{r}{n}$ \right ) ^{nt} = $$\frac{P}{P_{0}$$ }$

The money doubles, so substitute 2 for . Also, set and (bimonthly), and calculate as follows:

$\left (1+ $\frac{0.07}{6}$ \right ) ^{6t} = 2$

$\ln \left [ \left (1+ $\frac{0.07}{6}$ \right ) ^{6t} \right ]= \ln 2$

$6t \ln \left (1+ $\frac{0.07}{6}$ \right ) = \ln 2$

$t = $\frac{\ln 2}{ 6 \ln\left(1+ \frac{0.07}${6} \right ) }$

$\approx $\frac{\ln 2}{ 6 \ln\left(1.011666667 \right) }$$

$\approx $\frac{0.693147181 }{ 6 (0.011599136) }$$

$\approx 9.960$

$0.960 \times 12 \approx 11.52$ , so this is 9 years and 11.52 months; rounding up to the next bimonth, this is 10 years even.

← Didn't Know|Knew It →

Question

Money is deposited in a bank account at 7% interest per year, compounded bimonthly. If no money is deposited into or withdrawn from the account, then how long will it take for the money to double?

Tap to reveal answer

Answer

If is deposited in a bank account with interest rate ( converted to a decimal) compounded times a year for years, then the final balance of the account is

$P = $P_{0}$ \left (1+ $\frac{r}{n}$ \right ) ^{nt}$

Dividing both sides of the formula by , this is

$\left (1+ $\frac{r}{n}$ \right ) ^{nt} = $$\frac{P}{P_{0}$$ }$

The money doubles, so substitute 2 for . Also, set and (bimonthly), and calculate as follows:

$\left (1+ $\frac{0.07}{6}$ \right ) ^{6t} = 2$

$\ln \left [ \left (1+ $\frac{0.07}{6}$ \right ) ^{6t} \right ]= \ln 2$

$6t \ln \left (1+ $\frac{0.07}{6}$ \right ) = \ln 2$

$t = $\frac{\ln 2}{ 6 \ln\left(1+ \frac{0.07}${6} \right ) }$

$\approx $\frac{\ln 2}{ 6 \ln\left(1.011666667 \right) }$$

$\approx $\frac{0.693147181 }{ 6 (0.011599136) }$$

$\approx 9.960$

$0.960 \times 12 \approx 11.52$ , so this is 9 years and 11.52 months; rounding up to the next bimonth, this is 10 years even.

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