Mathematics of Finance - Finite Mathematics

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Question

What is the total interest made after 4 years on a simple interest loan that accumulates 13% each year and has original amount was $1350.

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Answer

To solve this problem, first recall the formula to calculate simple interest.

$I=A_0\times r\times t$

where

$\A_0=\text{initial amount} \r=\text{rate} \t=\text{time} \I=\text{interest accumulated}$

From the question it's known that,

$\A_0=1350\r=0.13 \t=4$

Thus the equation becomes,

$\I=1350\times 0.13\times 4 \I=702$

Therefore $702 in interested accumulated after four years.

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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 If no money is deposited into or withdrawn from the account, then how long will it take for the money to double?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

What is the total interest made after 4 years on a simple interest loan that accumulates 13% each year and has original amount was $1350.

Tap to reveal answer

Answer

To solve this problem, first recall the formula to calculate simple interest.

$I=A_0\times r\times t$

where

$\A_0=\text{initial amount} \r=\text{rate} \t=\text{time} \I=\text{interest accumulated}$

From the question it's known that,

$\A_0=1350\r=0.13 \t=4$

Thus the equation becomes,

$\I=1350\times 0.13\times 4 \I=702$

Therefore $702 in interested accumulated after four years.

← Didn't Know|Knew It →

Question

What is the total interest made after 4 years on a simple interest loan that accumulates 13% each year and has original amount was $1350.

Tap to reveal answer

Answer

To solve this problem, first recall the formula to calculate simple interest.

$I=A_0\times r\times t$

where

$\A_0=\text{initial amount} \r=\text{rate} \t=\text{time} \I=\text{interest accumulated}$

From the question it's known that,

$\A_0=1350\r=0.13 \t=4$

Thus the equation becomes,

$\I=1350\times 0.13\times 4 \I=702$

Therefore $702 in interested accumulated after four years.

← 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

What is the total interest made after 4 years on a simple interest loan that accumulates 13% each year and has original amount was $1350.

Tap to reveal answer

Answer

To solve this problem, first recall the formula to calculate simple interest.

$I=A_0\times r\times t$

where

$\A_0=\text{initial amount} \r=\text{rate} \t=\text{time} \I=\text{interest accumulated}$

From the question it's known that,

$\A_0=1350\r=0.13 \t=4$

Thus the equation becomes,

$\I=1350\times 0.13\times 4 \I=702$

Therefore $702 in interested accumulated after four years.

← 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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