Mathematics of Finance

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Questions 1 - 4

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.

$\$702$

$\$648$

$\$576$

$\$2150$

$\$750$

Explanation

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.

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.

$\$702$

$\$648$

$\$576$

$\$2150$

$\$750$

Explanation

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.

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 If no money is deposited into or withdrawn from the account, then how long will it take for the money to double?

10 years 0 months

10 years 2 months

9 years 8 months

9 years 10 months

9 years 6 months

Explanation

If $P_{0}$ 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 $P_{0}$ , this is

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

The money doubles, so substitute 2 for $\frac{P}{P_{0} }$ . 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.

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?

10 years 0 months

10 years 2 months

9 years 8 months

9 years 10 months

9 years 6 months

Explanation

If $P_{0}$ 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 $P_{0}$ , this is

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

The money doubles, so substitute 2 for $\frac{P}{P_{0} }$ . 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.