Number & Algebra

QUESTION 1 [5 marks] Easy

Ananya invests \$3000 in an account that pays 5.8\% compound interest per year. She does not invest any further amount or withdraw any money from the account.

a. Find the amount in her account after 8 years.

[2]

Show Solution

$$A = 3000(1+5.8/100)^{8} = 4,709.84$$

b. Find the number of years until the amount in her account exceeds \$5500.

[3]

Show Solution

$$3000(1+5.8/100)^n > 5500 \Rightarrow n > \dfrac{\ln(5500/3000)}{\ln(1+5.8/100)} = 10.75$$ So after 11 years.

QUESTION 2 [3 marks] Medium

An investment of \$650 is made at the beginning of every year for 10 years. Interest accumulates at a rate of 3.2\% per year, compounded annually. Find the value of the investment at the end of the 10 years.

Show Solution

Using the annuity-due formula: $$FV = 650\times\dfrac{(1+0.032)^{10}-1}{0.032}\times(1+0.032) = 7,761.18$$

QUESTION 3 [6 marks] Hard

A family takes out a loan of \$180000 to purchase a house. They agree to pay the bank \$1450 at the end of every month to amortise the loan, and interest accumulates on the balance at a rate of 1.1\% per year, compounded monthly.

a. Find how long it takes to pay back the loan, in years and months.

[4]

Show Solution

$$n = \dfrac{-\ln\left(1-\dfrac{Pi}{PMT}\right)}{\ln(1+i)} = 131.8 \text{ months} \approx 11\text{ years, } 0\text{ months}$$

b. Calculate how much they have paid in total in amortising the loan.

[2]

Show Solution

$$\text{Total} = 1450\times 132 = 191400$$

QUESTION 4 [4 marks] Easy

At her son's birth, Meera put \$800 into a savings account. The interest earned is 3.6\% per year, compounded quarterly. How much money will he have on his 18th birthday?

Show Solution

$$A = 800\left(1+\dfrac{3.6/100}{4}\right)^{4\times18} = 1,524.94$$

QUESTION 5 [6 marks] Hard

A bank offers loans of \$P at the beginning of a particular month at a monthly interest rate of $I\%$. The interest is calculated at the end of each month and added to the amount outstanding. A repayment of \$R is required at the end of each month. Let $S_n$ denote the amount outstanding, in dollars, immediately after the $n$th monthly repayment.

a. Find an expression for $S_1$.

[2]

Show Solution

After one month, interest adds $\dfrac{I}{100}P$, then a repayment of $R$ is made: $$S_1 = P\left(1+\dfrac{I}{100}\right) - R$$

b. Find a similar expression for $S_2$, and hence show the general pattern for $S_n$.

[4]

Show Solution

$$S_2 = S_1\left(1+\dfrac{I}{100}\right) - R = P\left(1+\dfrac{I}{100}\right)^2 - R\left(1+\left(1+\dfrac{I}{100}\right)\right)$$ Continuing this pattern (the $R$ terms form a geometric series): $$S_n = P\left(1+\dfrac{I}{100}\right)^n - \dfrac{100R}{I}\left(\left(1+\dfrac{I}{100}\right)^n - 1\right)$$

QUESTION 6 [5 marks] Hard

Kavya takes out a bank loan of \$200000 to buy a house, at an annual interest rate of 3.8\%, compounded annually and added to the balance. To pay off the loan, she makes annual deposits of \$P at the end of every year, into a savings account paying 2.4\% per year, compounded annually, making her first deposit at the end of year 1. Her goal is to have saved enough to pay off the loan after 20 years.

a. Find the amount Kavya would owe the bank after 20 years.

[2]

Show Solution

$$\text{Owed} = 200000(1+3.8/100)^{20} = 421,674.24$$

b. Find the value of $P$ required, to the nearest dollar.

[3]

Show Solution

Using the ordinary annuity future value formula: $$\text{Owed} = P\times\dfrac{(1+i)^{20}-1}{i}, \quad i=0.024$$ Solving: $$P = 421,674.24 \Big/ \dfrac{(1+0.024)^{20}-1}{0.024} = 16674$$

QUESTION 7 [4 marks] Medium

A bank pays interest at a rate of 3.9\% compounded monthly. If the inflation rate is 1.6\%, calculate the real rate of return per year.

Show Solution

First find the effective annual rate: $$\left(1+\dfrac{3.9/100}{12}\right)^{12} - 1 = 3.97\%$$ Then the real rate: $$\left(\dfrac{1+3.97/100}{1+1.6/100} - 1\right)\times 100 = 2.33\%$$

QUESTION 8 [7 marks] Medium

In 2020, the population growth rate in Country X was 1.4\% and the population was 42.3 million. The population growth rate in Country Y was 0.3\% and the population was 51.8 million. Assume both growth rates remain constant.

a. Find the size of the population in Country X in 2024.

[2]

Show Solution

$$42.3(1+1.4/100)^{4} = 44.72 \text{ million}$$

b. Find the size of the population in Country Y in 2024.

[2]

Show Solution

$$51.8(1+0.3/100)^{4} = 52.42 \text{ million}$$

c. Find when the population of Country X will exceed that of Country Y.

[3]

Show Solution

Solving $42.3(1.014)^n = 51.8(1.003)^n$ gives $n \approx 18.57$ years after 2020, i.e. during 2039.

QUESTION 9 [3 marks] Easy

A bank pays interest at a rate of 4.8\% compounded annually. If the inflation rate is 2.1\%, calculate the real rate of return per year.

Show Solution

$$\text{Real rate} = \left(\dfrac{1+4.8/100}{1+2.1/100} - 1\right)\times 100 = 2.64\%$$

QUESTION 10 [5 marks] Medium

Farhan invests \$6000 in an account that pays 4.2\% interest compounded monthly. He does not invest any further amount or withdraw any money.

a. Find the amount in his account after 6 years.

[2]

Show Solution

$$A = 6000\left(1+\dfrac{4.2/100}{12}\right)^{12\times6} = 7,716.18$$

b. Find the number of months until the amount in his account doubles.

[3]

Show Solution

$$\left(1+\dfrac{4.2/100}{12}\right)^n = 2 \Rightarrow n = \dfrac{\ln 2}{\ln(1+4.2/1200)} = 198.4$$ So 199 months.

QUESTION 11 [6 marks] Hard

A teacher invests \$90 at the beginning of every month into a fund earning 4.5\% per year, compounded monthly.

a. Find the value of the investment at the end of 20 years.

[3]

Show Solution

$$FV = 90\times\dfrac{(1+i)^{240}-1}{i}\times(1+i), \quad i=\dfrac{4.5/100}{12}$$ $$FV = 35,062.18$$

b. Another teacher wants to have \$35000 at the end of 20 years, investing monthly at the same rate. Calculate how much she must invest per month.

[3]

Show Solution

Rearranging the same formula for the payment: $$PMT = \dfrac{35000}{\left(\dfrac{(1+i)^{240}-1}{i}\right)(1+i)} = 89.84$$

Financial and Growth Models