Compound interest questions with answers

Questions

1. Question 1Total amount, yearly

   Daniel invests $20 000 at 2% per annum compound interest, compounded annually. Find the total amount after 3 years, correct to the nearest cent.

   Show the worked solution

   1. Step 1: Use the compound interest formula$A = P\left(1 + \dfrac{r}{100}\right)^n$$A = 20000\left(1 + \dfrac{2}{100}\right)^{3}$
   2. Step 2: Work it out$A = 21224.16$
2. Question 2Interest earned

   Priya invests $2500 at 5% per annum compound interest, compounded annually. Find the interest earned after 4 years, correct to the nearest cent.

   Show the worked solution

   1. Step 1: Use the compound interest formula$A = P\left(1 + \dfrac{r}{100}\right)^n$$A = 2500\left(1 + \dfrac{5}{100}\right)^{4}$
   2. Step 2: Work it out$A = 3038.7656\ldots$Interest $= A - P$$= 3038.7656\ldots - 2500$
   3. Step 3: Round to the nearest cent$538.77
3. Question 3Half-yearly, quarterly, monthly

   Priya invests $3000 at 2% per annum compound interest, compounded monthly. Find the total amount after 2 years, correct to the nearest cent.

   Show the worked solution

   1. Step 1: Use the compound interest formula, with the rate and the number of periods for each compounding period$A = P\left(1 + \dfrac{r}{100}\right)^n$$A = 3000\left(1 + \dfrac{2}{100 \times 12}\right)^{24}$
   2. Step 2: Work it out$A = 3122.3284\ldots$
   3. Step 3: Round to the nearest cent$3122.33
4. Question 4Depreciation

   The value of a car bought for $25 000 depreciates at 8% per annum. Find its value after 2 years, correct to the nearest cent.

   Show the worked solution

   1. Step 1: Use the compound interest formula$A = P\left(1 - \dfrac{r}{100}\right)^n$$A = 25000\left(1 - \dfrac{8}{100}\right)^{2}$
   2. Step 2: Work it out$A = 21160.00$
5. Question 5Work backwards to the amount invested

   Mr Lim invests some money at 5% per annum compound interest, compounded half-yearly. After 3 years the investment is worth $12 500. Find the amount invested, correct to the nearest cent.

   Show the worked solution

   1. Step 1: Write the formula with the unknown amount$12500 = P\left(1 + \dfrac{5}{100 \times 2}\right)^{6}$
   2. Step 2: Divide to find P$P = \dfrac{12500}{\left(1 + \dfrac{5}{100 \times 2}\right)^{6}}$$\phantom{P} = 10778.7108\ldots$Amount invested: $10778.71
6. Question 6Find the interest rate

   Wei Ling invests $5000 at a fixed rate of compound interest per annum, compounded yearly. After 4 years the investment is worth $5671.38. Find the rate of interest, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Put the numbers into the formula$A = P\left(1 + \dfrac{r}{100}\right)^n$$5671.38 = 5000\left(1 + \dfrac{r}{100}\right)^{4}$
   2. Step 2: Make the bracket the subject$\left(1 + \dfrac{r}{100}\right)^{4} = \dfrac{5671.38}{5000}$$1 + \dfrac{r}{100} = \sqrt[4]{\dfrac{5671.38}{5000}}$$\phantom{1 + \dfrac{r}{100}} = 1.032000\ldots$
   3. Step 3: Solve for r and round$r = 100 \times (1.032000\ldots - 1)$$\phantom{r} = 3.2000\ldots$Rate $= 3.2\%$ per annum
7. Question 7A different rate in the first year

   Ahmad invests $12 000 in a savings plan. The plan pays 1.5% interest for the first year, then 3.8% per annum compound interest, compounded yearly, for the next 3 years. Find the total amount at the end of the 4 years, correct to the nearest cent.

   Show the worked solution

   1. Step 1: After the first year: 1.5% more$12000 \times \left(1 + \dfrac{1.5}{100}\right) = 12180$
   2. Step 2: Then 3 years at 3.8% compound interest$A = 12180 \times \left(1 + \dfrac{3.8}{100}\right)^{3}$$A = 13621.9521\ldots$
   3. Step 3: Round to the nearest cent$13621.95
8. Question 8Find the later rate

   Mr Lim invests $25 000 in a savings plan. The plan pays 3% interest for the first year, then $r$% per annum compound interest, compounded yearly, for the next 2 years. At the end of the 3 years the investment is worth $26 895.46. Find $r$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: The amount after the first year$25000 \times \left(1 + \dfrac{3}{100}\right) = 25750$
   2. Step 2: That amount grows for 2 years at r%$25750\left(1 + \dfrac{r}{100}\right)^{2} = 26895.46$
   3. Step 3: Divide, then take the square root$\left(1 + \dfrac{r}{100}\right)^{2} = \dfrac{26895.46}{25750}$$1 + \dfrac{r}{100} = 1.022000\ldots$
   4. Step 4: Solve for r and round to 1 decimal place$r = 100 \times (1.022000\ldots - 1)$$r = 2.2000\ldots$$r = 2.2$
9. Question 9The least number of whole years

   Priya invests $20 000 at 2.5% per annum compound interest, compounded yearly. Find the least number of whole years after which the investment is worth more than $28 000.

   Show the worked solution

   1. Step 1: The value after n years$20000 \times 1.025^{n}$
   2. Step 2: Try whole numbers of years (to the nearest dollar): is it more than 28000?$20000 \times 1.025^{12} \approx 26898$$20000 \times 1.025^{13} \approx 27570$$20000 \times 1.025^{14} \approx 28259$$20000 \times 1.025^{15} \approx 28966$
   3. Step 3: The first whole year it is more than 28000After $13$ years it is $27570$; after $14$ years it is $28259$, more than $28000$.$14$ years