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Year 10AC9M10A04

Calculate Investment Interest

Use mathematical modelling to solve applied problems involving growth and decay, including financial contexts; formulate problems, choosing to apply linear, quadratic or exponential models; evaluate and modify models and report assumptions, methods and findings.

Worked example

e.g. an amount of $24200 is invested. What interest rate grows it to $30356.48 in 2 years?\text{an amount of } \$24\,200 \text{ is invested. What interest rate grows it to } \$30\,356.48 \text{ in 2 years?}
  1. Write down list of informationP=24200,A=30356.48,n=2P = 24\,200,\quad A = 30\,356.48,\quad n = 2
  2. Find a formula (memorise it)A=P(1+i)nA = P(1 + i)^{n}
  3. Substitute correct values into formula30356.48=24200(1+i)230\,356.48 = 24\,200(1 + i)^{2}
  4. Remember i=%times per yeari = \frac{\%}{\text{times per year}} as a decimal, and n=times per year×number of yearsn = \text{times per year} \times \text{number of years}
  5. Solve for the asked value(1+i)2=1.2544  1+i=1.12  i=12%(1+i)^{2} = 1.2544 \ \therefore \ 1 + i = 1.12 \ \therefore \ i = 12\%

Practice

Complete the following using the compound interest formula:

1
A=$3200, i=6.8%, n=7 years. Find PA = \$3\,200,\ i = 6.8\%,\ n = 7 \text{ years. Find } P
AnswerP=$2019.07P = \$2\,019.07
2
P=$7500, A=$9447.84, n=3 years. Find iP = \$7\,500,\ A = \$9\,447.84,\ n = 3 \text{ years. Find } i
Answeri=8%i = 8\%
3
P=$20500, i=12%, n=4 years. Find AP = \$20\,500,\ i = 12\%,\ n = 4 \text{ years. Find } A
AnswerA=$32257.15A = \$32\,257.15
4
P=$5000, i=6%, n=10 years. Find AP = \$5\,000,\ i = 6\%,\ n = 10 \text{ years. Find } A
AnswerA=$8954.24A = \$8\,954.24
5
P=$12000, i=4.5%, n=6 years. Find AP = \$12\,000,\ i = 4.5\%,\ n = 6 \text{ years. Find } A
AnswerA=$15627.12A = \$15\,627.12
6
P=$800, A=$1000, i=4.7%. Find n (nearest year)P = \$800,\ A = \$1\,000,\ i = 4.7\%. \text{ Find } n \text{ (nearest year)}
Answern=5 yearsn = 5 \text{ years}
7
P=$30000, i=7%, n=3 years. Find AP = \$30\,000,\ i = 7\%,\ n = 3 \text{ years. Find } A
AnswerA=$36751.29A = \$36\,751.29
8
P=$1000 at 10% compounded for 2 years. How much MORE than simple interest?P = \$1\,000 \text{ at } 10\% \text{ compounded for 2 years. How much MORE than simple interest?}
AnswerCompound $1210 vs simple $1200=$10 more\text{Compound } \$1\,210 \text{ vs simple } \$1\,200 = \$10 \text{ more}
Next step
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