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Year 10AC9M10A04

Investments with Multiple Rates

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. $20 000 invested for 5 years: 10% p.a. for the first 2 years, then 12% p.a. for the next 3\$20\,000 \text{ invested for 5 years: } 10\% \text{ p.a. for the first 2 years, then } 12\% \text{ p.a. for the next 3}
  1. Write down list of information AND A TIMELINE
    Five year timeline split into two interest periods10% p.a.2 years12% p.a.3 yearsP = $20 000A = ?year 0year 5
  2. Find a formula — one bracket per rateA=P(1+i1)n1(1+i2)n2A = P(1+i_{1})^{n_{1}}(1+i_{2})^{n_{2}}
  3. SubstituteA=20 000(1.1)2(1.12)3A = 20\,000(1.1)^{2}(1.12)^{3}
  4. SolveA=$33 999.26A = \$33\,999.26
    The same timeline showing the multiplier applied in each period× (1.1)²2 years× (1.12)³3 yearsP = $20 000A = $33 999.26year 0year 5

Practice

Calculate the final amount:

1
$10 000 for 4 years: 5% for 2 years, then 8% for 2 years\$10\,000 \text{ for 4 years: } 5\% \text{ for 2 years, then } 8\% \text{ for 2 years}
AnswerA=$12 859.56A = \$12\,859.56
2
$5 000 for 6 years: 6% for 3 years, then 9% for 3 years\$5\,000 \text{ for 6 years: } 6\% \text{ for 3 years, then } 9\% \text{ for 3 years}
AnswerA=$7 712.00A = \$7\,712.00
3
$50 000 for 5 years: 4% for 1 year, then 7% for 4 years\$50\,000 \text{ for 5 years: } 4\% \text{ for 1 year, then } 7\% \text{ for 4 years}
AnswerA=$68 161.39A = \$68\,161.39
4
$2 500 for 4 years: 10% for 2 years, then 5% for 2 years\$2\,500 \text{ for 4 years: } 10\% \text{ for 2 years, then } 5\% \text{ for 2 years}
AnswerA=$3 335.06A = \$3\,335.06
5
$100 000 for 6 years: 8% for 4 years, then 3% for 2 years\$100\,000 \text{ for 6 years: } 8\% \text{ for 4 years, then } 3\% \text{ for 2 years}
AnswerA=$144 334.27A = \$144\,334.27
Next step
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