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Compound Interest: Effective Rates and Inflation — Answers

Answer sheet with worked solutions for the Year 12 financial mathematics worksheet.

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Success TutoringWorksheetsAnswer sheetYear 12 · Mathematics · Financial Mathematics

Compound Interest: Effective Rates and Inflation — Answers

Part A: Compound interest with different compounding periods

A = P(1 + r)[sup n], where P is the principal, r is the interest rate per compounding period (as a decimal) and n is the number of compounding periods. For a nominal rate of R% p.a. compounded m times a year, r = and n = m × years.

Rearrange for other unknowns: P = , n = and r = (A ÷ P)1/n − 1.

Worked example. $5000 at 4% p.a. compounded quarterly for 2 years: r = 0.01, n = 8, A = 5000 × 1.018 = $5414.28, so the interest is $414.28.

1.$8000 is invested at 6% p.a. compounded monthly for 5 years. Find the final amount and the interest earned.

Answer: $10 790.80; interest $2790.80

  1. r = 0.06 ÷ 12 = 0.005 per month; n = 5 × 12 = 60
  2. A = 8000 × 1.00560 = 8000 × 1.348850… = $10 790.80
  3. Interest = 10 790.80 − 8000 = $2790.80
2.$12 000 is invested at 5.2% p.a. compounded quarterly for 3 years. Find the final amount and the interest earned.

Answer: $14 011.82; interest $2011.82

  1. r = 0.052 ÷ 4 = 0.013 per quarter; n = 3 × 4 = 12
  2. A = 12 000 × 1.01312 = $14 011.82
  3. Interest = 14 011.82 − 12 000 = $2011.82
3.How much must be invested now at 4.8% p.a. compounded monthly to have $20 000 in 4 years?

Answer: $16 512.46

  1. r = 0.048 ÷ 12 = 0.004; n = 48
  2. P = = 20 000 ÷ 1.211206… = $16 512.46
4.How many whole years does it take for an investment to at least double at 7% p.a. compounded annually?

Answer: 11 years

  1. Solve 1.07n = 2: n = = 10.24…
  2. After 10 years the factor is 1.0710 = 1.967 (not yet doubled); after 11 years it is 1.0711 = 2.105.
  3. So it takes 11 years (to the nearest whole year above).
5.$4000 grows to $5000 in 5 years with interest compounded annually. Find the annual interest rate, as a percentage correct to 2 decimal places.

Answer: 4.56% p.a.

  1. 4000(1 + r)5 = 5000, so (1 + r)5 = 1.25
  2. 1 + r = 1.251/5 = 1.045639…
  3. r = 0.045639… ≈ 4.56% p.a.

Part B: Effective annual rates

Nominal rates with different compounding periods cannot be compared directly. The effective annual rate is the single yearly rate that gives the same growth: i = (1 + [frac R/m])[sup m] − 1, where R is the nominal annual rate (as a decimal) and m is the number of compounding periods per year.

Worked example. 12% p.a. compounded monthly: i = 1.0112 − 1 = 0.126825…, so the effective rate is 12.68% p.a. — the same as 12.68% compounded once a year.

6.Find the effective annual rate for 6% p.a. compounded monthly, as a percentage correct to 2 decimal places.

Answer: 6.17% p.a.

  1. i = (1 + )12 − 1 = 1.00512 − 1 = 0.061678…
  2. Effective rate ≈ 6.17% p.a.
7.Find the effective annual rate for 8% p.a. compounded quarterly, as a percentage correct to 2 decimal places.

Answer: 8.24% p.a.

  1. i = (1 + )4 − 1 = 1.024 − 1 = 0.082432
  2. Effective rate ≈ 8.24% p.a.
8.Which savings account gives the highest effective annual rate?
  1. A6.00% p.a. compounded annually
  2. B5.90% p.a. compounded monthly
  3. C5.95% p.a. compounded quarterly
  4. D5.80% p.a. compounded daily (365 days)

Answer: 5.95% p.a. compounded quarterly

Effective rates: 6.00%; (1 + 0.059 ÷ 12)12 − 1 ≈ 6.06%; (1 + 0.0595 ÷ 4)4 − 1 ≈ 6.08%; (1 + 0.058 ÷ 365)365 − 1 ≈ 5.97%. The quarterly account is best, even though its nominal rate is not the highest.

9.For a positive nominal rate compounded more than once a year, the effective annual rate is always greater than the nominal rate.

Answer: True

Interest earned during the year itself earns interest, so the balance grows by more than the nominal rate. With yearly compounding the two rates are equal.

10.A nominal rate of 4.8% p.a. compounded monthly is a rate of (answer space)% per month.

Answer: 0.4

4.8 ÷ 12 = 0.4% per month.

Part C: Inflation and real returns

Inflation is the general rise in prices over time. With an average inflation rate f per year, a cost C today becomes C(1 + f)[sup n] after n years, and an amount A received in n years is worth [frac A/(1 + f)[sup n]] in today's dollars.

The real rate of return measures growth in buying power: real rate = [frac 1 + nominal rate/1 + inflation rate] − 1. A quick estimate is nominal rate − inflation rate.

Worked example. A $40 bill today with 2% inflation: in 3 years it costs 40 × 1.023 = $42.45.

11.An item costs $250 today. If inflation averages 3% p.a., what will a similar item cost in 10 years?

Answer: $335.98

  1. Prices grow like compound interest: 250 × 1.0310 = 250 × 1.343916… = $335.98
12.You will receive $10 000 in 5 years. If inflation averages 2.5% p.a., what is this amount worth in today's dollars?

Answer: $8838.54

  1. Divide by the inflation factor: = 10 000 ÷ 1.131408… = $8838.54
13.$5000 is invested at 6% p.a. compounded annually for 5 years, while inflation averages 3% p.a. Find the final amount and its value in today's dollars.

Answer: $6691.13; $5771.83 in today's dollars

  1. A = 5000 × 1.065 = $6691.13
  2. In today's dollars: = $5771.83 (calculated with the unrounded amount)
  3. The real increase in buying power is about $771.83, not $1691.13.
14.An investment earns 6% p.a. while inflation is 3% p.a. Find the exact real rate of return, correct to 2 decimal places, and compare it with the quick estimate.

Answer: 2.91% p.a. (the estimate 6% − 3% = 3% is slightly too high)

  1. Real rate = − 1 = 0.029126…
  2. Real rate ≈ 2.91% p.a.
  3. The quick estimate 6% − 3% = 3% is close but slightly overstates the real return.
15.Savings earn 2% p.a. while inflation is 4% p.a. Which statement is true?
  1. AThe balance grows but its buying power falls
  2. BThe balance and its buying power both grow
  3. CThe balance falls
  4. DBuying power is unchanged because interest is earned

Answer: The balance grows but its buying power falls

The dollar balance grows by 2% a year, but prices grow by 4% a year, so each year the savings buy less. The real rate is about 1.02 ÷ 1.04 − 1 ≈ −1.9%.

Part D: Extended problem

Ali has $15 000 to save for 4 years and is comparing three options. Option A: 5.4% p.a. compounded monthly. Option B: 5.5% p.a. compounded half-yearly. Option C: 5.6% p.a. simple interest.

(a) Find the value of each option after 4 years.

(b) Find the effective annual rates of options A and B.

(c) Which option should Ali choose? Explain why option C gives the least even though its rate is the highest.

(d) Inflation is expected to average 3% p.a. over the 4 years. Find the value of the best option in today's dollars and the real gain in buying power.

16.Answer parts (a) to (d), showing full working. Give dollar amounts to the nearest cent and rates as percentages correct to 2 decimal places.[10 marks]

Model answer: (a) A: 15 000 × (1 + )48 = 15 000 × 1.004548 = $18 607.52. B: 15 000 × (1 + )8 = 15 000 × 1.02758 = $18 635.71. C: 15 000 × (1 + 0.056 × 4) = 15 000 × 1.224 = $18 360.00. (b) A: 1.004512 − 1 ≈ 5.54% p.a.; B: 1.02752 − 1 ≈ 5.58% p.a. B has the higher effective rate, which agrees with (a). (c) Option B gives the most ($28.19 more than A and $275.71 more than C). Simple interest does not earn interest on interest, so C is lowest despite the highest nominal rate. (d) = $16 557.59 in today's dollars, so the real gain is about $1557.59, much less than the $3635.71 of interest.

Success criteria:

  • (a) correct final amounts for A, B and C with correct r and n for each (3 marks)
  • (b) effective rates 5.54% and 5.58% (2 marks)
  • (c) chooses B, with a reason referring to the amounts or effective rates and why simple interest is lowest (2 marks)
  • (d) divides by 1.034 to get $16 557.59 and states the real gain (3 marks)
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