Mathematics Standard • Year 11 • Module 3 • Lesson 12
Compound Interest
Practise HSC Mathematics Standard 2-style writing on compound interest — short answers and a structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 $9,500 is invested at 5.6% per annum compounding annually for 7 years. Calculate the final amount and the interest earned. 3 marks Band 3
1.2 A loan of $22,000 compounds quarterly at 7.2% per annum for 3 years. No repayments are made. Calculate the amount owed at the end of the 3 years. 3 marks Band 3-4
1.3 $16,000 is invested for 4 years.
(a) Calculate the interest earned under simple interest at 6% per annum.
(b) Calculate the interest earned under compound interest at 6% per annum compounding annually.
(c) State by how much compound interest exceeds simple interest over the 4 years. 4 marks Band 4
2. Extended response
2.1 Suri has $24,000 to invest for 6 years. Her broker presents two options.
Option P: 5.4% per annum compounding annually.
Option Q: 5.1% per annum compounding monthly.
(a) Calculate the final amount under Option P.
(b) Calculate the final amount under Option Q. (Show the adjusted r and n.)
(c) State which option produces the larger final amount and by how much. Then comment on whether monthly compounding is enough to overcome Option Q's lower nominal rate in this case. 6 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct final amount with (1.054)6 shown explicitly.
Part (b) — 2 marks
• 1 mark — correctly states r = 0.051 ÷ 12 and n = 6 × 12 = 72.
• 1 mark — correct final amount with (1 + r)72 shown explicitly.
Part (c) — 3 marks
• 1 mark — correct numerical comparison of the two amounts.
• 1 mark — explicit conclusion sentence naming the winning option and the dollar difference.
• 1 mark — accurate comment on whether the more frequent compounding was enough (i.e. compares the gain from monthly compounding versus the loss from a lower nominal rate).
Your response:
Stuck on (c)? Subtract the two final amounts, name the higher one, then write whether the gain from monthly compounding under Option Q was bigger or smaller than the gap created by Option P's higher nominal rate.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $9,500 at 5.6% p.a. annual for 7 years (3 marks)
Sample response.
P = $9,500, r = 0.056, n = 7.
(1.056)7 = 1.464706…
A = $9,500 × 1.464706 = $13,914.71.
I = A − P = $13,914.71 − $9,500 = $4,414.71.
Marking notes. 1 mark — correctly identifies P, r, n. 1 mark — calculates A correctly with (1+r)n shown. 1 mark — correctly calculates interest using I = A − P (with units). Writing A as the answer to the interest part = lose the third mark.
1.2 — Loan $22,000 at 7.2% p.a. quarterly for 3 years (3 marks)
Sample response.
r = 0.072 ÷ 4 = 0.018 per quarter; n = 3 × 4 = 12 quarters.
(1.018)12 = 1.238472…
A = $22,000 × 1.238472 = $27,246.39.
Marking notes. 1 mark — correctly adjusts both r and n for quarterly. 1 mark — shows (1.018)12 as a separate line. 1 mark — correct A with units. Common error: leaving r = 0.072 and using n = 12 → gives an inflated answer; or adjusting only one of r/n.
1.3 — Simple vs compound on $16,000 at 6% for 4 years (4 marks)
(a) Sample response. Isimple = Prn = 16,000 × 0.06 × 4 = $3,840.00.
(b) Sample response. (1.06)4 = 1.262477…; A = 16,000 × 1.262477 = $20,199.63; Icompound = $20,199.63 − $16,000 = $4,199.63.
(c) Sample response. Compound exceeds simple by $4,199.63 − $3,840.00 = $359.63.
Marking notes. (a) 1 mark — correct simple interest. (b) 1 mark — correct A; 1 mark — correct I via A − P. (c) 1 mark — correct difference with conclusion sentence. A bare number "$359.63" without identifying which method exceeds the other costs the final mark.
2.1 — Option P vs Option Q on $24,000 for 6 years (6 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Option P — 5.4% p.a. compounding annually.
P = $24,000, r = 0.054, n = 6.
(1.054)6 = 1.370185…
AP = $24,000 × 1.370185 = $32,884.44. [1 mark — Option P correct.]
(b) Option Q — 5.1% p.a. compounding monthly.
r = 0.051 ÷ 12 = 0.00425 per month; n = 6 × 12 = 72 months. [1 mark — adjusted r and n.]
(1.00425)72 = 1.357144…
AQ = $24,000 × 1.357144 = $32,571.46. [1 mark — Option Q correct with (1+r)72 shown.]
(c) Compare.
Difference = $32,884.44 − $32,571.46 = $312.98. [1 mark — correct numerical comparison.]
Conclusion: Option P (5.4% annually) gives the larger final amount, by $312.98 over 6 years. [1 mark — explicit conclusion naming the winning option, dollar difference, time unit.]
Comment: monthly compounding under Option Q does add value (it lifts an effective rate of about 5.22% from the nominal 5.1%), but it is not enough to overcome the 0.3 percentage-point gap in nominal rate — Option P's higher headline rate wins. More frequent compounding typically helps, but only by a small amount per year, so it cannot beat a clearly higher nominal rate. [1 mark — accurate comparison comment.]
Total: 6/6.
Band descriptors for marker.
Band 3: Calculates one option correctly (probably annual). Cannot or does not adjust r and n for monthly compounding. ≈ 2-3 marks.
Band 4: Both options calculated, but either (1+r)n is not shown explicitly, or the answer to (c) is a bare number with no conclusion sentence. ≈ 4 marks.
Band 5: Full numerical solution including correct adjustment of r and n and an explicit conclusion sentence, but the comparison comment in (c) is missing or wrong. ≈ 5 marks.
Band 6: Complete, correctly adjusted, explicit conclusion naming the winner and the dollar difference, AND an accurate comparison comment explaining the trade-off between compounding frequency and nominal rate. 6/6.