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Year 9 Simple and Compound Interest Worksheet — Level 2 — Answers

Answer sheet with worked solutions for the Year 9 financial mathematics worksheet.

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Success TutoringWorksheetsAnswer sheetYear 9 · Mathematics · Financial Mathematics

Year 9 Simple and Compound Interest Worksheet — Level 2 — Answers

Show your working. Write the rate as a decimal (5% = 0.05). Round money to the nearest cent.

1.$12,750.00 is invested at 4.5% per year, compounded yearly, for 4 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $15,204.61; interest $2,454.61

  1. A = P(1 + r)n = $12,750.00 × (1 + 0.045)4
  2. = $12,750.00 × 1.192519
  3. ≈ $15,204.61
  4. Interest = $15,204.61 − $12,750.00 = $2,454.61
2.Find the simple interest on $13,750.00 invested at 8% per year for 4 years. What is the total value at the end?

Answer: Interest $4,400.00; total $18,150.00

  1. I = P × r × n = $13,750.00 × 0.08 × 4
  2. = $4,400.00
  3. Total = $13,750.00 + $4,400.00 = $18,150.00
3.$9,000.00 is invested at 4% per year, compounded yearly, for 3 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $10,123.78; interest $1,123.78

  1. A = P(1 + r)n = $9,000.00 × (1 + 0.04)3
  2. = $9,000.00 × 1.124864
  3. ≈ $10,123.78
  4. Interest = $10,123.78 − $9,000.00 = $1,123.78
4.$10,250.00 is invested at 4.5% per year, compounded yearly, for 4 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $12,223.32; interest $1,973.32

  1. A = P(1 + r)n = $10,250.00 × (1 + 0.045)4
  2. = $10,250.00 × 1.192519
  3. ≈ $12,223.32
  4. Interest = $12,223.32 − $10,250.00 = $1,973.32
5.$3,250.00 is invested at 6% per year, compounded yearly, for 3 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $3,870.80; interest $620.80

  1. A = P(1 + r)n = $3,250.00 × (1 + 0.06)3
  2. = $3,250.00 × 1.191016
  3. ≈ $3,870.80
  4. Interest = $3,870.80 − $3,250.00 = $620.80
6.Find the simple interest on $3,250.00 invested at 4.5% per year for 2 years. What is the total value at the end?

Answer: Interest $292.50; total $3,542.50

  1. I = P × r × n = $3,250.00 × 0.045 × 2
  2. = $292.50
  3. Total = $3,250.00 + $292.50 = $3,542.50
7.Find the simple interest on $13,000.00 invested at 3% per year for 5 years. What is the total value at the end?

Answer: Interest $1,950.00; total $14,950.00

  1. I = P × r × n = $13,000.00 × 0.03 × 5
  2. = $1,950.00
  3. Total = $13,000.00 + $1,950.00 = $14,950.00
8.$13,500.00 is invested at 6% per year, compounded yearly, for 4 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $17,043.44; interest $3,543.44

  1. A = P(1 + r)n = $13,500.00 × (1 + 0.06)4
  2. = $13,500.00 × 1.262477
  3. ≈ $17,043.44
  4. Interest = $17,043.44 − $13,500.00 = $3,543.44
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