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Year 11 Simple and Compound Interest Worksheet — Level 2 — Practice Set 4 — Answers

Answer sheet with worked solutions for the Year 11 financial mathematics worksheet.

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Success TutoringWorksheetsAnswer sheetYear 11 · Mathematics · Financial Mathematics

Year 11 Simple and Compound Interest Worksheet — Level 2 — Practice Set 4 — Answers

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

1.$10,000.00 is invested at 2.5% per year, compounded yearly, for 2 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $10,506.25; interest $506.25

  1. A = P(1 + r)n = $10,000.00 × (1 + 0.025)2
  2. = $10,000.00 × 1.050625
  3. ≈ $10,506.25
  4. Interest = $10,506.25 − $10,000.00 = $506.25
2.$4,000.00 is invested at 2% per year, compounded yearly, for 4 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $4,329.73; interest $329.73

  1. A = P(1 + r)n = $4,000.00 × (1 + 0.02)4
  2. = $4,000.00 × 1.082432
  3. ≈ $4,329.73
  4. Interest = $4,329.73 − $4,000.00 = $329.73
3.$11,750.00 is invested at 3% per year, compounded yearly, for 3 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $12,839.54; interest $1,089.54

  1. A = P(1 + r)n = $11,750.00 × (1 + 0.03)3
  2. = $11,750.00 × 1.092727
  3. ≈ $12,839.54
  4. Interest = $12,839.54 − $11,750.00 = $1,089.54
4.$10,250.00 is invested at 2% per year, compounded yearly, for 3 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $10,877.38; interest $627.38

  1. A = P(1 + r)n = $10,250.00 × (1 + 0.02)3
  2. = $10,250.00 × 1.061208
  3. ≈ $10,877.38
  4. Interest = $10,877.38 − $10,250.00 = $627.38
5.Find the simple interest on $13,250.00 invested at 3% per year for 4 years. What is the total value at the end?

Answer: Interest $1,590.00; total $14,840.00

  1. I = P × r × n = $13,250.00 × 0.03 × 4
  2. = $1,590.00
  3. Total = $13,250.00 + $1,590.00 = $14,840.00
6.Find the simple interest on $6,750.00 invested at 5% per year for 5 years. What is the total value at the end?

Answer: Interest $1,687.50; total $8,437.50

  1. I = P × r × n = $6,750.00 × 0.05 × 5
  2. = $1,687.50
  3. Total = $6,750.00 + $1,687.50 = $8,437.50
7.$9,000.00 is invested at 3% per year, compounded yearly, for 4 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $10,129.58; interest $1,129.58

  1. A = P(1 + r)n = $9,000.00 × (1 + 0.03)4
  2. = $9,000.00 × 1.125509
  3. ≈ $10,129.58
  4. Interest = $10,129.58 − $9,000.00 = $1,129.58
8.$8,000.00 is invested at 5% per year, compounded yearly, for 6 years. Find the final amount and the interest earned, to the nearest cent.

Answer: Amount $10,720.77; interest $2,720.77

  1. A = P(1 + r)n = $8,000.00 × (1 + 0.05)6
  2. = $8,000.00 × 1.340096
  3. ≈ $10,720.77
  4. Interest = $10,720.77 − $8,000.00 = $2,720.77
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