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Compound interest practice problems for calculating final balances over different time periods and compounding frequencies.

A worksheet with ten math problems involving compound interest calculations for various investment amounts, interest rates, and time periods.

A worksheet with ten math problems involving compound interest calculations for various investment amounts, interest rates, and time periods.

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Show Answer Key & Explanations Step-by-step solution for: Compound Interest Worksheet by Hannah Douglas worksheets library
To solve these problems, we use the compound interest formula:

A = P(1 + r/n)^(nt)

Where:
- A = ending balance (what we’re solving for)
- P = principal (starting amount)
- r = annual interest rate (as a decimal)
- n = number of times interest is compounded per year
- t = time in years

Since all problems say “compounded annually” or “semiannually”, we’ll adjust n accordingly:
- Annually → n = 1
- Semiannually → n = 2

Let’s go one by one.

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1) $20,000 at 5.4% compounded annually for 3 years

P = 20000
r = 5.4% = 0.054
n = 1
t = 3

A = 20000 × (1 + 0.054/1)^(1×3)
= 20000 × (1.054)^3
Calculate 1.054^3:
1.054 × 1.054 = 1.110916
1.110916 × 1.054 ≈ 1.170905

A ≈ 20000 × 1.170905 ≈ $23,418.10

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2) $1,250 at 8% compounded annually for 2 years

P = 1250
r = 0.08
n = 1
t = 2

A = 1250 × (1.08)^2
1.08² = 1.1664
A = 1250 × 1.1664 = $1,458.00

---

3) $650 at 7% compounded annually for 5 years

P = 650
r = 0.07
n = 1
t = 5

A = 650 × (1.07)^5
1.07^5 = 1.4025517 (use calculator or step-by-step multiplication)
A ≈ 650 × 1.4025517 ≈ $911.66

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4) $1,340 at 8% compounded annually for 2 years

P = 1340
r = 0.08
n = 1
t = 2

A = 1340 × (1.08)^2 = 1340 × 1.1664 = $1,562.98

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5) $9,000 at 11% compounded annually for 2 years

P = 9000
r = 0.11
n = 1
t = 2

A = 9000 × (1.11)^2 = 9000 × 1.2321 = $11,088.90

---

6) $190 invested at 11% compounded annually for 6 months

Note: 6 months = 0.5 years
P = 190
r = 0.11
n = 1
t = 0.5

A = 190 × (1.11)^0.5
√1.11 ≈ 1.053565
A ≈ 190 × 1.053565 ≈ $200.18

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7) $1,500 at 7% compounded annually for 3 years

P = 1500
r = 0.07
n = 1
t = 3

A = 1500 × (1.07)^3
1.07^3 = 1.225043
A ≈ 1500 × 1.225043 ≈ $1,837.56

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8) $5,000 at 16% compounded semiannually for 2 years

Semiannually → n = 2
P = 5000
r = 0.16
n = 2
t = 2

A = 5000 × (1 + 0.16/2)^(2×2) = 5000 × (1.08)^4
1.08^4 = 1.36048896
A ≈ 5000 × 1.36048896 ≈ $6,802.44

---

9) $28,000 at 7.9% compounded semiannually for 2 years

P = 28000
r = 0.079
n = 2
t = 2

A = 28000 × (1 + 0.079/2)^(4) = 28000 × (1.0395)^4
1.0395^2 = 1.08056025
1.08056025^2 ≈ 1.167610
A ≈ 28000 × 1.167610 ≈ $32,693.08

---

10) $12,700 at 8.8% compounded semiannually for 1 year

P = 12700
r = 0.088
n = 2
t = 1

A = 12700 × (1 + 0.088/2)^2 = 12700 × (1.044)^2
1.044^2 = 1.089936
A ≈ 12700 × 1.089936 ≈ $13,842.19

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Final Answer:
1) $23,418.10
2) $1,458.00
3) $911.66
4) $1,562.98
5) $11,088.90
6) $200.18
7) $1,837.56
8) $6,802.44
9) $32,693.08
10) $13,842.19
Parent Tip: Review the logic above to help your child master the concept of compound interest word problems worksheet.
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