Compound Interest Formula Practice Worksheet with Word Problems

Let’s solve each problem one by one using the Compound Interest Formula:

> A = P(1 + r/n)^(nt)
Where:
- A = final amount (balance or total value)
- P = principal (starting amount)
- r = annual interest rate (as a decimal)
- n = number of times compounded per year
- t = time in years

Since all problems say “compounded semiannually”, that means n = 2.

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Problem 1:

You take out a loan for $672 at an interest rate of 8% compounded semiannually for eight years. What is the total amount you will owe?

→ P = 672, r = 0.08, n = 2, t = 8

A = 672 × (1 + 0.08/2)^(2×8)
= 672 × (1 + 0.04)^16
= 672 × (1.04)^16

Calculate (1.04)^16:
≈ 1.87298 (using calculator)

A ≈ 672 × 1.87298 ≈ 1258.64

✔ Final Answer for #1: $1,258.64

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Problem 2:

If a principal of $695 was invested at a rate of 5% compounded semiannually and terminates with a balance of $861.35, how long was the money invested for?

→ P = 695, A = 861.35, r = 0.05, n = 2 → find t

Use formula:
861.35 = 695 × (1 + 0.05/2)^(2t)
Divide both sides by 695:
861.35 / 695 ≈ 1.23935 = (1.025)^(2t)

Take log of both sides:
log(1.23935) = 2t × log(1.025)

log(1.23935) ≈ 0.0932
log(1.025) ≈ 0.01072

So:
0.0932 = 2t × 0.01072
→ 2t = 0.0932 / 0.01072 ≈ 8.694
→ t ≈ 4.347 years

But let’s check if it’s exact — maybe try t = 4.5? Let’s test:

(1.025)^(2×4.5) = (1.025)^9 ≈ 1.24886
695 × 1.24886 ≈ 867.96 → too high

Try t = 4.3: (1.025)^8.6 → use calculator: approx 1.237 → 695×1.237≈859.7 → close to 861.35

Actually, better to do precise calculation:

From earlier:
2t = log(861.35/695) / log(1.025)
= log(1.2393525) / log(1.025)
≈ 0.093207 / 0.010723 ≈ 8.692

So t = 8.692 / 2 = 4.346 years

But since this is likely expecting whole or half years, perhaps we made rounding error.

Wait — let’s compute exactly:

861.35 ÷ 695 = 1.2393525179856115

Now, (1.025)^x = 1.2393525 → x = ln(1.2393525)/ln(1.025) ≈ 0.2145 / 0.02469 ≈ 8.687 → so 2t = 8.687 → t = 4.3435

Hmm — but maybe the problem expects us to round to nearest half-year? Or perhaps I should check if t=4.5 gives closer?

At t=4.5: A = 695*(1.025)^9 = 695 * 1.248863 = 867.96 → too big

At t=4: A = 695*(1.025)^8 = 695 * 1.218403 = 846.79 → too small

Difference: 861.35 - 846.79 = 14.56
Total jump from t=4 to t=4.5: 867.96 - 846.79 = 21.17

So fraction: 14.56 / 21.17 ≈ 0.688 → so t ≈ 4 + 0.688*0.5 ≈ 4.344 → same as before.

But perhaps the problem has a typo or expects exact match? Wait — let me recalculate with more precision.

Actually, let’s assume the answer is meant to be 4.5 years, but our calc shows it’s not matching. Alternatively, maybe the balance is approximate.

Wait — let’s reverse: what if t=4.346, then A=695*(1.025)^(8.692)=?

Using calculator: 1.025^8.692 ≈ e^(8.692 * ln(1.025)) ≈ e^(8.692 * 0.0246926) ≈ e^0.2146 ≈ 1.2393 → yes, matches.

So t ≈ 4.35 years — but since the problem says “how long”, and doesn’t specify format, we can leave as decimal.

But looking at other problems, they might expect rounded answers. However, let’s keep it accurate.

Alternatively — perhaps I miscalculated division.

861.35 / 695 = let's do exact: 861.35 ÷ 695 = 1.2393525179856115

Now, solve (1.025)^(2t) = 1.2393525

Take natural log:
2t * ln(1.025) = ln(1.2393525)
2t = ln(1.2393525) / ln(1.025)
ln(1.2393525) ≈ 0.214507
ln(1.025) ≈ 0.0246926
2t ≈ 0.214507 / 0.0246926 ≈ 8.687
t ≈ 4.3435 years

So approximately 4.34 years

But let’s see if the problem expects integer or half — maybe it’s 4.5? But 4.5 gives higher amount.

Perhaps there’s a mistake in the problem? Or maybe I need to report as 4.3 years? But let’s move on and come back.

Actually, let’s calculate with t=4.34:

2t=8.68, (1.025)^8.68 = ?

Use: 1.025^8 = 1.218403, 1.025^0.68 = ?

Better to use direct calculator: 1.025^8.68 ≈ 1.2385 → 695*1.2385≈860.76 — close to 861.35

t=4.35: 2t=8.7, 1.025^8.7≈1.2398 → 695*1.2398≈861.66 — very close to 861.35

So t≈4.35 years is good.

But perhaps the intended answer is 4.5 years? No, because that gives 867.96 which is off by $6.61.

Maybe the balance is wrong? Or perhaps I should accept 4.35.

Looking at the context, maybe they want it in years and months? 0.35 years * 12 = 4.2 months — so about 4 years and 4 months.

But since the problem doesn't specify, I'll go with 4.35 years.

Wait — let me check online or standard way — actually, in many textbooks, they expect you to solve logarithmically and report decimal.

So I’ll put 4.35 years

But let’s double-check with another method.

Set up equation:

695 * (1.025)^(2t) = 861.35
(1.025)^(2t) = 861.35 / 695 = 1.2393525
Take log base 10:
2t * log(1.025) = log(1.2393525)
2t = log(1.2393525) / log(1.025) = 0.093207 / 0.010723 = 8.692
t = 4.346

So 4.35 years when rounded to two decimals.

✔ Final Answer for #2: 4.35 years

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Problem 3:

How much interest is earned on a principal of $168 invested at an interest rate of 10% compounded semiannually for two years?

First, find A, then subtract P to get interest.

P = 168, r = 0.10, n = 2, t = 2

A = 168 × (1 + 0.10/2)^(2×2) = 168 × (1.05)^4

(1.05)^4 = 1.21550625

A = 168 × 1.21550625 ≈ 204.205

Interest = A - P = 204.205 - 168 = 36.205 → $36.21

✔ Final Answer for #3: $36.21

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Problem 4:

$202.18 is earned on funds invested at a rate of 8% compounded semiannually over five years. What was the amount of the original investment?

Here, "earned" means interest, so I = 202.18

We know: I = A - P, and A = P(1 + r/n)^(nt)

So: I = P[(1 + r/n)^(nt) - 1]

Given: I = 202.18, r = 0.08, n = 2, t = 5

So: 202.18 = P[(1 + 0.04)^10 - 1] = P[(1.04)^10 - 1]

(1.04)^10 ≈ 1.480244

So: 202.18 = P[1.480244 - 1] = P × 0.480244

Thus, P = 202.18 / 0.480244 ≈ ?

Calculate: 202.18 ÷ 0.480244 ≈ 421.00 (let me compute)

0.480244 × 421 = 0.480244 × 400 = 192.0976, 0.480244 × 21 = 10.085124, total 202.182724 — perfect!

So P = $421.00

✔ Final Answer for #4: $421.00

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Problem 5:

If you put $287 into a savings account and after six years the balance is $247.48, what was the interest rate if it was compounded semiannually?

Wait — balance is less than principal? That can’t be right for positive interest.

$287 → $247.48? That’s a loss. Probably a typo.

Perhaps it’s $347.48? Or $287 became $347.48?

Let me read: "after six years the balance is $247.48" — but 247 < 287, so negative interest? Unlikely.

Perhaps it’s $347.48? Let me assume that’s a typo and it’s $347.48.

Because otherwise, it doesn’t make sense.

In many worksheets, typos happen. Let me check the numbers.

If P=287, A=247.48, t=6, n=2, find r.

Then A = P(1 + r/2)^(12)

247.48 = 287 × (1 + r/2)^12

(1 + r/2)^12 = 247.48 / 287 ≈ 0.8623

Then 1 + r/2 = (0.8623)^(1/12) ≈ ?

0.8623^(1/12) = e^(ln(0.8623)/12) = e^(-0.1482/12) = e^(-0.01235) ≈ 0.9877

So 1 + r/2 = 0.9877 → r/2 = -0.0123 → r = -0.0246 or -2.46%

Negative interest? Not typical for such problems.

Probably a typo, and it should be $347.48.

Let me assume A = 347.48

Then: 347.48 = 287 × (1 + r/2)^12

(1 + r/2)^12 = 347.48 / 287 ≈ 1.21073

Take 12th root: (1.21073)^(1/12) = ?

Or take log: 12 * ln(1 + r/2) = ln(1.21073) ≈ 0.1912

So ln(1 + r/2) = 0.1912 / 12 ≈ 0.015933

Then 1 + r/2 = e^0.015933 ≈ 1.01606

So r/2 = 0.01606 → r = 0.03212 or 3.212%

Approximately 3.21%

But let’s verify: (1 + 0.03212/2)^12 = (1.01606)^12 ≈ ?

1.01606^12 = (1.01606^2)^6 ≈ (1.03238)^6

1.03238^2 ≈ 1.0658, ^3 ≈ 1.0658*1.03238≈1.099, better use calculator.

1.01606^12 = exp(12*ln(1.01606)) = exp(12*0.01593) = exp(0.19116) ≈ 1.2107 — yes, matches 347.48/287=1.21073

So r ≈ 3.21%

But the problem says $247.48, which is probably a typo. Perhaps it’s $347.48.

Maybe it’s $287 became $347.48? That makes sense.

I think it's safe to assume it's $347.48.

Otherwise, the answer is negative, which is unlikely.

So I'll proceed with A=347.48

r = 2 * [ (A/P)^(1/(nt)) - 1 ] = 2 * [ (347.48/287)^(1/12) - 1 ]

= 2 * [1.21073^(1/12) - 1]

1.21073^(1/12) = 1.01606 as above

So r = 2 * 0.01606 = 0.03212 → 3.21%

✔ Final Answer for #5: 3.21%

(Note: If the balance is indeed $247.48, then rate is negative, but I believe it's a typo.)

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Problem 6:

$60.36 is earned on funds invested at a rate of 3% compounded semiannually over three years. What was the amount of the original investment?

Again, "earned" means interest I = 60.36

I = P[(1 + r/n)^(nt) - 1]

r = 0.03, n = 2, t = 3, so nt = 6

(1 + 0.03/2)^6 = (1.015)^6

(1.015)^6 = 1.093443 (approx)

So I = P[1.093443 - 1] = P * 0.093443

60.36 = P * 0.093443

P = 60.36 / 0.093443 ≈ ?

Calculate: 0.093443 * 646 = ? First, 0.093443 * 600 = 56.0658, 0.093443 * 46 = 4.298378, total 60.364178 — very close to 60.36

So P ≈ $646.00

✔ Final Answer for #6: $646.00

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Problem 7:

If an investment over nine years at a rate of 7% compounded semiannually results in a final balance of $787.58, what was the original investment?

A = 787.58, r = 0.07, n = 2, t = 9, find P

A = P(1 + r/n)^(nt) = P(1 + 0.035)^18

(1.035)^18 = ? Calculate:

1.035^2 = 1.071225
1.035^4 = (1.071225)^2 ≈ 1.147523
1.035^8 = (1.147523)^2 ≈ 1.3168
1.035^16 = (1.3168)^2 ≈ 1.734
Then 1.035^18 = 1.035^16 * 1.035^2 ≈ 1.734 * 1.071225 ≈ 1.857

More accurately: use calculator or known value.

1.035^18 = (1.035^9)^2, but better: 1.035^10 = 1.410598, 1.035^8 = 1.316809, so 1.035^18 = 1.410598 * 1.316809 ≈ 1.8574

Actual value: 1.035^18 = 1.857489 (from calculator)

So A = P * 1.857489 = 787.58

P = 787.58 / 1.857489 ≈ ?

787.58 ÷ 1.857489 ≈ 424.00 (let me compute)

1.857489 * 424 = 1.857489 * 400 = 742.9956, 1.857489 * 24 = 44.579736, total 787.575336 — very close to 787.58

So P = $424.00

✔ Final Answer for #7: $424.00

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Problem 8:

If an investment over three years at a rate of 3% compounded semiannually results in a final balance of $640.76, what was the original investment?

A = 640.76, r = 0.03, n = 2, t = 3, nt = 6

(1 + 0.03/2)^6 = (1.015)^6 = 1.093443 (as before)

A = P * 1.093443 = 640.76

P = 640.76 / 1.093443 ≈ ?

1.093443 * 586 = ? 1.093443 * 500 = 546.7215, 1.093443 * 86 = 94.036098, total 640.757598 — very close to 640.76

So P = $586.00

✔ Final Answer for #8: $586.00

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Problem 9:

If you put $661 in a savings account that pays 10% compounded semiannually for six years, what is the amount of money you will have at the end of the six years?

P = 661, r = 0.10, n = 2, t = 6, nt = 12

A = 661 * (1 + 0.05)^12 = 661 * (1.05)^12

(1.05)^12 = 1.795856

A = 661 * 1.795856 ≈ ?

661 * 1.795856 = 661 * 1.7 = 1123.7, 661 * 0.095856 ≈ 63.36, total approx 1187.06

Compute: 661 * 1.795856

First, 600 * 1.795856 = 1077.5136
61 * 1.795856 = 109.547216
Total = 1077.5136 + 109.547216 = 1187.060816

So A ≈ $1,187.06

✔ Final Answer for #9: $1,187.06

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Problem 10:

If you put money into a savings account that earns $6.39 over one year at a rate of 6% compounded semiannually, how much money did you put into the account?

"Earms $6.39" means interest I = 6.39

r = 0.06, n = 2, t = 1, nt = 2

I = P[(1 + r/n)^(nt) - 1] = P[(1 + 0.03)^2 - 1] = P[(1.03)^2 - 1] = P[1.0609 - 1] = P * 0.0609

So 6.39 = P * 0.0609

P = 6.39 / 0.0609 ≈ ?

0.0609 * 105 = 6.3945 — very close to 6.39

So P ≈ $105.00

Check: 105 * 0.0609 = 6.3945 ≈ 6.39 — yes, probably rounded.

So P = $105.00

✔ Final Answer for #10: $105.00

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## Summary of Answers:

1. $1,258.64
2. 4.35 years
3. $36.21
4. $421.00
5. 3.21% (assuming balance is $347.48, not $247.48)
6. $646.00
7. $424.00
8. $586.00
9. $1,187.06
10. $105.00

Note: For problem 5, if the balance is indeed $247.48, the rate would be negative, which is unusual. I assumed a typo and used $347.48. If not, please clarify.

Final Answer:
1. $1,258.64
2. 4.35 years
3. $36.21
4. $421.00
5. 3.21%
6. $646.00
7. $424.00
8. $586.00
9. $1,187.06
10. $105.00