Simple & Compound Interest Problems worksheet - Free Printable

Let's solve each problem step by step using the formulas for simple interest and compound interest.

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1. Lisa invested $8000 at 9.25% for 30 months. Calculate the interest earned and the final amount.

- Principal (P) = $8000
- Rate (r) = 9.25% = 0.0925
- Time (t) = 30 months = 30/12 = 2.5 years

Simple Interest Formula:
$$
I = P \times r \times t
$$

$$
I = 8000 \times 0.0925 \times 2.5 = 8000 \times 0.23125 = 1850
$$

Final Amount (A):
$$
A = P + I = 8000 + 1850 = 9850
$$

✔ Find = Interest Earned → Answer = $1850
✔ Find = Final Amount → Answer = $9850

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2. Dylan has an investment that earns him $300 each year. If the annual rate is 8%, what is the principal?

- Interest (I) = $300
- Rate (r) = 8% = 0.08
- Time (t) = 1 year (since it’s annual)

$$
I = P \times r \times t \Rightarrow 300 = P \times 0.08 \times 1
$$

$$
P = \frac{300}{0.08} = 3750
$$

✔ Find = Principal → Answer = $3750

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3. David invested in a 15-week term deposit that earned 7¾% of simple interest. When it matured he received $1250 in interest annually. He then reinvested all his money in a 40-week term deposit so that it would earn 8.5% annually.

#### a) How much was the original investment?

- Interest (I) = $1250
- Rate (r) = 7¾% = 7.75% = 0.0775
- Time (t) = 15 weeks = 15/52 ≈ 0.28846 years

$$
I = P \times r \times t \Rightarrow 1250 = P \times 0.0775 \times \frac{15}{52}
$$

First compute:
$$
0.0775 \times \frac{15}{52} = 0.0775 \times 0.28846 ≈ 0.02234
$$

$$
P = \frac{1250}{0.02234} ≈ 55950.76
$$

Wait — this seems too high. Let's double-check: Is the $1250 the annual interest, or the interest from the 15-week deposit?

> The problem says: "When it matured he received $1250 in interest annually." This phrasing is ambiguous.

But likely, he received $1250 as interest from the 15-week deposit, not annually.

So re-read: “earned 7¾% of simple interest. When it matured he received $1250 in interest” — probably $1250 total interest over the 15-week period.

So:

$$
I = 1250,\quad r = 0.0775,\quad t = \frac{15}{52} \approx 0.28846
$$

$$
1250 = P \times 0.0775 \times \frac{15}{52}
\Rightarrow P = \frac{1250}{0.0775 \times 15 / 52}
= \frac{1250}{(0.0775 \times 15)/52}
= \frac{1250}{1.1625 / 52}
= \frac{1250}{0.022346} ≈ 55950.76
$$

That still gives ~$55,950 — but let's check if we misinterpreted.

Alternatively, perhaps the $1250 is the annual interest, meaning the interest rate is 7.75% per year, and he earned $1250 per year, so:

Then:
$$
I = P \times r \Rightarrow 1250 = P \times 0.0775 \Rightarrow P = \frac{1250}{0.0775} = 16153.85
$$

But then the time doesn't matter because it's annual interest.

But the deposit is only for 15 weeks — so it can’t be annual interest unless it's a yearly yield.

Wait — the phrase is: “earned 7¾% of simple interest. When it matured he received $1250 in interest annually.”

This is poorly worded. But likely, the $1250 is the interest earned over the 15-week period.

So go back to:

$$
I = P \cdot r \cdot t
\Rightarrow 1250 = P \cdot 0.0775 \cdot \left(\frac{15}{52}\right)
$$

$$
P = \frac{1250}{0.0775 \cdot 15 / 52} = \frac{1250}{(1.1625)/52} = \frac{1250}{0.022346} ≈ 55950.76
$$

So original investment ≈ $55,950.76

But let's do exact calculation:

$$
P = \frac{1250 \times 52}{0.0775 \times 15} = \frac{65000}{1.1625} = 55950.76
$$

✔ a) Find = Original Investment → Answer ≈ $55,950.76

#### b) How much will David have when the second term deposit matures?

He reinvests the total amount from first deposit:
$$
A_1 = P + I = 55950.76 + 1250 = 57200.76
$$

Now reinvested in a 40-week term deposit at 8.5% annually, simple interest.

- P = $57,200.76
- r = 8.5% = 0.085
- t = 40/52 ≈ 0.76923 years

$$
I_2 = P \times r \times t = 57200.76 \times 0.085 \times \frac{40}{52}
$$

First:
$$
0.085 \times \frac{40}{52} = 0.085 \times 0.76923 ≈ 0.06538
$$

$$
I_2 = 57200.76 \times 0.06538 ≈ 3747.00
$$

Total amount after second deposit:
$$
A_2 = 57200.76 + 3747.00 = 60947.76
$$

✔ b) Find = Final Amount → Answer ≈ $60,947.76

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4. Alexie invested $5000 at 6% compounded annually for 17 years. Meja invested $5000 at 6% compounded daily for 17 years.

Use compound interest formula:

$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$

#### a) Which investment earned more money?

Alexie (Annually):
- $ P = 5000 $
- $ r = 0.06 $
- $ n = 1 $
- $ t = 17 $

$$
A_A = 5000 \left(1 + \frac{0.06}{1}\right)^{1 \times 17} = 5000 (1.06)^{17}
$$

Calculate $ (1.06)^{17} $:

Using calculator:
$$
(1.06)^{17} ≈ 2.69277
\Rightarrow A_A ≈ 5000 \times 2.69277 = 13,463.85
$$

Meja (Daily):
- $ n = 365 $
- $ r = 0.06 $
- $ t = 17 $

$$
A_M = 5000 \left(1 + \frac{0.06}{365}\right)^{365 \times 17}
= 5000 \left(1 + \frac{0.06}{365}\right)^{6295}
$$

$$
\frac{0.06}{365} ≈ 0.000164384
\Rightarrow (1.000164384)^{6295}
$$

Take natural log:
$$
\ln(A_M / 5000) = 6295 \times \ln(1.000164384) ≈ 6295 \times 0.00016437 ≈ 1.034
\Rightarrow A_M / 5000 = e^{1.034} ≈ 2.799
\Rightarrow A_M ≈ 5000 \times 2.799 = 13,995.00
$$

More accurately, use:
$$
A_M = 5000 \times e^{0.06 \times 17} = 5000 \times e^{1.02} ≈ 5000 \times 2.773 = 13,865
$$

Actually, better to use direct calculation:

$$
(1 + 0.06/365)^{6295} ≈ e^{0.06 \times 17} = e^{1.02} ≈ 2.773
\Rightarrow A_M ≈ 5000 \times 2.773 = 13,865
$$

But actual value with daily compounding is slightly higher than continuous.

Better: Use financial calculator approximation:

$$
(1 + 0.06/365)^{6295} ≈ 2.774
\Rightarrow A_M ≈ 5000 × 2.774 = 13,870
$$

Compare:
- Alexie: $13,463.85
- Meja: ~$13,870

✔ a) Meja earned more

✔ Answer Alexie = $13,463.85
✔ Answer Meja = $13,870

#### b) How much more?

$$
13,870 - 13,463.85 = 406.15
$$

✔ Answer = $406.15

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5. Jessica wants to have $15,000 in 7 years. Her bank will pay her 5% interest compounded quarterly. How much does she have to invest now?

Use present value formula:

$$
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}
$$

- $ A = 15000 $
- $ r = 0.05 $
- $ n = 4 $ (quarterly)
- $ t = 7 $

$$
P = \frac{15000}{\left(1 + \frac{0.05}{4}\right)^{4 \times 7}} = \frac{15000}{(1 + 0.0125)^{28}} = \frac{15000}{(1.0125)^{28}}
$$

Compute $ (1.0125)^{28} $:

$$
(1.0125)^{28} ≈ e^{28 \times \ln(1.0125)} ≈ e^{28 \times 0.01242} ≈ e^{0.34776} ≈ 1.392
$$

Or directly: $ (1.0125)^{28} ≈ 1.392 $

$$
P = \frac{15000}{1.392} ≈ 10,775.86
$$

✔ Find = Present Investment → Answer ≈ $10,775.86

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6. Phil wants to buy a car worth $20,000 in 5 years.

#### a) How much should he invest in a GIC earning 6% compounded semi-annually?

- $ A = 20000 $
- $ r = 0.06 $
- $ n = 2 $
- $ t = 5 $

$$
P = \frac{20000}{\left(1 + \frac{0.06}{2}\right)^{2 \times 5}} = \frac{20000}{(1.03)^{10}}
$$

$$
(1.03)^{10} ≈ 1.343916
\Rightarrow P = \frac{20000}{1.343916} ≈ 14,881.17
$$

✔ Answer = $14,881.17

#### b) How much interest did he earn?

$$
I = A - P = 20000 - 14881.17 = 5118.83
$$

✔ Answer = $5,118.83

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7. Megan will receive $5000 on her 21st birthday from her parents who invested when she was born. If the interest rate was 4¾% compounded monthly, how much did her parents invest when she was born?

- $ A = 5000 $
- $ r = 4.75\% = 0.0475 $
- $ n = 12 $
- $ t = 21 $

$$
P = \frac{5000}{\left(1 + \frac{0.0475}{12}\right)^{12 \times 21}} = \frac{5000}{\left(1 + 0.0039583\right)^{252}}
$$

$$
(1.0039583)^{252} ≈ e^{252 \times \ln(1.0039583)} ≈ e^{252 \times 0.003954} ≈ e^{0.9959} ≈ 2.707
$$

Or calculate directly:
$$
(1.0039583)^{252} ≈ 2.707
\Rightarrow P = \frac{5000}{2.707} ≈ 1847.44
$$

✔ Answer = $1,847.44

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## ✔ Final Answers Summary:

1. Interest Earned: $1850, Final Amount: $9850
2. Principal: $3750
3. a) Original Investment: $55,950.76
b) Final Amount: $60,947.76
4. a) Alexie: $13,463.85, Meja: $13,870
b) Difference: $406.15
5. Investment Needed: $10,775.86
6. a) Investment: $14,881.17
b) Interest Earned: $5,118.83
7. Original Investment: $1,847.44

Let me know if you'd like these rounded differently or formatted neatly!