Simple & Compound Interest Problems worksheet for math practice.

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)

$$
I = P \times r \times t \Rightarrow 300 = P \times 0.08 \times 1
\Rightarrow P = \frac{300}{0.08} = 3750
$$

✔ 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 earned annually = $1250
- Annual rate = 7¾% = 7.75% = 0.0775
- Time = 15 weeks = 15/52 ≈ 0.28846 years

But wait: the interest earned annually is $1250, but this is from a 15-week investment. So we need to clarify: does “$1250 in interest annually” mean that the annual interest is $1250, or that the total interest over 15 weeks is $1250?

Looking closely: "When it matured he received $1250 in interest annually." This wording is ambiguous, but likely means that the interest earned over the 15-week period was $1250, not annually.

So:
- Interest (I) = $1250
- Rate (r) = 7.75% = 0.0775
- Time (t) = 15/52 years

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

First compute:
$$
0.0775 \times \frac{15}{52} = 0.0775 \times 0.28846 \approx 0.02233
$$

Now:
$$
P = \frac{1250}{0.02233} \approx 55977.74
$$

Wait — this seems very high. Let’s re-evaluate.

Alternatively, maybe the annual interest is $1250, meaning the interest earned over one year would be $1250, but the deposit was only for 15 weeks.

But the sentence says: "he received $1250 in interest annually" — possibly a misstatement. It might mean: he received $1250 as interest when it matured, and the interest rate is 7.75% per year.

Let’s assume: Total interest earned over 15 weeks = $1250, rate = 7.75% per year.

Then:

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

Compute:
$$
\frac{15}{52} \approx 0.28846
\Rightarrow 0.0775 \times 0.28846 \approx 0.02233
\Rightarrow P = \frac{1250}{0.02233} \approx 55977.74
$$

This seems too large. Let's check: if $55,977.74 @ 7.75% for 1 year → $4334 interest, so for 15 weeks → ~$1250. Yes, correct.

But let's double-check with exact fractions.

$$
t = \frac{15}{52},\quad r = \frac{31}{400} = 0.0775
$$

$$
I = P \cdot \frac{31}{400} \cdot \frac{15}{52} = P \cdot \frac{465}{20800}
= P \cdot 0.022355...
$$

Set equal to 1250:
$$
P = \frac{1250}{0.022355} \approx 55925.93
$$

Let’s keep it precise.

But perhaps there's a better interpretation.

Wait — maybe the interest rate is 7.75% per annum, and the interest earned over 15 weeks is $1250, which is not annual.

Yes, that must be it.

So:
$$
I = P \cdot r \cdot t \Rightarrow 1250 = P \cdot 0.0775 \cdot \frac{15}{52}
\Rightarrow P = \frac{1250 \cdot 52}{0.0775 \cdot 15}
= \frac{65000}{1.1625} = 55925.93
$$

✔ a) Original Investment = $55,925.93

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

He reinvests the entire amount (principal + interest) into a 40-week term deposit at 8.5% annually.

So:
- New principal = $55,925.93 + $1,250 = $57,175.93
- Rate = 8.5% = 0.085
- Time = 40 weeks = 40/52 ≈ 0.76923 years

Use simple interest again:
$$
I = 57175.93 \times 0.085 \times \frac{40}{52}
$$

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

$$
I = 57175.93 \times 0.06538 \approx 3745.86
$$

Final amount:
$$
A = 57175.93 + 3745.86 = 60,921.79
$$

✔ b) Final amount = $60,921.79

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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.

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

Alexie (annually):
$$
A = P(1 + r)^t = 5000(1 + 0.06)^{17} = 5000(1.06)^{17}
$$

Calculate $ (1.06)^{17} $:

$$
(1.06)^{17} \approx 2.69277
\Rightarrow A = 5000 \times 2.69277 \approx 13,463.85
$$

Meja (daily compounding):

$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
- $ n = 365 $
- $ r = 0.06 $
- $ t = 17 $

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

$$
\left(1 + \frac{0.06}{365}\right)^{6295} \approx e^{0.06 \times 17} = e^{1.02} \approx 2.773
$$

More accurately:
$$
\left(1 + \frac{0.06}{365}\right)^{6295} \approx 2.7731
\Rightarrow A \approx 5000 \times 2.7731 = 13,865.50
$$

✔ Alexie = $13,463.85
✔ Meja = $13,865.50
→ Meja earned more

#### b) How much more?

$$
13,865.50 - 13,463.85 = 401.65
$$

✔ Answer: $401.65 more

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

We use present value formula:

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

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

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

Calculate $ (1.0125)^{28} $:

$$
(1.0125)^{28} \approx 1.4160
\Rightarrow P = \frac{15000}{1.4160} \approx 10,600.00
$$

✔ Answer: $10,600.00

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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 $

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

$$
(1.03)^{10} \approx 1.3439
\Rightarrow P = \frac{20000}{1.3439} \approx 14,881.40
$$

✔ Answer: $14,881.40

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

$$
I = 20000 - 14881.40 = 5118.60
$$

✔ Answer: $5,118.60

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

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

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

$$
(1.0039583)^{252} \approx e^{0.0475 \times 21} = e^{0.9975} \approx 2.713
$$

More accurately:
$$
(1.0039583)^{252} \approx 2.713
\Rightarrow P = \frac{5000}{2.713} \approx 1842.90
$$

✔ Answer: $1,842.90

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✔ Final Answers Summary:

1. Interest = $1850, Final Amount = $9850
2. Principal = $3750
3. a) Original Investment = $55,925.93
b) Final Amount = $60,921.79
4. a) Alexie = $13,463.85, Meja = $13,865.50
b) Meja earned $401.65 more
5. Investment needed = $10,600.00
6. a) Invest $14,881.40
b) Interest earned = $5,118.60
7. Parents invested = $1,842.90

Let me know if you'd like these rounded differently or formatted for a worksheet!