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Simple Interest Math Worksheet with problems on calculating interest for loans and real-world scenarios.

Math worksheet titled "Simple Interest" from EffortlessMath, featuring ten problems to determine simple interest for various loans and six word problems involving simple interest calculations.

Math worksheet titled "Simple Interest" from EffortlessMath, featuring ten problems to determine simple interest for various loans and six word problems involving simple interest calculations.

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Show Answer Key & Explanations Step-by-step solution for: Simple Interest: Math Worksheets | PDF | Interest | Political Economy
Absolutely! Let’s solve each problem step by step using the Simple Interest Formula:

> I = P × r × t

Where:
- I = Interest (what we’re solving for in most cases)
- P = Principal (initial amount)
- r = annual interest rate (as a decimal — divide % by 100)
- t = time in years (convert months to years by dividing by 12)

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## Part 1: Determine the simple interest for these loans.

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1) $450 at 7% for 2 years

I = 450 × 0.07 × 2 = $63.00

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2) $5,200 at 4% for 3 years

I = 5200 × 0.04 × 3 = $624.00

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3) $1,300 at 5% for 6 years

I = 1300 × 0.05 × 6 = $390.00

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4) $5,400 at 3.5% for 6 months

→ 6 months = 6/12 = 0.5 years

I = 5400 × 0.035 × 0.5 = $94.50

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5) $600 at 4% for 9 months

→ 9 months = 9/12 = 0.75 years

I = 600 × 0.04 × 0.75 = $18.00

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6) $24,000 at 5.5% for 5 years

I = 24000 × 0.055 × 5 = $6,600.00

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7) $15,600 at 3% for 2 years

I = 15600 × 0.03 × 2 = $936.00

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8) $1,200 at 5.5% for 4 years

I = 1200 × 0.055 × 4 = $264.00

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9) $1,600 at 4.5% for 9 months

→ 9/12 = 0.75 years

I = 1600 × 0.045 × 0.75 = $54.00

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10) $12,000 at 2.2% for 5 years

I = 12000 × 0.022 × 5 = $1,320.00

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## Part 2: Solve each simple interest word problem.

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11) A new car, valued at $28,000, depreciates at 9% per year. What is the value of the car one year after purchase?

⚠️ This is depreciation, not interest, but we use the same math.

Depreciation amount = 28000 × 0.09 × 1 = $2,520

Value after 1 year = 28000 - 2520 = $25,480

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12) Sara puts $4,000 into an investment yielding 5% annual simple interest; she left the money in for five years. How much interest does Sara get at the end of those five years?

I = 4000 × 0.05 × 5 = $1,000

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13) A bank is offering 3.5% simple interest on a savings account. If you deposit $7,500, how much interest will you earn in two years?

I = 7500 × 0.035 × 2 = $525

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14) $400 interest is earned on a principal of $2,000 at a simple interest rate of 5% interest per year. For how many years was the principal invested?

We rearrange formula:
t = I / (P × r)

t = 400 / (2000 × 0.05) = 400 / 100 = 4 years

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15) In how many years will $1,200 yield an interest of $180 at 3% simple interest?

t = I / (P × r) = 180 / (1200 × 0.03) = 180 / 36 = 5 years

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16) Jim invested $4,000 in a bond at a yearly rate of 4.5%. He earned $540 in interest. How long was the money invested?

t = I / (P × r) = 540 / (4000 × 0.045) = 540 / 180 = 3 years

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## 📝 Final Answers Summary:

Part 1:


1) $63.00
2) $624.00
3) $390.00
4) $94.50
5) $18.00
6) $6,600.00
7) $936.00
8) $264.00
9) $54.00
10) $1,320.00

Part 2:


11) $25,480
12) $1,000
13) $525
14) 4 years
15) 5 years
16) 3 years

All problems solved using the simple interest formula and careful unit conversion (months → years). Let me know if you’d like to see any step explained further!
Parent Tip: Review the logic above to help your child master the concept of simple interest math problems worksheet.
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