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Practice Problems

Problems:


Problem 1

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Sarah invests €3,000 at an annual interest rate of 5% compounded yearly. Question: How much will she have after 4 years?


Problem 2

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Liam deposits €1,200 in a savings account that earns 7% interest compounded yearly. Question: How much will his deposit be worth after 3 years?


Problem 3

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A bank account has a final amount of €8,245.12 after 5 years of being compounded annually at an interest rate of 6%. Question: What was the initial deposit?


Problem 4

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A car was purchased for €30,000 and depreciates at a rate of 15% per year. Question: What will the car's value be after 3 years?


Solutions:


Problem 1

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Sarah invests €3,000 at an annual interest rate of 5% compounded yearly. Question: How much will she have after 4 years?

Step 1: Identify the Variables

  • PP (Principal) = €3,000 (This is the initial amount Sarah invests.)

  • ii (Interest Rate) = 5% or 0.05 (This is the rate at which interest is compounded annually.)

  • tt (Time) = 4 years (This is the duration for which Sarah's money is invested.)

  • FF (Final Amount) is what we are looking for (the amount Sarah will have after 4 years). Step 2: Substitute the Values into the Formula

  • The formula for compound interest is: F=P(1+i)tF = P(1 + i)^t

  • Substituting the values we have: F=3,000(1+0.05)4F = 3,000(1 + 0.05)^4 Step 3: Add 1 to the Interest Rate

  • Inside the brackets, add 11 to the interest rate: 1+0.05=1.051 + 0.05 = 1.05 Step 4: Raise This Number to the Power of 44

  • Raise 1.051.05 to the power of 44: 1.054=1.215506251.05^4 = 1.21550625 Step 5: Multiply by the Principal

  • Finally, multiply by the principal (€3,000): F=3,000×1.21550625=3,646.52F = 3,000 \times 1.21550625 = €3,646.52 Final Answer: After 4 years, Sarah will have €3,646.52.


Problem 2

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Liam deposits €1,200 in a savings account that earns 7% interest compounded yearly. Question: How much will his deposit be worth after 3 years?

Step 1: Identify the Variables

  • PP (Principal) = €1,200 (This is the initial amount Liam deposits.)

  • ii (Interest Rate) = 7% or 0.07 (This is the rate at which interest is compounded annually.)

  • tt (Time) = 3 years (This is the duration for which Liam's money is invested.)

  • FF (Final Amount) is what we are looking for (the amount Liam will have after 3 years). Step 2: Substitute the Values into the Formula

  • The formula for compound interest is: F=P(1+i)tF = P(1 + i)^t

  • Substituting the values we have: F=1,200(1+0.07)3F = 1,200(1 + 0.07)^3 Step 3: Add 1 to the Interest Rate

  • Inside the brackets, add 11 to the interest rate: 1+0.07=1.071 + 0.07 = 1.07 Step 4: Raise This Number to the Power of 33

  • Raise 1.071.07 to the power of 33: 1.073=1.2250431.07^3 = 1.225043 Step 5: Multiply by the Principal

  • Finally, multiply by the principal (€1,200): F=1,200×1.225043=1,470.05F = 1,200 \times 1.225043 = €1,470.05 Final Answer: After 3years3 years, Liam's deposit will be worth €1,470.05.


Problem 3

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A bank account has a final amount of €8,245.12 after 5 years of being compounded annually at an interest rate of 6%. Question: What was the initial deposit?

Step 1: Identify the Variables

  • FF (Final Amount) = €8,245.12 (This is the amount after 5 years.)

  • ii (Interest Rate) = 6% or 0.06 (This is the rate at which interest is compounded annually.)

  • tt (Time) = 5 years (This is the duration for which the money was invested.)

  • PP (Principal) is what we are looking for (the initial amount deposited). Step 2: Rearrange and Substitute the Values into the Formula

  • We need to rearrange the formula to solve for PP: P=F(1+i)tP = \frac{F}{(1 + i)^t}

  • Substituting the values we have: P=8,245.12(1+0.06)5P = \frac{8,245.12}{(1 + 0.06)^5} Step 3: Add 1 to the Interest Rate

  • Inside the brackets, add 11 to the interest rate: 1+0.06=1.061 + 0.06 = 1.06 Step 4: Raise This Number to the Power of 55

  • Raise 1.061.06 to the power of 55: 1.065=1.338225581.06^5 = 1.33822558 Step 5: Divide the Final Amount by This Number

  • Finally, divide the final amount (€8,245.12) by 1.33822558: P=8,245.121.33822558=6,161.23P = \frac{8,245.12}{1.33822558} = €6,161.23 Final Answer: The initial deposit was €6,161.23.


Problem 4

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A car was purchased for €30,000 and depreciates at a rate of 15% per year. Question: What will the car's value be after 3 years?

Step 1: Identify the Variables

  • PP (Initial Value) = €30,000 (This is the starting value of the car.)

  • ii (Depreciation Rate) = 15% or 0.15 (This is the rate at which the car loses value each year.)

  • tt (Time) = 3 years (This is the duration over which the car depreciates.)

  • FF (Final Value) is what we are looking for (the car's value after 3 years). Step 2: Substitute the Values into the Formula

  • The formula for depreciation is: F=P(1i)tF = P(1 - i)^t

  • Substituting the values we have: F=30,000(10.15)3F = 30,000(1 - 0.15)^3 Step 3: Subtract the Depreciation Rate from 11

  • Inside the brackets, subtract 0.15 from 11: 10.15=0.851 - 0.15 = 0.85 Step 4: Raise This Number to the Power of 33

  • Raise 0.85 to the power of 33: 0.853=0.6141250.85^3 = 0.614125 Step 5: Multiply by the Initial Value

  • Finally, multiply by the initial value (€30,000): F=30,000×0.614125=18,423.75F = 30,000 \times 0.614125 = €18,423.75 Final Answer: After 3 years, the car will be worth €18,423.75.


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