Practice Problems (Junior Cert Mathematics): Revision Notes

Practice Problems

Problems:

---

Problem 1

---

Problem 2

---

Problem 3

---

Problem 4

---

Solutions:

---

Problem 1

Step 1: Identify the Variables

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

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

• $t$ (Time) = 4 years (This is the duration for which Sarah's money is invested.)

• $F$ (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: $F = P(1 + i)^t$

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

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

• Raise $1.05$ to the power of $4$: $1.05^4 = 1.21550625$ Step 5: Multiply by the Principal

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

---

Problem 2

Step 1: Identify the Variables

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

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

• $t$ (Time) = 3 years (This is the duration for which Liam's money is invested.)

• $F$ (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: $F = P(1 + i)^t$

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

• Inside the brackets, add $1$ to the interest rate: $1 + 0.07 = 1.07$ Step 4: Raise This Number to the Power of $3$

• Raise $1.07$ to the power of $3$: $1.07^3 = 1.225043$ Step 5: Multiply by the Principal

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

---

Problem 3

Step 1: Identify the Variables

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

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

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

• $P$ (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 $P$: $P = \frac{F}{(1 + i)^t}$

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

• Inside the brackets, add $1$ to the interest rate: $1 + 0.06 = 1.06$ Step 4: Raise This Number to the Power of $5$

• Raise $1.06$ to the power of $5$: $1.06^5 = 1.33822558$ Step 5: Divide the Final Amount by This Number

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

---

Problem 4

Step 1: Identify the Variables

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

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

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

• $F$ (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: $F = P(1 - i)^t$

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

• Inside the brackets, subtract 0.15 from $1$: $1 - 0.15 = 0.85$ Step 4: Raise This Number to the Power of $3$

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

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

---