https://file.aiquickdraw.com/mathbot/file/1729258681521v5r22de9.jpg

Let's solve each of the questions in the image one by one. These are compound interest-related problems, and we'll use the compound interest formula for solving them:

### Compound Interest Formula:
\[
A = P \left(1 + \frac{r}{100}\right)^n
\]
Where:
- \(A\) = Amount after interest
- \(P\) = Principal (Initial amount)
- \(r\) = Rate of interest (per annum)
- \(n\) = Number of years

The interest can be calculated by:
\[
\text{Interest} = A - P
\]

### 1. **Question 1:**
Find the amount at the end of 2 years on ₹1000 at 8% p.a. compounded annually.

#### Solution:
- Principal \(P = ₹1000\)
- Rate \(r = 8\%\)
- Time \(n = 2\) years

Using the formula:
\[
A = 1000 \left(1 + \frac{8}{100}\right)^2 = 1000 \times \left(1.08\right)^2
\]
\[
A = 1000 \times 1.1664 = 1166.40
\]

Thus, the amount after 2 years is **₹1166.40**.

---

### 2. **Question 2:**
Find the interest on ₹2048 for 3 years, interest compounded annually at the rate of \(6 \frac{1}{4}\%\) p.a.

#### Solution:
- Principal \(P = ₹2048\)
- Rate \(r = 6 \frac{1}{4}\% = 6.25\%\)
- Time \(n = 3\) years

Using the formula:
\[
A = 2048 \left(1 + \frac{6.25}{100}\right)^3 = 2048 \times \left(1.0625\right)^3
\]
\[
A = 2048 \times 1.1989 = 2454.56
\]

Interest = \(A - P = 2454.56 - 2048 = 406.56\)

Thus, the interest is **₹406.56**.

---

### 3. **Question 3:**
Madhu borrowed a sum of ₹24000 for 2 years at the rate of 10% p.a., interest compounded annually. Find how much money will she have to pay to clear her debt?

#### Solution:
- Principal \(P = ₹24000\)
- Rate \(r = 10\%\)
- Time \(n = 2\) years

Using the formula:
\[
A = 24000 \left(1 + \frac{10}{100}\right)^2 = 24000 \times \left(1.1\right)^2
\]
\[
A = 24000 \times 1.21 = 29040
\]

Thus, Madhu will have to pay **₹29040** to clear her debt.

---

### 4. **Question 4:**
Vimal obtains a loan of ₹64000 against his fixed deposits. If the rate of interest be 2.5 paise per rupee per annum, calculate the compound interest payable after 3 years.

#### Solution:
- Principal \(P = ₹64000\)
- Rate \(r = 2.5 \text{ paise per rupee per annum} = 2.5\%\)
- Time \(n = 3\) years

Using the formula:
\[
A = 64000 \left(1 + \frac{2.5}{100}\right)^3 = 64000 \times \left(1.025\right)^3
\]
\[
A = 64000 \times 1.0769 = 68841.60
\]

Interest = \(A - P = 68841.60 - 64000 = 4841.60\)

Thus, the compound interest payable after 3 years is **₹4841.60**.

---

Do you need any further details on these calculations or have any questions?

### Further Questions:
1. What happens to the total interest if we increase the number of compounding periods per year?
2. How would the interest change if Madhu borrowed ₹50000 instead of ₹24000 at the same rate and time?
3. Can you calculate the amount after 5 years if ₹5000 is invested at a rate of 7% p.a. compounded annually?
4. What would the compound interest be if Vimal's loan had a higher interest rate of 3% instead of 2.5%?
5. How does compound interest differ from simple interest for the same principal and rate?

**Tip:** Always ensure that you convert all rates and units properly before applying formulas. This helps avoid common calculation errors.

Solve the compound interest problems as shown in the image.

Learn how to solve compound interest problems involving principal, interest rate, and time. This example covers four distinct scenarios, including finding the amount and interest using the compound interest formula. Suitable for students in Grades 8-10.

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