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.