A certain sum of money amounts to Rs. 6600 in 4 years at a certain rate percent simple interest. If the rate of interest be increased by its 25%, the same sum would amount to Rs. 7000 during the same period. Find the sum.

Given:

Amount (A₁) = ₹6600

Amount (A₂) = ₹7000

Time (t) = 4 years

New Rate = r + 25% of r = r + (r × 25/100) = 1.25r

Formula used:

Simple Interest (SI) = \( \frac{P \times R \times T}{100} \)

Amount = Principal + SI

Calculation:

Let P = x, r = R

A₁ = x + \( \frac{x \times R \times 4}{100} \) = 6600

⇒ x(1 + \( \frac{4R}{100} \)) = 6600

⇒ x(1 + \( \frac{R}{25} \)) = 6600.  —(1)

A₂ = x + \( \frac{x \times (1.25R) \times 4}{100} \) = 7000

⇒ x(1 + \( \frac{1.25R \times 4}{100} \)) = 7000

⇒ x(1 + \( \frac{5R}{100} \)) = 7000

⇒ x(1 + \( \frac{R}{20} \)) = 7000  —(2)

Divide (2) by (1):

⇒ \( \frac{1 + \frac{R}{20}}{1 + \frac{R}{25}} = \frac{7000}{6600} = \frac{35}{33} \)

Let’s simplify:

⇒ \( \frac{20+R}{20} \div \frac{25+R}{25} = \frac{35}{33} \)

⇒ \( \frac{(20+R)\times25}{(25+R)\times20} = \frac{35}{33} \)

⇒ \( \frac{500 + 25R}{500 + 20R} = \frac{35}{33} \)

⇒ Cross multiply:

33(500 + 25R) = 35(500 + 20R)

⇒ 16500 + 825R = 17500 + 700R

⇒ 825R - 700R = 17500 - 16500

⇒ 125R = 1000

⇒ R = 8%

Now put R in eq (1):

x(1 + \( \frac{8}{25} \)) = 6600

⇒ x × \( \frac{33}{25} \) = 6600

⇒ x = \( \frac{6600 \times 25}{33} = 5000 \)

∴ The correct answer is ₹5000.