Assertion (A): A certain sum of money is borrowed at 10% compound interest for two years, it amounts to ₹ 5,000. The sum borrowed = ₹ 5,000 $\Big(1 -\dfrac{10}{100}\Big)\Big(1 -\dfrac{10}{100}\Big)$.

Reason (R): Since, amount = sum borrowed x $\Big(1+\dfrac{10}{100}\Big)\Big(1+\dfrac{10}{100}\Big)$

∴ Sum borrowed = Amount x $\Big(1 -\dfrac{10}{100}\Big)\Big(1 -\dfrac{10}{100}\Big)$.

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Both A and R are false.

Explanation

Given,

A = ₹ 5,000, R = 10 %, T = 2 years

A = P x $\Big(1 + \dfrac{R}{100}\Big)^T$

$⇒ 5,000 = P \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] ⇒ 5,000 = P \Big(1 + \dfrac{1}{10}\Big)^2 \\[1em] ⇒ 5,000 = P \Big(\dfrac{11}{10}\Big)^2 \\[1em] ⇒ 5,000 = P \Big(\dfrac{121}{100}\Big)\\[1em] ⇒ P = \Big(\dfrac{100 \times 5,000}{121}\Big)\\[1em] ⇒ P = \Big(\dfrac{5,00,000}{121}\Big)\\[1em] ⇒ P = ₹ 4,132.23$

According to assertion :

The sum borrowed,

$= ₹ 5,000 \Big(1 -\dfrac{10}{100}\Big)\Big(1 -\dfrac{10}{100}\Big)\\[1em] = ₹ 5,000 \Big(1 -\dfrac{1}{10}\Big)\Big(1 -\dfrac{1}{10}\Big)\\[1em] = ₹ 5,000 \Big(\dfrac{9}{10}\Big)\Big(\dfrac{9}{10}\Big)\\[1em] = ₹ 50 \times 9 \times 9\\[1em] = ₹ 4050$

This is incorrect because the actual sum borrowed is ₹ 4,132.23, not ₹ 4,050.

∴ Assertion(A) is false.

Given,

Amount = sum borrowed x $\Big(1+\dfrac{10}{100}\Big)\Big(1+\dfrac{10}{100}\Big)$

⇒ A = sum borrowed x $\Big(1+\dfrac{1}{10}\Big)\Big(1+\dfrac{1}{10}\Big)$

⇒ A = sum borrowed x $\Big(\dfrac{11}{10}\Big)\Big(\dfrac{11}{10}\Big)$

⇒ Sum borrowed = $\dfrac{A}{\Big(\dfrac{11}{10}\Big)\Big(\dfrac{11}{10}\Big)}$

⇒ Sum borrowed = A x ${\Big(\dfrac{10}{11}\Big)\Big(\dfrac{10}{11}\Big)}$

According to reason:

Sum borrowed = Amount x $\Big(1 -\dfrac{10}{100}\Big)\Big(1 -\dfrac{10}{100}\Big)$

∴ Reason(R) is false.

Hence, both Assertion (A) and Reason (R) are false.