For a recurring deposit account in a bank, the deposit is ₹1,000 per month for 2 years at 10% p.a. rate of interest.

Statement (1): The interest earned is 10% of ₹(24 x 1,000).

Statement (2): For monthly instalment = ₹P, number of instalment = n and rate of interest r% p.a.; the interest earned = $\dfrac{P \times n \times (n + 1)}{12} \times \dfrac{r}{100}$.

1. Both statements are true.

2. Both statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Both statements are false.

Reason

Given, P = ₹1,000, n = 2 years = 24 months and r = 10%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ 1,000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{10}{100} \\[1em] = ₹ 1,000 \times \dfrac{600}{24} \times \dfrac{1}{10} \\[1em] = ₹ 100 \times 25 \\[1em] = ₹ 2,500$

According to statement 1, the interest earned is 10% of ₹(24 x 1,000) = $\dfrac{10}{100} \times 24000$ = ₹ 2,400.

∵ ₹ 2,400 ≠ ₹ 2,500.

So, statement 1 is false.

According to statement 2:

Given, monthly instalment = ₹P, number of instalment = n and rate of interest r% p.a.

the interest earned = $P \times \dfrac{n(n + 1)}{12} \times \dfrac{r}{100}$

But the correct formula is:

the interest earned = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

So, statement 2 is false.

Hence, Both statements are false.