Mr. Bajaj needs ₹ 30000 after 2 years. What least money (in multiple of ₹ 5) must be deposit every month in a recurring deposit account to get required money at the end of 2 years, the rate of interest being 8% p.a.?

Let money deposited per month be ₹ x.

So,

P = x, n = (2 × 12) = 24 months, r = 8%.

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

$\therefore I = ₹ x \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{8}{100} \\[1em] = 2x$

Maturity value = Sum deposited + Interest

⇒ 30000 = x × 24 + 2x

⇒ 30000 = 24x + 2x

⇒ 30000 = 26x

x = $\dfrac{30000}{26} = 1153.84$

Rounding off to nearest multiple of 5 = ₹ 1155.

Hence, the money that must be deposited every month = ₹ 1155.