Case study:
Manish, a bank employee, purchased a plot (15 m × 18 m) in Ghaziabad. He paid ₹ 2,00,000 at the beginning as down payment and agreed to pay the remaining ₹ 6,00,000 at the end of 2 years from the date of purchase.

In order to pay ₹ 6,00,000 at the end of two years, he opened an R.D. account in his bank, with ₹ 20,000 per month at 8% rate of interest.

(i) Find the maturity value of this account at end of 2 years.

(ii) Is the M.V. of the above R.D. account equal to ₹ 6,00,000 ?

If not, how much more/less should the monthly instalment be so that Manish gets the required money (₹ 6,00,000) at the end of two years ?

(i) Given,

n = 2 years = 24 months, P = ₹ 20,000, r = 8%

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

Substituting values we get :

$\therefore I = 20000 \times \dfrac{24(25)}{24} \times \dfrac{8}{100} \\[1em] = 20000 \times 25 \times \dfrac{8}{100} \\[1em] = 20000 \times 2 \\[1em] = ₹ 40,000$

Maturity Value (M.V.) = Sum deposited + Interest

= ₹ 20,000 × 24 + ₹ 40,000 = ₹ 5,20,000.

Hence, the maturity value of this account at end of 2 years = ₹ 5,20,000.

(ii) Maturity value of the R.D. account is ₹ 5,20,000.

So the maturity value is not equal to ₹ 6,00,000.

₹ 6,00,000 - ₹ 5,20,000 = ₹ 80,000

So, Manish gets ₹ 80,000 less.

Thus, the M.V. of the above R.D. account is not equal to ₹ 6,00,000.

Let required monthly instalment be x.

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

Substituting values we get :

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

Maturity value = 24x + 2x = 26x

Since required M.V. = 6,00,000

26x = 6,00,000

x = $\dfrac{6,00,000}{26}$ = 23,076.92 ≈ ₹ 23,077.

Difference in monthly instalment = ₹ 23,077 − ₹ 20,000

= ₹ 3,077.

Hence, monthly instalment should be ₹ 3,077 more per month to get ₹ 6,00,000 at the end of two years.