Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets ₹ 33 more than Ashok in 18 months, calculate the money invested.

Given,

r = 20%

n = 18 months or 1.5 years

Let sum of money invested by both be ₹ x.

For Ashok interest is compounded annually :

For 1st year :

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x \times \Big(1 + \dfrac{20}{100}\Big)^1 \\[1em] = x \times \Big(\dfrac{120}{100}\Big)^1 \\[1em] = x \times \dfrac{6}{5} \\[1em] = \dfrac{6x}{5}.$

For next $\dfrac{1}{2}$ year :

P = $\dfrac{6x}{5}$

$A = P\Big(1 + \dfrac{r}{100 \times 2}\Big)^{n \times 2} \\[1em] = \dfrac{6x}{5} \times \Big(1 + \dfrac{20}{100 \times 2}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = \dfrac{6x}{5} \times \Big(1 + \dfrac{20}{200}\Big)^1 \\[1em] = \dfrac{6x}{5} \times \Big(\dfrac{220}{200}\Big) \\[1em] = \dfrac{66x}{50}.$

For Geeta interest is compounded half-yearly :

$A = P\Big(1 + \dfrac{r}{100 \times 2}\Big)^{n \times 2} \\[1em] = x \times \Big(1 + \dfrac{20}{100 \times 2}\Big)^{1.5 \times 2} \\[1em] = x \times \Big(1 + \dfrac{20}{200}\Big)^3 \\[1em] = x \times \Big(\dfrac{220}{200}\Big)^3 \\[1em] = x \times \Big(\dfrac{11}{10}\Big)^3 \\[1em] = x \times \dfrac{1331}{1000} \\[1em] = \dfrac{1331x}{1000}.$

Given, Geeta receives ₹ 33 more :

$\therefore \dfrac{1331x}{1000} - \dfrac{66x}{50} = 33 \\[1em] \Rightarrow \dfrac{1331x - 1320x}{1000} = 33 \\[1em] \Rightarrow \dfrac{11x}{1000} = 33 \\[1em] \Rightarrow x = \dfrac{33 \times 1000}{11} \\[1em] \Rightarrow x = ₹ 3000.$

Hence, sum invested by both = ₹ 3000.