$\dfrac{3}{5}$ part of certain sum is lent at S.I. and the remaining is lent at C.I. If the rate of interest in both the cases is 20%. On the whole the total interest in 1 year is ₹ 1,000 then the sum is:

1. ₹ 2,000

2. ₹ 4,000

3. ₹ 2,500

4. ₹ 5,000

Let the principal amount be ₹ x.

Let the first part be ₹ $\dfrac{3}{5}$ of $x$ and the second part be ₹ $x - \dfrac{3}{5}x$ = $\dfrac{5}{5}x - \dfrac{3}{5}x$ = $\dfrac{2}{5}x$.

Hence

P₁ = ₹ $\dfrac{3x}{5}$

R₁ = 20%

T₁ = 1 year

$\text{S.I.} = \Big(\dfrac{P \times R \times T}{100}\Big)\\[1em] = \Big(\dfrac{3x \times 20 \times 1}{5 \times 100}\Big)\\[1em] = \dfrac{60x}{500}\\[1em] = \dfrac{3x}{25}$

P₂ = ₹ $\dfrac{2x}{5}$

R₂ = $20\%$

T₂ = 1 year

$\text{A} = P\Big[1 + \dfrac{R}{100}\Big]^n\\[1em] = \dfrac{2x}{5} \Big[1 + \dfrac{20}{100}\Big]^1\\[1em] = \dfrac{2x}{5} \Big[1 + \dfrac{1}{5}\Big]\\[1em] = \dfrac{2x}{5} \Big[\dfrac{5}{5} + \dfrac{1}{5}\Big]\\[1em] = \dfrac{2x}{5} \Big[\dfrac{(5 + 1)}{5}\Big]\\[1em] = \dfrac{2x}{5} \Big[\dfrac{6}{5}\Big]\\[1em] = \dfrac{12x}{25} \\[1em]$

And

$\text{C.I. = A - P}\\[1em] = \dfrac{12x}{25} - \dfrac{2x}{5} \\[1em] = \dfrac{12x}{25} - \dfrac{5\times 2x}{5 \times 5} \\[1em] = \dfrac{12x}{25} - \dfrac{10x}{25} \\[1em] = \dfrac{(12 - 10)x}{25} \\[1em] = \dfrac{2x}{25} \\[1em]$

Total amount = ₹ 1,000

$\dfrac{3x}{25} + \dfrac{2x}{25} = 1,000\\[1em] \Rightarrow\dfrac{(3x + 2x)}{25} = 1,000\\[1em] \Rightarrow\dfrac{5x}{25} = 1,000\\[1em] \Rightarrow\dfrac{x}{5} = 1,000\\[1em] \Rightarrow x = 1,000 \times 5\\[1em] \Rightarrow x = 5,000$

Hence, option 4 is the correct option.