Answer : Option A

Explanation :

Let the parts be x, y and z
R = 5%
x + interest on x for 2 years = y + interest on y for 3 years = z + interest on z for 4 years
$MF#%\begin{align}&\left(x + \dfrac{x \times 5 \times 2}{100}\right) = \left(y + \dfrac{y \times 5 \times 3}{100}\right) = \left(z + \dfrac{z \times 5 \times 4}{100}\right)\\\\ &\left(x + \dfrac{x}{10}\right) = \left(y + \dfrac{3y}{20}\right) = \left(z + \dfrac{z}{5}\right)\\\\ &\dfrac{11x}{10} = \dfrac{23y}{20} = \dfrac{6z}{5}\\\\\\\\ &\text{ Let }\dfrac{11x}{10} = \dfrac{23y}{20} = \dfrac{6z}{5} = \text{ k} \quad \text{ (where k is a constant)}\\\\ &\text{Then, }x = \dfrac{10k}{11}, \quad y = \dfrac{20k}{23}, \quad z = \dfrac{5k}{6}\\\\ &\text{ we know that }x + y + z = 2379 \\\\ &\dfrac{10k}{11} + \dfrac{20k}{23} + \dfrac{5k}{6} = 2379\\\\ &10k \times 23 \times 6 + 20k \times 11 \times 6 + 5k \times 11 \times 23 = 2379 \times11 \times23 \times 6\\\\ &1380k + 1320k + 1265k = 2379 \times11 \times23 \times 6\\\\ &3965k = 2379 \times11 \times23 \times 6\\\\ &k = \dfrac{2379 \times11 \times 23 \times 6}{3965}\\\\ &\text{First part, x = }\dfrac{10k}{11} = \dfrac{10}{11} \times \dfrac{2379 \times11 \times 23 \times 6}{3965} = \dfrac{10 \times 2379\times 23 \times 6}{3965} \\\\&= \dfrac{2\times 2379\times 23 \times 6}{793} = 2 \times 3 \times 23 \times 6 = 828\end{align}$MF#%