# A sum of Rs. 1440 is lent out in three parts in such away that the interests on first part at 2% for 3 years, second part at 3% for 4 years and third part at 4% for 5 years are equal. Then the difference between the largest and the smallest sum is –

A. Rs. 200 B. Rs. 400 C. Rs. 460 D. Rs. 560 Answer: Option D
Let the parts be Rs. x, Rs. y and Rs. [1440 – (x + y)]. Then, $\begin{array}{l}=\frac{x\times2\times3}{100}=\frac{y\times3\times4}{100}\\ =\frac{[1440-(x+y)]\times4\times5}{100}\\ \therefore6x=12y\text{ or }x=2y.\\ \text{ So },\frac{x\times2\times3}{100}=\frac{[1440-(x+y)]\times4\times5}{100}\\ \Leftrightarrow12y=(1440-3y)\times20\\ \Rightarrow72y=28800\\ \Rightarrow y=400\end{array}$ $\begin{array}{l} \text { First part }=x=2 y=\text { Rs. } 800, \\ \text { Second part }=\text { Rs. } 400 \\ \text { Third part } \\ =\text { Rs. }[1440-(800+400)] \\ =\text { Rs. } 240 . \\ \therefore \text { Required difference } \\ =\text { Rs. }(800-240) \\ =\text { Rs. } 560 \end{array}$