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