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 -

Solution(By Examveda Team)

Let the parts be Rs. x, Rs. y and Rs. [1440 - (x + y)].
Then,
$$\eqalign{ & = \frac{{x \times 2 \times 3}}{{100}} = \frac{{y \times 3 \times 4}}{{100}} \cr & = \frac{{\left[ {1440 - \left( {x + y} \right)} \right] \times 4 \times 5}}{{100}} \cr & \therefore 6x = 12y\,or\,x = 2y. \cr & So,\frac{{x \times 2 \times 3}}{{100}} = \frac{{\left[ {1440 - \left( {x + y} \right)} \right] \times 4 \times 5}}{{100}} \cr & \Leftrightarrow 12y = \left( {1440 - 3y} \right) \times 20 \cr & \Rightarrow 72y = 28800 \cr & \Rightarrow y = 400 \cr & {\text{First part}} = x = 2y = {\text{Rs}}{\text{. }}800, \cr & {\text{Second part}} = {\text{Rs}}{\text{. }}400 \cr & {\text{Third part}} \cr & = {\text{Rs}}.\left[ {1440 - \left( {800 + 400} \right)} \right] \cr & = {\text{Rs}}{\text{. }}240. \cr & \therefore {\text{Required difference}} \cr & {\text{ = Rs}}{\text{.}}\left( {800 - 240} \right) \cr & = {\text{Rs}}.560 \cr} $$