Question

A sum of Rs. 1,440 is lent out in three parts in such a way 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. 400
• B. Rs. 560
• C. Rs. 460
• D. Rs. 200

Answer 1

Answer: (B)

Rs. 560

Answer 2

Let the three parts be Rs x, Rs y, Rs z, Then,
$\frac{x\times2\times\times}{100}=\frac{y\times3\times4}{100}=\frac{y\times4\times5}{100}$
⇒ $\frac{6x}{100}=\frac{12y}{100}=\frac{20z}{100}=k$ (say)
⇒ $x=\frac{100k}{6},y=\frac{100k}{12}, z=\frac{100k}{20}$
⇒ $x=\frac{50k}{3},y=\frac{25k}{3},z=5k$
Given, $x+y+z=1440$
⇒ $\frac{50k}{3}+\frac{25k}{3}+5k=1440\times3$
⇒ $50k+25k+15k=1440\times3$
⇒ $90k=1440\times3$
⇒ $k=\frac{1440\times3}{90}=48$
$\therefore$ Largest share $x=\frac{50\times48}{3}=Rs800$
Smallest share $=z=5\times48=Rs240$
Required difference $=Rs800-Rs240=Rs560$
The correct option is (B): Rs. 560