Question

A man divided two sums of money among 4 sons R, S, T, U the first in the ratio 4 : 3 : 2 : 1 and the second in the ration 5 : 6 : 7 : 8 If the second sum of money is twice the first the first sum which son receives the largest part?

• A. R
• B. S
• C. T
• D. U

Answer 1

The correct option is A R

Let the first sum=x.

Then the second sum=2x when x is in Rs.

Let us divide x in the ratio 4:3:2:1

and 2x in the ratio 5:6:7:8.

$\therefore$ In the first case, the shares of R, S, T & U are,

$R⟶\frac{x}{4+3+2+1}\times 4=\frac{4x}{10},S⟶\frac{x}{4+3+2+1}\times 3=\frac{3x}{10},T⟶\frac{x}{4+3+2+1}\times 2=\frac{2x}{10}\&U⟶\frac{x}{4+3+2+1}\times 1=\frac{x}{10}.$

In the second case

$R⟶\frac{2x}{5+6+7+8}\times 5=\frac{5x}{13},S⟶\frac{2x}{5+6+7+8}\times 6=\frac{6x}{13},T⟶\frac{2x}{5+6+7+8}\times 7=\frac{7x}{13}\&U⟶\frac{2x}{5+6+7+8}\times 8=\frac{8x}{13}.$

$\therefore$ The total sum of

$R=\frac{4x}{10}+\frac{5x}{13}=\frac{102x}{130},S=\frac{3x}{10}+\frac{6x}{13}=\frac{99x}{130},T=\frac{2x}{10}+\frac{7x}{13}=\frac{96x}{130}\&U=\frac{x}{10}+\frac{8x}{13}=\frac{93x}{130}.\therefore \frac{93x}{130}<\frac{96x}{130}<\frac{99x}{130}<\frac{102x}{130}.$

So R gets the highest share.

Ans- Option A.