Question

A father divided a rectangular piece of land between his 2 sons in ratio 4 : 5. The length of the field was twice it's breadth.

If breadth is 6 m, then find the area of land which the 1st son got. (The lesser share)

• A. 40 $m^2$
• B. 30 $m^2$
• C. 32 $m^2$
• D. 36 $m^2$

Answer 1

The correct option is C

32 $m^2$

Given breadth = 6 m

Length = Twice the breadth = 2 $\times$ 6 = 12 m

Area of the Land = Length $\times$ Breadth = 6 $\times$ 12 = 72 $m^2$

Let the 1st son get 4x and 2nd son get 5x.

4x + 5x = 9x = 72

$\Rightarrow$ x = $\frac{72}{9}$ = 8 $m^2$

$\therefore$ The area of land which the 1st son gets = 4x = 4$\times$8 = 32 $m^2$