Question

Ravish has three boxes whose total weight is $60\frac{1}{2}kg.$ Box B weighs $3\frac{1}{2}kg$ more than box A and box C weighs $5\frac{1}{3}$ more than box B. Find the weight of box A.

• A. $\frac{289}{18}kg$
• B. $\frac{229}{18}kg$
• C. $\frac{241}{18}kg$
• D. $\frac{265}{18}kg$

Answer 1

The correct option is A $\frac{289}{18}kg$

Total weight of three boxes $=60\frac{1}{2}kg.$

Let weight of box A = x kg.

Then weight of box $B=(x+3\frac{1}{2})kg.$ and weight of box C = $(x+3\frac{1}{2}+5\frac{1}{3})kg.$

$\therefore$ According to the condition,

$x+(x+3\frac{1}{2})+(x+3\frac{1}{2}+5\frac{1}{3})=60\frac{1}{2}$

$\Rightarrow x+x+\frac{7}{2}+x+\frac{7}{2}+\frac{16}{3}=\frac{121}{2}$

$\Rightarrow 3x+(\frac{7}{2}+\frac{7}{2}+\frac{16}{3})$

$\Rightarrow 3x=\frac{121}{2}-(\frac{7}{2}+\frac{7}{2}+\frac{16}{3})$

$\Rightarrow 3x=\frac{363-(21+21+32)}{6}=\frac{363-74}{6}$

$\Rightarrow 3x=\frac{289}{6}$

$\Rightarrow x=\frac{289}{6\times 3}=\frac{289}{18}kg.$

$\therefore$ Weight to Box $A=\frac{289}{18}kg$