Ravish has three boxes whose total weight is 60
Question:

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

Solution:

Let the weight of box A be $\mathrm{x} \mathrm{kg}$.

Therefore, the weights of box B and box C will be $\left(\mathrm{x}+3 \frac{1}{2}\right) \mathrm{kg}$ and $\left(\mathrm{x}+3 \frac{1}{2}+5 \frac{1}{3}\right) \mathrm{kg}$, respectively.

According to the question,

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

or $3 \mathrm{x}=\frac{121}{2}-\frac{7}{2}-\frac{7}{2}-\frac{16}{3}$

or $3 \mathrm{x}=\frac{363-21-21-32}{6}$

or $3 \mathrm{x}=\frac{289}{6}$

or $\mathrm{x}=\frac{289}{18}$

Thus, weight of box $\mathrm{A}=\frac{289}{18} \mathrm{~kg}$

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