Hamid has three boxes of different fruits.
Question:

Hamid has three boxes of different fruits. Box A weighs $2 \frac{1}{2} \mathrm{~kg}$ more than box B and Box C

weighs $10 \frac{1}{4} \mathrm{~kg}$ more than box $\mathrm{B}$. The total weight of the three boxes is $48 \frac{3}{4} \mathrm{~kg}$. How many kilograms

does box A weigh?

Solution:

Let the weight of box A be $x \mathrm{~kg}$. According to the question,

Weight of Box $A=$ Weight of Box $B+\frac{5}{2} \mathrm{~kg}$ [given]

$\Rightarrow \quad$ Weight of box $B=\left(x-\frac{5}{2}\right) \mathrm{kg}$

and weight of $B 0 \times C=$ Weight of box $B+\frac{41}{4} \mathrm{~kg}$ [given]

$\Rightarrow \quad$ Weight of Box $C=x-\frac{5}{2}+\frac{41}{4} \mathrm{~kg}=x+\left(\frac{-10+41}{4}\right) \mathrm{kg}=\left(x+\frac{31}{4}\right) \mathrm{kg}$

As, total weight of three boxes $=48 \frac{3}{4} \mathrm{~kg}=\frac{195}{4} \mathrm{~kg}$ [given]

$\because$ Total weight of three boxes $=$ Weight of box $A+$ Weight of box $B+$ Weight of box $C$

$\Rightarrow$ $\frac{195}{4}=3 x+\frac{31-10}{4}$

$\Rightarrow$ $\frac{195}{4}=3 x+\frac{21}{4} \Rightarrow 3 x=\frac{195}{4}-\frac{21}{4}$

$\Rightarrow$ $3 x=\frac{195-21}{4} \Rightarrow 3 x=\frac{174}{4}$

$\Rightarrow$ $x=\frac{174}{3 \times 4} \Rightarrow x=\frac{29}{2}$

$\therefore$ $x=14 \frac{1}{2} \mathrm{~kg}$

Hence, box $A$ weighs $14 \frac{1}{2} \mathrm{~kg}$.

 

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