Question

Question 95

Solve the following:

Hamid has three boxes of different fruits. Box A weighs 2 $\frac{1}{2}$ kg more than box B and Box C weighs 10 $\frac{1}{4}$ more than box B. The total weight of the three boxes is 48$\frac{3}{4}kg$. How many kilograms does box A weigh?

Answer 1

Let the weight of box A be x kg.
According to the equation,
Weight of Box A=Weight of Box B + $\frac{5}{2}$ kg
$\Rightarrow WeightofboxB=(x-\frac{5}{2})kg$
and weight of Box C=Weight of Box B+ $\frac{41}{4}$kg
$\Rightarrow WeightofBoxC=x-\frac{5}{2}+\frac{41}{4}kg=x+\left(\frac{-10+41}{4}\right)kg=(x+\frac{31}{4})kgAs,totalweightofthreeboxes=48\frac{3}{4}kg=\frac{195}{4}kg$
$\because$Total weight of three boxes=Weight of box A+Weight of Box B+ Weight of Box C
$\therefore \frac{195}{4}=x+x-\frac{5}{2}+x+\frac{31}{4}\Rightarrow \frac{195}{4}=3x+\frac{31-10}{4}\Rightarrow \frac{195}{4}=3x+\frac{21}{4}\Rightarrow 3x=\frac{195}{4}-\frac{21}{4}\Rightarrow 3x=\frac{195-21}{4}\Rightarrow 3x=\frac{174}{4}\Rightarrow x=\frac{174}{3\times 4}\Rightarrow x=\frac{29}{2}\therefore x=14\frac{1}{2}kg.$
Hence, box A weighs $14\frac{1}{2}kg.$