Question

A basket contains three types of fruits weighing $19\frac{1}{3}\mathrm{kg}$ in all. If $8\frac{1}{9}\mathrm{kg}$ of these be apples, $3\frac{1}{6}\mathrm{kg}$ be oranges and the rest pears, what is the weight of the pears in the basket?

Answer 1

Weight of pears in the basket = Weight of the basket containing three types of fruits - (Weight of apples + Weight of oranges)
=$19\frac{1}{3}-\left(8\frac{1}{9}+3\frac{1}{6}\right)$
Now,

$$\begin{gathered} \left(8\frac{1}{9}+3\frac{1}{6}\right)\Rightarrow \left(8+\frac{1}{9}\right)+\left(3+\frac{1}{6}\right) \\ =\frac{73}{9}+\frac{19}{6} \end{gathered}$$

LCM of 9 and 6 is 18, that is, $\left(3\times 3\times 2\right)$.

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \frac{\left(73\times 2\right)+\left(19\times 3\right)}{18} \\ =\frac{146+57}{18} \\ =\frac{203}{18} \end{gathered}$$

∴​ $8\frac{1}{9}+3\frac{1}{6}=\frac{203}{18}$
Now,
Weight of pears in the basket = $19\frac{1}{3}-\frac{203}{18}$
$$\begin{gathered} =\left(19+\frac{1}{3}\right)-\frac{203}{18} \\ =\frac{58}{3}-\frac{203}{18} \\ =\frac{58}{3}+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{203}{18}\right) \\ =\frac{348-203}{18} \\ =\frac{145}{18} \\ =8\frac{1}{18}\text{kg} \end{gathered}$$
​Therefore, the weight of the pears in the basket is $8\frac{1}{18}\text{kg}$.