# A basket contains three types of fruits weighing

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?

Solution:

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,

$\left(8 \frac{1}{9}+3 \frac{1}{6}\right) \Rightarrow\left(8+\frac{1}{9}\right)+\left(3+\frac{1}{6}\right)$

$=\frac{73}{9}+\frac{19}{6}$

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

We have:

$\frac{(73 \times 2)+(19 \times 3)}{18}$

$=\frac{146+57}{18}$

$=\frac{203}{18}$

$\therefore 8 \frac{1}{9}+3 \frac{1}{6}=\frac{203}{18}$

Now,

Weight of pears in the basket $=19 \frac{1}{3}-\frac{203}{18}$

$=\left(19+\frac{1}{3}\right)-\frac{203}{18}$

$=\frac{58}{3}-\frac{203}{18}$

$=\frac{58}{3}+\left(\right.$ Additive inverse of $\left.\frac{203}{18}\right)$

$=\frac{348-203}{18}$

$=\frac{145}{18}$

$=8 \frac{1}{18} \mathrm{~kg}$

Therefore, the weight of the pears in the basket is $8 \frac{1}{18} \mathrm{~kg}$.