Question

In a bag, there are $20kg$ of fruits. If $7\dfrac{1}{6}kg$ of these fruits are oranges and $8\dfrac{2}{3}kg$ of these are apples and rest are grapes. Find the mass of the grapes in the bag.

Answer 1

Hint: This is a very simple question, in this question first find the sum of the masses of oranges and apples and then take the difference of the obtained sum from total mass of the fruits.

Complete step-by-step answer:
Since, there are $3$ kinds of fruits, the sum of the individual mass of all the three fruits must be equal to the given total mass of the fruits, that is, $20kg$.
Here, clearly we can see that the masses are given in mixed fraction. But to carry out algebraic operations like addition and subtraction, we must convert them into improper fraction.
Total mass of fruits is $ = 20kg$ (1)
Mass of oranges $ = 7\dfrac{1}{6}kg$
Improper fraction $ = \dfrac{{43}}{6}$
Hence, mass of oranges $ = \dfrac{{43}}{6}kg$ (2)
Similarly, mass of apples is
$
   = 8\dfrac{2}{3}kg \\
    \\
 $
$ = \left( {\dfrac{{8 \times 3 + 2}}{3}} \right) = \dfrac{{26}}{3}kg$ (3)
Let mass of grapes be $x kg$ (4)
Now,
Mass of oranges + mass of apples + mass of grapes = total mass of fruits
Substituting the values from the equation (1), (2), (3) and (4)
$
  \dfrac{{43}}{6} + \dfrac{{26}}{3} + x = 20 \\
  x = \dfrac{{20}}{1} - \left( {\dfrac{{43}}{6} + \dfrac{{26}}{3}} \right) \\
  x = 20 - \left( {\dfrac{{43 + 26 \times 2}}{6}} \right) \\
  x = 20 - \dfrac{{95}}{6} \\
  x = \dfrac{{20 \times 6 - 95}}{6} \\
  x = \dfrac{{25}}{6}kg \\
 $ (LCM of $3$ and $6$ is $6$)
Mass of grapes $ = \dfrac{{25}}{6} = 4\dfrac{1}{6}kg$

Note: In this type of questions, students often tend to make calculation mistakes. So one must concentrate hard so that you may avoid any calculation mistake. Moreover, the first step is always to convert mixed fraction into improper fraction.