Question

Average weight of 25 persons is increased by 1 kg when one man weighing 60 kg is replaced by a new person. Weight of new person is
(a) 50 kg
(b) 60 kg
(c) 56 kg
(d) 85 kg

Answer 1

Hint: We will assume that the sum of the original 25 persons is W and the weight of the new person is x. Then we will subtract the 60 kg from W and add x. Then we will find the average of the new sum of weight which will be one more than the original average. We will then find the value of x from the derived equation.

Complete step by step solution:
If W is the sum of the weights of the original 25 persons, with 60 kg person included, then the average of the weights of these 25 persons will be $\dfrac{W}{25}$.
After the person with weight 60 is removed, the sum of the weights of 24 people change to (W – 60).
In this changed weight, we add a person with weight x. The sum of weight of 25 people now is (W – 60 + x) and the average weight changes to $\dfrac{W-60+x}{25}$.
This changed average is 1 more than the original average.
Therefore,
\[\dfrac{W}{25}+1=\dfrac{W-60+x}{25}\]
To solve this equation, first we split the terms based on operation signs.
$\dfrac{W}{25}+1=\dfrac{W}{25}-\dfrac{60}{25}+\dfrac{x}{25}$
Now, $\dfrac{W}{25}$ cancels out from both sides. We will take $\dfrac{60}{25}$ to the other side of equals to sign.
\[\begin{align}
  & \Rightarrow \dfrac{W}{25}+1=\dfrac{W}{25}-\dfrac{60}{25}+\dfrac{x}{25} \\
 & \Rightarrow \dfrac{x}{25}=1+\dfrac{60}{25}
\end{align}\]
To solve further, we make the denominators the same by taking LCM of 1 and 25 on the right side of the equals to sign. 25 in the denominator now can be cancelled out from both sides.
\[\begin{align}
  & \Rightarrow \dfrac{x}{25}=\dfrac{1\left( 25 \right)}{25}+\dfrac{60}{25} \\
 & \Rightarrow \dfrac{x}{25}=\dfrac{60+25}{25} \\
 & \Rightarrow x=60+25
\end{align}\]
Therefore, x = 60 + 25 = 85.
Hence, option (d) is the correct option.

Note: It is advisable to be careful while taking LCM of 1 and 25 to solve the equations. Students can also solve by option verification method instead of solving the equation for getting the value of x.