Question

The average weight of students in a class of $35$ students is $40 kg$. If the weight of the teacher is included, the average rises by $\dfrac{1}{2} kg$. Then what is the weight of the teacher?
A. $40 kg$
B. $50 kg$
C. $41 kg$
D. $58 kg$

Answer 1

Hint: First, use the formula of mean and calculate the sum of weight of $35$ students. Then apply the formula of mean, and calculate the sum of the weight of $35$ students and a teacher. In the end, substitute the sum of the weight of $35$ students in the mean of $35$ students and a teacher to get the required answer.

Formula Used:
$\text{Mean} = \dfrac{{\text{Sum of observations}}}{{\text{Total number of observations}}}$

Complete step by step solution:
Given:
The average weight of $35$ students is: $40 kg$
The average weight of $35$ students and a teacher is: $40\dfrac{1}{2} kg$

Let's consider, $x$ to be the sum of weight of $35$ students.
Apply the formula of the mean.
We get,
$\text{Average weight} = \dfrac{{\text{Sum of weight}}}{{\text{Total number of students}}}$
Substitute the values in the above formula.
$40 kg = \dfrac{x}{{35}}$
Solve the above equation.
$x = 35\left( {40 kg} \right)$
$x = 1400 kg$
Thus, the sum of the weight of $35$ students is $1400 kg$.
Let's consider, $y$ to be the weight of a teacher.
Then the sum of the weight of $35$ students and a teacher is: $\left( {1400 + y} \right) kg$
Apply the formula of the mean.
We get,
$\text{Average weight} = \dfrac{{\text{Sum of weight 35 students and a teacher}}}{{\text{Total number of students and a teacher}}}$
Substitute the values in the above formula.
$40\dfrac{1}{2} kg = \dfrac{{\left( {1400 + y} \right)kg}}{{35 + 1}}$
$ \Rightarrow \dfrac{{81}}{2} kg = \dfrac{{\left( {1400 + y} \right)kg}}{{36}}$
Cross multiply.
$36\left( {81} \right) kg = 2\left( {1400 + y} \right)kg$
$ \Rightarrow 2916 kg = 2y + 2800 kg$
Subtract $2800 kg$ from both sides.
$2y = \left( {2916 - 2800} \right) kg$
$ \Rightarrow 2y = 116 kg$
Divide both sides by 2.
$y = 58 kg$
Thus, the weight of a teacher is $58 kg$.

Option ‘D’ is correct

Note: The average of a mean is total sum upon total number. Every precision value matters in this type of question, note that if all numbers of a list are multiplied by the same positive number, then its average changes by the same factor.