Question

If the average weight of $6$ students is $50$kg that of $2$ students is $51$ kg and that of rest of $2$ students is $55$kg. Then the average weight of all the students is
${\text{(A) 61 kg}}$
${\text{(B) 51}}{\text{.5 kg}}$
${\text{(C) 52 kg}}$
${\text{(D) 51}}{\text{.2 kg}}$

Answer 1

Hint: Here we have to find the average weight of all the students by using the formula. We will first make the appropriate equations for all the cases given to us and then using that, we will find the average weight of all the students. Finally we get the required answer.

Formula used: ${\text{Mean = }}\dfrac{{{\text{Sum of terms}}}}{{{\text{Number of terms}}}}$

Complete step-by-step solution:
From the question, we know that the mean of $6$ students is $50kg$.
Therefore, let us consider the students to be ${x_1},{x_2},{x_3},{x_4},{x_5},{x_6},{x_7},{x_8},{x_9},{x_{10}}$
Since we know the mean is $50kg$, it can be mathematically being written using the mean formula as:
$ \Rightarrow 50 = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6}$
Now on cross multiplication we get:
$ \Rightarrow 50 \times 6 = {x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}$
On simplifying we get:
$ \Rightarrow 300 = {x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} \to (1)$
Now, we know the mean of $2$ other student is $51kg$.
Therefore, let us consider the $2$ students to be ${x_7}$ and ${x_8}$
Since we know the mean is $51kg$, it can be mathematically being written using the mean formula as:
$ \Rightarrow 51 = \dfrac{{{x_7} + {x_8}}}{2}$
Now on cross multiplication we get:
$ \Rightarrow 51 \times 2 = {x_7} + {x_8}$
On simplifying we get:
$ \Rightarrow 102 = {x_7} + {x_8} \to (2)$
Now, we know the mean of the remaining $2$ other student is $55kg$.
Therefore, let us consider the $2$ students to be ${x_9}$ and${x_{10}}$.
Since we know the mean is $55kg$, it can be mathematically being written using the mean formula as:
$ \Rightarrow 55 = \dfrac{{{x_9} + {x_{10}}}}{2}$
Now on cross multiplication we get:
$ \Rightarrow 55 \times 2 = {x_9} + {x_{10}}$
On simplifying we get:
$ \Rightarrow 102 = {x_7} + {x_8} \to (3)$
Now since we have to find the mean of all the $10$ students, this could be calculated using the formula of mean as:
$Mean = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} + {x_7} + {x_8} + {x_9} + {x_{10}}}}{{10}}$
On grouping the terms for simplification, we get:
$\Rightarrow$$Mean = \dfrac{{({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}) + ({x_7} + {x_8}) + ({x_9} + {x_{10}})}}{{10}}$
Now using equations $(1),(2)$and $(3)$we substitute the value of the sum, we get:
$\Rightarrow$$Mean = \dfrac{{300 + 102 + 110}}{{10}}$
On simplifying the numerator, we get:
$\Rightarrow$$Mean = \dfrac{{512}}{{10}}$
On simplifying we get:
$\Rightarrow$$Mean = 51.2$, which is the required final answer.

Therefore, the correct option is option $(D)$.

Note: The term average used in the question statement is nothing other than the mean of the distribution.
Mean is the most commonly used measure of central tendency, there also exists other central tendencies such as median and mode which is used in statistics.