Question

The mean wage of $150$ laborers working in a factory running three shifts with $60$,$40$ and $50$ laborers is Rs $114.0$ The mean wage of $60$ laborers working in the first shift is ${\text{Rs 121}}{\text{.50}}$ and that of $40$ laborers working in the second shift is ${\text{Rs 107}}{\text{.75}}$ then the mean wage of the laborers working in the third shift is
\[{\text{(A) Rs 110}}\]
${\text{(B) Rs 100}}$
${\text{(C) Rs 100}}$
${\text{(D) Rs 114}}{\text{.75}}$

Answer 1

Hint: In this question we will use calculate the total wages of all the laborers in the company and then subtract the total wages of laborers working in the first and second shift to find the total wage of the laborers working in the third shift, from the total wages we will find the mean wage.

Formula used: ${\text{Mean wage of laborers = }}\dfrac{{{\text{Total wage of laborers}}}}{{{\text{Total number of laborers}}}}$

Complete step-by-step solution:
Let the total wages of all the laborers working in the first shift is $x$
Let the total wages of all the laborers working in the second shift is $y$
Let the total wages of all the laborers working in the third shift is $z$
Number of laborers working in shift $a = 60$
Number of laborers working in shift $b = 40$
Number of laborers working in shift $c = 50$
We know, ${\text{Mean Wage = }}\dfrac{{{\text{x + y + z}}}}{{{\text{a + b + c}}}}$
Since, $x + y + z$ represents the total wage of all the laborers and $a + b + c$ represents the total number of laborers.
We know, total number of laborers in the factory are $150$
Therefore $a + b + c = 150$
Therefore, mean wage = $\dfrac{{x + y + z}}{{150}} \to (1)$
Now given, mean wage of $150$ laborers $ = 114$
Therefore, $\dfrac{{x + y + z}}{{150}} = 114$
On Cross multiplying we get:
$ \Rightarrow x + y + z = 114 \times 150$
On multiplying the Right-Hand Side, we get:
$ \Rightarrow x + y + z = 17100 \to (1)$
The total wages of all the laborers are $17100$
Now, mean wage of laborers in shift \[1\] is $\dfrac{x}{a} = 121.50$
Putting the value of \[a\]and we get,
$ \Rightarrow \dfrac{x}{{60}} = 121.50$
On Cross Multiplying we get,
$x = 121.50 \times 60$
$\therefore $ Total wage of laborers in shift \[1\] is $7290$
Also, mean wage of laborers in shift \[2\] is $\dfrac{y}{b} = 107.75$
Putting the value of \[b\] and we get,
$ \Rightarrow \dfrac{y}{{40}} = 107.75$
On Cross Multiplying we get,
$y = 107.75 \times 40$
$\therefore $ Total wage of laborers in shift \[2\] is $4310$
Now, putting the values of \[x\]and \[y\] in equation $(1)$ we get
$7290 + 4310 + z = 17100$
Taking \[z\] as LHS and remaining as RHS on the negative sign because of the equal sign we get,
$z = 17100 - 7290 - 4310$
On doing some add and subtract the values and we get:
$z = 5500$
Now, we have to find out the mean wage of laborers in shift $3$
So we use the formula and putting the values we get,
Mean wage of laborers in shift $3$ is $\dfrac{z}{c} = \dfrac{{5500}}{{50}}$
On dividing the values we get,
$ \Rightarrow 110$
Hence the mean wage of laborers in shift $3$ is $110$

$\therefore $ The correct option is $(A)$ which is $110$

Note: Mean is called average in layman terms and it is always the total of a value of a property in a distribution divided by the total number of terms in that distribution.
A common place to make mistakes is Cross multiplying, always cross multiplying the denominator of the fraction in R.H.S with the numerator of the fraction in the L.H.S.