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Question: The mean weight of \(6\) boys in a group is \(48\)kg. The individual weights of \(5\) of them are \(...

The mean weight of 66 boys in a group is 4848kg. The individual weights of 55 of them are 50kg,45kg,49kg,46kg50kg,45kg,49kg,46kg and 44kg44kg. Find the weight of the sixth boy.

Explanation

Solution

Here we have to find the weight of the sixth boy. First we have to group the mean weight of 66 boys and then assume the sixth boy. By using the formula to solve them and us get the required answer.

Formula used: Mean weight of the boys =Total weight of boysNo. of boys = \dfrac{{{\text{Total weight of boys}}}}{{{\text{No}}{\text{. of boys}}}}

Complete step-by-step solution:
It is given that the total number of boys in a group is 48kg48kg
Let the weight of the 55 boys from the group are as 50kg,45kg,49kg,46kg50kg,45kg,49kg,46kg and 44kg.44kg.
Now we have to find the weight of the sixth boy.
It is given that the mean weight of 66 boys in the group is 48kg.48kg.
That means weight of the 66 boys group =48kg = 48kg
We know mean weight of boys =Total weight of all boysNo. of boys....(1) = \dfrac{{{\text{Total weight of all boys}}}}{{{\text{No}}{\text{. of boys}}}}....(1)
Let the weight of the sixth boy is aa kg.
Substitute the values in the above mean formula, we get
50+45+49+46+44+a6=48\Rightarrow \dfrac{{50 + 45 + 49 + 46 + 44 + a}}{6} = 48
Now adding the numerator values
234+a6=48\Rightarrow \dfrac{{234 + a}}{6} = 48
We just cross multiply the terms
234+a=48×6\Rightarrow 234 + a = 48 \times 6
Now we are multiplying the right handed term and
We get,
234+a=288\Rightarrow 234 + a = 288
Now keeping left handed side and take the numerical value to the right hand side
We can write it as,
a=288234\Rightarrow a = 288 - 234
Now by subtraction, we get
a=54a = 54

Therefore Weight of the 6th6^{th} boy is 54kg54kg.

Note: The mean of the series of numbers or observations a1,a2,a3,a4,a5..............an{a_1},{a_2},{a_3},{a_4},{a_5}..............{a_n} Is given as a1+a2+a3+.......+ann\dfrac{{{a_1} + {a_2} + {a_3} + ....... + {a_n}}}{n} =i=1nain = \dfrac{{\sum\limits_{i = 1}^n {{a_i}} }}{n}, where nn is the number of the value in the series.
We need to remember that to have a good understanding of how to compute the average of some numbers.
Also we should be able to solve linear equations in one variable which helps us in simplification of the problem and reach the correct answer.