Question

Mean weight of 25 students in a class is 60 kg. The mean weight of the first 13 students is 57 kg and that of the last 13 students is 63 kg. Then the weight of the ${{13}^{th}}$ student is
A. 60
B. 40
C. 30
D. None of these

Answer 1

Hint: We first find the general form of deviations of n observations and the form of mean for those n observations. We find the formula for the mean using the values of number of observations and the deviation value. We place the values and get the solution.

Complete step by step answer:
We take the weight of the students as ${{x}_{i}}$. The mean of the set of the observations will be \[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\].
It’s given that the mean weight of 25 students in a class is 60 kg. Putting the values, we get
\[60=\dfrac{1}{25}\sum\limits_{i=1}^{25}{{{x}_{i}}}\Rightarrow \sum\limits_{i=1}^{25}{{{x}_{i}}}=60\times 25=1500\].
For mean weight of first 13 students is 57 kg, we can write
\[57=\dfrac{1}{13}\sum\limits_{i=1}^{13}{{{x}_{i}}}\Rightarrow \sum\limits_{i=1}^{13}{{{x}_{i}}}=57\times 13=741\]
For mean weight of last 13 students is 63 kg, we can write
\[63=\dfrac{1}{13}\sum\limits_{i=13}^{25}{{{x}_{i}}}\Rightarrow \sum\limits_{i=13}^{25}{{{x}_{i}}}=63\times 13=819\].
When we are considering the two divisions of first 13 students and last 13 students out of 25 students, we are considering the middle student, the ${{13}^{th}}$ student twice.
Therefore, we can write \[\sum\limits_{i=1}^{13}{{{x}_{i}}}+\sum\limits_{i=13}^{25}{{{x}_{i}}}=\sum\limits_{i=1}^{25}{{{x}_{i}}}+{{x}_{13}}\].
We get $1500=741+819+{{x}_{13}}\Rightarrow {{x}_{13}}=1560-1500=60$.

So, the correct answer is “Option A”.

Note: We need to remember that the value of m in the equation of \[\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-m \right)}=\sum\limits_{i=1}^{n}{{{x}_{i}}}-mn\] was constant and the summation wouldn’t have worked. That’s why we multiplied the number of iterations of n with m to find the summation.