Question

The average of marks of 28 students in Mathematics was 50. Eight students left the school and then this average increased by 5. What is the average of marks obtained by the students who left the school? \[\]
A.37.5\[\]
B.42.5\[\]
C.45\[\]
D. 50.5\[\]

Answer 1

Hint: We denote the number of students, average marks in Mathematics, sum of marks before the departure of 8 students as ${{n}_{old}},{{\left( \overline{x} \right)}_{old}},{{S}_{old}}$ an after departure of 8 students as ${{n}_{new}},{{\left( \overline{x} \right)}_{new}},{{S}_{new}}$. We use the formula for sum of data values $S$ with any average $\overline{x}$, number of data values $n$ that is $S=n\times \overline{x}$ to find ${{S}_{old}},{{S}_{new}}$. We find the sum of marks of 8 leaving students ${{S}_{l}}={{S}_{old}}-{{S}_{new}}$ and average as $\dfrac{{{S}_{l}}}{8}$.\[\]

Complete step-by-step solution:
We know that mean or average is the expected value of the data sample set. It is the value we expect without going detailed into data and it is a measure of central tendency. It is denoted by $\overline{x}$. If there are $n$ number of data values with equal weights in the samples say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then the mean is calculated by first finding the sum of data values and then by dividing the sum by $n.$ So the sample mean is given by
\[\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}=\dfrac{S}{n}\]
So the sum of the values is given by
\[S=n\overline{x}\]
We are given the question that the average of marks of 28 students in Mathematics was 50. Let the number of students initially be denoted as ${{n}_{old}}$ and the average of marks of students in Mathematics be denoted as $ {{\left( \overline{x} \right)}_{old}}$. We have${{n}_{old}}=28,{{\left( \overline{x} \right)}_{old}}=50$. So the sum of marks ${{S}_{i}}$of students in Mathematics is
\[{{S}_{old}}=n\times {{\left( \overline{x} \right)}_{old}}=28\times 50=1400\]
We are also given in the question that 8 students left the school and then this average increased by 5. Let us denote the new number of students as ${{n}_{new}}$, the new average marks as ${{\left( \overline{x} \right)}_{new}}$ and the new of sum of marks as ${{S}_{new}}$. We have 8 students leaving the school. So we have new number of students as,
\[{{n}_{new}}={{n}_{old}}-8=28-8=20\]
The new average increased by 5. So we have new average marks as,
\[{{\left( \overline{x} \right)}_{old}}={{\left( \overline{x} \right)}_{old}}+5=50+5=55\]
So the new of sum of marks is,
\[{{S}_{new}}={{n}_{new}}\times {{\left( \overline{x} \right)}_{new}}=20\times 55=1100\]
Let the number of students, average marks and sum of marks who left the school be denoted as ${{n}_{l}},{{\left( \overline{x} \right)}_{l}},{{S}_{l}}$ respectively. We have ${{n}_{l}}=8$.The sum of marks 8 students who left the school is the difference between and old sum of marks of class which is
\[{{S}_{l}}={{S}_{old}}-{{S}_{new}}=1400-1100=300\]
So the average marks of 8 students who left the school is
\[{{\left( \overline{x} \right)}_{l}}=\dfrac{{{S}_{l}}}{{{n}_{l}}}=\dfrac{300}{8}=37.5\]
So the correct option is A.

Note: We can alternatively solve by first finding find new average with formula y ${{\left( \overline{x} \right)}_{new}}-{{\left( \overline{x} \right)}_{old}}=k$ where $k$ is the change in average. Then we use the formula the formula $\dfrac{{{S}_{old}}-{{S}_{new}}}{n-h}$ where $h$ is the decrease in number of values. Here the mean is the arithmetic means AM. Other types of mean are geometric mean (GM) and harmonic mean (HM). The relation among mean is $AM\ge GM\ge HM$.