
A student obtained 60, 75, 85 marks respectively in three monthly examinations and 95 in the final examinations. The three-monthly examinations are of equal weightage whereas the final examination is weighted twice as much as a monthly examination. The mean marks of mathematics are.
A. 80
B. 81
C. 82
D. 85
Answer
544.2k+ views
Hint: First we will note the formula to find the weighted mean of $n$ numbers which are ${{a}_{1}},{{a}_{2}},{{a}_{3}}...,{{a}_{n}}$ and their weights are given by ${{w}_{1}},{{w}_{2}},{{w}_{3}}...{{w}_{n}}$. So, the weighted mean is given by the formula $\dfrac{{{a}_{1}}{{w}_{1}}+{{a}_{2}}{{w}_{2}}+{{a}_{3}}{{w}_{3}}...\,+{{a}_{n}}{{w}_{n}}}{{{w}_{1}}+{{w}_{2}}+{{w}_{3}}...+{{w}_{n}}}$. Then we will replace the variable given in the formula by the values given in the question to get our result.
Complete step-by-step solution:
The weighted mean of n numbers which are ${{a}_{1}},{{a}_{2}},{{a}_{3}}...,{{a}_{n}}$ and their weights are given by ${{w}_{1}},{{w}_{2}},{{w}_{3}}...{{w}_{n}}$. So, the weighted mean is given by the formula
$\dfrac{{{a}_{1}}{{w}_{1}}+{{a}_{2}}{{w}_{2}}+{{a}_{3}}{{w}_{3}}...+{{a}_{n}}{{w}_{n}}}{{{w}_{1}}+{{w}_{2}}+{{w}_{3}}...+{{w}_{n}}}$ …. (i), where ${{w}_{1}},{{w}_{2}},{{w}_{3}}...{{w}_{n}}$ are the weights of the corresponding values ${{a}_{1}},{{a}_{2}},{{a}_{3}}...,{{a}_{n}}$.
Let the weightage of the monthly examinations be $x$.
So, the weightage of the final examination according to the question is $2x$ as the final examination is weighted twice as much as a monthly examination.
Now according to question the value of $n$ is 4, similarly the corresponding values and their weights are given by ${{a}_{1}}=60,{{a}_{2}}=75,{{a}_{3}}=85,{{a}_{4}}=95$ and their corresponding weights are given by ${{w}_{1}}=x,{{w}_{2}}=x,{{w}_{3}}=x,{{w}_{4}}=2x$.
Now replacing the above values in the corresponding equation (i) we get.
$=\dfrac{60x+75x+85x+95\times 2x}{x+x+x+2x}$
$=\dfrac{60x+75x+85x+190x}{5x}$
$=\dfrac{410x}{5x}$
$=82$
So, the mean mark of the student in subject mathematics is 82. So, option C is correct.
Note: While calculating weighted mean we must keep in mind the difference between the arithmetic mean and the weighted mean, while the former is just the normal average that we calculate in day to day life while the latter is calculated by multiplying the corresponding weights with the values and then dividing with the total weight.
Complete step-by-step solution:
The weighted mean of n numbers which are ${{a}_{1}},{{a}_{2}},{{a}_{3}}...,{{a}_{n}}$ and their weights are given by ${{w}_{1}},{{w}_{2}},{{w}_{3}}...{{w}_{n}}$. So, the weighted mean is given by the formula
$\dfrac{{{a}_{1}}{{w}_{1}}+{{a}_{2}}{{w}_{2}}+{{a}_{3}}{{w}_{3}}...+{{a}_{n}}{{w}_{n}}}{{{w}_{1}}+{{w}_{2}}+{{w}_{3}}...+{{w}_{n}}}$ …. (i), where ${{w}_{1}},{{w}_{2}},{{w}_{3}}...{{w}_{n}}$ are the weights of the corresponding values ${{a}_{1}},{{a}_{2}},{{a}_{3}}...,{{a}_{n}}$.
Let the weightage of the monthly examinations be $x$.
So, the weightage of the final examination according to the question is $2x$ as the final examination is weighted twice as much as a monthly examination.
Now according to question the value of $n$ is 4, similarly the corresponding values and their weights are given by ${{a}_{1}}=60,{{a}_{2}}=75,{{a}_{3}}=85,{{a}_{4}}=95$ and their corresponding weights are given by ${{w}_{1}}=x,{{w}_{2}}=x,{{w}_{3}}=x,{{w}_{4}}=2x$.
Now replacing the above values in the corresponding equation (i) we get.
$=\dfrac{60x+75x+85x+95\times 2x}{x+x+x+2x}$
$=\dfrac{60x+75x+85x+190x}{5x}$
$=\dfrac{410x}{5x}$
$=82$
So, the mean mark of the student in subject mathematics is 82. So, option C is correct.
Note: While calculating weighted mean we must keep in mind the difference between the arithmetic mean and the weighted mean, while the former is just the normal average that we calculate in day to day life while the latter is calculated by multiplying the corresponding weights with the values and then dividing with the total weight.
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