
In a class, the average marks obtained by the number of students in English is \[52.25\]. \[25\% \] students placed in the C category made the average of 31 marks, while \[20\% \] who were placed in a category made the average of 80 marks. Find the average marks of the remaining students?
Answer
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Hint: Average can be defined as the ratio of the sum of the observations in the given set of data to the total number of the observations. For example: The given data is \[11,{\text{ }}12,{\text{ }}13,{\text{ }}14\].
Average \[ = {\text{ }}\dfrac{{\left( {11 + 12 + 13 + 14} \right)}}{4}{\text{ }} = {\text{ }}12.5\]
In this particular problem, the average is associated with the percentages. So, we need to assume that the total number of students in the class is 100 just to make the calculation easier.
Complete step-by-step answer:
Let the total number of students in the class be \[100\].
Average marks of all the students in English is given to be \[52.25\]
According to the problem,
Number of the students in the C category = \[25\% \] of the total students \[ = {\text{ }}25\]
Average marks of the students in C category in English \[ = {\text{ }}31\]
Number of the students in the A category = \[20\% \] of the total students = \[20\]
Average marks of the students in A category in English = 80
Hence, the number of the remaining students \[ = {\text{ }}100{\text{ }}-{\text{ }}\left( {25{\text{ }} + {\text{ }}20} \right){\text{ }} = {\text{ }}100{\text{ }}-{\text{ }}45{\text{ }} = {\text{ }}55\]
Let Average marks of the remaining students in English = x
On solving,
\[ \Rightarrow \] \[52.25 \times 100{\text{ }} = {\text{ }}\left( {31 \times 25} \right){\text{ }} + {\text{ }}\left( {80 \times 20} \right){\text{ }} + {\text{ }}\left( {55 \times x} \right)\]
\[ \Rightarrow \] \[5225{\text{ }} = {\text{ }}775{\text{ }} + {\text{ }}160{\text{ }} + {\text{ }}55x\]
\[ \Rightarrow \] \[55x{\text{ }} = {\text{ }}5225{\text{ }}-{\text{ }}775{\text{ }}-{\text{ }}160\]
\[ \Rightarrow \] 55x = 2850
\[ \Rightarrow \] \[x{\text{ }} = {\text{ }}\dfrac{{2850}}{{55}}{\text{ }} = {\text{ }}51.8\]
Hence, the average marks of the remaining students in English is \[51.8\].
Note: The average is commonly known as the Mean or the Expected value.
To find the averages associated with the percentages, we always consider the total number of the observations to be 100 so that the calculations become easier.
So, on solving this problem by using the same method, 25% of the total students i.e. \[100\] becomes \[25\] in the C category and similarly, 20% of the total students i.e. \[100\] becomes \[20\] in A category and hence, we get the average marks of the remaining students in English, \[51.8\].
Average \[ = {\text{ }}\dfrac{{\left( {11 + 12 + 13 + 14} \right)}}{4}{\text{ }} = {\text{ }}12.5\]
In this particular problem, the average is associated with the percentages. So, we need to assume that the total number of students in the class is 100 just to make the calculation easier.
Complete step-by-step answer:
Let the total number of students in the class be \[100\].
Average marks of all the students in English is given to be \[52.25\]
According to the problem,
Number of the students in the C category = \[25\% \] of the total students \[ = {\text{ }}25\]
Average marks of the students in C category in English \[ = {\text{ }}31\]
Number of the students in the A category = \[20\% \] of the total students = \[20\]
Average marks of the students in A category in English = 80
Hence, the number of the remaining students \[ = {\text{ }}100{\text{ }}-{\text{ }}\left( {25{\text{ }} + {\text{ }}20} \right){\text{ }} = {\text{ }}100{\text{ }}-{\text{ }}45{\text{ }} = {\text{ }}55\]
Let Average marks of the remaining students in English = x
On solving,
\[ \Rightarrow \] \[52.25 \times 100{\text{ }} = {\text{ }}\left( {31 \times 25} \right){\text{ }} + {\text{ }}\left( {80 \times 20} \right){\text{ }} + {\text{ }}\left( {55 \times x} \right)\]
\[ \Rightarrow \] \[5225{\text{ }} = {\text{ }}775{\text{ }} + {\text{ }}160{\text{ }} + {\text{ }}55x\]
\[ \Rightarrow \] \[55x{\text{ }} = {\text{ }}5225{\text{ }}-{\text{ }}775{\text{ }}-{\text{ }}160\]
\[ \Rightarrow \] 55x = 2850
\[ \Rightarrow \] \[x{\text{ }} = {\text{ }}\dfrac{{2850}}{{55}}{\text{ }} = {\text{ }}51.8\]
Hence, the average marks of the remaining students in English is \[51.8\].
Note: The average is commonly known as the Mean or the Expected value.
To find the averages associated with the percentages, we always consider the total number of the observations to be 100 so that the calculations become easier.
So, on solving this problem by using the same method, 25% of the total students i.e. \[100\] becomes \[25\] in the C category and similarly, 20% of the total students i.e. \[100\] becomes \[20\] in A category and hence, we get the average marks of the remaining students in English, \[51.8\].
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