Question

A school has 4 sections of Chemistry in class X having 40, 35, 45 and 42 students. The mean marks obtained in the Chemistry test are 50, 60, 55, and 45 respectively for the 4 sections. Determine the overall average of marks per student.

Answer 1

Hint: First we will find total marks obtained in each section. Then add all marks and divide the sum by the total number of students. The quotient is the required solution.

Formula Used:
${\rm{Mean = }}\dfrac{{{\rm{Sum}}\,{\rm{of}}\,{\rm{terms}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{terms}}}}$

Complete step by step solution:
Given that in class X, there are four sections of Chemistry, each with 40, 35, 45, and 42 students.
The average scores for the four portions of the Chemistry test were 50, 60, 55, and 45, respectively.
Total marks obtained by students in the first section is $ = \left( {40 \times 50} \right) = 2000$.
Total marks obtained by students in the second section is $ = \left( {35 \times 60} \right) = 2100$.
Total marks obtained by students in the third section is $ = \left( {45 \times 55} \right) = 2475$.
Total marks obtained by students in the fourth section is $ = \left( {42 \times 45} \right) = 1890$.
Total number of marks obtained by all students is $ = 2000 + 2100 + 2475 + 1890 = 8465$
The total number of students in 4 sections is $40 + 35 + 45 + 42 = 162$
Divide total number of marks by total number of students
${\rm{Mean}} = \dfrac{{8465}}{{162}}$
$ \Rightarrow {\rm{Mean}} \approx {\rm{52}}{\rm{.25}}$
The average marks per student is 52.25.

Note: There is a formula to calculate the mean $\overline X = \dfrac{{{n_1}\overline {{X_1}} + {n_2}\overline {{X_2}} + {n_3}\overline {{X_3}} + {n_4}\overline {{X_4}} }}{{{n_1} + {n_2} + {n_3} + {n_4}}}$ where ${n_1},{n_2},{n_3},{n_4}$ are the number of students and $\overline {{X_1}} $ ,$\overline {{X_2}} $, $\overline {{X_3}} $, $\overline {{X_4}} $ are means of marks.