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There are total 120 students in three sections A, B and C of Class 10. The average marks of the three sections is 85. The average marks of section B and C is 88 and the average marks of section A is 78. Find the number of students in section A.
36
28
33
24
36
Use the weighted average formula or alligation method. By total marks balance: 120×85=(120−nA)×88+nA×78, where nA is the number of students in section A.
Total students = 120, Average marks of all sections (A, B, C) = 85, Average marks of sections B and C = 88, Average marks of section A = 78.
Total Marks=Total Students×Overall Average
Use the weighted average formula or alligation method. By total marks balance: 120×85=(120−nA)×88+nA×78, where nA is the number of students in section A.
Students often mistakenly assume that the number of students in section A is simply calculated by taking the simple average of the averages (88+78) / 2, ignoring the weighted contribution based on student strength.
Calculate Total Marks of All Students
Using the overall average of all 120 students, find the sum of all marks: 120×85=10200.
120×85=10200
Express Section A and Sections B+C Strength
Let the number of students in section A be nA. Then the combined number of students in sections B and C is (120−nA).
nB+nC=120−nA
Set Up Total Marks Equation
The sum of the total marks of section A and the total marks of sections B and C equals the grand total marks: 78nA+88(120−nA)=10200.
78nA+88(120−nA)=10200
Solve for nA
Expanding the equation: 78nA+10560−88nA=10200⟹−10nA=10200−10560⟹−10nA=−360⟹nA=36.
nA=36
A is correct because the number of students in section A calculated via the weighted average equation is 36.
Weighted average problems frequently appear in mixture and allegation questions in quantitative aptitude tests.
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