Question

The average marks of the students in four sections A, B, C and D of a school is 60%. The average marks of the students of A, B, C and D individually are 45%, 50%, 72% and 80%, respectively. If the average marks of the students of sections A and B together is 48% and that of the students of B and C together is 60%, what is the ratio of the number of students in sections A and D?

• A. 2 : 3
• B. 4 : 3
• C. 5 : 3
• D. 3 : 5

Answer 1

The Correct Option is B

Let \(a,b,c,d\) be the number of students in A, B, C and D respectively

Given, average marks of the students in section A, B, C and D of the school = 60

Then \(\frac{45a+50b+72c+80d}{a+b+c+d}=60\%\)

\(=45a+50b+72c+80d=60a+60b+60c+60d\)

\(=12c+20d=15a+10b..............(1)\)

Average marks of the students of sections A and B together is \(48\%\)

\(=\frac{45a+50b}{a+b}=48\%\)

\(=454a+50b=48a+48b\)

\(=3a=2b\) (or) \(15a=10b.................(2)\)

Average marks of the students of sections B and C together is \(60\%\)

\(=\frac{72c+80d}{c+d}=60\%\)

\(=\$\$72c+80d=60c+60d\)

\(=12c=12d..............(3)\)

Substitute equations (2) and (3) in equation (1)

⇒ \(20d+20d=15a+15a\)

⇒ \(40d=30a\)

⇒ \(a:d=4:3\)

Hence, option B is the correct answer.The correct option is (B): 4 : 3