Question

The average marks of section A are 65 and that of section B are 70. The average of both the sections combined is 67. The ratio of the number of students of section A to B is :

• A. 1 : 3
• B. 2 : 3
• C. 3 : 1
• D. 3 : 2

Hint:

Use the rule of alligation: the ratio of group sizes is inverse to the ratio of their deviations from the combined mean.

Answer 1

The Correct Option is D

Step 1: Let the number of students in section A be \(n_1\) and in section B be \(n_2\). The combined mean formula for two groups is\[\bar{X} = \frac{n_1 \bar{X_1} + n_2 \bar{X_2}}{n_1+n_2}\]

Step 2: Substitute the given values \(\bar{X_1} = 65\), \(\bar{X_2} = 70\) and \(\bar{X} = 67\).\[67 = \frac{65 n_1 + 70 n_2}{n_1+n_2}\]

Step 3: Multiply both sides by \((n_1+n_2)\).\[67 n_1 + 67 n_2 = 65 n_1 + 70 n_2\]

Step 4: Collect the \(n_1\) terms on one side and the \(n_2\) terms on the other.\[67 n_1 - 65 n_1 = 70 n_2 - 67 n_2\]\[2 n_1 = 3 n_2\]

Step 5: Solve for the ratio.\[\frac{n_1}{n_2} = \frac{3}{2}\]So the ratio of students of section A to B is \(3:2\), which is option (D).