Question

In an engineering college of 10,000 students, 1,500 like neither their core branches nor other branches. The number of students who like their core branches is \( \frac{1}{4} \) of the number of students who like other branches. The number of students who like both their core and other branches is 500. The number of students who like their core branches is:

• A. 1,800
• B. 3,500
• C. 1,600
• D. 1,500

Hint:

In set theory problems, use union and intersection relations systematically, and leverage De Morgan's laws for complements.

Answer 1

The Correct Option is A

Let \( \Omega \), \( C \), and \( O \) be the set of total students (universal set), students who like their core branch, and students who like other branches, respectively. We are given: \[ |\Omega| = 10,000, \quad |C^c \cap O^c| = 1,500, \quad |O| = 4|C|, \quad |C \cap O| = 500. \] We need to find \( |C| \). Step 1: Calculate \( |C \cup O| \).
Using De Morgan's law: \[ |C \cup O| = |\Omega| - |C^c \cap O^c| = 10,000 - 1,500 = 8,500. \] Step 2: Relation between \( |C| \) and \( |O| \).
From set theory: \[ |C \cup O| = |C| + |O| - |C \cap O|. \] Substitute \( |C \cup O| = 8,500 \), \( |O| = 4|C| \), and \( |C \cap O| = 500 \): \[ 8,500 = |C| + 4|C| - 500. \] Simplify: \[ 5|C| = 8,500 + 500 = 9,000 \implies |C| = \frac{9,000}{5} = 1,800. \] Final Answer: \[ \boxed{1,800} \]