GATE CS 2024 Set 2 Q.3: Quantitative Aptitude

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 $1/4^{\text{th}}$ 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

1,800

3,500

1,600

1,500

Official answer and solution

Official answer: A
From: Official GATE 2024 answer key

Solution

Apply the principle of inclusion-exclusion to the sets of students liking core branches ( $C$ ) and other branches ( $O$ ).

Let $C$ be the set of students who like their core branches and $O$ be the set of students who like other branches.

Total students $= 10,000$ , and students who like neither $= 1,500$ , so: $|C \cup O| = 10,000 - 1,500 = 8,500$
Given $|C| = \frac{1}{4}|O| \implies |O| = 4|C|$ , and $|C \cap O| = 500$ .
By the inclusion-exclusion principle: $|C \cup O| = |C| + |O| - |C \cap O|$ $8,500 = |C| + 4|C| - 500 \implies 5|C| = 9,000 \implies |C| = 1,800$

Final answer

Option (A)

Common trap

Subtracting 500 instead of adding 500 when solving $5|C| - 500 = 8500$ , which gives $1600$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2024 Computer Science and Information Technology - Set 2 (CS2)
Question: 3 of 65
Section: General Aptitude

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