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

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

From the problem statement:

• Total number of students, $N = 10{,}000$

• Students who like neither: $|(C \cup O)'| = 1{,}500$

• Students who like at least one branch: $|C \cup O| = 10{,}000 - 1{,}500 = 8{,}500$

• Students who like both: $|C \cap O| = 500$

• Relation between $|C|$ and $|O|$: $|C| = \frac{1}{4}|O| \implies |O| = 4|C|$

Using the principle of inclusion-exclusion:

Substitute the values into the equation:

Therefore, the number of students who like their core branches is 1,800, which corresponds to option (A).