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:

When solving problems involving inclusion-exclusion, carefully account for overlaps and use algebraic expressions for unknown quantities.

Answer 1

The Correct Option is A

Step 1: Total students and given information.

The total number of students in the college is \( 10,000 \). Out of these, \( 1,500 \) students like neither their core branches nor other branches. Hence, the remaining students who like at least one of the branches is:

\[ 10,000 - 1,500 = 8,500 \] Step 2: Let the number of students who like other branches be \( x \).

The number of students who like their core branches is given as \( \frac{1}{4}x \), and the number of students who like both branches is 500.

Using the principle of inclusion-exclusion for the students who like at least one branch:

\[ \text{Students liking at least one branch} = \text{Students liking core branches} + \text{Students liking other branches} - \text{Students liking both branches} \]

Substituting the values:

\[ 8,500 = \frac{1}{4}x + x - 500 \] Step 3: Simplify the equation to find \( x \).

Combine terms:

\[ 8,500 = \frac{5}{4}x - 500 \]

Add 500 to both sides:

\[ 9,000 = \frac{5}{4}x \]

Multiply through by 4 and divide by 5:

\[ x = \frac{9,000 \times 4}{5} = 7,200 \] Step 4: Find the number of students who like core branches.

The number of students who like their core branches is:

\[ \frac{1}{4}x = \frac{1}{4} \times 7,200 = 1,800 \]

Thus, the correct answer is \( 1,800 \).