Question

In a group of students, 10 students like Mathematics, 12 students like English, 4 students like both Mathematics and English, and 6 students like neither Mathematics nor English. The number of students in the group is ____

• A. 18
• B. 20
• C. 24
• D. 32

Hint:

First find how many students like at least one subject using the union formula, then add the students who like neither.

Answer 1

The Correct Option is C

Step 1: Define the sets.
Let \(M\) be the set of students who like Mathematics and \(E\) be the set of students who like English. We are given \(n(M) = 10\), \(n(E) = 12\), \(n(M \cap E) = 4\) (students who like both), and the number of students who like neither subject is 6.

Step 2: Apply the formula for the union of two sets.
The number of students who like Mathematics or English (or both) is given by \(n(M \cup E) = n(M) + n(E) - n(M \cap E)\).

Step 3: Substitute the given values.
\(n(M \cup E) = 10 + 12 - 4 = 18\). So 18 students like at least one of the two subjects.

Step 4: Add the students who like neither subject.
The total group size is the number who like at least one subject plus the number who like neither, since these two groups do not overlap and together cover everyone: \(\text{Total} = n(M \cup E) + \text{neither} = 18 + 6 = 24\).

Step 5: Compare with the options.
The total of 24 matches option (C). Option (A) 18 is only the number who like at least one subject, without adding the "neither" group. Option (B) 20 and option (D) 32 do not correspond to any correct combination of the given numbers.

\[ \boxed{24} \]