In a class of 35 students, 18 students opt Maths, 12 students opt Maths but not Science. Find the number of students who opt both subjects. Also find the number of students who opt Science but not Maths. It is given that each student opt atleast one subject.

Answer

Let's solve the problem step by step.

### Step 1: Identify the given information
- Total number of students in the class = 35
- Number of students who opt for Maths (M) = 18
- Number of students who opt for Maths but not Science = 12

### Step 2: Find the number of students who opt for both Maths and Science
Let:
- \( n(M) \) = Number of students who opt for Maths = 18
- \( n(M \cap S) \) = Number of students who opt for both Maths and Science
- \( n(M \setminus S) \) = Number of students who opt for Maths but not Science = 12

Using the relationship:
\[ n(M) = n(M \setminus S) + n(M \cap S) \]

We can substitute the known values:
\[ 18 = 12 + n(M \cap S) \]

Now, solving for \( n(M \cap S) \):
\[ n(M \cap S) = 18 - 12 = 6 \]

### Step 3: Find the number of students who opt for Science but not Maths
Let:
- \( n(S) \) = Number of students who opt for Science
- \( n(S \setminus M) \) = Number of students who opt for Science but not Maths

We know that:
\[ n(M \cup S) = n(M) + n(S) - n(M \cap S) \]
Where \( n(M \cup S) \) is the total number of students, which is 35.

Substituting the known values:
\[ 35 = 18 + n(S) - 6 \]

Now, solving for \( n(S) \):
\[ 35 = 12 + n(S) \]
\[ n(S) = 35 - 12 = 23 \]

Now, we can find the number of students who opt for Science but not Maths:
\[ n(S \setminus M) = n(S) - n(M \cap S) \]
\[ n(S \setminus M) = 23 - 6 = 17 \]

### Final Answers
- Number of students who opt for both subjects (Maths and Science) = 6
- Number of students who opt for Science but not Maths = 17

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