In a school there are 100 students 60 of them don't like Chocolate and 50 don’t like Biscuit and 10 of them like none then how many of them like both?

Options

A. 20
B. 30
C. 40
D. None of these

Answer

To solve the problem step by step, we will denote:

- Let \( C \) be the set of students who like chocolate.
- Let \( B \) be the set of students who like biscuits.

Given data:
- Total students, \( n = 100 \)
- Students who don't like chocolate, \( n(C') = 60 \)
- Students who don't like biscuits, \( n(B') = 50 \)
- Students who like neither, \( n(N) = 10 \)

### Step 1: Calculate the number of students who like chocolate

The number of students who like chocolate can be calculated as follows:

\[ n(C) = n - n(C') = 100 - 60 = 40 \]

### Step 2: Calculate the number of students who like biscuits

The number of students who like biscuits can be calculated similarly:

\[ n(B) = n - n(B') = 100 - 50 = 50 \]

### Step 3: Calculate the number of students who like at least one of the two (chocolate or biscuits)

The number of students who like at least one of the two can be calculated by subtracting those who like neither from the total number of students:

\[ n(C \cup B) = n - n(N) = 100 - 10 = 90 \]

### Step 4: Use the principle of inclusion-exclusion to find the number of students who like both

According to the principle of inclusion-exclusion, we have:

\[ n(C \cup B) = n(C) + n(B) - n(C \cap B) \]

Rearranging this formula to find \( n(C \cap B) \):

\[ n(C \cap B) = n(C) + n(B) - n(C \cup B) \]

Substituting the values we calculated:

\[ n(C \cap B) = 40 + 50 - 90 = 0 \]

### Conclusion

Thus, the number of students who like both chocolate and biscuits is \( 0 \).

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