If in a class there are 200 students in which 120 take Mathematics, 90 take Physics, 60 take Chemistry, 50 take Mathematics & Physics, 50 take Mathematics & Chemistry, 43 take Physics & Chemistry and 38 take Mathematics Physics & Chemistry, then the number of students who hace taken exactly one subject is

Options

A. 42
B. 56
C. 270
D. 98

Answer

To solve the problem of finding the number of students who have taken exactly one subject, we can use the principle of inclusion-exclusion. Let's denote:

- \( n(M) \): Number of students taking Mathematics = 120
- \( n(P) \): Number of students taking Physics = 90
- \( n(C) \): Number of students taking Chemistry = 60
- \( n(M \cap P) \): Number of students taking both Mathematics and Physics = 50
- \( n(M \cap C) \): Number of students taking both Mathematics and Chemistry = 50
- \( n(P \cap C) \): Number of students taking both Physics and Chemistry = 43
- \( n(M \cap P \cap C) \): Number of students taking all three subjects = 38

### Step 1: Calculate the number of students taking at least one subject

Using the principle of inclusion-exclusion, the total number of students taking at least one subject is given by:

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

Substituting the values:

\[ n(M \cup P \cup C) = 120 + 90 + 60 - 50 - 50 - 43 + 38 \]

Calculating this gives:

\[ n(M \cup P \cup C) = 120 + 90 + 60 - 50 - 50 - 43 + 38 = 165 \]

### Step 2: Calculate the number of students taking exactly one subject

The number of students taking exactly one subject can be calculated as follows:

\[ n(\text{exactly one}) = n(M) + n(P) + n(C) - 2(n(M \cap P) + n(M \cap C) + n(P \cap C)) + 3n(M \cap P \cap C) \]

Substituting the values:

\[ n(\text{exactly one}) = 120 + 90 + 60 - 2(50 + 50 + 43) + 3(38) \]

Calculating this step by step:

1. Calculate the total for the first part:
\[ 120 + 90 + 60 = 270 \]

2. Calculate the total for the intersections:
\[ 50 + 50 + 43 = 143 \]
Then multiply by 2:
\[ 2 \times 143 = 286 \]

3. Calculate the total for the intersection of all three subjects:
\[ 3 \times 38 = 114 \]

4. Now substitute these results back into the equation:
\[ n(\text{exactly one}) = 270 - 286 + 114 \]

5. Finally, calculate:
\[ n(\text{exactly one}) = 270 - 286 + 114 = 98 \]

### Conclusion

Thus, the number of students who have taken exactly one subject is **98**.