Read the following information carefully and answer the questions given below-
1000 students appeared in an examination Out of which 275 students passed in physics, 320 students passed in chemistry and 350 students passed in mathematics. 84 students passed in both chemistry and mathematics 60 students passed in both physics and mathematics, 50 students passed in both physics and chemistry and 34 students passed in all the three subjects.
What is the number of students who passed in at least two subjects ?

Options

A. A) 126
B. B) 127
C. C) 128
D. D) 129

Answer

To solve the problem, we will use the principle of inclusion-exclusion to find the number of students who passed in at least two subjects.

Let's define the sets:
- Let \( P \) be the set of students who passed in Physics.
- Let \( C \) be the set of students who passed in Chemistry.
- Let \( M \) be the set of students who passed in Mathematics.

From the information given:
- \( |P| = 275 \) (students passed in Physics)
- \( |C| = 320 \) (students passed in Chemistry)
- \( |M| = 350 \) (students passed in Mathematics)
- \( |C \cap M| = 84 \) (students passed in both Chemistry and Mathematics)
- \( |P \cap M| = 60 \) (students passed in both Physics and Mathematics)
- \( |P \cap C| = 50 \) (students passed in both Physics and Chemistry)
- \( |P \cap C \cap M| = 34 \) (students passed in all three subjects)

We need to find the number of students who passed in at least two subjects. This can be calculated using the formula:

\[ |P \cap C| + |C \cap M| + |P \cap M| - 2|P \cap C \cap M| \]

Now, substituting the values we have:

\[ = |P \cap C| + |C \cap M| + |P \cap M| - 2|P \cap C \cap M| \]
\[ = 50 + 84 + 60 - 2 \times 34 \]
\[ = 50 + 84 + 60 - 68 \]
\[ = 94 + 60 \]
\[ = 154 \]

So, the number of students who passed in at least two subjects is **154**.