In an examination, 31 candidates passed chemistry, 29 passed physics and 3 failed both subjects.If 50 candidates sat for the examination, how many of them passed chemistry only?

Answer 1

To solve this problem, let's denote the following:

• Let \( A \) be the set of candidates who passed chemistry.
• Let \( B \) be the set of candidates who passed physics.
• The number of candidates who passed chemistry \( |A| = 31 \).
• The number of candidates who passed physics \( |B| = 29 \).
• The number of candidates who failed both subjects is 3.
• Therefore, the total number of candidates who sat for the examination is 50.

From this, we can find the total number of candidates who passed at least one of the subjects:

\[ \text{Candidates passed at least one subject} = \text{Total candidates} - \text{Failed both} \] \[ = 50 - 3 = 47 \]

Using the principle of inclusion-exclusion for the sets \( A \) and \( B \):

\[ |A \cup B| = |A| + |B| - |A \cap B| \]

Where \( |A \cup B| \) is the total number of candidates who passed at least one subject (which we found to be 47). Thus, we write:

\[ 47 = 31 + 29 - |A \cap B| \]

Now, simplifying this expression:

\[ 47 = 60 - |A \cap B| \]

So, we can solve for \( |A \cap B| \):

\[ |A \cap B| = 60 - 47 = 13 \]

Now we have:

• \( |A| = 31 \) (candidates passed chemistry)
• \( |B| = 29 \) (candidates passed physics)
• \( |A \cap B| = 13 \) (candidates who passed both subjects)

To find the number of candidates who passed chemistry only, we use the formula:

\[ |A \text{ only}| = |A| - |A \cap B| \] \[ |A \text{ only}| = 31 - 13 = 18 \]

Thus, the number of candidates who passed chemistry only is \( \boxed{18} \).