For an examination consisting of three subjects -Maths Physics and Chemistry, 280 students appeared. When the results were declared, 185 students had passed in Maths, 210 had passed in Physics and 222 had passed in Chemistry.
All those except 5 students who passed in Maths, passed in Physics.
All those except 10 students who passed in Maths, passed in Chemistry.
47 students failed in all the three subjects.
200 students who passed in Physics also passed in Chemistry :
How many students passed in Chemistry only ?

Options

A. 18
B. 21
C. 25
D. 29

Explanation

To solve the problem step by step, we will use the information provided and apply set theory concepts, particularly focusing on the Venn diagram approach.

### Step 1: Identify the total number of students and those who passed in each subject
- Total students = 280
- Passed in Maths (M) = 185
- Passed in Physics (P) = 210
- Passed in Chemistry (C) = 222
- Failed in all subjects = 47

### Step 2: Calculate the number of students who passed at least one subject
To find the number of students who passed at least one subject, we subtract those who failed in all subjects from the total number of students:
\[ \text{Students who passed at least one subject} = 280 - 47 = 233 \]

### Step 3: Use the information about students passing in multiple subjects
From the problem:
- All except 5 students who passed in Maths passed in Physics. This means:
\[ M \cap P = 185 - 5 = 180 \]
- All except 10 students who passed in Maths passed in Chemistry. This means:
\[ M \cap C = 185 - 10 = 175 \]
- 200 students who passed in Physics also passed in Chemistry:
\[ P \cap C = 200 \]

### Step 4: Set up the equation using the principle of inclusion-exclusion
Using the formula for the union of three sets:
\[ N(M \cup P \cup C) = N(M) + N(P) + N(C) - N(M \cap P) - N(M \cap C) - N(P \cap C) + N(M \cap P \cap C) \]
Substituting the known values:
\[ 233 = 185 + 210 + 222 - 180 - 175 - 200 + N(M \cap P \cap C) \]

### Step 5: Simplify the equation
Calculating the right side:
\[ 233 = 185 + 210 + 222 - 180 - 175 - 200 + N(M \cap P \cap C) \]
\[ 233 = 185 + 210 + 222 - 555 + N(M \cap P \cap C) \]
\[ 233 = 617 - 555 + N(M \cap P \cap C) \]
\[ 233 = 62 + N(M \cap P \cap C) \]
Now, solving for \( N(M \cap P \cap C) \):
\[ N(M \cap P \cap C) = 233 - 62 = 171 \]

### Step 6: Analyze the Venn diagram
Now we can fill in the Venn diagram:
- Let \( x \) be the number of students who passed only in Chemistry.
- We know:
- \( N(M \cap P) = 180 \)
- \( N(M \cap C) = 175 \)
- \( N(P \cap C) = 200 \)
- \( N(M \cap P \cap C) = 171 \)

### Step 7: Calculate the number of students passing only in Chemistry
From the intersections:
- Students passing in both Maths and Chemistry but not Physics:
\[ N(M \cap C) - N(M \cap P \cap C) = 175 - 171 = 4 \]
- Students passing in both Physics and Chemistry but not Maths:
\[ N(P \cap C) - N(M \cap P \cap C) = 200 - 171 = 29 \]

### Step 8: Calculate the total number of students passing in Chemistry
Total students passing in Chemistry:
\[ N(C) = N(M \cap C) + N(P \cap C) + x + N(M \cap P \cap C) \]
Substituting the known values:
\[ 222 = 4 + 29 + x + 171 \]
\[ 222 = 204 + x \]
Thus, solving for \( x \):
\[ x = 222 - 204 = 18 \]

### Final Answer
The number of students who passed only in Chemistry is **18**.