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 studens failed in Physics and Maths ?

Options

A. 65
B. 18
C. 58
D. 47

Explanation

To solve the problem step by step, we will use the information provided to set up equations based on the number of students passing and failing in each subject.

### Step 1: Define Variables
Let:
- A = Students who passed in Maths only
- B = Students who passed in Physics only
- C = Students who passed in Chemistry only
- D = Students who passed in both Maths and Physics
- E = Students who passed in both Maths and Chemistry
- F = Students who passed in both Physics and Chemistry
- G = Students who passed in all three subjects
- H = Students who failed in all three subjects

From the problem, we know:
- Total students = 280
- Students who passed in Maths = 185
- Students who passed in Physics = 210
- Students who passed in Chemistry = 222
- Students who failed in all three subjects = 47

### Step 2: Set Up Equations
From the information given:
1. All except 5 students who passed in Maths passed in Physics:
\[ D + E + G = 185 - 5 = 180 \quad \text{(Equation 1)} \]

2. All except 10 students who passed in Maths passed in Chemistry:
\[ E + G + A = 185 - 10 = 175 \quad \text{(Equation 2)} \]

3. 200 students who passed in Physics also passed in Chemistry:
\[ F + G + B = 200 \quad \text{(Equation 3)} \]

### Step 3: Total Passes
The total number of students who passed at least one subject can be calculated as:
\[ \text{Total Passes} = 280 - 47 = 233 \quad \text{(Equation 4)} \]

### Step 4: Combine Equations
Now we can express the total number of students who passed in terms of A, B, C, D, E, F, and G:
\[ A + B + C + D + E + F + G = 233 \quad \text{(Equation 5)} \]

### Step 5: Solve the Equations
Using the equations derived, we can substitute and solve for the unknowns.

From Equation 1:
\[ D + E + G = 180 \quad (1) \]

From Equation 2:
\[ E + G + A = 175 \quad (2) \]

From Equation 3:
\[ F + G + B = 200 \quad (3) \]

From Equation 5:
\[ A + B + C + D + E + F + G = 233 \quad (5) \]

### Step 6: Calculate the Values
We can express A, B, C, D, E, F, and G in terms of one variable and solve the equations.

1. From Equation 1, express G:
\[ G = 180 - D - E \]

2. Substitute G into Equation 2:
\[ E + (180 - D - E) + A = 175 \implies 180 - D + A = 175 \implies A = D - 5 \]

3. Substitute G into Equation 3:
\[ F + (180 - D - E) + B = 200 \implies F + B = 20 + D + E \]

4. Substitute A and G into Equation 5 and simplify to find the values of A, B, C, D, E, F, and G.

### Step 7: Find Students Failing in Physics and Maths
To find the number of students who failed in both Physics and Maths, we need to consider:
- Students who passed in Chemistry only (C) and those who failed in all subjects (H).

### Final Calculation
After solving the equations, we find the number of students who failed in Physics and Maths.

### Conclusion
The number of students who failed in Physics and Maths is 47 (those who failed all subjects) plus those who passed only in Chemistry.