In an examination `67%` of the students passed in mathis `53%` of the student passed in english, `25%` of the students passed in both subjects. If 200 students failled in both subjects. Then how many students passed in only maths.

Options

A. `1880`
B. `1780`
C. `1680`
D. `1980`

Answer

To solve the problem step by step, we will use the information given about the percentages of students passing in Maths and English, as well as the number of students who failed in both subjects.

### Step 1: Define the total number of students
Let the total number of students be \( N \).

### Step 2: Calculate the percentage of students who passed in each subject
- Students who passed in Maths = \( 67\% \) of \( N \) = \( 0.67N \)
- Students who passed in English = \( 53\% \) of \( N \) = \( 0.53N \)
- Students who passed in both subjects = \( 25\% \) of \( N \) = \( 0.25N \)

### Step 3: Use the principle of inclusion-exclusion to find the total number of students who passed in at least one subject
The formula for the number of students who passed in at least one subject is:
\[ \text{Passed in at least one subject} = (\text{Passed in Maths}) + (\text{Passed in English}) - (\text{Passed in both}) \]
Substituting the values:
\[ \text{Passed in at least one subject} = 0.67N + 0.53N - 0.25N = 0.95N \]

### Step 4: Calculate the number of students who failed in both subjects
According to the problem, \( 200 \) students failed in both subjects. Therefore, the number of students who passed in at least one subject can also be expressed as:
\[ N - 200 \]

### Step 5: Set up the equation
Since both expressions represent the number of students who passed in at least one subject, we can set them equal to each other:
\[ 0.95N = N - 200 \]

### Step 6: Solve for \( N \)
Rearranging the equation gives:
\[ 0.95N - N = -200 \]
\[ -0.05N = -200 \]
\[ N = \frac{-200}{-0.05} = 4000 \]

### Step 7: Calculate the number of students who passed in Maths
Now that we know \( N = 4000 \), we can find the number of students who passed in Maths:
\[ \text{Passed in Maths} = 0.67N = 0.67 \times 4000 = 2680 \]

### Step 8: Calculate the number of students who passed in only Maths
To find the number of students who passed only in Maths, we need to subtract those who passed in both subjects from those who passed in Maths:
\[ \text{Passed only in Maths} = \text{Passed in Maths} - \text{Passed in both} \]
\[ \text{Passed only in Maths} = 2680 - 0.25N = 2680 - 0.25 \times 4000 = 2680 - 1000 = 1680 \]

### Final Answer
The number of students who passed in only Maths is \( 1680 \).

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