In a class, `80%` of students pass in maths exam and `70%` of students pass in English exam. `10%` of students fail in Both subjects and if 144 students pass in Both subejects. Find the total number of students.

Answer

To solve the problem step by step, we will use the information provided in the question and apply the concept of percentages.

### Step 1: Assume the Total Number of Students
Let's assume the total number of students in the class is \( x \).

### Step 2: Calculate the Number of Students Passing in Each Subject
- **Students passing in Maths**: Since \( 80\% \) of students pass in Maths, the number of students passing in Maths is:
\[ \text{Students passing in Maths} = 0.80 \times x \]

- **Students passing in English**: Since \( 70\% \) of students pass in English, the number of students passing in English is:
\[ \text{Students passing in English} = 0.70 \times x \]

### Step 3: Calculate the Number of Students Failing in Both Subjects
- **Students failing in both subjects**: According to the problem, \( 10\% \) of students fail in both subjects, so the number of students failing in both is:
\[ \text{Students failing in both} = 0.10 \times x \]

### Step 4: Calculate the Number of Students Passing in At Least One Subject
- The number of students passing in at least one subject can be calculated as:
\[ \text{Students passing in at least one subject} = x - \text{Students failing in both} \]
Substituting the value from Step 3:
\[ \text{Students passing in at least one subject} = x - 0.10x = 0.90x \]

### Step 5: Set Up the Equation Using the Inclusion-Exclusion Principle
Using the inclusion-exclusion principle, we can express the number of students passing in at least one subject as:
\[ \text{Students passing in at least one subject} = \text{Students passing in Maths} + \text{Students passing in English} - \text{Students passing in both} \]
Substituting the values from Steps 2 and 4:
\[ 0.90x = (0.80x + 0.70x - \text{Students passing in both}) \]
Let \( y \) be the number of students passing in both subjects. From the problem, we know \( y = 144 \). Thus, we can rewrite the equation as:
\[ 0.90x = 0.80x + 0.70x - 144 \]

### Step 6: Solve for \( x \)
Combine like terms:
\[ 0.90x = 1.50x - 144 \]
Rearranging gives:
\[ 144 = 1.50x - 0.90x \]
\[ 144 = 0.60x \]
Now, solving for \( x \):
\[ x = \frac{144}{0.60} = 240 \]

### Conclusion
The total number of students in the class is \( 240 \).

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