In an examination a student got `32%` marks and failed by `4` marks . While an another student got `35%` marks and got 5 marks more than marks. Find the max. marks in the examination ?

Options

A. `300`
B. `400`
C. `350`
D. `250`

Answer

To solve the problem step by step, we will denote the maximum marks in the examination as \( x \).

### Step 1: Set up the equations based on the information given.

1. The first student got \( 32\% \) of the maximum marks and failed by \( 4 \) marks. This means that if he had \( 4 \) more marks, he would have passed.
- Marks obtained by the first student = \( \frac{32}{100} \times x \)
- Passing marks = Marks obtained + 4
- Therefore, we can write the equation:
\[ \frac{32}{100} \times x + 4 = \text{Passing Marks} \]

2. The second student got \( 35\% \) of the maximum marks and scored \( 5 \) marks more than the passing marks.
- Marks obtained by the second student = \( \frac{35}{100} \times x \)
- Therefore, we can write the equation:
\[ \frac{35}{100} \times x - 5 = \text{Passing Marks} \]

### Step 2: Equate the two expressions for passing marks.

From the two equations we have:
\[ \frac{32}{100} \times x + 4 = \frac{35}{100} \times x - 5 \]

### Step 3: Simplify the equation.

1. Rearranging the equation gives:
\[ \frac{32}{100} \times x + 4 + 5 = \frac{35}{100} \times x \]
\[ \frac{32}{100} \times x + 9 = \frac{35}{100} \times x \]

2. Now, subtract \( \frac{32}{100} \times x \) from both sides:
\[ 9 = \frac{35}{100} \times x - \frac{32}{100} \times x \]
\[ 9 = \left(\frac{35 - 32}{100}\right) \times x \]
\[ 9 = \frac{3}{100} \times x \]

### Step 4: Solve for \( x \).

1. To isolate \( x \), multiply both sides by \( \frac{100}{3} \):
\[ x = 9 \times \frac{100}{3} \]
\[ x = \frac{900}{3} \]
\[ x = 300 \]

### Conclusion

The maximum marks in the examination is \( \boxed{300} \).

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