In an exam , pass marks are 37.5%. A got 33(1/3)% more marks than passing marks .B failed by 50 marks. B's marks are 62.5% of A's marks . Find the passing marks .

Let the passing marks be denoted by \( P \).

### Step 1: A's marks calculation
A got \( 33 \frac{1}{3}\% \) more than the passing marks, so A's marks are:
\[
A = P \times \left(1 + \frac{33 \frac{1}{3}}{100}\right) = P \times \left(1 + \frac{1}{3}\right) = P \times \frac{4}{3}
\]
Thus, A's marks are \( A = \frac{4}{3}P \).

### Step 2: B's marks calculation
B's marks are \( 62.5\% \) of A's marks. Converting \( 62.5\% \) to a fraction gives \( \frac{5}{8} \), so:
\[
B = \frac{5}{8}A = \frac{5}{8} \times \frac{4}{3}P = \frac{20}{24}P = \frac{5}{6}P
\]
Thus, B's marks are \( B = \frac{5}{6}P \).

### Step 3: B failed by 50 marks
We are told that B failed by 50 marks, meaning:
\[
B = P - 50
\]
Substitute \( B = \frac{5}{6}P \) into this equation:
\[
\frac{5}{6}P = P - 50
\]

### Step 4: Solving for \( P \)
Rearrange the equation to solve for \( P \):
\[
P - \frac{5}{6}P = 50
\]
\[
\frac{1}{6}P = 50
\]
Multiply both sides by 6:
\[
P = 50 \times 6 = 300
\]

Thus, the passing marks are \( \boxed{300} \).

Let me know if you want further details on any step! 

### Additional Questions:
1. What would A's marks be if the passing marks were 400?
2. How much did B score in this exam based on the calculated passing marks?
3. If the passing marks increased by 10%, how much would A score?
4. What percentage did B score relative to the passing marks?
5. What would B's score be if he failed by 70 marks instead?

### Tip:
To solve percentage-based problems efficiently, always convert percentages to fractions or decimals to simplify the equations.

In an exam, pass marks are 37.5%. A got 33(1/3)% more marks than passing marks. B failed by 50 marks. B's marks are 62.5% of A's marks. Find the passing marks.

This problem helps you calculate the passing marks in an exam where A scored 33(1/3)% more than the passing marks and B's marks were 62.5% of A's, with B failing by 50 marks. Learn how to use algebra and percentage calculations to solve this problem. Suitable for students in Grades 9-12.

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