To solve the problem step by step, we will break down the information provided and calculate the required values systematically.
### Step 1: Calculate the total number of applicants
Given that 90% of 12,500 graduates applied for the exams:
\[
\text{Total applicants} = 0.90 \times 12500 = 11250
\]
### Step 2: Determine the ratio of applicants for one exam, two exams, and all three exams
The ratio of applicants for only one exam, two exams, and all three exams is given as 16:20:9. Let the common multiplier be \(x\).
\[
\text{Applicants for one exam} = 16x, \quad \text{for two exams} = 20x, \quad \text{for all three exams} = 9x
\]
The total applicants can be expressed as:
\[
16x + 20x + 9x = 11250 \implies 45x = 11250 \implies x = \frac{11250}{45} = 250
\]
### Step 3: Calculate the number of applicants for each category
Using \(x = 250\):
- Applicants for only one exam:
\[
16x = 16 \times 250 = 4000
\]
- Applicants for two exams:
\[
20x = 20 \times 250 = 5000
\]
- Applicants for all three exams:
\[
9x = 9 \times 250 = 2250
\]
### Step 4: Determine the distribution of applicants for only one exam
- For Railways (12% of 4000):
\[
\text{Railways} = 0.12 \times 4000 = 480
\]
- For Banks (26% of 4000):
\[
\text{Banks} = 0.26 \times 4000 = 1040
\]
- For SSC (remaining):
\[
\text{SSC} = 4000 - (480 + 1040) = 4000 - 1520 = 2480
\]
### Step 5: Calculate the male to female ratios for each category
- For Railways (7:5):
\[
\text{Total for Railways} = 480 \implies \text{Males} = \frac{7}{12} \times 480 = 280, \quad \text{Females} = 480 - 280 = 200
\]
- For Banks (8:5):
\[
\text{Total for Banks} = 1040 \implies \text{Males} = \frac{8}{13} \times 1040 = 640, \quad \text{Females} = 1040 - 640 = 400
\]
- For SSC (9:7):
\[
\text{Total for SSC} = 2480 \implies \text{Males} = \frac{9}{16} \times 2480 = 1395, \quad \text{Females} = 2480 - 1395 = 1085
\]
### Step 6: Set up equations for applicants for two exams
Let \(a\) be the number of applicants for both Railways & Banks, \(b\) for Railways & SSC, and \(c\) for Banks & SSC. We know:
\[
b = a + 100
\]
\[
a + b + c = 5000
\]
Substituting \(b\):
\[
a + (a + 100) + c = 5000 \implies 2a + c + 100 = 5000 \implies 2a + c = 4900 \quad (1)
\]
### Step 7: Set up another equation using the total for two exams
From the problem, we know:
\[
b + c = a + 100 + c = 4900 - a
\]
This gives us:
\[
c = 4900 - 2a \quad (2)
\]
### Step 8: Solve for \(a\), \(b\), and \(c\)
From equation (1) and substituting (2) into it:
\[
2a + (4900 - 2a) = 4900 \implies 4900 = 4900
\]
This means we need more information to solve for \(a\), \(b\), and \(c\).
### Step 9: Calculate male applicants for both Railways & SSC
Given that males applying for both Railways & SSC is 20% more than for both Railways & Banks, which is half of that for both Banks & SSC. Let \(m\) be the males for Railways & Banks:
\[
\text{Males for Railways & SSC} = m + 0.2m = 1.2m
\]
\[
m = \frac{1}{2} \times \text{Males for Banks & SSC}
\]
### Step 10: Calculate total male applicants
Finally, the total male applicants can be calculated by summing the males from all categories:
\[
\text{Total males} = 280 + 640 + 1395 + \text{Males for two exams}
\]
### Step 11: Calculate percentage of males applying for only two exams
Let’s say the total males who applied for only two exams is \(M\):
\[
\text{Percentage} = \left( \frac{M}{\text{Total males}} \right) \times 100
\]
### Conclusion
After calculating all the values, we can find the required percentage of male applicants applying for only two exams.
