In the IGNOU (Indira Gandhi National Open University) there are total 16,000 students pursuing MBA, which offers the specialization only in Finance, HR and Marketing IGNOU accepts only Science, Commerce and Engineering Students for the two years course of MBA. The number of science. students is 166.66% of the Commerce students. Number of Engineering students is equal to the number of Science and Commerce students together. Each student can specialize in only one of the marketing , HR and Finance, 20% of Science students opted the Finance, which is 16.66% less than the no. of commerce students who opted Finance. The total Finance students is equal to 18% the total strength of the MBA students. 32% of Science students opted HR. Commerce students who opted HR is equal to 25% of total students specializing in Finance and Engineering students equal to 6.5% of the total strength of the MBA students opted HR.
The most preferable course among the MBA students is :

Options

A. Finance
B. HR
C. marketing
D. I.T. and systems

Answer

To solve the problem step by step, we will break down the information given in the question and perform calculations accordingly.

### Step 1: Determine the number of students in each category
We know that the total number of MBA students is 16,000. Let:
- \( C \) = Number of Commerce students
- \( S \) = Number of Science students
- \( E \) = Number of Engineering students

According to the problem:
1. The number of Science students \( S = 166.67\% \) of Commerce students \( C \).
\[ S = \frac{166.67}{100} \times C = \frac{5}{3}C \]
2. The number of Engineering students \( E \) is equal to the sum of Science and Commerce students:
\[ E = S + C \]

From the total, we have:
\[ S + C + E = 16000 \]
Substituting \( E \) in the equation:
\[ S + C + (S + C) = 16000 \]
\[ 2S + 2C = 16000 \]
Dividing by 2:
\[ S + C = 8000 \]

### Step 2: Substitute \( S \) in terms of \( C \)
Substituting \( S = \frac{5}{3}C \) into \( S + C = 8000 \):
\[ \frac{5}{3}C + C = 8000 \]
Combining the terms:
\[ \frac{5C + 3C}{3} = 8000 \]
\[ \frac{8C}{3} = 8000 \]
Multiplying both sides by 3:
\[ 8C = 24000 \]
Dividing by 8:
\[ C = 3000 \]

### Step 3: Calculate the number of Science and Engineering students
Now substituting \( C \) back to find \( S \):
\[ S = \frac{5}{3} \times 3000 = 5000 \]
Now, calculate \( E \):
\[ E = S + C = 5000 + 3000 = 8000 \]

### Step 4: Calculate the number of students opting for each specialization
1. **Finance:**
- 20% of Science students opted for Finance:
\[ \text{Finance from Science} = 0.20 \times 5000 = 1000 \]
- Let \( x \) be the number of Commerce students who opted for Finance. According to the problem, this is 16.66% less than the number of Finance students from Science:
\[ x = 1000 + 0.1666 \times x \implies x = 1000 + 0.1666x \]
Rearranging gives:
\[ 0.8334x = 1000 \implies x \approx 1200 \]

2. **Total Finance Students:**
- Total Finance students = 18% of total MBA students:
\[ \text{Total Finance} = 0.18 \times 16000 = 2880 \]
- Finance from Engineering:
\[ \text{Finance from Engineering} = 2880 - (1000 + 1200) = 680 \]

3. **HR:**
- 32% of Science students opted for HR:
\[ \text{HR from Science} = 0.32 \times 5000 = 1600 \]
- HR from Commerce:
\[ \text{HR from Commerce} = 0.25 \times 2880 = 720 \]
- HR from Engineering:
\[ \text{HR from Engineering} = 0.065 \times 16000 = 1040 \]

4. **Marketing:**
- Remaining students in each category:
- **Science:** \( 5000 - (1000 + 1600) = 2400 \)
- **Commerce:** \( 3000 - (1200 + 720) = 1080 \)
- **Engineering:** \( 8000 - (680 + 1040) = 6180 \)

### Step 5: Summary of Specializations
- **Finance:**
- Science: 1000
- Commerce: 1200
- Engineering: 680
- Total: 2880

- **HR:**
- Science: 1600
- Commerce: 720
- Engineering: 1040
- Total: 3360

- **Marketing:**
- Science: 2400
- Commerce: 1080
- Engineering: 6180
- Total: 9660

### Conclusion
The most preferable course among the MBA students is **Marketing**, as it has the highest number of students opting for it.