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 number of Engineering students who opted marketing is :

Options

A. 7850
B. 7500
C. 8850
D. none of these

Answer

To solve the problem step by step, we will break down the information given and use it to find the number of Engineering students who opted for Marketing.

### Step 1: Define Variables
Let:
- \( SC \) = Number of Science students
- \( CO \) = Number of Commerce students
- \( E \) = Number of Engineering students

Given:
- Total students = 16,000
- \( SC + CO + E = 16,000 \)

### Step 2: Establish Relationships
From the problem, we know:
1. The number of Science students is 166.66% of the Commerce students:
\[ SC = 1.6666 \times CO \]
2. The number of Engineering students is equal to the number of Science and Commerce students combined:
\[ E = SC + CO \]

### Step 3: Substitute and Solve
Substituting \( E \) in the total students equation:
\[ SC + CO + (SC + CO) = 16,000 \]
This simplifies to:
\[ 2(SC + CO) = 16,000 \implies SC + CO = 8,000 \]

### Step 4: Substitute for \( SC \)
Now substitute \( SC \) in terms of \( CO \):
\[ 1.6666 \times CO + CO = 8,000 \]
Combining terms:
\[ 2.6666 \times CO = 8,000 \]
Solving for \( CO \):
\[ CO = \frac{8,000}{2.6666} \approx 3,000 \]

### Step 5: Calculate \( SC \)
Now, substituting \( CO \) back to find \( SC \):
\[ SC = 1.6666 \times 3,000 \approx 5,000 \]

### Step 6: Calculate \( E \)
Now, substituting \( SC \) and \( CO \) to find \( E \):
\[ E = SC + CO = 5,000 + 3,000 = 8,000 \]

### Step 7: Determine Specializations
Now we need to find how many Engineering students opted for Marketing.

1. **Finance Students**:
- 20% of Science students opted for Finance:
\[ \text{Finance from Science} = 0.20 \times 5,000 = 1,000 \]
- Let \( x \) be the number of Commerce students who opted for Finance. The problem states that this is 16.66% less than the Finance students from Science:
\[ x = 1,000 \div (1 - 0.1666) \approx 1,200 \]
- Total Finance students:
\[ \text{Total Finance} = 1,000 + 1,200 = 2,200 \]
- Total Finance students is also given as 18% of total students:
\[ 0.18 \times 16,000 = 2,880 \]

2. **HR Students**:
- 32% of Science students opted for HR:
\[ \text{HR from Science} = 0.32 \times 5,000 = 1,600 \]
- Commerce students who opted for HR is 25% of total Finance students:
\[ \text{HR from Commerce} = 0.25 \times 2,200 = 550 \]
- Engineering students who opted for HR is 6.5% of total students:
\[ \text{HR from Engineering} = 0.065 \times 16,000 = 1,040 \]

### Step 8: Calculate Marketing Students
Now, we can find the number of Engineering students who opted for Marketing:
- Total Engineering students = 8,000
- Total Engineering students who opted for Finance = 680 (from previous calculations)
- Total Engineering students who opted for HR = 1,040

Thus, the number of Engineering students who opted for Marketing:
\[ \text{Marketing from Engineering} = E - (\text{Finance from Engineering} + \text{HR from Engineering}) = 8,000 - (680 + 1,040) = 8,000 - 1,720 = 6,280 \]

### Final Answer
The number of Engineering students who opted for Marketing is **6,280**.