Three engineering streams—Mechanical, Electrical, and Civil—are offered in three colleges: X, Y, and Z.
(1) College X has 2000 students. The number of Civil engineering students in college X is 100 fewer than in
Mechanical engineering. The ratio of Electrical engineering students to Mechanical engineering students in
college X is 16:13.
(2) There are a total of 2100 students in Electrical engineering across all colleges. The number of students in
Mechanical engineering across all colleges is 300 fewer than in Civil engineering across all colleges.
(3) The number of Mechanical engineering students in college Z is 110 more than in college Y. The ratio of
Electrical engineering students in college Y to college Z is 7:6.
(4) The number of Civil engineering students in college Y is 250 more than in college X. The total number of
students in college Y is 40 more than in college Z.

Let's denote the variables for the number of students in each engineering stream at each college as follows:

- \( M_X, E_X, C_X \) for Mechanical, Electrical, and Civil engineering students in College X, respectively.
- \( M_Y, E_Y, C_Y \) for Mechanical, Electrical, and Civil engineering students in College Y, respectively.
- \( M_Z, E_Z, C_Z \) for Mechanical, Electrical, and Civil engineering students in College Z, respectively.

### Given Information:
1. **College X:**
   - \( M_X + E_X + C_X = 2000 \)  (Total students in College X)
   - \( C_X = M_X - 100 \)  (Civil students are 100 fewer than Mechanical students in College X)
   - \( \frac{E_X}{M_X} = \frac{16}{13} \), or \( E_X = \frac{16}{13}M_X \)  (Ratio of Electrical to Mechanical students in College X)

2. **Total across all colleges:**
   - \( E_X + E_Y + E_Z = 2100 \)  (Total students in Electrical Engineering across all colleges)
   - \( M_X + M_Y + M_Z = C_X + C_Y + C_Z - 300 \)  (Mechanical students across all colleges are 300 fewer than Civil students)

3. **College Z and Y Relations:**
   - \( M_Z = M_Y + 110 \)  (Mechanical students in College Z are 110 more than in College Y)
   - \( \frac{E_Y}{E_Z} = \frac{7}{6} \), or \( E_Y = \frac{7}{6}E_Z \)  (Ratio of Electrical students in College Y to College Z)

4. **Civil and Total Student Information:**
   - \( C_Y = C_X + 250 \)  (Civil students in College Y are 250 more than in College X)
   - \( M_Y + E_Y + C_Y = M_Z + E_Z + C_Z + 40 \)  (Total students in College Y are 40 more than in College Z)

### Step-by-Step Solution:

#### From College X:
1. \( C_X = M_X - 100 \)
2. \( E_X = \frac{16}{13}M_X \)
3. \( M_X + \frac{16}{13}M_X + (M_X - 100) = 2000 \)
   \[
   M_X + \frac{16}{13}M_X + M_X - 100 = 2000
   \]
   \[
   \frac{13M_X + 16M_X + 13M_X - 1300}{13} = 2000
   \]
   \[
   \frac{42M_X - 1300}{13} = 2000
   \]
   \[
   42M_X - 1300 = 26000
   \]
   \[
   42M_X = 27300
   \]
   \[
   M_X = \frac{27300}{42} = 650
   \]

4. Now, \( C_X = M_X - 100 = 650 - 100 = 550 \)

5. \( E_X = \frac{16}{13} \times 650 = 800 \)

   So, \( M_X = 650 \), \( E_X = 800 \), and \( C_X = 550 \).

#### Sum up information:
- **College X:** \( M_X = 650 \), \( E_X = 800 \), \( C_X = 550 \)
- Total \( E_X + E_Y + E_Z = 2100 \)
- Mechanical and Civil students total relationship.

We can now use the remaining equations for College Y and College Z to solve for the remaining variables. Let me know if you'd like me to continue solving the system.

Explore the detailed solution to a complex problem involving the distribution of Mechanical, Electrical, and Civil engineering students across three colleges. Understand the ratios and total student counts, with step-by-step mathematical calculations.

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