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The problem provides information about three students—Vidhi, Sumit, and Varun—and their marks based on a percentage of maximum marks. Let's extract the relevant information and translate it into mathematical terms to solve for \(x\).

### Information given:
1. **Vidhi's marks**:
   - Vidhi scores \((x - 14)\)% of the maximum marks.
   - She fails by 25 marks, meaning her score is 25 marks less than the passing marks.

2. **Sumit's marks**:
   - Sumit scores \((x + 12)\)% of the maximum marks.
   - He scores 40 marks more than the passing marks.

3. **Varun's marks**:
   - Varun scores 42% of the maximum marks.
   - He gets 5 marks more than the passing marks.

### Let the maximum marks be \( M \), and let the passing marks be \( P \).

### Mathematical expressions:
1. Vidhi's marks = \(\frac{(x - 14)}{100} \times M = P - 25\)
2. Sumit's marks = \(\frac{(x + 12)}{100} \times M = P + 40\)
3. Varun's marks = \(\frac{42}{100} \times M = P + 5\)

Now, we have the system of three equations:
1. \[\frac{(x - 14)}{100} \times M = P - 25\]
2. \[\frac{(x + 12)}{100} \times M = P + 40\]
3. \[\frac{42}{100} \times M = P + 5\]

We can solve this system of equations to find \(x\).

### Step-by-step solution:

From equation (3):
\[
P = \frac{42}{100} \times M - 5
\]
Substitute this expression for \(P\) into equations (1) and (2):

**Equation (1):**
\[
\frac{(x - 14)}{100} \times M = \left(\frac{42}{100} \times M - 5\right) - 25
\]
Simplifying:
\[
\frac{(x - 14)}{100} \times M = \frac{42}{100} \times M - 30
\]
Multiply both sides by 100 to eliminate the denominator:
\[
(x - 14) \times M = 42 \times M - 3000
\]
\[
x \times M - 14M = 42M - 3000
\]
\[
x \times M = 56M - 3000 \quad \text{(Equation A)}
\]

**Equation (2):**
\[
\frac{(x + 12)}{100} \times M = \left(\frac{42}{100} \times M - 5\right) + 40
\]
Simplifying:
\[
\frac{(x + 12)}{100} \times M = \frac{42}{100} \times M + 35
\]
Multiply both sides by 100:
\[
(x + 12) \times M = 42 \times M + 3500
\]
\[
x \times M + 12M = 42M + 3500
\]
\[
x \times M = 30M + 3500 \quad \text{(Equation B)}
\]

### Solving for \(x\):

From Equation A and Equation B:
1. \(x \times M = 56M - 3000\)
2. \(x \times M = 30M + 3500\)

Equating the two expressions for \(x \times M\):
\[
56M - 3000 = 30M + 3500
\]
Simplifying:
\[
56M - 30M = 3500 + 3000
\]
\[
26M = 6500
\]
\[
M = 250
\]

Now, substitute \(M = 250\) into one of the equations to find \(x\). Using Equation A:
\[
x \times 250 = 56 \times 250 - 3000
\]
\[
x \times 250 = 14000 - 3000
\]
\[
x \times 250 = 11000
\]
\[
x = \frac{11000}{250} = 44
\]

### Final Answer:
The value of \(x\) is \(44\).

Would you like further clarification or additional details on any step?

### Relative questions:
1. What happens if \(x\) were a different value? Would the system of equations still hold?
2. How would the solution change if the maximum marks were different?
3. Can we set up a similar problem but with a different set of students' scores?
4. How do we verify the correctness of the value \(x = 44\)?
5. Could we solve this problem using a different mathematical method?

### Tip:
Always double-check your substituted values to ensure consistent solutions across all equations in a system!

Vidhi gets (x-14)% of maximum marks and failed by 25 marks, whereas Sumit gets (x+12)% of maximum marks and scored 40 marks more than minimum passing marks. If Varun gets 42% marks and gets 5 more marks than minimum passing marks, then find the value of 'x'?

Solve for x in a system of equations where students Vidhi, Sumit, and Varun score percentages of maximum marks and have differences from passing marks. This problem covers algebraic concepts and systems of equations, ideal for students in Grades 8-10.

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