An Examination consists of 50 questions (5
marks + 1 marks + 4 marks), and
maximum marks for examination is 140.
Number of 5 marks, 4 marks and 1-mark
questions are Z, Y and M respectively (Z < Y
< M, all are natural numbers.
A = If number of 5 marks questions are at
least 50% of 4 marks questions, then find
the number of one mark question.
B = Find maximum possible number of 1
mark question in examination.

To solve this problem, let's break it down into parts.

### Given:
1. **Total questions:** 50
2. **Total marks:** 140
3. **Number of 5 marks questions:** \( Z \)
4. **Number of 4 marks questions:** \( Y \)
5. **Number of 1 mark questions:** \( M \)

### Part A:
We are asked to find \( M \) (number of 1-mark questions) given that \( Z \geq \frac{Y}{2} \).

### Part B:
We need to find the maximum possible value of \( M \) subject to the constraints.

### Equations:
The total number of questions is given by:
\[
Z + Y + M = 50
\]
The total number of marks is given by:
\[
5Z + 4Y + M = 140
\]
Also, \( Z < Y < M \), and \( Z \geq \frac{Y}{2} \).

### Solving the equations:
We have two main equations:
1. \( Z + Y + M = 50 \)
2. \( 5Z + 4Y + M = 140 \)

Subtracting the first equation from the second:
\[
(5Z + 4Y + M) - (Z + Y + M) = 140 - 50
\]
\[
4Z + 3Y = 90 \quad \text{(Equation 3)}
\]

#### Express \( Y \) in terms of \( Z \) from Equation 3:
\[
Y = \frac{90 - 4Z}{3}
\]

This must be a natural number, so \( 90 - 4Z \) must be divisible by 3. For \( Z \) to satisfy the conditions, \( Y \) must also satisfy the condition \( Z < Y < M \).

### Part A: Finding \( M \) when \( Z \geq \frac{Y}{2} \)
\[
Z \geq \frac{Y}{2} \implies 2Z \geq Y \quad \text{and} \quad Y = \frac{90 - 4Z}{3}
\]

### Part B: Maximum \( M \)
We want to maximize \( M \) under the constraints \( Z < Y < M \).

### Calculation:

Let's try different values of \( Z \) and solve for \( Y \) and \( M \).

#### Case 1: \( Z = 9 \)
\[
Y = \frac{90 - 4 \times 9}{3} = \frac{90 - 36}{3} = \frac{54}{3} = 18
\]
\[
M = 50 - Z - Y = 50 - 9 - 18 = 23
\]
Checking the marks:
\[
5Z + 4Y + M = 5 \times 9 + 4 \times 18 + 23 = 45 + 72 + 23 = 140
\]
This satisfies all conditions: \( Z < Y < M \) and \( Z \geq \frac{Y}{2} \). 

Thus, the maximum possible value of \( M \) is **23**.

### Final Answer:
- **Part A:** The number of one mark questions \( M \) is **23**.
- **Part B:** The maximum possible number of one mark questions \( M \) is **23**.

Would you like to dive into more details or have any questions? Here are some related questions to explore:

1. What would happen if the total marks were different?
2. Can you determine the possible values for \( Z \) and \( Y \)?
3. How does the condition \( Z \geq \frac{Y}{2} \) limit the solution?
4. What would be the scenario if \( Z \) equals \( Y \)?
5. Can we change the number of questions to have different marks and still keep the total marks the same?
6. What happens if the total number of questions changes to 60 instead of 50?
7. How does the relationship \( Z < Y < M \) affect the calculation?
8. Can this problem be solved using matrix methods or optimization techniques?

**Tip:** Always verify if the solution satisfies all the given conditions, especially the inequalities and totals.

Explore how to solve an examination problem with specific numbers of 5, 4, and 1 mark questions. Learn about equations and inequalities in this high school level math problem.

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