Quick Solution

We can solve this puzzle by setting up equations and an inequality based on the given conditions for the number of questions and marks in each part.

1. Set Up Equations for Questions and Marks

• Let N_A, N_B, and N_C be the number of questions in parts A, B, and C.
• Let M_A, M_B, and M_C be the marks for each part.

From the problem statement, we can establish the following:

• Question Count: N_A + N_B + N_C = 100. Since we know N_B = 23, this simplifies to N_A + N_C = 77.
• Mark Calculation: M_A = 1 × N_A, M_B = 2 × 23 = 46, and M_C = 3 × N_C.
• Total Marks: M_Total = M_A + M_B + M_C = N_A + 46 + 3N_C.

2. Use the 60% Marks Condition

The most important condition is that the marks for Part A must be at least 60% of the total marks for the entire exam:

M_A ≥ 0.60 × M_Total

Substituting our expressions for marks and questions:

N_A ≥ 0.6 × (N_A + 46 + 3N_C)

To solve this, we'll use our first equation (N_A = 77 - N_C) to get an inequality with only one variable, N_C:

(77 - N_C) ≥ 0.6 × ( (77 - N_C) + 46 + 3N_C )
77 - N_C ≥ 0.6 × (123 + 2N_C)
77 - N_C ≥ 73.8 + 1.2N_C
77 - 73.8 ≥ 1.2N_C + N_C
3.2 ≥ 2.2N_C
N_C ≤ 3.2 / 2.2
N_C ≤ 1.45...

3. Find the Final Answer

- Our calculation shows that the number of questions in Part C (N_C) must be less than or equal to approximately 1.45.
- The problem also states that "each part should contain at least one question," so N_C must be 1 or more.
- The only integer that satisfies both conditions (1 ≤ N_C ≤ 1.45...) is 1.

Final Answer: There must be 1 question in part C.