Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there are three persons ahead of C and 21 behind A, then what could be the minimum number of persons in the queue?
Step-by-step Solution:
Let's carefully analyze the correct solution.
Given Information:
1. A and B: 5 persons between them → B is 6 places after A.
2. B and C: 8 persons between them → C is 9 places after B.
3. C has 3 persons ahead → C is in 4th position.
4. A has 21 persons behind → Total persons = Position of A + 21.
Possible Arrangements:
Since C is **4th in the queue**, we consider valid arrangements:
Case 1: C → B → A
- C at position 4
.
- B at position = 4 + 9 = 13
.
- A at position = 13 - 6 = 7
.
- There are 21 persons behind A
.
- Total persons = 7 (A’s position) + 21 = 28
Thus, the correct answer is:
B. 28 ✅