Two buses, Bus X and Bus Y, travel through five stops in order: A, B, C, D, and E. At each stop, a certain number of passengers may board or alight from each bus. Initially, both buses start empty at Stop A. The number of passengers in each bus changes throughout the journey according to the facts given below:
- At Stop A, 8 passengers board Bus X, and 5 passengers board Bus Y.
- At Stop B, 3 passengers alight from Bus X, and 4 more board it. On Bus Y, 2 passengers alight, and no one boards.
- At Stop C, Bus X is recorded to have exactly 9 passengers on board after all boarding and alighting are done. Bus Y has 3 passengers board at Stop C, and no one alights.
- At Stop D, 5 passengers alight from Bus X. The number of passengers boarding Bus Y at this stop is equal to the number of passengers who alighted from Bus X at Stop D.
- At Stop E, all remaining passengers in both buses alight, and both buses become empty.
- The total number of passengers who boarded Bus Y across all stops is known to be 13.
Questions based on the above instructions Scheduling Puzzle in NMAT 2024 Actual Questions
1. How many passengers boarded Bus X at Stop C?
A) 3
B) 4
C) 5
D) Cannot be determined
E) 2
2. How many passengers alighted from Bus Y at Stop E?
A) 6
B) 9
C) 11
D) 14
E) 13
3. How many passengers were in Bus Y before reaching Stop D?
A) 3
B) 6
C) 9
D) 11
E) Cannot be determined
4. What is the total number of passengers who alighted from Bus X over all five stops?
A) 10
B) 12
C) 14
D) Cannot be determined
E) 9
5. How many passengers boarded Bus Y at Stop D?
A) 3
B) 4
C) 5
D) 6
E) Cannot be determined
Answers and Explanation Scheduling Puzzle in NMAT 2024 Actual Questions
Step 1: Both buses start empty. At Stop A, 8 passengers board Bus X and 5 board Bus Y → Bus X = 8, Bus Y = 5.
Step 2: At Stop B, 3 passengers alight and 4 board Bus X → Bus X = 8 – 3 + 4 = 9. Bus Y has 2 alight and no one board → Bus Y = 5 – 2 = 3.
Step 3: After Stop C, Bus X is stated to have 9 passengers → net change at C is zero (x board and x alight). Bus Y has 3 passengers board and no alighting → Bus Y = 3 + 3 = 6.
Step 4: At Stop D, 5 passengers alight from Bus X → Bus X = 9 – 5 = 4. The same number (5) board Bus Y → Bus Y = 6 + 5 = 11.
Step 5: At Stop E, all remaining passengers alight → Bus X = 4 – 4 = 0, Bus Y = 11 – 11 = 0.
| Stop | Bus X: Board | Bus X: Alight | Bus X: Total After Stop | Bus Y: Board | Bus Y: Alight | Bus Y: Total After Stop |
|---|---|---|---|---|---|---|
| Start | 0 | 0 | 0 | 0 | 0 | 0 |
| Stop A | 8 | 0 | 8 | 5 | 0 | 5 |
| Stop B | 4 | 3 | 9 | 0 | 2 | 3 |
| Stop C | ? | ? | 9 | 3 | 0 | 6 |
| Stop D | 0 | 5 | 4 | 5 | 0 | 11 |
| Stop E | 0 | 4 | 0 | 0 | 11 | 0 |
Answer: D) Cannot be determined – Bus X had 9 passengers after Stop B and also 9 after Stop C, but the exact number boarding or alighting at Stop C is not provided.
Answer: C) 11 – Bus Y had 6 passengers after Stop C and 5 more boarded at Stop D, making 11. All 11 alighted at Stop E.
Answer: B) 6 – Before reaching Stop D, Bus Y had 3 passengers after Stop B and 3 boarded at Stop C, totaling 6.
Answer: D) Cannot be determined – Bus X alighted 3 at B, 5 at D, and 4 at E, but alighting at Stop C is unknown, so total cannot be calculated.
Answer: C) 5 – At Stop D, Bus Y boarded as many passengers as alighted from Bus X, which was 5.
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