Comprehension
Alex Company has its o ce at the third oor of a multi - storied building in Mumbai. There are 5 rooms to be allotted to 5 managers (designated M1 to M5), each of whom will occupy one room. Each room has its own advantages and disadvantages. Some have the sea view, while others are closer to either the lift or the dining room, while some are more spacious. Each of the ve managers was asked to rank the room preferences amongst the rooms 301, 302, 303, 304 and 305. Their preferences were recorded as follows:
Table:
| Preference | M1 | M2 | M3 | M4 | M5 |
|---|---|---|---|---|---|
| 1st | 302 | 302 | 303 | 302 | 301 |
| 2nd | 303 | 304 | 301 | 305 | 302 |
| 3rd | 304 | 305 | 304 | 304 | 305 |
| 4th | 301 | 305 | 303 | ||
| 5th | 302 |
Question: 1
How many managers would get their rooms as per their best preference?
How many managers would get their rooms as per their best preference?
Show Hint
In allocation problems, first resolve conflicts by spotting rooms with only one claimant. Then handle conflicts systematically to maximize satisfaction.
Updated On: Aug 23, 2025
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The Correct Option is C
Solution and Explanation
Step 1: Recall the preference table.
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
Preference &
M1 &
M2 &
M3 &
M4 &
M5
\hline 1^\text{st} & 302 & 302 & 303 & 302 & 301
2^\text{nd} & 303 & 304 & 301 & 305 & 302
3^\text{rd} & 304 & 305 & 304 & 304 & 305
4^\text{th} & 301 & 305 & 305 & 303 &
5^\text{th} & & & 302 & &
\hline \end{array} \]
Step 2: Identify unique allocations by first preference.
- Room 301: preferred only by M5. Hence, M5 surely gets Room 301. ✓
- Room 303: preferred only by M3. Hence, M3 surely gets Room 303. ✓
- Room 302: preferred by M1, M2, M4 (conflict; only one can get).
Step 3: Assign conflict-free rooms.
- M5 gets Room 301 (best choice). ✓
- M3 gets Room 303 (best choice). ✓
Step 4: Handle Room 302 conflict.
Among M1, M2, and M4, only one can be allotted Room 302 as per best choice. ✓
Step 5: Count managers satisfied with best choice.
Thus, M3, M5, and one among \{M1, M2, M4\} get their best choice = total 3. \[ \boxed{3} \]
Preference &
M1 &
M2 &
M3 &
M4 &
M5
\hline 1^\text{st} & 302 & 302 & 303 & 302 & 301
2^\text{nd} & 303 & 304 & 301 & 305 & 302
3^\text{rd} & 304 & 305 & 304 & 304 & 305
4^\text{th} & 301 & 305 & 305 & 303 &
5^\text{th} & & & 302 & &
\hline \end{array} \]
Step 2: Identify unique allocations by first preference.
- Room 301: preferred only by M5. Hence, M5 surely gets Room 301. ✓
- Room 303: preferred only by M3. Hence, M3 surely gets Room 303. ✓
- Room 302: preferred by M1, M2, M4 (conflict; only one can get).
Step 3: Assign conflict-free rooms.
- M5 gets Room 301 (best choice). ✓
- M3 gets Room 303 (best choice). ✓
Step 4: Handle Room 302 conflict.
Among M1, M2, and M4, only one can be allotted Room 302 as per best choice. ✓
Step 5: Count managers satisfied with best choice.
Thus, M3, M5, and one among \{M1, M2, M4\} get their best choice = total 3. \[ \boxed{3} \]
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Question: 2
If manager X gets his/her 1st choice, then his/her preference ranking is 1 and so on. Management decided to allot rooms so that the sum of the preference ranking of all the managers is minimized. What is the total preference ranking for the rooms allotted to all the managers?
If manager X gets his/her 1st choice, then his/her preference ranking is 1 and so on. Management decided to allot rooms so that the sum of the preference ranking of all the managers is minimized. What is the total preference ranking for the rooms allotted to all the managers?
Show Hint
When minimizing total preference ranking, treat allocation as an optimization problem. First assign rooms with unique claimants, then resolve conflicts by distributing based on next-best available choices.
Updated On: Aug 23, 2025
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The Correct Option is C
Solution and Explanation
Step 1: Start with certain allocations.
- M5 gets Room 301 (1st preference, rank = 1).
- M3 gets Room 303 (1st preference, rank = 1).
Step 2: Deal with Room 302 conflict (M1, M2, M4).
Only one of them can get Room 302 as 1st preference. Suppose M1 gets it (rank = 1).
Now, M2 and M4 cannot get 302; they move to next available preferences.
Step 3: Assign second preferences.
- M2’s 2nd choice is 304 → assign Room 304 (rank = 2).
- M4’s 2nd choice is 305 → assign Room 305 (rank = 2).
Step 4: Verify uniqueness of allocations.
Now each room 301–305 is allocated uniquely:
M1 → 302, M2 → 304, M3 → 303, M4 → 305, M5 → 301.
Step 5: Compute total ranking sum.
M1 (1) + M2 (2) + M3 (1) + M4 (2) + M5 (1) = 7. \[ \boxed{7} \]
- M5 gets Room 301 (1st preference, rank = 1).
- M3 gets Room 303 (1st preference, rank = 1).
Step 2: Deal with Room 302 conflict (M1, M2, M4).
Only one of them can get Room 302 as 1st preference. Suppose M1 gets it (rank = 1).
Now, M2 and M4 cannot get 302; they move to next available preferences.
Step 3: Assign second preferences.
- M2’s 2nd choice is 304 → assign Room 304 (rank = 2).
- M4’s 2nd choice is 305 → assign Room 305 (rank = 2).
Step 4: Verify uniqueness of allocations.
Now each room 301–305 is allocated uniquely:
M1 → 302, M2 → 304, M3 → 303, M4 → 305, M5 → 301.
Step 5: Compute total ranking sum.
M1 (1) + M2 (2) + M3 (1) + M4 (2) + M5 (1) = 7. \[ \boxed{7} \]
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Question: 3
Suppose that Manager M2 decides not to join the new zonal office and Manager M6 takes his place. Manager M6 has the following preference ranking in decreasing order: 301, 302, 303, 304. In this case what would be the sum of the preference ranking allotted to all five managers?
Suppose that Manager M2 decides not to join the new zonal office and Manager M6 takes his place. Manager M6 has the following preference ranking in decreasing order: 301, 302, 303, 304. In this case what would be the sum of the preference ranking allotted to all five managers?
Show Hint
When preference conflicts occur, allocate unique rooms first, then resolve overlaps by distributing choices so that the total ranking sum is minimized.
Updated On: Dec 6, 2025
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The Correct Option is
Solution and Explanation
Step 1: Write down the new preference table (M1, M3, M4, M5, M6).
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
Preference &
M1 &
M3 &
M4 &
M5 &
M6
\hline 1^\text{st} & 302 & 303 & 302 & 301 & 301
2^\text{nd} & 303 & 301 & 305 & 302 & 302
3^\text{rd} & 304 & 304 & 304 & 305 & 303
4^\text{th} & 301 & 305 & 303 & & 304
5^\text{th} & 302 & & & &
\hline \end{array} \]
Step 2: Handle clear allocations.
- M3’s 1st preference is Room 303 → no conflict (allocate 303). Rank = 1. ✓
Step 3: Deal with conflicts.
- Room 301 is preferred 1st by M5 and M6. Only one of them can get 301.
- Room 302 is preferred 1st by M1 and M4 (and also 2nd by M6). Only one of M1/M4 can get it.
Step 4: Assign optimal allocation to minimize total.
- Assign M6 → Room 301 (1st preference). Rank = 1. ✓
- Assign M5 → Room 305 (3rd preference, since 301 and 302 are taken). Rank = 3. ✓
- Assign M1 → Room 302 (1st preference). Rank = 1. ✓
- Assign M4 → Room 304 (3rd preference, since 302 taken). Rank = 3. ✓
Step 5: Add up preference rankings.
M1 (1) + M3 (1) + M4 (3) + M5 (3) + M6 (1) = 9. \[ \boxed{9} \]
Preference &
M1 &
M3 &
M4 &
M5 &
M6
\hline 1^\text{st} & 302 & 303 & 302 & 301 & 301
2^\text{nd} & 303 & 301 & 305 & 302 & 302
3^\text{rd} & 304 & 304 & 304 & 305 & 303
4^\text{th} & 301 & 305 & 303 & & 304
5^\text{th} & 302 & & & &
\hline \end{array} \]
Step 2: Handle clear allocations.
- M3’s 1st preference is Room 303 → no conflict (allocate 303). Rank = 1. ✓
Step 3: Deal with conflicts.
- Room 301 is preferred 1st by M5 and M6. Only one of them can get 301.
- Room 302 is preferred 1st by M1 and M4 (and also 2nd by M6). Only one of M1/M4 can get it.
Step 4: Assign optimal allocation to minimize total.
- Assign M6 → Room 301 (1st preference). Rank = 1. ✓
- Assign M5 → Room 305 (3rd preference, since 301 and 302 are taken). Rank = 3. ✓
- Assign M1 → Room 302 (1st preference). Rank = 1. ✓
- Assign M4 → Room 304 (3rd preference, since 302 taken). Rank = 3. ✓
Step 5: Add up preference rankings.
M1 (1) + M3 (1) + M4 (3) + M5 (3) + M6 (1) = 9. \[ \boxed{9} \]
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