CAT 2024 Slot 3 · DILR · Set 5: Air-Conditioner and Room Temperature
The full set and all 5 questions (Q18–Q22) from CAT 2024 Slot 3, with answers and solutions.
The air-conditioner (AC) in a large room can be operated either in REGULAR mode or in POWER mode to reduce the temperature.
If the AC operates in REGULAR mode, then it brings down the temperature inside the room (called inside temperature) at a constant rate to the set temperature in 1 hour. If it operates in POWER mode, then this is achieved in 30 minutes.
If the AC is switched off, then the inside temperature rises at a constant rate so as to reach the temperature outside at the time of switching off in 1 hour.
The temperature outside has been falling at a constant rate from 7 pm onward until 3 am on a particular night. The following graph shows the inside temperature between 11 pm (23:00) and 2 am (2:00) that night.
The following facts are known about the AC operation that night.
• The AC was turned on for the first time that night at 11 pm (23:00).
• The AC setting was changed (including turning it on/off, and/or setting different temperatures) only at the beginning of the hour or at 30 minutes after the hour.
• The AC was used in POWER mode for longer duration than in REGULAR mode during this 3-hour period.
Question 18 · Reasoning Based DI
- Acannot be determined
- B2
- C0
- D1
Answer & solution
Answer: Option B
Solving the set
Step 1 – Read the graph and decide when the AC was on or off. A fall in inside temperature means the AC is cooling; a rise means it is off.
Interval | Inside temperature | Change | AC status | What it tells us |
23:00–23:30 | 38 → 32 | −6 | On | Cooling |
23:30–0:00 | 32 → 26 | −6 | On | Cooling |
0:00–0:30 | 26 → 31 | +5 | Off (at 0:00) | Rises 10 per hour ⇒ outside at 0:00 = 26 + 10 = 36 |
0:30–1:00 | 31 → 26 | −5 | On (at 0:30) | Cooling |
1:00–1:30 | 26 → 30 | +4 | Off (at 1:00) | Rises 8 per hour ⇒ outside at 1:00 = 26 + 8 = 34 |
1:30–2:00 | 30 → 28 | −2 | On (at 1:30) | Cooling |
Step 2 – Outside temperature. When the AC is switched off, the inside temperature rises to the outside temperature (at the moment of switching off) in exactly 1 hour. From 0:00: 26 → 31 in 30 minutes, so the full hour would give 26 + 10 = 36 = outside at 0:00. From 1:00: 26 → 30 in 30 minutes, so outside at 1:00 = 26 + 8 = 34. The outside temperature falls at a constant rate: 2° per hour (1° per 30 minutes).
Time | 21:00 | 22:00 | 23:00 | 23:30 | 0:00 | 0:30 | 1:00 | 1:30 | 2:00 |
Outside | 42 | 40 | 38 | 37 | 36 | 35 | 34 | 33 | 32 |
Inside | — | — | 38 | 32 | 26 | 31 | 26 | 30 | 28 |
Outside − Inside | — | — | 0 | 5 | 10 | 4 | 8 | 3 | 4 |
Step 3 – Modes. In REGULAR mode the drop in 30 minutes is half the gap to the set temperature; in POWER mode the set temperature is reached in 30 minutes. The AC runs for 2 hours in total (23:00–0:00, 0:30–1:00, 1:30–2:00), and POWER must be used for more than 1 hour.
• If 23:00–0:00 were one REGULAR setting (38 → 26 in 1 hour), POWER could be at most 1 hour (the other two half-hours). Not allowed.
• A single POWER setting cannot produce the straight 38 → 26 line over a whole hour, because POWER reaches its set temperature in 30 minutes and then holds it.
• So the setting must be changed at 23:30 (for example, POWER to 32 at 23:00, then POWER to 26 at 23:30).
Time | Event | Counts as 'turned on or setting altered'? |
23:00 | AC turned on (outside the 11:01–1:59 window) | Not counted |
23:30 | Setting changed (needed to get enough POWER time) | Yes |
0:00 | AC turned off | No (switch-off) |
0:30 | AC turned on | Yes |
1:00 | AC turned off | No (switch-off) |
1:30 | AC turned on | Yes |
This question
From Step 1, the AC was switched off at 0:00 and at 1:00 (the only two rising intervals). Both lie between 11:01 pm and 1:59 am.
Hence, the answer is option 2.
Question 19 · Reasoning Based DI
Answer & solution
Answer: 34
Solving the set
Step 1 – Read the graph and decide when the AC was on or off. A fall in inside temperature means the AC is cooling; a rise means it is off.
Interval | Inside temperature | Change | AC status | What it tells us |
23:00–23:30 | 38 → 32 | −6 | On | Cooling |
23:30–0:00 | 32 → 26 | −6 | On | Cooling |
0:00–0:30 | 26 → 31 | +5 | Off (at 0:00) | Rises 10 per hour ⇒ outside at 0:00 = 26 + 10 = 36 |
0:30–1:00 | 31 → 26 | −5 | On (at 0:30) | Cooling |
1:00–1:30 | 26 → 30 | +4 | Off (at 1:00) | Rises 8 per hour ⇒ outside at 1:00 = 26 + 8 = 34 |
1:30–2:00 | 30 → 28 | −2 | On (at 1:30) | Cooling |
Step 2 – Outside temperature. When the AC is switched off, the inside temperature rises to the outside temperature (at the moment of switching off) in exactly 1 hour. From 0:00: 26 → 31 in 30 minutes, so the full hour would give 26 + 10 = 36 = outside at 0:00. From 1:00: 26 → 30 in 30 minutes, so outside at 1:00 = 26 + 8 = 34. The outside temperature falls at a constant rate: 2° per hour (1° per 30 minutes).
Time | 21:00 | 22:00 | 23:00 | 23:30 | 0:00 | 0:30 | 1:00 | 1:30 | 2:00 |
Outside | 42 | 40 | 38 | 37 | 36 | 35 | 34 | 33 | 32 |
Inside | — | — | 38 | 32 | 26 | 31 | 26 | 30 | 28 |
Outside − Inside | — | — | 0 | 5 | 10 | 4 | 8 | 3 | 4 |
Step 3 – Modes. In REGULAR mode the drop in 30 minutes is half the gap to the set temperature; in POWER mode the set temperature is reached in 30 minutes. The AC runs for 2 hours in total (23:00–0:00, 0:30–1:00, 1:30–2:00), and POWER must be used for more than 1 hour.
• If 23:00–0:00 were one REGULAR setting (38 → 26 in 1 hour), POWER could be at most 1 hour (the other two half-hours). Not allowed.
• A single POWER setting cannot produce the straight 38 → 26 line over a whole hour, because POWER reaches its set temperature in 30 minutes and then holds it.
• So the setting must be changed at 23:30 (for example, POWER to 32 at 23:00, then POWER to 26 at 23:30).
Time | Event | Counts as 'turned on or setting altered'? |
23:00 | AC turned on (outside the 11:01–1:59 window) | Not counted |
23:30 | Setting changed (needed to get enough POWER time) | Yes |
0:00 | AC turned off | No (switch-off) |
0:30 | AC turned on | Yes |
1:00 | AC turned off | No (switch-off) |
1:30 | AC turned on | Yes |
This question
From Step 2, the outside temperature at 1 am was 26 + 2 × 4 = 34 °C.
Hence, the answer is 34.
Question 20 · Reasoning Based DI
Answer & solution
Answer: 42
Solving the set
Step 1 – Read the graph and decide when the AC was on or off. A fall in inside temperature means the AC is cooling; a rise means it is off.
Interval | Inside temperature | Change | AC status | What it tells us |
23:00–23:30 | 38 → 32 | −6 | On | Cooling |
23:30–0:00 | 32 → 26 | −6 | On | Cooling |
0:00–0:30 | 26 → 31 | +5 | Off (at 0:00) | Rises 10 per hour ⇒ outside at 0:00 = 26 + 10 = 36 |
0:30–1:00 | 31 → 26 | −5 | On (at 0:30) | Cooling |
1:00–1:30 | 26 → 30 | +4 | Off (at 1:00) | Rises 8 per hour ⇒ outside at 1:00 = 26 + 8 = 34 |
1:30–2:00 | 30 → 28 | −2 | On (at 1:30) | Cooling |
Step 2 – Outside temperature. When the AC is switched off, the inside temperature rises to the outside temperature (at the moment of switching off) in exactly 1 hour. From 0:00: 26 → 31 in 30 minutes, so the full hour would give 26 + 10 = 36 = outside at 0:00. From 1:00: 26 → 30 in 30 minutes, so outside at 1:00 = 26 + 8 = 34. The outside temperature falls at a constant rate: 2° per hour (1° per 30 minutes).
Time | 21:00 | 22:00 | 23:00 | 23:30 | 0:00 | 0:30 | 1:00 | 1:30 | 2:00 |
Outside | 42 | 40 | 38 | 37 | 36 | 35 | 34 | 33 | 32 |
Inside | — | — | 38 | 32 | 26 | 31 | 26 | 30 | 28 |
Outside − Inside | — | — | 0 | 5 | 10 | 4 | 8 | 3 | 4 |
Step 3 – Modes. In REGULAR mode the drop in 30 minutes is half the gap to the set temperature; in POWER mode the set temperature is reached in 30 minutes. The AC runs for 2 hours in total (23:00–0:00, 0:30–1:00, 1:30–2:00), and POWER must be used for more than 1 hour.
• If 23:00–0:00 were one REGULAR setting (38 → 26 in 1 hour), POWER could be at most 1 hour (the other two half-hours). Not allowed.
• A single POWER setting cannot produce the straight 38 → 26 line over a whole hour, because POWER reaches its set temperature in 30 minutes and then holds it.
• So the setting must be changed at 23:30 (for example, POWER to 32 at 23:00, then POWER to 26 at 23:30).
Time | Event | Counts as 'turned on or setting altered'? |
23:00 | AC turned on (outside the 11:01–1:59 window) | Not counted |
23:30 | Setting changed (needed to get enough POWER time) | Yes |
0:00 | AC turned off | No (switch-off) |
0:30 | AC turned on | Yes |
1:00 | AC turned off | No (switch-off) |
1:30 | AC turned on | Yes |
This question
From Step 2, the outside temperature falls 2° per hour and is 36 °C at midnight. At 9 pm (3 hours earlier): 36 + 6 = 42 °C.
Hence, the answer is 42.
Question 21 · Reasoning Based DI
- AMore than 3
- BEither 2 or 3
- CExactly 2
- DExactly 3
Answer & solution
Answer: Option D
Solving the set
Step 1 – Read the graph and decide when the AC was on or off. A fall in inside temperature means the AC is cooling; a rise means it is off.
Interval | Inside temperature | Change | AC status | What it tells us |
23:00–23:30 | 38 → 32 | −6 | On | Cooling |
23:30–0:00 | 32 → 26 | −6 | On | Cooling |
0:00–0:30 | 26 → 31 | +5 | Off (at 0:00) | Rises 10 per hour ⇒ outside at 0:00 = 26 + 10 = 36 |
0:30–1:00 | 31 → 26 | −5 | On (at 0:30) | Cooling |
1:00–1:30 | 26 → 30 | +4 | Off (at 1:00) | Rises 8 per hour ⇒ outside at 1:00 = 26 + 8 = 34 |
1:30–2:00 | 30 → 28 | −2 | On (at 1:30) | Cooling |
Step 2 – Outside temperature. When the AC is switched off, the inside temperature rises to the outside temperature (at the moment of switching off) in exactly 1 hour. From 0:00: 26 → 31 in 30 minutes, so the full hour would give 26 + 10 = 36 = outside at 0:00. From 1:00: 26 → 30 in 30 minutes, so outside at 1:00 = 26 + 8 = 34. The outside temperature falls at a constant rate: 2° per hour (1° per 30 minutes).
Time | 21:00 | 22:00 | 23:00 | 23:30 | 0:00 | 0:30 | 1:00 | 1:30 | 2:00 |
Outside | 42 | 40 | 38 | 37 | 36 | 35 | 34 | 33 | 32 |
Inside | — | — | 38 | 32 | 26 | 31 | 26 | 30 | 28 |
Outside − Inside | — | — | 0 | 5 | 10 | 4 | 8 | 3 | 4 |
Step 3 – Modes. In REGULAR mode the drop in 30 minutes is half the gap to the set temperature; in POWER mode the set temperature is reached in 30 minutes. The AC runs for 2 hours in total (23:00–0:00, 0:30–1:00, 1:30–2:00), and POWER must be used for more than 1 hour.
• If 23:00–0:00 were one REGULAR setting (38 → 26 in 1 hour), POWER could be at most 1 hour (the other two half-hours). Not allowed.
• A single POWER setting cannot produce the straight 38 → 26 line over a whole hour, because POWER reaches its set temperature in 30 minutes and then holds it.
• So the setting must be changed at 23:30 (for example, POWER to 32 at 23:00, then POWER to 26 at 23:30).
Time | Event | Counts as 'turned on or setting altered'? |
23:00 | AC turned on (outside the 11:01–1:59 window) | Not counted |
23:30 | Setting changed (needed to get enough POWER time) | Yes |
0:00 | AC turned off | No (switch-off) |
0:30 | AC turned on | Yes |
1:00 | AC turned off | No (switch-off) |
1:30 | AC turned on | Yes |
This question
From Step 3, the events that turn the AC on or change its setting between 11:01 pm and 1:59 am are: the change at 23:30 (needed so that POWER runs more than 1 hour), turning on at 0:30, and turning on at 1:30. Switch-offs at 0:00 and 1:00 are not counted, and no other change is possible within the on-periods (changes only on the hour or half hour).
Exactly 3.
Hence, the answer is option 4.
Question 22 · Reasoning Based DI
Answer & solution
Answer: 10
Solving the set
Step 1 – Read the graph and decide when the AC was on or off. A fall in inside temperature means the AC is cooling; a rise means it is off.
Interval | Inside temperature | Change | AC status | What it tells us |
23:00–23:30 | 38 → 32 | −6 | On | Cooling |
23:30–0:00 | 32 → 26 | −6 | On | Cooling |
0:00–0:30 | 26 → 31 | +5 | Off (at 0:00) | Rises 10 per hour ⇒ outside at 0:00 = 26 + 10 = 36 |
0:30–1:00 | 31 → 26 | −5 | On (at 0:30) | Cooling |
1:00–1:30 | 26 → 30 | +4 | Off (at 1:00) | Rises 8 per hour ⇒ outside at 1:00 = 26 + 8 = 34 |
1:30–2:00 | 30 → 28 | −2 | On (at 1:30) | Cooling |
Step 2 – Outside temperature. When the AC is switched off, the inside temperature rises to the outside temperature (at the moment of switching off) in exactly 1 hour. From 0:00: 26 → 31 in 30 minutes, so the full hour would give 26 + 10 = 36 = outside at 0:00. From 1:00: 26 → 30 in 30 minutes, so outside at 1:00 = 26 + 8 = 34. The outside temperature falls at a constant rate: 2° per hour (1° per 30 minutes).
Time | 21:00 | 22:00 | 23:00 | 23:30 | 0:00 | 0:30 | 1:00 | 1:30 | 2:00 |
Outside | 42 | 40 | 38 | 37 | 36 | 35 | 34 | 33 | 32 |
Inside | — | — | 38 | 32 | 26 | 31 | 26 | 30 | 28 |
Outside − Inside | — | — | 0 | 5 | 10 | 4 | 8 | 3 | 4 |
Step 3 – Modes. In REGULAR mode the drop in 30 minutes is half the gap to the set temperature; in POWER mode the set temperature is reached in 30 minutes. The AC runs for 2 hours in total (23:00–0:00, 0:30–1:00, 1:30–2:00), and POWER must be used for more than 1 hour.
• If 23:00–0:00 were one REGULAR setting (38 → 26 in 1 hour), POWER could be at most 1 hour (the other two half-hours). Not allowed.
• A single POWER setting cannot produce the straight 38 → 26 line over a whole hour, because POWER reaches its set temperature in 30 minutes and then holds it.
• So the setting must be changed at 23:30 (for example, POWER to 32 at 23:00, then POWER to 26 at 23:30).
Time | Event | Counts as 'turned on or setting altered'? |
23:00 | AC turned on (outside the 11:01–1:59 window) | Not counted |
23:30 | Setting changed (needed to get enough POWER time) | Yes |
0:00 | AC turned off | No (switch-off) |
0:30 | AC turned on | Yes |
1:00 | AC turned off | No (switch-off) |
1:30 | AC turned on | Yes |
This question
Both temperatures change linearly between the half-hour marks, so the largest gap occurs at one of these marks. From the table in Step 2, the differences are 5, 10, 4, 8, 3 and 4. The maximum is 10 at midnight (outside 36, inside 26).
Hence, the answer is 10.
Take this paper as a timed test
Real exam timing, instant score, rank and topic-wise analysis.