There are four bulbs A, B, C and D in a room. A glows for 2 minutes, then goes off for 1 minute and the process continues in this manner. Similarly, B glows for 3 minutes and then goes off for 1 minute and the process continues. For C, it is 4 minutes and 1 minute, while for D, it is 5 minutes and 1 minute respectively for glowing and going off. A man enters the room and comes out of it as soon as he completes an hour. What is the maximum number of times that he may face complete darkness in the room?

Q: What is the answer to CAPF 2025 Paper I Q68?

B. Correct — B, 2. Each bulb repeats on its own cycle: A every 3 minutes, B every 4, C every 5 and D every 6, with the off-minute at the end of each cycle. Number the minutes from a common start. A is off in the minutes numbered 2 mod 3, B in those numbered 3 mod 4, C in 4 mod 5 and D in 5 mod 6. Complete darkness needs all four at once. The D condition already forces the A condition, since a number that leaves 5 on division by 6 leaves 2 on division by 3, and combining the rest gives one solution in every sixty minutes — the sixtieth minute, numbered 59. So darkness lasts one minute and returns exactly an hour later. Now the stay: it is exactly sixty minutes long, and the gaps between dark minutes are also sixty. A man who walks in part-way through one dark minute is in darkness at once, and sixty minutes later he is still inside for the first part of the next one. That is two spells of darkness, which is the most that any hour-long stay can catch. (Provisional — pending the official UPSC key.)

(a)3
(b)2
(c)0
(d)1

Answer

Why

Correct — B, 2. Each bulb repeats on its own cycle: A every 3 minutes, B every 4, C every 5 and D every 6, with the off-minute at the end of each cycle. Number the minutes from a common start. A is off in the minutes numbered 2 mod 3, B in those numbered 3 mod 4, C in 4 mod 5 and D in 5 mod 6. Complete darkness needs all four at once. The D condition already forces the A condition, since a number that leaves 5 on division by 6 leaves 2 on division by 3, and combining the rest gives one solution in every sixty minutes — the sixtieth minute, numbered 59. So darkness lasts one minute and returns exactly an hour later. Now the stay: it is exactly sixty minutes long, and the gaps between dark minutes are also sixty. A man who walks in part-way through one dark minute is in darkness at once, and sixty minutes later he is still inside for the first part of the next one. That is two spells of darkness, which is the most that any hour-long stay can catch.

Why the others are wrong

(a)3 — Would need three dark minutes inside sixty. The four cycles have a common period of sixty minutes and yield a single dark minute in each, so three cannot occur however the visit is timed.
(c)0 — Only true if the bulbs never go dark together. With a common start they do, once an hour, so an hour inside the room cannot miss it entirely.
(d)1 — The count for a visit that starts on a whole minute, and it is what the word 'maximum' is there to exclude. A stay that straddles the boundary catches the end of one dark minute and the start of the next.

Concept

Repeating on-off patterns are handled with cycle lengths and remainders. A bulb glowing for g minutes and resting for 1 has a cycle of g + 1, and it is dark in the last minute of every cycle. Several such bulbs are dark together only at instants satisfying all the remainder conditions at once, and the pattern of coincidences repeats with the lowest common multiple of the cycles — here 3, 4, 5 and 6 give 60.

The word doing the work in this stem is 'maximum'. The events are one minute long and sixty minutes apart, and the visit is sixty minutes long, so the count depends on where the visit begins. Start it on a clean minute boundary and you see one dark spell; start it in the middle of a dark minute and you catch the second half of that one on the way in and the first half of the next on the way out. Whenever an interval and the gap between events are equal, that straddle is worth checking — it is the difference between one and two here.

Key facts

A bulb glowing g minutes and resting 1 minute has a cycle of g + 1 minutes, dark in the final minute.
The cycles here are 3, 4, 5 and 6 minutes, with a common period of 60 minutes.
The condition for D, a remainder of 5 on division by 6, already implies A's condition of remainder 2 on division by 3.
Combining all four leaves one dark minute in every sixty — the sixtieth.
A sixty-minute visit can overlap two consecutive dark minutes if it begins part-way through one.

When are all four dark at once

Bulb A→Glows 2, rests 1 — cycle 3 minutes, dark when the minute number leaves 2 on division by 3

Bulb B→Glows 3, rests 1 — cycle 4 minutes, dark when it leaves 3 on division by 4

Bulb C→Glows 4, rests 1 — cycle 5 minutes, dark when it leaves 4 on division by 5

Bulb D→Glows 5, rests 1 — cycle 6 minutes, dark when it leaves 5 on division by 6

All four together→One minute in every 60 — the last minute of each hour-long cycle

A stay of exactly 60 minutes→Entering part-way through a dark minute catches that one and the start of the next: 2

The interval between dark minutes equals the length of the visit, which is what makes the straddle possible.

Study next

Lowest common multiple applied to repeating events and traffic-light problemsRemainder conditions and the Chinese remainder theoremCounting events in a window when the window equals the gapBells, blinking lights and other periodic-coincidence questionsReading 'maximum' and 'at least' as instructions to choose the timing

Common traps

Assuming the bulbs never coincide and answering zero.
Answering one by silently starting the visit at a minute boundary.
Taking the cycle of a bulb as its glowing time rather than glowing plus resting.

A periodic-coincidence item whose answer depends on the word 'maximum' as much as on the arithmetic.

Related PYQs

No directly related past PYQ was found.

Practice

practice — not a real PYQ
Three bells ring at intervals of 6, 9 and 15 minutes. If they ring together at 10:00 AM, when will they next ring together?
- (a)10:30 AM
- (b)11:00 AM
- (c)11:30 AM
- (d)12:00 noon
Answer(c) 11:30 AM — the lowest common multiple of 6, 9 and 15 is 90 minutes.
practice — not a real PYQ
A lamp is on for 4 minutes and off for 1 minute, repeating. Over one hour, for how many minutes is it off?
- (a)10
- (b)12
- (c)15
- (d)20
Answer(b) 12 — the cycle is 5 minutes long, so there are 12 cycles in an hour and one off-minute in each.

There are four bulbs A, B, C and D in a room. A glows for 2 minutes, then goes off for 1 minute and the process continues in this manner. Similarly, B glows for 3 minutes and then goes off for 1 minute and the process continues. For C, it is 4 minutes and 1 minute, while for D, it is 5 minutes and 1 minute respectively for glowing and going off. A man enters the room and comes out of it as soon as he completes an hour. What is the maximum number of times that he may face complete darkness in the room?

(a)3
(b)2
(c)0
(d)1