2024 CSAT — Q37

The hands coincide $11$ times in every $12$ hours — once every $\tfrac{12}{11}$ hours $\approx 65$ min $27$ s, not once an hour. Crucially, in the stretch from $11$ o’clock to $12$ o’clock there is no separate coincidence: the “$11$-o’clock” overlap is exactly $12{:}00$.

List the coincidences in the window $10{:}00$ a.m. $\to 2{:}00$ p.m.:

• $\approx 10{:}54{:}33$ — the overlap in the $10$–$11$ hour,
• $12{:}00{:}00$ — exact overlap (this absorbs the $11$-o’clock one),
• $\approx 1{:}05{:}27$ — the overlap in the $1$–$2$ hour.

That is $3$ coincidences. (Between $11$ and $12$ there is no extra one; $2{:}00$ itself is not a coincidence.)

Answer: (a) 3 times.

The whole item rests on one fact: $11$ coincidences in $12$ hours, so they are spaced $\tfrac{12}{11}$ hr apart and there is a “missing” hour near $11$–$12$ where the only overlap is $12{:}00$. Over a $4$-hour window you would naively expect $4$, but the $11$-o’clock gap drops it to $3$. The safe exam move is to mark the actual overlap times ($10{:}54$, $12{:}00$, $1{:}05$) rather than count hour-boundaries — that is exactly where the off-by-one (b) creeps in.