CSAT Solved Papers/ 2023/Q50

2023 CSAT — Q50

Quant Number theory 2.5 marks Medium

There are three traffic signals. Each signal changes colour from green to red and then from red to green. The first signal takes $25$ seconds, the second signal takes $39$ seconds and the third signal takes $60$ seconds to change the colour from green to red. The durations for green and red colours are same. At $2{:}00$ p.m, they together turn green. At what time will they change to green next, simultaneously?

A 4:00 p.m.
B 4:10 p.m. Answer
C 4:20 p.m.
D 4:30 p.m.

Worked rationale

“Green and red durations are same,” and the green $\to$ red time is given, so each signal’s full green-to-green cycle is twice that time:

T_1 = 2\times 25 = 50,\quad T_2 = 2\times 39 = 78,\quad T_3 = 2\times 60 = 120 \ \text{seconds}.

They are all green together again after $\operatorname{lcm}(50, 78, 120)$ seconds. Factorise:

50 = 2\cdot 5^2,\quad 78 = 2\cdot 3\cdot 13,\quad 120 = 2^3\cdot 3\cdot 5.

\operatorname{lcm} = 2^3\cdot 3\cdot 5^2\cdot 13 = 8\cdot 3\cdot 25\cdot 13 = 7800 \ \text{seconds}.

Convert: $7800 / 60 = 130$ minutes $= 2$ h $10$ min. So $2{:}00$ p.m. $+\ 2$ h $10$ min $=$ $4{:}10$ p.m.

Answer: (b) 4:10 p.m.

Why the other options miss

A

wrong formula: takes $\operatorname{lcm}(25,39,60)$ (the green $\to$ red times) without doubling for the full cycle, or rounds to a “nice” $2$ -hour gap.
C

an arithmetic slip: a slip in the LCM (e.g. $2^2$ instead of $2^3$ , or dropping the $13$ ), giving a wrong period near $140$ minutes.
D

missed a case: uses $\operatorname{lcm}(50,78,120)$ but mis-factors $78$ (omits $13$ or adds a stray factor), overshooting to $150$ minutes.

Specialist insight

Two traps stacked: (1) the full cycle is double the stated green $\to$ red time (green and red, equal durations), so use $50, 78, 120$ — not $25, 39, 60$ ; (2) the keystone is $78 = 2\cdot 3\cdot 13$ — the factor $13$ is what makes the LCM $7800$ (not a round $7200$ ). Forgetting the doubling gives $3900$ s and the wrong “4:00/4:05” family; mishandling the $13$ gives the other decoys. Convert seconds to h:min at the end ( $7800\text{s} = 130\text{ min}$ ).

The trap, in one line

Full cycle $=2\times$ given (green $=$ red): $\operatorname{lcm}(50,78,120) = 7800$ s $= 2$ h $10$ min $\Rightarrow$ 4:10 p.m. (b).

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