CSAT Solved Papers/ 2022/Q46

2022 CSAT — Q46

Quant Arithmetic & numeracy 2.5 marks Hard

A man started from home at $14{:}30$ hours and drove to a village, arriving there when the village clock indicated $15{:}15$ hours. After staying for $25$ minutes, he drove back by a different route of length $1.25$ times the first route at a rate twice as fast reaching home at $16{:}00$ hours. As compared to the clock at home, the village clock is

A 10 minutes slow
B 5 minutes slow
C 10 minutes fast
D 5 minutes fast Answer

Worked rationale

Work in real (home-clock) time first; the home clock is the reliable reference.

Total elapsed on the home clock: $14{:}30 \to 16{:}00 = 90$ minutes. This covers outbound trip $+$ $25$ -min stay $+$ return trip.

Let the outbound time be $t$ . The return route is $1.25\times$ as long at $2\times$ the speed, so the return time is $\dfrac{1.25}{2}\,t = 0.625\,t$ . Then:

t + 25 + 0.625t = 90 \;\Rightarrow\; 1.625t = 65 \;\Rightarrow\; t = 40 \text{ min}.

So the real outbound arrival is $14{:}30 + 40 = 15{:}10$ . But the village clock read $15{:}15$ at that moment — it is ahead by $5$ minutes.

(Check: return time $= 0.625 \times 40 = 25$ min; real departure from village $15{:}10 + 25 = 15{:}35$ , home arrival $15{:}35 + 25 = 16{:}00$ ✓.)

Answer: (d) 5 minutes fast.

Why the other options miss

A

an arithmetic slip: mis-solves $1.625t = 65$ (e.g. $t = 50$ ), placing real arrival at $15{:}20$ and inverting/doubling the gap.
B

the direction reversed: gets the $5$ -minute gap but reverses which way it points — the village clock reads later than real, hence fast, not slow.

Specialist insight

The whole problem hinges on the return-time factor $\dfrac{\text{distance}}{\text{speed}} = \dfrac{1.25}{2} = 0.625$ of the outbound time. Solve $1.625t = 65$ to get $t = 40$ , then compare the real arrival ( $15{:}10$ ) to the village clock reading ( $15{:}15$ ): the village clock is ahead, i.e. fast by $5$ min. The sign trap (slow vs fast) is the examiner’s real test.

The trap, in one line

Return time $= 0.625t$ ; $1.625t = 65 \Rightarrow t = 40$ , real arrival $15{:}10$ vs clock $15{:}15$ $\Rightarrow$ $5$ min fast $=$ (d).

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