CSAT Solved Papers/ 2024/Q38

2024 CSAT — Q38

Quant Logical & quantitative reasoning 2.5 marks Medium

The calendar for the year $2025$ is same for

A 2029
B 2030
C 2031 Answer
D 2033

Worked rationale

A year’s calendar repeats when (i) the total odd days elapsed since it is a multiple of $7$ (same weekday start) and (ii) the new year has the same leap status. $2025$ is an ordinary year ( $2025 = 4\times506 + 1$ , not divisible by $4$ ).

A common year contributes $1$ odd day, a leap year $2$ . Accumulate odd days from $2025$ onward (each year’s contribution moves the next year’s start):

year	odd days	cumulative
2025	1	1
2026	1	2
2027	1	3
2028 (leap)	2	5
2029	1	6
2030	1	7 ≡ 0

The cumulative reaches a multiple of $7$ at the end of $2030$ , so the calendar repeats in $2031$ . And $2031$ is an ordinary year (matching $2025$ ‘s leap status). Both conditions met.

Why the other options miss

A

counted one too few: stops after $3$ years (cumulative $3$ ), guessing the usual ” $+6$ years” shortcut without accounting for the $2028$ leap year’s extra odd day.
B

a convention slip: counts the start year or shifts the boundary by one, landing on the year before the true repeat.
D

an arithmetic slip: over-accumulates past $7$ (e.g. miscounts $2032$ ‘s leap day), overshooting the cycle.

Specialist insight

Calendar-repeat problems are pure odd-day accounting mod $7$ , and the leap year in the run is the trap: $2028$ adds $2$ odd days, breaking the naive “calendars repeat every $6$ or $11$ years” reflex. Tabulate cumulative odd days year by year and stop at the first multiple of $7$ — here end of $2030$ , giving repeat year $2031$ , which is conveniently also non-leap (so no further adjustment). Mishandling the leap-year contribution is the single biggest source of error.

The trap, in one line

Accumulate odd days mod $7$ with the $2028$ leap year counting as $2$ ; the cumulative hits $7$ at end of $2030$ , so $2025$ 's calendar repeats in $2031$ .

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