CSAT Solved Papers/ 2021/Q48

2021 CSAT — Q48

Quant Logical & quantitative reasoning 2.5 marks Hard

Which day is $10$ th October, $2027$ ?

A Sunday Answer
B Monday
C Tuesday
D Saturday

Worked rationale

Anchor on a known day: $1$ January $2021$ was a Friday. Count odd days forward.

$1$ Jan $2021 \to 1$ Jan $2027$ : the years $2021,2022,2023,2024,2025,2026$ contribute $365\times 5 + 366 = 2191$ days ( $2024$ is a leap year). Since $2191 = 7\times 313$ , that is $0$ odd days. So $1$ Jan $2027$ is also a Friday.

$1$ Jan $2027 \to 10$ Oct $2027$ ( $2027$ is not a leap year): day-of-year of Oct $10$ is

31+28+31+30+31+30+31+31+30+10 = 283.

The offset from Jan $1$ is $283 - 1 = 282$ days, and $282 \equiv 2 \pmod 7$ .

So $10$ Oct $2027 = \text{Friday} + 2 = \textbf{Sunday}$ .

Answer: (a) Sunday.

Why the other options miss

B

counts one day too many: uses offset $283$ instead of $282$ (counts Jan $1$ itself), shifting one day.
C

a leap-year convention slip: treats $2027$ as a leap year (adds a spurious Feb $29$ ) or mis-handles the $2024$ leap year, adding an extra odd day.
D

an arithmetic slip: a one-day slip in the year-block odd-day total ( $2191 \bmod 7$ ).

Specialist insight

Two independent fencepost traps live here: the leap-year count over $2021$ – $2026$ (exactly one, $2024$ ) and the offset within $2027$ (days since Jan $1$ is day-of-year $-1$ ). Anchoring on a memorised day ( $1$ Jan $2021 =$ Friday) and reducing each leg mod $7$ keeps the arithmetic tiny: the six-year block is $0$ odd days, and $282 \equiv 2$ gives Friday $\to$ Sunday. Always subtract $1$ when converting a day-of-year to an offset from the anchor.

The trap, in one line

$1$ Jan $2027 =$ Friday; Oct $10$ is offset $282\equiv 2$ , so Friday $+2 =$ Sunday $\Rightarrow$ (a).

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