CSAT Solved Papers/ 2021/Q47

2021 CSAT — Q47

Quant Data sufficiency 2.5 marks Medium

Consider two Statements and a Question:

Statement-1: The last day of the month is a Wednesday.

Statement-2: The third Saturday of the month was the seventeenth day.

Question: What day is the fourteenth of the given month?

Which one of the following is correct in respect of the Statements and the Question?

A Statement-1 alone is sufficient to answer the Question
B Statement-2 alone is sufficient to answer the Question Answer
C Both Statement-1 and Statement-2 are required to answer the Question
D Neither Statement-1 alone nor Statement-2 alone is sufficient to answer the Question

Worked rationale

Statement-1 alone: “Last day is Wednesday” — but the month length is unknown ( $28,29,30$ or $31$ days), so the last date could be the $28$ th– $31$ st. Without knowing how many days the month has, the $14$ th cannot be pinned to a weekday. Not sufficient.

Statement-2 alone: the $3$ rd Saturday is the $17$ th, so Saturdays fall on $17-14 = 3$ rd, $10$ th, $17$ th, $24$ th, $31$ st. From “the $3$ rd is a Saturday,” the $14$ th is $14 - 3 = 11$ days later; $11 \equiv 4 \pmod 7$ , so the $14$ th is $\text{Saturday}+4 = \text{Wednesday}$ . Determines the day. Sufficient.

Answer: (b) Statement-2 alone is sufficient.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2021 CSAT Q47 — Data sufficiency — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

a convention slip: assumes a fixed month length, treating “last day Wednesday” as if it fixed every earlier date.
C

thought both were needed when S2 alone suffices: thinks S2 needs S1’s anchor, missing that “ $3$ rd Saturday $=17$ th” alone fixes the whole calendar of weekdays.
D

misreads the clue: fails to convert “ $3$ rd Saturday” into the first Saturday ( $3$ rd), which is what makes S2 self-contained.

Specialist insight

A single dated weekday fixes the whole month’s days-of-week; an undated one (S1’s “last day,” with unknown length) does not. From S2, back out the first Saturday ( $17 - 14 = 3$ rd), then the $14$ th is a fixed offset away ( $+11 \equiv +4$ ), landing on Wednesday. The trap is S1 looking decisive — but “last day” is anchored to an unknown date, so it’s a decoy. In day-of-week DS, always ask: is the given weekday tied to a known date?

The trap, in one line

S2 fixes the $3$ rd as Saturday $\Rightarrow$ $14$ th $=$ Sat $+11 \equiv$ Wednesday; S1's unknown month length is a decoy $\Rightarrow$ (b).

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