CSAT Solved Papers/ 2023/Q20

2023 CSAT — Q20

Quant Statement validity 2.5 marks Hard

$A, B, C$ working independently can do a piece of work in $8, 16$ and $12$ days respectively. $A$ alone works on Monday, $B$ alone works on Tuesday, $C$ alone works on Wednesday; $A$ alone, again works on Thursday and so on. Consider the following statements:

The work will be finished on Thursday.
The work will be finished in $10$ days.

Which of the above statements is/are correct?

A 1 only Answer
B 2 only
C Both 1 and 2
D Neither 1 nor 2

Worked rationale

Use a common denominator so each turn adds an integer of “work units.” LCM $(8,16,12) = 48$ , so total work $= 48$ units and the daily outputs are

A = \tfrac{48}{8} = 6,\quad B = \tfrac{48}{16} = 3,\quad C = \tfrac{48}{12} = 4 \ \text{units/day}.

The turn order cycles $A, B, C, A, B, C, \dots$ One full cycle (3 days) does $6 + 3 + 4 = 13$ units.

After $3$ cycles ( $9$ days): $3 \times 13 = 39$ units. Remaining $48 - 39 = 9$ .
Day $10$ ( $A$ ‘s turn): $+6 \Rightarrow 45$ units. Remaining $3$ .
Day $11$ ( $B$ ‘s turn): $B$ does $3$ units $\Rightarrow 45 + 3 = 48$ . Work completes on day $11$ .

Now the weekday of day $11$ : day $1 =$ Monday, so day $11 =$ Monday $+ 10 =$ Thursday (Mon, Tue, Wed, Thu, Fri, Sat, Sun, Mon, Tue, Wed, Thu).

Statement 1 (finished on Thursday): day $11$ is Thursday. True.
Statement 2 (finished in $10$ days): it finishes on day $11$ , not day $10$ . False.

Answer: (a) 1 only.

Visual solution

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

Hand-drawn worked solution for UPSC 2023 CSAT Q20 — Statement validity — Tap the drawing to open it full size for the fine detail.

Why the other options miss

B

stops one day early: stops at day $10$ (where $45 < 48$ units), declaring completion a day early and mismatching the weekday.
C

gets the Thursday finish but also asserts ” $10$ days,” not noticing the $3$ remaining units force an $11$ th day.
D

mis-tracks the cycle (e.g. wrong daily units or a bad cycle sum), landing on a different day and rejecting both.

Specialist insight

Scaling to LCM units ( $48$ , with daily outputs $6,3,4$ ) turns fractions into clean integer bookkeeping — essential under the clock. The decisive subtlety is who works the finishing day: after $9$ days ( $39$ ) and day $10$ ( $A$ , $+6 = 45$ ), the last $3$ units fall to $B$ on day $11$ , so it is not a $10$ -day job, but day $11$ is a Thursday. The two statements are deliberately split — one true (weekday), one false (day-count) — punishing anyone who conflates “Thursday” with ” $10$ days.”

The trap, in one line

In LCM units ( $6,3,4$ /day, cycle $13$ ): $39$ by day $9$ , $45$ by day $10$ ( $A$ ), last $3$ on day $11$ ( $B$ ) $=$ Thursday $\Rightarrow$ St-1 true, St-2 false $=$ (a).

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