CSAT Solved Papers/ 2023/Q38

2023 CSAT — Q38

Hybrid Logical & quantitative reasoning 2.5 marks Medium

Consider the following statements:

$A$ is older than $B$ .
$C$ and $D$ are of the same age.
$E$ is the youngest.
$F$ is younger than $D$ .
$F$ is older than $A$ .

How many statements given above are required to determine the oldest person/persons?

A Only two
B Only three
C Only four
D All five Answer

Worked rationale

Build the order chain and check which statements are load-bearing for naming the oldest.

From (5) $F > A$ , (1) $A > B$ : $F > A > B$ . From (4) $D > F$ : $D > F > A > B$ . From (2) $C = D$ : so $C = D$ sit at the top. From (3) $E$ is youngest. Full order:

C = D \;>\; F \;>\; A \;>\; B \;>\; E.

The oldest are $C$ and $D$ (tied). Now test necessity — drop each statement and ask “can someone else be oldest?”:

Drop (2): $C$ ‘s age is unknown $\Rightarrow$ can’t say $C = D$ are both oldest. Needed.
Drop (4): $D$ vs $F$ unknown $\Rightarrow$ $F$ could top $D$ . Needed.
Drop (5): $A$ vs the rest is unanchored $\Rightarrow$ $A$ could be oldest. Needed.
Drop (1): $B$ is unconstrained $\Rightarrow$ $B$ could be oldest. Needed.
Drop (3): $E$ is unconstrained $\Rightarrow$ $E$ could be oldest. Needed.

Every statement removes a candidate that could otherwise top the list. All five are required.

Answer: (d) All five.

Visual solution

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

Hand-drawn worked solution for UPSC 2023 CSAT Q38 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

missed a case: uses just (4)+(2) to crown $C, D$ , ignoring that $A, B, E$ are then unbounded and could exceed $D$ .
B

solved the wrong question: stops once a top pair appears, not testing whether the unmentioned people are ruled out.
C

counted one too few: drops one statement (commonly (1), thinking $B$ “obviously” young) without checking that nothing else then bounds $B$ .

Specialist insight

“How many statements are required” is a minimal-sufficient-set test, not “can I find the oldest.” The disciplined check is to remove each statement and ask whether some person becomes a possible oldest. Here the chain $C{=}D > F > A > B > E$ is a single thread in which every link is needed to pin the top — drop any one and a different person could rise. Recognising that all five are independently load-bearing is the whole item.

The trap, in one line

Order is $C{=}D > F > A > B > E$ ; removing *any* statement frees a person to possibly be oldest, so all five are required $=$ (d).

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