CSAT Solved Papers/ 2021/Q63

2021 CSAT — Q63

Quant Logical & quantitative reasoning 2.5 marks Medium

There are three points $P$ , $Q$ and $R$ on a straight line such that $PQ:QR = 3:5$ . If $n$ is the number of possible values of $PQ:PR$ , then what is $n$ equal to?

A 1
B 2 Answer
C 3
D 4

Worked rationale

Distances $PQ = 3$ and $QR = 5$ (in proportion). On a line, the value of $PR$ depends on which point lies between the other two — enumerate the three orderings:

$Q$ between $P$ and $R$ ( $P\text{–}Q\text{–}R$ ): $PR = PQ + QR = 3 + 5 = 8$ , so $PQ:PR = 3:8$ .
$P$ between $Q$ and $R$ ( $Q\text{–}P\text{–}R$ ): $QR = QP + PR \Rightarrow PR = 5 - 3 = 2$ , so $PQ:PR = 3:2$ .
$R$ between $P$ and $Q$ ( $P\text{–}R\text{–}Q$ ): $PQ = PR + RQ \Rightarrow PR = 3 - 5 = -2$ , impossible (a length can’t be negative).

So $PQ:PR$ takes exactly two values, $3:8$ and $3:2$ .

Answer: (b) 2.

Visual solution

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

Hand-drawn worked solution for UPSC 2021 CSAT Q63 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

misses a case: considers only $Q$ in the middle ( $3:8$ ), forgetting that $P$ can lie between $Q$ and $R$ .
C

misses a case: counts the third ordering ( $R$ in the middle) as valid, missing that it forces a negative length.
D

misses a case: double-counts orderings or treats reversed labellings as new ratios.

Specialist insight

Collinear-point ratio problems hinge on which point is between which — there are three orderings, but the constraint $QR = 5 > PQ = 3$ rules out $R$ being between $P$ and $Q$ (that would need $PR = 3 - 5 < 0$ ). The discipline is to test feasibility, not just list cases: an ordering survives only if every segment length stays positive. Two survive, so $n = 2$ .

The trap, in one line

Only $Q$ -middle ( $3:8$ ) and $P$ -middle ( $3:2$ ) give positive $PR$ ; $R$ -middle needs $PR<0 \Rightarrow n=2 \Rightarrow$ (b).

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