CSAT Solved Papers/ 2025/Q50

2025 CSAT — Q50

Quant Statement validity 2.5 marks Hard

Three teams $P, Q, R$ participated in a tournament in which the teams play with one another exactly once. A win fetches a team $2$ points and a draw $1$ point. A team gets no point for a loss. Each team scored exactly one goal in the tournament. The team $P$ got $3$ points, $Q$ got $2$ points and $R$ got $1$ point. Which of the following statements is/are correct?

I. The result of the match between $P$ and $Q$ is a draw with the score $0 - 0$ .

II. The number of goals scored by $R$ against $Q$ is $1$ .

A I only
B II only
C Both I and II Answer
D Neither I nor II

Worked rationale

There are $3$ matches ( $P$ – $Q$ , $P$ – $R$ , $Q$ – $R$ ); each match awards $2$ points total, so points sum to $6 = 3+2+1$ ✓. Decode the points table.

Points ⇒ results. $P=3$ over two matches forces one win $+$ one draw ( $2+1$ ; no other split). $R=1$ forces one draw $+$ one loss ( $1+0$ ). $Q=2$ is either two draws or one win $+$ one loss. The points table alone admits two result sets:

(A) $P$ – $Q$ draw, $P$ beat $R$ , $Q$ – $R$ draw — check: $P=2{+}1=3$ , $Q=1{+}1=2$ , $R=0{+}1=1$ ✓.
(B) $P$ beat $Q$ , $P$ – $R$ draw, $Q$ beat $R$ — check: $P=2{+}1=3$ , $Q=0{+}2=2$ , $R=1{+}0=1$ ✓.

The goal constraint then eliminates (B): under (B), $P$ won vs $Q$ so (as $P$ scored only $1$ goal total) $P$ – $Q$ is $1$ – $0$ and the $P$ – $R$ draw is $0$ – $0$ ; $Q$ won vs $R$ scoring $\ge1$ there but $Q$ scored $0$ vs $P$ , so $Q$ ‘s goal is vs $R$ , forcing $R$ to score $0$ vs $Q$ ; but then $R$ ‘s total is $0$ (draw vs $P$ at $0$ – $0$ ) $+\,0 = 0 \ne 1$ — contradiction. So only set (A) survives:

P\text{–}Q:\ \text{draw},\qquad P\text{ beat }R,\qquad Q\text{–}R:\ \text{draw}.

Goal accounting. Each team scored exactly $1$ goal in total (across its two matches). Let the $P$ – $Q$ draw be $d\!-\!d$ and the $Q$ – $R$ draw be $e\!-\!e$ (draws are level). $P$ won $P$ – $R$ , so $P$ ‘s goals there $> R$ ‘s.

$P$ total: $d + (P\text{'s goals vs }R) = 1$ . Since $P$ won, $P$ ‘s goals vs $R \ge 1$ , forcing $d = 0$ and $P$ scored $1$ vs $R$ . So $P$ – $Q$ is $0$ – $0$ — Statement I is TRUE.
$Q$ total: $d + e = 0 + e = 1 \Rightarrow e = 1$ .
$R$ total: $(R\text{'s goals vs }P) + e = 1$ . As $P$ won $1$ – $0$ , $R$ ‘s goals vs $P = 0$ , consistent with $e=1$ .

So the $Q$ – $R$ draw is $1$ – $1$ : $R$ scored $e = 1$ goal against $Q$ — Statement II is TRUE.

Answer: (c) Both I and II.

Why the other options miss

A

stops one deduction short: nails the $0$ – $0$ draw but stops before the goal total forces the $Q$ – $R$ draw to be $1$ – $1$ , missing II.
B

reasons the goal totals but never deduces that the $P$ – $Q$ draw must be scoreless, missing I.
D

stops at the points table (which alone leaves two result sets) and declares the scenario under-determined, never applying the goal-total constraint that eliminates set (B) and pins both scorelines.

Specialist insight

This is a two-layer deduction: the points table alone narrows the results to two sets (3 = win+draw, 1 = draw+loss), and the “exactly one goal each” constraint then both eliminates one result set and pins the scorelines. The decisive insight is that $P$ won a match yet scored only one goal in the whole tournament — so that win is $1$ – $0$ and its draw must be $0$ – $0$ , which cascades to make the $Q$ – $R$ draw $1$ – $1$ . Treat the points and the goal-totals as two separate accounting books that must both balance; that is exactly the structured reasoning CSAT rewards here.

The trap, in one line

Points leave two result sets; "one goal each" eliminates one and forces $P$ 's win to be $1$ – $0$ , the $P$ – $Q$ draw $0$ – $0$ , and the $Q$ – $R$ draw $1$ – $1$ — so both I and II hold.

← All 2025 CSAT questions