CSAT Solved Papers/ 2024/Q70

2024 CSAT — Q70

Quant Data sufficiency 2.5 marks Hard

A Question is given followed by two Statements I and II. Consider the Question and the Statements.

$P$ , $Q$ , $R$ and $S$ appeared in a test.

Question: Has $P$ scored more marks than $Q$ ?

Statement-I: The sum of the marks scored by $P$ and $Q$ is equal to the sum of the marks scored by $R$ and $S$ .

Statement-II: The sum of the marks scored by $P$ and $S$ is more than the sum of the marks scored by $Q$ and $R$ .

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

A The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
B The Question can be answered by using either Statement alone
C The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
D The Question cannot be answered even by using both the Statements together Answer

Worked rationale

Question: is $P > Q$ ? Statements: I: $P+Q = R+S$ . II: $P+S > Q+R$ .

Each alone clearly says nothing about $P$ vs $Q$ directly (each mixes in $R,S$ ), so the live question is whether both together decide it. The fast algebra: add I and II.

(P+Q) + (P+S) \;>\; (R+S) + (Q+R) \ \Rightarrow\ 2P + Q + S > 2R + Q + S \ \Rightarrow\ P > R.

So together they force $P > R$ — but that is a fact about $P$ vs $R$ , not $P$ vs $Q$ . To settle $P$ vs $Q$ , hunt for two cases that both satisfy I and II yet disagree:

“Yes” case ( $P > Q$ ): $P=10,\,Q=2,\,R=5,\,S=7$ . I: $10+2 = 5+7$ ✓. II: $10+7=17 > 2+5=7$ ✓. Here $P = 10 > 2 = Q$ .
“No” case ( $P < Q$ ): $P=2,\,Q=10,\,R=1,\,S=11$ . I: $2+10 = 1+11$ ✓. II: $2+11=13 > 10+1=11$ ✓. Here $P = 2 < 10 = Q$ .

Both statements hold in each case, yet $P>Q$ is true in one and false in the other.

Answer: (d) cannot be answered even by using both the Statements together.

Why the other options miss

A

read one sum as ordering two of four people: imagines a single sum condition orders two of the four people; ignores that $R,S$ are free.
B

the same over-reach, symmetrised.
C

answered the wrong comparison: the engineered trap. The student adds the statements, correctly derives $P > R$ , and mistakes $P>R$ for $P>Q$ — answering the wrong comparison. The algebra is right; the target is wrong. The deduction $P>R$ is a true consequence, but the question asked about $Q$ , and $Q,S$ are still free to flip the order.

Specialist insight

The structural read: from one equality (I) and one strict inequality (II) you can derive exactly one clean consequence by adding them — and here it lands on $P > R$ , not the $P$ vs $Q$ the question wants. That mismatch is the entire trap: the examiner gives you a derivable-but-irrelevant fact to reward the over-eager algebraist with a wrong (c). The gold DS habit defeats it cold: before committing to “sufficient,” construct one yes and one no case under all the given conditions. Two minutes of counterexample-hunting (swap $P$ and $Q$ , then choose $R,S$ to keep I and II true) proves (d) and is exactly the move the $-\tfrac13$ negative marking pays you to make rather than guessing (c). Note $R,S$ act as free “ballast” — that freedom is what keeps $P$ and $Q$ un-orderable.

The trap, in one line

Both statements together prove $P > R$ , not $P > Q$ — answering the wrong comparison is the (c)-vs-(d) trap; swap $P,Q$ to build a counterexample.

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