CSAT Solved Papers/ 2024/Q64

2024 CSAT — Q64

Quant Data sufficiency 2.5 marks Medium

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

Question: What are the unique values of $x$ and $y$ , where $x,\ y$ are distinct natural numbers?

Statement-I: $x/y$ is odd.

Statement-II: $xy = 12$ .

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 Answer
D The Question cannot be answered even by using both the Statements together

Worked rationale

Decide, don’t just compute — and read ” $x/y$ is odd” precisely: an odd value is an odd integer, so $y \mid x$ and the quotient is one of $1,3,5,\dots$

Statement-I alone ( $x/y$ odd): infinitely many distinct pairs work — $(3,1),(5,1),(6,2),(15,3),\dots$ all give an odd integer quotient. Not unique.

Statement-II alone ( $xy = 12$ ): the distinct factor pairs are $(1,12),(2,6),(3,4),(4,3),(6,2),(12,1)$ . Not unique.

Both together. Write $x = (\text{odd})\cdot y$ , so $xy = (\text{odd})\cdot y^2 = 12$ . Test $y$ :

$y = 1$ : $\text{odd} = 12$ — not odd. ✗
$y = 2$ : $\text{odd} = 12/4 = 3$ — odd ✓, giving $x = 6$ .
$y = 3$ : $\text{odd} = 12/9$ — not an integer. ✗

Only $(x,y) = (6,2)$ survives: $xy = 12$ ✓ and $x/y = 3$ (odd) ✓. Unique.

Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

Why the other options miss

A

thought a statement was enough when it wasn’t: imagines $x/y$ odd plus the small search space already pins the pair, missing that Statement I alone admits infinitely many quotients.
B

thought either statement was enough when neither is: treats $xy = 12$ as if a single “natural” factor pair were intended, ignoring the six ordered possibilities.
D

stopped short of the combined condition: stops at “many factor pairs” without imposing $(\text{odd})\cdot y^2 = 12$ , which eliminates every pair except $(6,2)$ .

Specialist insight

The decisive move is rewriting the two statements as one Diophantine condition: $(\text{odd})\cdot y^2 = 12$ . Squaring the denominator is what shrinks the search to $y\in\{1,2,3\}$ and the oddness of the cofactor kills $y=1$ and $y=3$ instantly. This is the general DS discipline — fold both statements into a single equation and test for a second admissible solution; here there is exactly one, so “together” suffices and neither alone does. Note the language trap: “odd” silently forces integrality ( $y\mid x$ ); a reader who allows $x/y$ to be a non-integer would wrongly think Statement I says almost nothing.

The trap, in one line

" $x/y$ odd" means an odd integer, so $(\text{odd})\cdot y^2 = 12$ forces $(x,y)=(6,2)$ only when paired with $xy=12$ — neither statement alone is unique $=(c)$ .

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