CSAT Solved Papers/ 2024/Q73

2024 CSAT — Q73

Quant Data sufficiency 2.5 marks Hard

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

Age of each of $P$ and $Q$ is less than $100$ years but more than $10$ years. If you interchange the digits of the age of $P$ , the number represents the age of $Q$ .

Question: What is the difference of their ages?

Statement-I: The age of $P$ is greater than the age of $Q$ .

Statement-II: The sum of their ages is $11/6$ times their difference.

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 Answer
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

Worked rationale

Encode the digit-reversal structure first. Let $P$ ‘s age $= 10a + b$ and $Q$ ‘s age $= 10b + a$ , with digits $a, b \in \{1,\dots,9\}$ (both ages are two-digit numbers strictly between $10$ and $100$ ). Then:

\text{sum} = 11(a+b), \qquad \text{difference} = 9\,|a-b|.

Statement-I alone ( $P > Q$ , i.e. $a > b$ ): the difference is $9(a-b)$ , but $a-b$ can be anything from $1$ to $8$ , so the difference is not pinned. Not sufficient.

Statement-II alone ( $\text{sum} = \tfrac{11}{6}\,\text{difference}$ ):

11(a+b) = \frac{11}{6}\cdot 9\,|a-b| \;\Longrightarrow\; a+b = \frac{3}{2}\,|a-b| \;\Longrightarrow\; 2(a+b) = 3\,|a-b|.

Take $a > b$ (a non-zero difference is required for the equation to make sense): $2(a+b) = 3(a-b) \Rightarrow 5b = a$ . With single digits, $b = 1,\ a = 5$ is the only solution. Hence the difference is $9(a-b) = 9 \times 4 = 36$ — uniquely determined, and the answer is the same whichever of $P,Q$ is older (difference is symmetric). Sufficient.

So the Question is answered by Statement-II alone but not by Statement-I alone.

Answer: (a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

Why the other options miss

B

thought the ordering statement was enough when it isn’t: thinks " $P > Q$ " alone fixes the difference, missing that it only orders the ages without pinning $a-b$ .
C

imported a redundant tie-breaker: solves $2(a+b)=3|a-b|$ but fails to notice the solution $(5,1)$ is unique, so it wrongly imports Statement-I as a tie-breaker that isn’t needed.
D

never reduced the ratio to a digit equation: never reduces $\text{sum}=\tfrac{11}{6}\, \text{difference}$ to $5b = a$ , so it never sees that the digits are forced.

Specialist insight

The reversal identity is the whole game: for $\overline{ab}$ and $\overline{ba}$ , the sum is always a multiple of $11$ and the difference is always a multiple of $9$ . Substituting these turns Statement-II into $11(a+b) = \tfrac{11}{6}\cdot 9|a-b|$ , where the $11$ ‘s cancel and you are left with the clean linear digit equation $5b = a$ — which has a single one-digit solution. Because difference is symmetric, the ordering in Statement-I adds nothing to a difference question. The deadliest trap is defaulting to “(c) both needed” out of DS habit; the reversal algebra makes Statement-II self-sufficient.

The trap, in one line

For $\overline{ab},\overline{ba}$ the sum is a multiple of $11$ and the difference a multiple of $9$ , so Statement-II reduces to $5b=a \Rightarrow$ difference $=36$ alone $=(a)$ — Statement-I's ordering is redundant for a difference.

← All 2024 CSAT questions