UPSC CDS 2 2022 Math Paper

Concept:The sum of a two-digit number and its reverse is always a multiple of 11.Explanation:Let the two-digit number be $10x + y$, where $x$ is the tens digit ($1 \le x \le 9$) and $y$ is the units digit ($0 \le y \le 9$).The reversed number is $10y + x$.Their sum is $(10x + y) + (10y + x) = 11(x + y)$.Statement‑1 says the sum is divisible by $9$.Since $11$ is not divisible by $9$, $x + y$ must be divisible by $9$.Possible values for $x + y$ are $9$ or $18$.For $x+y = 9$ there are many possible digits (e.g. $1+8$, $2+7$, … , $9+0$).For $x+y = 18$ only $x=y=9$ works.Thus Statement‑1 alone does not give a unique number.Statement‑2 says the sum is divisible by $2$ (even).$11(x+y)$ is even only when $x+y$ is even.There are many even sums from $2$ to $18$, so many numbers satisfy this.Thus Statement‑2 alone is not sufficient.Combining both statements: $x+y$ must be divisible by $9$ and even.The only possibility is $x+y = 18$, which forces $x = y = 9$.So the number is $99$.Hence both statements together are sufficient.Answer:Both Statement‑1 and Statement‑2 are sufficient to answer the question. (Option C)