2024 CSAT — Q65

Let the total be $T$. The trick on “fraction of what the others got” is to convert each statement into a fraction of the whole.

Statement-I: $X = \tfrac45(Y+Z) = \tfrac45(T - X) \Rightarrow 5X = 4T - 4X \Rightarrow 9X = 4T \Rightarrow X = \tfrac49 T.$ This fixes $X$, but leaves $Y+Z = \tfrac59 T$ split unknown — $Y$ or $Z$ could be the smallest, or even smaller than… we can’t rank. I alone insufficient.

Statement-II: $Y = \tfrac27(X+Z) = \tfrac27(T-Y) \Rightarrow 9Y = 2T \Rightarrow Y = \tfrac29 T.$ Fixes $Y$, but $X+Z = \tfrac79 T$ is split unknown ($X$ or $Z$ could be below $\tfrac29 T$). II alone insufficient.

Both together: $X = \tfrac49 T$, $Y = \tfrac29 T$, so $Z = T - \tfrac49 T - \tfrac29 T = \tfrac39 T$. Now all three are pinned: $X = \tfrac49,\ Z = \tfrac39,\ Y = \tfrac29$ (of $T$). The least is $Y$. Determined.

Answer: (c) both together, but neither alone.

The decisive move — and the one a generalist fumbles — is ”$X$ is $\tfrac45$ of the rest” $\Rightarrow$ ”$X = \tfrac45(T-X) \Rightarrow X = \tfrac{4}{4+5}T = \tfrac49 T$.” In general, ”$A$ is $\tfrac pq$ of the others” makes $A = \tfrac{p}{p+q}$ of the whole. Once both statements are fractions of the same total $T$, the third share is forced by subtraction and the ranking is immediate — the unknown total $T$ cancels out entirely (you never need its value, only the fractions $\tfrac49,\tfrac29,\tfrac39$). The DS discipline: each statement alone fixes one share but leaves a two-way lump unranked (insufficient); only together do all three fractions sum to $1$ and the order locks. Recognising that the total is a red herring is what turns this from a “we need the amount!” panic into a clean (c).