UPSC Prelims 2026 · CSAT
Question 51: Logical Reasoning and Mental Ability
Seven cubes are identical in shape. Out of these, the weight of each of the six cubes is equal and the weight of the remaining cube is less than the weight of any other cube. A balance is used to identify the lightest cube. What is the minimum number of attempts required to distinguish the odd cube with certainty?
AnswerUPSC’s answer is (a).
Put 3 cubes on each pan and keep one aside. If the pans balance, the lighter cube is the one set aside: one weighing. If one pan is lighter, the odd cube is among its three; weigh any two of them against each other. The lighter one is the odd cube, and if they balance it is the third. So 2 weighings are enough in the worst case, and one weighing cannot settle the unbalanced case with three suspects.
Why each option is right or wrong
- (a) Right: 3 against 3, then 1 against 1, settles it.
- (b) Wrong: three weighings are more than needed.
- (c) Wrong: four weighings are more than needed.
- (d) Wrong: one weighing cannot separate three suspects after an unbalanced result.
Only the provisional key (27 May 2026) was found; the final key replaces it when found.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2026, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.