Weighing Puzzles— Concepts, Formulas & Shortcuts
- A balance has three outcomes per weighing, so n weighings distinguish up to 3ⁿ possibilities.
- Finding one heavier ball among n takes ⌈log₃ n⌉ weighings — 2 for up to 9 balls, 3 for up to 27.
- Always split into three roughly equal groups, not two.
- If the odd item may be heavier OR lighter, the capacity halves: 3 weighings handle up to 12 balls.
- With weights allowed on BOTH pans, powers of 3 (1, 3, 9, 27) let you weigh every integer up to 40 kg.
- With weights allowed on only ONE pan, you need powers of 2 (1, 2, 4, 8, 16, 32).
Weighing Puzzles Practice Questions with Answers
Attempt each question first, then open the explanation. All 6 questions below are free to read and require no signup.
Q1.You have 8 identical-looking balls, one of which is heavier. What is the minimum number of weighings on a balance scale to find it?
Moderate- A1
- B2
- C3
- D4
Weighing Puzzles question 1 of 6+Show Answer & Explanation
Answer: B. 2
Explanation: Weigh 3 against 3. If they balance, the heavy ball is among the remaining 2 and one more weighing settles it; otherwise weigh 1 against 1 from the heavier group.
Q2.With 9 balls, one of which is heavier, the minimum number of weighings needed is:
Moderate- A2
- B3
- C4
- D5
Weighing Puzzles question 2 of 6+Show Answer & Explanation
Answer: A. 2
Explanation: Split into three groups of 3. One weighing finds the heavy group; a second finds the ball. 3² = 9 exactly.
Q3.Among 12 balls one differs in weight but you do not know whether it is heavier or lighter. The minimum number of weighings is:
Difficult- A2
- B3
- C4
- D5
Weighing Puzzles question 3 of 6+Show Answer & Explanation
Answer: B. 3
Explanation: There are 24 possibilities (12 balls × 2 directions) and 3³ = 27 ≥ 24, so three carefully chosen weighings suffice.
Q4.What is the minimum number of weights needed to weigh any whole number of kilograms from 1 to 40 on a two-pan balance, if weights may be placed on both pans?
Difficult- A3
- B4
- C5
- D6
Weighing Puzzles question 4 of 6+Show Answer & Explanation
Answer: B. 4
Explanation: Weights of 1, 3, 9 and 27 kg work, because each may be added, omitted or placed opposite — balanced ternary covering 1 to 40.
Q5.With 27 balls, one heavier than the rest, the minimum number of weighings is:
Moderate- A2
- B3
- C4
- D9
Weighing Puzzles question 5 of 6+Show Answer & Explanation
Answer: B. 3
Explanation: 3³ = 27, so three weighings of three equal groups are sufficient.
Q6.If weights may be placed on only ONE pan, which set weighs every integer from 1 to 31 kg?
Difficult- A1, 3, 9, 27
- B1, 2, 4, 8, 16
- C1, 5, 10, 25
- D2, 4, 8, 16, 32
Weighing Puzzles question 6 of 6+Show Answer & Explanation
Answer: B. 1, 2, 4, 8, 16
Explanation: With one-sided placement each weight is either used or not, so binary applies: 1 + 2 + 4 + 8 + 16 = 31 covers every value in the range.
Weighing Puzzles — Frequently Asked Questions
Why divide into three groups instead of two?+
Because a balance returns three distinct outcomes, not two. Splitting into thirds extracts the maximum information per weighing, which is what makes 9 balls solvable in 2 weighings instead of 3.
Why are the weights 1, 3, 9 and 27 enough for 1–40 kg?+
Because weights may be placed on either pan, each weight can contribute +1, 0 or −1. That is balanced ternary, and four powers of 3 cover every integer from 1 to 40.
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