There are eight identical looking balls, but one of them is heavier than the others. Using a balance scale only twice, which strategy will always allow you to find the overweight ball?

Introduction / Context:
This classic reasoning puzzle involves eight identical looking balls where exactly one is heavier than the others. You are given a balance scale, which can compare the weights of two groups of balls, and you are allowed to use the scale only twice. The challenge is to find a strategy that guarantees you will identify the heavier ball in these two weighings. This type of problem is common in aptitude tests and measures logical planning and optimal use of limited information.

Given Data / Assumptions:

• There are 8 balls that look the same.
• Exactly one ball is heavier than the rest; all others have equal, lighter weight.
• You have a balance scale that indicates which side is heavier or whether the two sides balance.
• You may use the scale at most two times.
• You must always be able to identify the heavier ball, regardless of which one it is.

Concept / Approach:
The key concept is to divide the balls into groups so that each weighing gives maximum useful information. Because one ball is heavier, the first weighing should split the set in a way that either isolates the heavy ball in a smaller suspect group or shows that it lies outside the weighed balls. Dividing the balls into groups of 3, 3, and 2 allows you to use the first weighing to decide whether the heavy ball is among 6 balls on the scale or among 2 balls off the scale. The second weighing then narrows down the suspect group to a single ball.

Step-by-Step Solution:
Step 1: Label the balls conceptually as A, B, C, D, E, F, G, and H. Step 2: For the first weighing, place A, B, and C on the left pan and D, E, and F on the right pan. Step 3: Observe the result of the first weighing. There are two possibilities: the scale balances, or one side is heavier. Step 4: If the scale balances, then all six balls A through F are of normal weight, and the heavier ball must be among G and H. Step 5: For the second weighing in this balanced case, compare G against H. The heavier side contains the overweight ball. Step 6: If, in the first weighing, one side is heavier (for example, left side heavier with A, B, C), then the heavy ball must be among those three balls because the other three on the lighter side must be normal. Step 7: For the second weighing in this unbalanced case, pick any two balls from the heavier group, say A and B, and compare A against B. Step 8: If A and B balance, then C is the heavier ball. If one of them is heavier, that ball is the overweight one.
Verification / Alternative check:
To verify the strategy, examine all possible locations of the heavy ball. If it is G or H, the first weighing will balance, and the second weighing between G and H will identify it. If it is A, B, C, D, E, or F, then the first weighing will tilt towards the side containing that ball. By limiting the suspect group to three and then doing a simple comparison between two of them, the second weighing always isolates the heavy one. In every scenario, you have an unconditional method to find the overweight ball in no more than two weighings.

Why Other Options Are Wrong:
Weighing 4 balls against 4 balls as in option A can tell you which group contains the heavy ball, but then you still have four candidates and only one weighing left, which is not enough to guarantee a unique identification. Option C with 2 against 2 then 2 against 2 splits information poorly and does not necessarily isolate a single ball in two weighings. Option D is not a correct use of a balance scale and would require far more than two weighings or would not give any clear comparison at all.

Common Pitfalls:
A common mistake is to divide the balls into two equal groups of four, which feels natural but does not optimise information for a second weighing. Another pitfall is to ignore the fact that the scale gives a trinary outcome (left heavy, right heavy, or balance) and to treat it as if it only gave one bit of information. Recognising that a 3, 3, and 2 split makes full use of the scale outcomes is essential. By planning carefully before touching the scale, you ensure that every possible result leads you closer to the solution.

Final Answer:
The correct strategy is: First weigh 3 balls against 3 balls. If they balance, weigh the remaining 2 against each other; if not, weigh 1 against 1 from the heavier group of 3.