IIM B UG (B.Sc-Hons.)- Sample Paper 2026 · Logical Reasoning · Question 9

Each test has only two possible outcomes — the faulty cable is in the group tested, or it is not.

So after n tests there are at most 2ⁿ different patterns of answers, and each pattern can point to at most one cable.

To separate 100 cables we therefore need 2ⁿ ≥ 100.

2⁶ = 64, which is too few, and 2⁷ = 128, which is enough.

So at least 7 tests are needed.

Seven also suffice: test half the cables, keep whichever half the answer points to, and repeat. Each test halves the number of candidates, so 100 → 50 → 25 → 13 → 7 → 4 → 2 → 1.

The least number of tests that guarantees success is 7.