You have 28 balls of which 27 weigh equal and 1 is heavier than the rest . you are given a weighing machine ?
How will you weigh the heavier ball and what is the min no of 3 chances required.Explain with logical reasoning
- NCSLv 71 month ago
Using @dixon's procedure you could identify the heavy ball in three weighings as long as it isn't the 27th or 28th ball. In that case it would take a fourth weighing to discover which one was heavier.
Specifically: weigh the first two sets of nine. They're equal. Discard those 18 balls, leaving 9 + 1. Weigh two sets of three. They're equal. Discard those 6 balls leaving 3 + 1. Weigh one against another. They're equal. You're left with two balls, requiring a fourth weighing.
To guarantee it can be done in n weighings, you can have no more than 3ⁿ balls. We have 3³+1 = 28 balls, so no guarantee of 3 weighings can be made. Might get lucky, might not.
- DixonLv 71 month ago
Split into three equal groups of 9 + 9 + 9
Weigh two groups against each other.
Looking at the outcome you can deduce which group of 9 includes the heavy ball.
Repeat by splitting that group into 3 + 3 + 3, then again with 1 + 1 + 1