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## Interview Question

Software Engineer Interview(Student Candidate) Bangalore (India)

# There are 100 balls out of which 99 are of the same weight

but one is heavier. In how many minimum attempts you can find the heavier ball?
Tags:

1

5

Mayank on Jun 23, 2012
0

The obvious way is to break them up in two groups, and weigh those against each other. Choose the heavier group and repeat. If there is an odd number, add one ball from a previous weigh. Each side will contain the following in consecutive weighs : 50 , 25, 13, 7, 4, 2, 1. so 7 weighs in total.

A less obvious way would be to divide each group in to 3 groups. Weigh two against each other. If one is heavier choose that group and continue. If both sides are equal, then the heavier ball is in the group that wasn't weighed.
The largest group in each weigh in the WCS - 34, 12, 4, 2, 1 - so 5 weighs.

Roman on Jul 18, 2012
0

5

Vinay on Aug 30, 2012
1

Minimum would be 2 attempts....

So you have 100 balls and only 1 ball weighs different from the rest of the 99.

Split the 100 balls into two groups. One group with 49 balls and the other with 49 balls. Weight those two groups and we are assume that they will be balanced on a scale (also assuming you could use a scale too). This leaves 2 balls left that needs to be weighed and 1 out of those 2 balls is the heavier one.

FooBar on Oct 30, 2012