Friday, 16 March 2012


There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is
(A)  2         (B)    3        (C)   4           (D)   8
Solution follows here:

This is a good one and needs extension of our thought process.
Most general mistake done here is dividing all the eight bags in to 2 lots of four each, weighing them by placing each lot on one of the two sides of the balance. This makes us to identify the lot with heavier bag. Now we came down to 4 bags from 8 bags. Repeat this process for two lots, each lot containing two bags. Then we will come down to two bags and one more iteration of the process gives us the final result of finding the odd bag out. This process involves three stages of weighing which is not the minimum possible one.
Right approach:
Keep two bags aside and consider two lots of three bags each and weigh them. Go to Case-I in case of unequal-weighing lest go to case-II.

Unequal weighing hints that the odd one is in a set of three bags. Then keeping one bag aside, and weighing the rest two, with one on each side, reveals the odd-bag. 
Altogether, this process involves two stages of weighing.

Equal weighing of three-three bags on each side hints that the odd one is in the set of two bags kept aside. Then proceed with weighing of those two bags and get the odd-bag out. 
Altogether, this process involves two stages of weighing only.

No comments:

Post a Comment