mathbyvemuri
Branch Unspecified
14 May 2012

# Vemuri's Brain Teaser-1

I want to present a series of brain teaser problems collected from various sources, for the benefit of GMAT takers looking for some tougher-level questions. Here it goes the first one.....

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) 5 (E) 8
6 years ago
mathbyvemuri : The approach to this problem changes depending upon the type of balance.

mathbyvemuri

Branch Unspecified
6 years ago
The_Big_K
mathbyvemuri : The approach to this problem changes depending upon the type of balance.
The_Big_K,
Yes, I got your point, it's a simple common balance

Ankita Katdare

Computer Science
6 years ago
The answer should be a = 2.

Process:
Divide the 8 bags into groups od 3, 3 and 2
First compare 3 and 3. If it is equal, then the heavy bag is in the next group i.e. of 2.
If it is not equal, then we take the group that weighs more. Keep one bag in it aside and weigh the rest of the 2. This will give the answer.

mathbyvemuri

Branch Unspecified
6 years ago
@ AbraKaDabra,you are right. For the sake of all I want to elaborate on the approach to solve it:
This 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.

Case-I
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.

Case-II
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.