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Vemuri's Brain Teaser-1

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by mathbyvemuri » Tue May 01, 2012 6:16 am
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

Check the Brain Teaser-2: https://www.beatthegmat.com/vemuri-s-bra ... 11675.html
Last edited by mathbyvemuri on Wed May 09, 2012 1:03 am, edited 1 time in total.
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by shantanu86 » Tue May 01, 2012 7:22 am
IMO [A]..

Will not discuss, to save it from being ruined ;)
If you feel like it, hit thanks :)
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by mathbyvemuri » Tue May 01, 2012 8:52 am
Shantanu, it's a good gesture from you
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by neelgandham » Sun May 06, 2012 12:21 pm
I concur with shantanu ! I go with A
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by klmehta03 » Sun May 06, 2012 12:31 pm
First take 3 bags on each side
If they have same weight than only one more weighing would be req. to identify the lighter bag.

second scenario
if the weight on both sides is not same than take one bag on each side and identify the lighter bag.
IMO A
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by Bill@VeritasPrep » Sun May 06, 2012 12:39 pm
I love brainteasers! I would also go with A
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by mathbyvemuri » Mon May 07, 2012 9:43 pm
Yes, as most said, option "A" is the right one. Thanks for the replies.
And here is the detailed explanation for the problem:

This one 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 it comes down to 4 bags from 8 bags. Repeat this process for two lots, each lot containing two bags. Then it comes 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.

Answer(A)
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