Who is behind QuickerMaths?
I am Vineet Patawari - PGDM (IIM Indore), ACA, B.Com(H). My passion for Mathematics, specially Vedic Maths encouraged me to start QuickerMaths
I believe that if trained properly using powerful tools like Vedic Maths, the immense intellect of human mind can be ignited instantly - find out more
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June 12th, 2010 - 13:27
4 WIEGHINGS REQUIRED
October 1st, 2009 - 21:34
But if a heavier coin is in a group of two then?
July 15th, 2009 - 12:52
Heyy Vishal congratulations!!!!!!!..you are absolutely correct.
and Ranjana your answer is also right but you did not give the solution.Thats why credit goes to Vishal.
July 14th, 2009 - 11:20
The answer to the above question is 3 weighing.
1) First divide the coins in 3 equal parts of 8 each.
2) Then compare the 2 parts on the beam balance and check any one of them is heavier or both are equal. You will come to know about that group of 8 coins which is heavier than other 2 groups.
3) Then, divide these eight coins in 3 groups of 3, 3 and 2 coins. Weigh 3 and 3 on the beam balance and if anyone of them is heavier, divide them in 3 parts of 1 each and compare to get a final heavy coin.
4) If the two groups of 3 coins each are of same weight then compare the remaining 2 coins, by putting one coin on each side of the balance.
5) Finally you will get the heavier coin in 3 weighing.
July 13th, 2009 - 17:21
3 weighings