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(#M40002268) TCS Numerical Ability question Keep an EYE

In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin

Asked In TCS xyz (13 years ago)

Solved Jai Read Solution (7)

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Read Solution (7) :

y.veera sekhar rao  (13 years ago)

80
=================================
27 27 26
====================================
9 9 9
====================================
3 3 3
===================================
1 1 1

so weighings are 4

Is this solution Helpfull? Yes (19) | No (10)  

lakshita  (13 years ago)

it will be 5.
make groups like
40,40
then 20,20
then 10 ,10
then 5,5
then in case to tell minimum no of cases...divide them in 2,2
they will weigh equal
the one left with us will be required coin.

Is this solution Helpfull? Yes (16) | No (7)  

Jai  (13 years ago)

minimum no. of weighings is 6
1)40 - 40 (lot with counterfeit coin is taken)
2) 20 - 20 (lot with counterfeit coin is taken)
3) 10 - 10 (lot with counterfeit coin is taken)
4) 5 - 5 (lot with counterfeit coin is taken)
5) 2 - 2 (lot with counterfeit coin is taken) (or if both the lots are same the left out is counterfeit coin)
6) 1 - 1 (one which lesser weight is counterfeit coin)

Is this solution Helpfull? Yes (9) | No (4)  

Rajeev Dubey  (13 years ago)

1st40-40
2nd20-20
3rd10-10
4th5-5
5th2-2
if needed 1-1

Is this solution Helpfull? Yes (3) | No (6)  

Dhandayuthapani  (13 years ago)

1st 39 39 2
2nd 13 13 13
3rd 5 5 3
4th 2 2 1
5th 1 1

Is this solution Helpfull? Yes (2) | No (7)  

Minal  (12 years ago)

make 4 lots of 20..den compare first two..and other two..one of these will bi unequal..say 20(1)=20(2) and 20(3)!=20(4). now compare 20(1) with 20(3), if they are equal 20(4) contains counterfeit coin otherwise 20(4) contains. Say 20(4) contains..till now v have done 3 weighings..now take 4 lots of 5 each out of 20..similarly you will get 5 coins in 3 weighings..till now v have done 6 weighings now compare 4 out of these 5 coins with any coin from other lots..1 showing imbalance is counterfeit otherwise 5th one is. this way it requires 10 weighings :)

Is this solution Helpfull? Yes (1) | No (4)  

Subhradip Biswas  (13 years ago)

Most probably 6 times but not sure.....

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