Latest Infosys Aptitude Question SOLUTION: There are 27 coins and a two-pan balance. All coins has same weight except for one, which is heavier than all others. All coins looks identical. What is the minimum number of weigh

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There are 27 coins and a two-pan balance. All coins has same weight except for one, which is heavier than all others. All coins looks identical. What is the minimum number of weighing required to certainly find out heavier coin?

Asked In Infosys AISHWARYA (10 years ago)

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Brati Biswas  (10 years ago)

divide 27 coins into 3 equal sections of 9 coins each.one of these sections will contain the heavy coin.Now put the 9 coins in eah of the two plan balance and weigh it.Whichever side will weigh more,contain the heavy coin.Or if the two sides weigh equal then the other section containing 9 coins will have the heavy coin.
continue this process of dividing into 3 sections until you get 1 coin.
In this way we will weigh 3 times.
27/3=9 , 9/3=3 , 3/3=1. hence the answer.

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Rahul Sharan  (10 years ago)

I think if u do this u ll b getting in 8 steps of weighing

13 13 1 (if both 13 weigh same ,the other one will be the heavier coin or take the heavier one)
6 6 1 (if both 6 weigh same ,the other one will be the heavier coin or take the heavier one)
3 3 0 (take the heavier one)
1 1 1 (now heavier ll be seperated)

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Nupur  (9 years ago)

Divide it in 3 sets of 9.
Out of the three sets weigh any two sets. If one of them i heavy, use it further. Otherwise the third set of 9 coins contains the defective coin.
Divide this set of 9 coins into three sets of 3 coins each.
Repeat the weighing procedure.
You will get one set of three coins left to decide from.
Among these three weigh any two, if they are equal then the third coin is heavy otherwise the heavier one will be evident from the weighing balance.

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Deven Patidar  (8 years ago)

Given 27 coins
dividing in 3 parts 13,13 (for 2 pan) and 1(put outside)
start with taking 13 coins in each pan and remaining 1 placed outside.
if pan is balanced, it means coin which has placed outside is heavier. count=1

if pan is unbalanced...heavier coin will be in heavier side
so take heavier side's 13 coin (other 14 put away outside)
now taking 6 coins in each pan and 1 placed outside (not mixed with 14 outside coins)
if pan is balanced, it means single coin which has placed outside is heavier. count=2

if pan is unbalanced...if pan is unbalanced...heavier coin will be in heavier side
so take heavier side's 6 coin (other 6 put away outside-we can put it with all outside coins bcz all are of same weight)
now heavier side's 6 coin is divided into 3,3
now take heavier side's 3 coins (lighter put away outside) in each pan count=3
(one side will be heavier bcz of heavier coin)
so now take heavier side's 3 coins (lighter put away outside) and divide into 3 parts 1,1(one in each pan) and 1 (placed outside{not with other outside coin)
if pan is balanced outside's single placed coin will be heavier
if pan is unbalanced then heavier pan side's coin will be heavier count=4


ans minimum 4 comparison has required to identify the heavier coin.

ANSWER: 4

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harsh shah  (8 years ago)

this is based on concept of binary tree
first we have to group 13 Vs 13
if plan balanced then 27th coin is heavier
else make group of 6 Vs 6 from group of coin that makes plan downward,
similary 3 Vs 3
1 Vs 1
and we get ans

So ans should be 4....
hope it might help . . . .

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Laxmi Srinivas  (10 years ago)

3. If we weigh once the maximum number of coins that can be eliminated is 2/3rd

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amaresh  (10 years ago)

26 min to six

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Shilpa  (8 years ago)

1st time - 13 vs 14.
2nd time - 7 vs 7
3rd time - 4 vs 3
4th time - 2 vs 2
5th time - 1 vs 1

5 times.

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mohan  (8 years ago)

cuberootof(27)
i.e,3
(or)
(9,9,9) and (3,3,3) and (1,1,1)

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