Latest TCS Aptitude Question SOLUTION: In a group of 80 coins,exactly one is counterfeit and weighs less than others. You are provided to weigh the coins. The minimum number of weighing required to determine counterfeit

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In a group of 80 coins,exactly one is counterfeit and weighs less than others. You are provided to weigh the coins. The minimum number of weighing required to determine counterfeit coin is?

Asked In TCS PRIYA NAMDEO (10 years ago)

Solved Mahesh Read Solution (5)

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Mahesh  (10 years ago)

The weighing method is,
Make the three groups as 27, 27 and 26
Compare two 27 groups, if they are equal, the wrong coin is in the third group.
If the first 27 group is heavier, the wrong coin is in the secong 27 group.
If the first 27 group is lighter , the wrong coin is in the first 27 group

After the first attempt, we could figure out, the group of 27 or 26 coins , where the wrong coin is.

Now, the selected 27 group should be made as 9, 9 and 9 groups(Make 9,9 and 8 if 26 is selected)

Of the three 9 groups, compare the first 9 group and the second 9 group, if they are the same, the wrong coin is in the third group.
If they are not the same, we can select the corresponding group of 9 coins, where the wrong coin is

So, after the second usage of the balance, we can finalise, the 9 coins group.

In the third usage, make them groups of three each and we can find the three coin group by using the balance for the third time.
For the fourth time, we are left with three coins.
Compare one each in each side of the balance, if they are the same, we know the third one is the wrong coin.If they are not the same , we know which one of them is the wrong coin.
Answer is 4 times.

Is this solution Helpfull? Yes (27) | No (2)  

Mahesh  (10 years ago)

This question follows a pattern,
Upto 3^n coins, we can find out the wrong one ( the heavier or the lighter coin as mentioned in the question) , by using the balance for (n) times.

For any number up to 3, we have to use the balance for once
For any number up to 9, we have to use the balance for 2 times.
For any number up to 81, we have to use the balance for 3 times

For any number up to 81, we have to use the balance for 4 times
For any number up to 243, we have to use the balance for five times

So, the answer is 4 times

Note: The pattern will be different if we do not know the coin as lighter or heavier

Is this solution Helpfull? Yes (10) | No   

kratika jain  (10 years ago)

6 turns....
1. 80 (divide in two parts 40 40)=take less one
2. 40(less) (divide into 2 parts 20 20)= take less one
3. 20(less) (divide into 2 parts 10 10)= take less one
4. 10(less) (divide into 2 parts 5 5)= take less one
5. 5(less) (divide into 2 parts 2 2)= equal then 5th one left is defected one
inequal then take less one
6. 2(less) (weigh 1 and 1) can detect defeted one

thus ans =6

Is this solution Helpfull? Yes (9) | No (6)  

pankaj bajaj  (10 years ago)

14 time weighing required
bcoz first we divides coin in 40+40 n weight them n find which 40 coins has that coin
now these 40 divided into 20+20
then in 10+10
then in 5+5
now we add a coin n divide 3+3
now again add one coin n 2+2
n finally we got that coin

Is this solution Helpfull? Yes | No (5)  

ansh tiwari  (10 years ago)

14 times , becouse any thin 80

Is this solution Helpfull? Yes | No (3)  

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