8 coins

Question:

Suppose you have 8 coins, all of equal size. 7 of them are of equal weight and one of them is heavier. If it takes minimum of x tries to find out the coin of a different weight then what is the value of x?

.

.

C

A

P

T

A

I

N

I

N

T

E

R

V

I

E

W

.

.

Solution:

Only 2 tries First divide 8 coins in 3 – 3 – 2 piles. compare the weight of the equal number of piles (3 – 3). now if they are of equal weight then the coin with a heavier weight will be in the third pile (of 2 quantity). so, in second try you can decide which one is heavier. Now, suppose in the first try, the two piles of 3 quantity are not of equal weight then the heavier coin will be in the heavier pile. so now you will have a pile of 3 coins. so, in second try you can take 2 coins from this pile and compare the weight. so, if the weights are equal them the heavier will be remaining one and if not equal then the heavier coin is whichever heavy on those two coins. So, you are done with 2 trials.

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