Accountancy, asked by rohitrkamble2003, 11 months ago

If you have 7 identical coins,out of which 6 coins are heavier than that of the remaining 1. How would you find that coin if the only thing you have is a weighing machine(not electrical)...which has place to keep objects on both sides!
But you can weigh only 2 times

Answers

Answered by sneharekhi14
0

Divide the coins in two equal groups(that is of 3 coins each) and separate the left out 7th coin

In your first turn

Check which group is heavier by using a physical balance

If the physical balance is in equilibrium then the left coin is the heavier one

Else

Take the heavier group and separate all the 3 coins

In your second turn

Place any two coins one on each side of physical balance

If the physical balance is in equilibrium then the coin kept out side is the heavier one

If the physical balance is not in equilibrium then the lower side of the physical balance has the heavier coin

Hope it HELPS

Plzzzz mark as BRAINLIEST !

Answered by ShibamNath2004
0

Answer:

First, divide the coins in three groups consisting of 2, 3 and 2 each.

Now weigh those two groups consisting 2 coins each. If any one group weighs more, then weigh those two coins of that group individually.

If both weigh same, then leave them and weigh any 2 coins from the third group. The one which will be heavier is the odd one out. Or if both weigh same, then the left over coin is the odd one out.

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