You have 8 balls. One of them is defective and weighs less
than others. You have a balance to measure balls against
each other. In 2 weighings how do you find the defective one?
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Answer / balbir singh
Firstly we take 6 balls, placing 3 on each pan of the
balance & check which pan goes up(1st weighing). If no
change it means the balls don't contain the defective
one,therefore the other two contain the defective
ball,which can be found out easily(2nd weighing).
Now if one of the pan goes up,it means it contains the
defective ball(out of 3 balls).We select these 3 & unload
the pans. Now we keep any of the 2 balls(from the selected
balls)on either side of the balance & weigh(2nd weighing).
One of these two will be the defective one and if not the
3rd one leftover is.
| Is This Answer Correct ? | 67 Yes | 6 No |
Answer / savitha shreyas
Firstly we take 6 balls, placing 3 on each pan of the
balance & check which pan goes up(1st weighing). If no
change it means the balls don't contain the defective
one,therefore the other two contain the defective
ball,which can be found out easily(2nd weighing).
Now if one of the pan goes up,it means it contains the
defective ball(out of 3 balls).We select these 3 & unload
the pans. Now we keep any of the 2 balls(from the selected
balls)on either side of the balance & weigh(2nd weighing).
One of these two will be the defective one and if not the
3rd one leftover is.
| Is This Answer Correct ? | 20 Yes | 2 No |
Answer / abi
Firstly we take 6 balls, placing 3 on each pan of the
balance & check which pan goes up(1st weighing). If no
change it means the balls don't contain the defective
one,therefore the other two contain the defective
ball,which can be found out easily(2nd weighing).
Now if one of the pan goes up,it means it contains the
defective ball(out of 3 balls).We select these 3 & unload
the pans. Now we keep any of the 2 balls(from the selected
balls)on either side of the balance & weigh(2nd weighing).
One of these two will be the defective one and if not the
3rd one leftover is.
| Is This Answer Correct ? | 8 Yes | 1 No |
Answer / aby
Firstly we take 4 balls, placing 2 on each pan of the
balance & check which pan goes up(1st weighing). If no
change it means the balls don't contain the defective
one,therefore the other Group of 4 balls contain the
defective
ball,Now if one of the pan goes up,it means it contains the
defective ball(out of 2 balls).We select these 2 & unload
the pans. Now we keep 1 ball(from the selected
balls)on either side of the balance & weigh(2nd weighing).
One of these two will be the defective one otherwise it
must b in the second group of balls.
| Is This Answer Correct ? | 3 Yes | 11 No |
Answer / rajeev kumar
Solution:
a. first take 4-4 group of balls and weight this. Then one
of group will be less.
b. next we take less weight group and take 2-2 group and
weight this and choose less weight group.
c. next take 1-1 group and weight choose less weight group.
and there is one ball.
So this ball is defective ball.
| Is This Answer Correct ? | 6 Yes | 25 No |
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