How can you find out the ball with slight difference out of 80 balls by measurin!
Question: How can you find out the ball with slight difference out of 80 balls by measuring 4 times only!?
- 80 balls are identical in shapes for naked eye!. 79 balls
are each 50 grams weight and 1 ball is 55 grams weight!.
- by using common balance, you have to identify 55 grams weight ball by measuring only 4 times!. you can also use measuring weights!.Www@Enter-QA@Com
are each 50 grams weight and 1 ball is 55 grams weight!.
- by using common balance, you have to identify 55 grams weight ball by measuring only 4 times!. you can also use measuring weights!.Www@Enter-QA@Com
Answers:
There are different outcome each time you measure!. It takes long to explain and with out pictures, you might not see it!. Here it goes:
I will try to explain by first, going with the straight way, then I will go over those if one side is wrong during measure!.
First, put 27--27 on the scale, that leaves 26
_____________________________________
(((Scenario 1)))
27--27 = balanced so one of the 26 is wrong!. Continue on 2nd Measure!.
(If 27--27 not balanced, see scenario 2 below)
((2nd Measure)) put 9--9 on the scale, that leaves 8 (26 from 1st step -9 -9 = 8)
9--9 balanced, then 8 has the wrong one!. Continue to 3rd measure!.
(If 9--9 is not balanced, the heavier side has the wrong one, goto scenario 2 measure #3)
((3rd Measure)) put 3--3 on the scale, that leaves 2
If 3-3 balanced, one of the other 2 is wrong!.
Then, in ((4th Measure)), you just need to measure 1--1 and the heavier one is wrong!.
If 3--3 is not balanced, the heavier side is wrong!.
Then on ((4th Measure)), from the wrong set of (3) put 1--1 that leaves 1
If 1--1 is balanced, the 1 that is left out is the wrong one!.
If it's not balanced, the heavier one is the wrong one!.
______________________________________!.!.!.
Now, if things doesn't go that way on the FIRST measure, here goes:
(((Scenario 2)))
If First measure yield an un balanced situation, then the heavier side of 27 has the wrong one!.
((2nd Measure)), do the same 9--9!.
That leave you 9 (27 - 9 - 9 = 9)
If 9--9 balanced, then the remaining 9 has the wrong one!.
If 9--9 is not balanced, heavier side has the wrong one!.
Use the set of 9 that is wrong and do
((3rd Measure)), 3--3 that leaves 3
if 3-3 balanced, one of the other 3 is wrong!.
if 3--3 is not balanced, the heavier side is wrong!.
((4th Measure)), from the wrong set of (3) put 1--1 that leaves 1
If 1--1 is balanced, the 1 that is left is the wrong one!.
If it's not balanced, the heavier one is the wrong one!.
______________________________________!.!.!.
You might have to draw the situations on a piece of paper to see for yourself!.
Ok, heres the diagram!. From first measure, follow to the right if balanced!.
Whenever not balanced, go down, and keep going down from there(balance or not)!.
http://www!.geocities!.com/gamblermaniac/8!.!.!.Www@Enter-QA@Com
I will try to explain by first, going with the straight way, then I will go over those if one side is wrong during measure!.
First, put 27--27 on the scale, that leaves 26
_____________________________________
(((Scenario 1)))
27--27 = balanced so one of the 26 is wrong!. Continue on 2nd Measure!.
(If 27--27 not balanced, see scenario 2 below)
((2nd Measure)) put 9--9 on the scale, that leaves 8 (26 from 1st step -9 -9 = 8)
9--9 balanced, then 8 has the wrong one!. Continue to 3rd measure!.
(If 9--9 is not balanced, the heavier side has the wrong one, goto scenario 2 measure #3)
((3rd Measure)) put 3--3 on the scale, that leaves 2
If 3-3 balanced, one of the other 2 is wrong!.
Then, in ((4th Measure)), you just need to measure 1--1 and the heavier one is wrong!.
If 3--3 is not balanced, the heavier side is wrong!.
Then on ((4th Measure)), from the wrong set of (3) put 1--1 that leaves 1
If 1--1 is balanced, the 1 that is left out is the wrong one!.
If it's not balanced, the heavier one is the wrong one!.
______________________________________!.!.!.
Now, if things doesn't go that way on the FIRST measure, here goes:
(((Scenario 2)))
If First measure yield an un balanced situation, then the heavier side of 27 has the wrong one!.
((2nd Measure)), do the same 9--9!.
That leave you 9 (27 - 9 - 9 = 9)
If 9--9 balanced, then the remaining 9 has the wrong one!.
If 9--9 is not balanced, heavier side has the wrong one!.
Use the set of 9 that is wrong and do
((3rd Measure)), 3--3 that leaves 3
if 3-3 balanced, one of the other 3 is wrong!.
if 3--3 is not balanced, the heavier side is wrong!.
((4th Measure)), from the wrong set of (3) put 1--1 that leaves 1
If 1--1 is balanced, the 1 that is left is the wrong one!.
If it's not balanced, the heavier one is the wrong one!.
______________________________________!.!.!.
You might have to draw the situations on a piece of paper to see for yourself!.
Ok, heres the diagram!. From first measure, follow to the right if balanced!.
Whenever not balanced, go down, and keep going down from there(balance or not)!.
http://www!.geocities!.com/gamblermaniac/8!.!.!.Www@Enter-QA@Com