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Guest
Guest
Hello,
I was recently reading a lateral thinking book, and I came about this problem. I’ve been racking my brain trying to figure it out. Here is the problem
There are two glasses of equal size, glass A and B. A is empty and B contains some water. Half of glass B is poured into A. Once again, half of what is in B is poured into A. Once more, half of glass B is poured into glass A. At the end of 3 pourings, A is half full. How much water is left in B?'
Here is what I have so far...
Say X is the original amount of water in glass B
1/2X + 1/4X + 1/8X = 7/8 X
You see, however much water was in glass B to start, half is left after one pouring, a quarter is left after the second pouring and an eighth is left after the third pouring. So seven eights has been transferred.
..BUT WHAT DO I DO KNOW?? Please help me solve this sucker, and please show your steps algebraically (as best you can) and provide reasoning. Thanks!!
-Ryan
I was recently reading a lateral thinking book, and I came about this problem. I’ve been racking my brain trying to figure it out. Here is the problem
There are two glasses of equal size, glass A and B. A is empty and B contains some water. Half of glass B is poured into A. Once again, half of what is in B is poured into A. Once more, half of glass B is poured into glass A. At the end of 3 pourings, A is half full. How much water is left in B?'
Here is what I have so far...
Say X is the original amount of water in glass B
1/2X + 1/4X + 1/8X = 7/8 X
You see, however much water was in glass B to start, half is left after one pouring, a quarter is left after the second pouring and an eighth is left after the third pouring. So seven eights has been transferred.
..BUT WHAT DO I DO KNOW?? Please help me solve this sucker, and please show your steps algebraically (as best you can) and provide reasoning. Thanks!!
-Ryan
