Puzzle: Of wine and water…
There is a wine glass filled with 1000 spoonful of wine. There is a water beaker with 2000 spoonful of water.
You take a spoon of water from the beaker and put it in the wine glass and mix it thoroughly. Then you take one spoon of the stuff (mostly wine with a little water) from the wine glass and put it in the water beaker and mix it thoroughly.
You do the whole thing over again. And one more time (so total of three times).
Of course increasingly the water beaker is getting more wine and vice versa.
Question is at the end of the third full round, do you have more wine (by volume) in the water beaker or more water (by volume) in the wine glass and by how much?
Here is a tougher version… what if after the above three cycles, you did the water beaker to wine glass one more time and stopped? (You did not do the wine glass to water beaker to complete the cycle; essentially three and a half cycles). Now, do you have more wine (by volume) in the water beaker or more water (by volume) in the wine glass and by how much?
Instead of putting your answer (and how you arrived at it) in the Comments section, write to me directly in messenger.
To begin with wine already has 85% of water in it, so water beaker will never be able to beat wine.
Abhishek is the first one to send the correct answer
Answer:
In the first case, the amount of wine in the water beaker is going to be exactly the same as water in the wine glass. In eighth grade, I remember our math teacher (Mr. Nandi) had given a similar problem in the Alligation chapter from K.C. Nag’s math book. I had done the whole ratio proportion thing and came to the same answer. Now I know there is a very simple way to find the answer:
First, let’s realize that after the three operations, the wine glass has the original volume of liquid (1000 spoonful) and the water beaker also has its original volume of liquid (2000 spoonful). (same volume as they started but of course the liquids they contain are neither pure wine nor pure water).
Let’s say the wine glass has “x” spoonful of water in it now. Then the wine in it has to be (1000-x) spoonful. (total in the glass is 1000).
So the amount of wine NOT present in the glass is 1000 – (1000-x) (original amount of wine was 1000) = x.
Where has that wine gone? Obviously it is sitting in the water beaker.
Thus the amount of water in the wine glass is exactly the same as the amount of wine in the water beaker.
Next part is a little tricky. After your “three and a half” round,
Wine glass has 1001 spoonful of liquid and water beaker has 1999 spoonful of liquid.
Let’s say the amount of water in the wine glass in “x” again.
Then the amount of wine is (1001-x).
The amount of wine missing is 1000 – (1001-x) = x-1 And that is sitting in the water beaker.
So the water in the wine glass will be 1 spoonful more than the wine in water glass.
The generalization is this:
1. Initially, when you put one spoonful of water in the wine glass, the amount of water in the wineglass (1 spoonful) is 1 spoon more than the wine in water (initially 0).
2. When you do the next exchange, the wine in water beaker and water in wine glass is same.
And as you keep doing, the answers toggle between 1 and 2 above!!
Answer:
In the first case, the amount of wine in the water beaker is going to be exactly the same as water in the wine glass. In eighth grade, I remember our math teacher (Mr. Nandi) had given a similar problem in the Alligation chapter from K.C. Nag’s math book. I had done the whole ratio proportion thing and came to the same answer. Now I know there is a very simple way to find the answer:
First, let’s realize that after the three operations, the wine glass has the original volume of liquid (1000 spoonful) and the water beaker also has its original volume of liquid (2000 spoonful). (same volume as they started but of course the liquids they contain are neither pure wine nor pure water).
Let’s say the wine glass has “x” spoonful of water in it now. Then the wine in it has to be (1000-x) spoonful. (total in the glass is 1000).
So the amount of wine NOT present in the glass is 1000 – (1000-x) (original amount of wine was 1000) = x.
Where has that wine gone? Obviously it is sitting in the water beaker.
Thus the amount of water in the wine glass is exactly the same as the amount of wine in the water beaker.
Next part is a little tricky. After your “three and a half” round,
Wine glass has 1001 spoonful of liquid and water beaker has 1999 spoonful of liquid.
Let’s say the amount of water in the wine glass in “x” again.
Then the amount of wine is (1001-x).
The amount of wine missing is 1000 – (1001-x) = x-1 And that is sitting in the water beaker.
So the water in the wine glass will be 1 spoonful more than the wine in water glass.
The generalization is this:
1. Initially, when you put one spoonful of water in the wine glass, the amount of water in the wineglass (1 spoonful) is 1 spoon more than the wine in water (initially 0).
2. When you do the next exchange, the wine in water beaker and water in wine glass is same.
And as you keep doing, the answers toggle between 1 and 2 above!!