There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Five cups of alcohol from the first container is taken out and is mixed well in the second container. Then, five cups of this mixture is taken out from the second container and put back into the first container. Let X and Y denote the proportion of alcohol in the first and the proportion of water in the second container. Then what is the relationship between X & Y? (Assume the size of the cups to be identical). Need answer with explain please.
Answers
Given : the first contains 500 ml of alcohol, while the second contains 500 ml of water. Five cups of alcohol from the first container is taken out and is mixed well in the second container. Then, five cups of this mixture is taken out from the second container and put back into the first container
To find : X and Y denote the proportion of alcohol in the first and the proportion of water in the second container. Then what is the relationship between X & Y
Solution:
Initially
1st Container = 500 ml alcohol
2nd container = 500 ml water
Five cups of alcohol from the first container is taken out and is mixed well in the second container
Let say cup size = C mL of course 0 ≤ C ≤ 100
1st container = 500 - 5C mL alcohol
2nd Container = 500 ml Water + 5C mL alcohol
Mixture in 2nd container = 500 + 5c mL
five cups of this mixture is taken out from the second container and put back into the first container.
Water taken out = (500/(500 + 5C)) * 5C = 500C/(100 + C)
Alcohol taken out = 5C - 500C/(100 + C) = 5C²/(100 + C)
1st container alcohol = 500 - 5C + 5C²/(100 + C)
= 5 ( 100² - C² + C²)/ (100 + C)
= 50000/ (100 + C)
X = (50000/ (100 + C)500
=> X = 100/(100 + C)
2nd container water = 500 - 500C/(100 + C)
= 50000/(100 + C)
Y = 50000/(100 + C)500
=> Y = 100/(100 + C)
X : Y = 100/(100 + C) : 100/(100 + C)
=> X : Y = 1 : 1
=> X & Y are equal
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