SOLUTION: How much water must be added to 400L of mixture that is 80% alcohol to reduce it to a 60% mixture?

Question 936149: How much water must be added to 400L of mixture that is 80% alcohol to reduce it to a 60% mixture?
Found 3 solutions by josgarithmetic, srinivas.g, TimothyLamb:
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
http://www.algebra.com/my/Two-Part-Mixture-with-one-material-quantity-unknown.lesson?content_action=show_dev

The lesson explains thoroughly how this works.
Answer by srinivas.g(540)   (Show Source): You can put this solution on YOUR website!
initial quantity of mixture = 400 L
alcohol % = 80 %
quantity of alcohol in the mixture = 80 % of 400 L
=
=
= L
Let X be the quantity of water added to the mixture
final quantity of the mixture = 400+ x
alcohol 5 in final mixture = 60 %
quantity of alcohol in final mixture = 60 % of (400+x)
=
=
quantity of alcohol in the initial mixture and final mixture is same

multiply with 5 on both sides

move 1200 to the left

divide with 3 on both sides

Result: Quantity of water added = 133.33 L
Answer by TimothyLamb(4379)   (Show Source): You can put this solution on YOUR website!
x = liters of 80% alcohol/water = liters of 20% water/alcohol
y = liters of pure water
z = liters of 60% alcohol/water = liters of 40% water/alcohol
---
x = 400
z = x + y
z = 400 + y
0.20x + 1.00y = 0.40z
0.20*400 + 1.00y = 0.40*(400 + y)
0.20*400 + y = 0.40*400 + 0.40y
y - 0.40y = 0.40*400 - 0.20*400
0.60y = 0.40*400 - 0.20*400
y = (0.40*400 - 0.20*400)/0.60
y = 133.333333333
---
answer:
y = liters of pure water = 133 1/3
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