SOLUTION: 8. A mixture with 60% orange content, what % of water should be added to make this mixture as 40%?

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Question 264388: 8. A mixture with 60% orange content, what % of water should be added to make this mixture as 40%?
Found 2 solutions by drk, vksarvepalli:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
This is a mixture problem. we can set up a table based on given information
liquid . . . . . . % . . . . . . L . . . . . . .%L
orange. . . . . .60 . . . . . x . . . . . . .60x
water. . . . . . . 0 . . . . . . y. . . . . . . 0y
mixture . . . . . 40 . . . . . x+y. . . . .40x + 40y
Looking at the third column, we get
60x+=+40x+%2B+40y
or
20x+=+40y
dividing, we get
x+=+2y
This means what ever the amount of orange juice is, the amount of water must be twice that.
we don't have enough information to answer with actual hard numbers.

Answer by vksarvepalli(154) About Me  (Show Source):
You can put this solution on YOUR website!
let there is 100 ml of original solution
in that there is 60% that is 60 ml of orange content
now suppose x ml of water is added
so total it is 100+x ml and orange content is 60 ml
now this 60 ml should constitute 40% of 100+x
60=(40/100)(100+x)
60=40+.4x
.4x=20
x=50 ml
so 50% of water should be added