SOLUTION: I'm a 66 year old man with a debate with my wife. I say the mixture would be 4 to 1 she disagrees. A 40% solution is to be mixed with water how much water is needed to make the so

Question 1011852: I'm a 66 year old man with a debate with my wife. I say the mixture would be 4 to 1 she disagrees. A 40% solution is to be mixed with water how much water is needed to make the solution 10%?
Found 5 solutions by ikleyn, Fombitz, stanbon, Alan3354, MathTherapy:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
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I'm a 66 year old man with a debate with my wife. I say the mixture would be 4 to 1 she disagrees.
A 40% solution is to be mixed with water how much water is needed to make the solution 10%?
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3 volumes of water per volume of 40% solution will give you 10% solution.

The wife is always right.

This is why. = 0.1. V i the volume of your original 40% solution. W is the volume of water to be added. Solve it: 0.4V = 0.1V + 0.1W, 0.3V = 0.1W = = 3.

Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Assume that you have 1 cup of 40% solution.
You add x cups of water (0% solution).
Look at the concentrations.

So it would be 3:1.

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A 40% solution is to be mixed with water how much water is needed to make the solution 10%?
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If it's a 40% solution, it is 60% water.
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Assume you start with 100 gallons of the 40% solution.
60 gallons of it must be water
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You are going to add "x" gallons of the water to the solution.
You now have (100+x) gallons of solution.
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You want that to be a 10% solution.
You have to solve for "x".
Equation:
water + water = water
60 gal + x gal = 0.90(100+x)
60 + x = 90 + 0.9x
0.1x = 30
x = 300 (number of gallons of water you would have to add)
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You started with 100 gallons of solution; 60 gallons was water
You added 300 gallons of water so you have 400 gallons of solution
40 gallons of active ingredient are still in the solution.
And 40 is 10% of 400.
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The original solution was 4 out of 10 or 2/5 active.
The new solution is 4 out of 40 or 1/10 active
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The mixture is now 10 inactive to 1 active.
Cheers,
Stan H.

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
If a man says something, and his wife doesn't hear it, is he still wrong?
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
I'm a 66 year old man with a debate with my wife. I say the mixture would be 4 to 1 she disagrees. A 40% solution is to be mixed with water how much water is needed to make the solution 10%?

Sorry, sir, you're wrong! I assume you merely did a ratio of 40%:10% to get 4:1, but it doesn't work like that
I don't know what your wife's ratio is, but yours is incorrect.
Let original amount of solution be S, and amount of water to add, W
Let's assume that the S units of solution contains .6S water
Water + water = water
Then we get: .6(S) + W = .9(S + W) ---- Amounts of water in original solution and new solution are: .6, and .9, respectively
.6S + W = .9S + .9W
W � .9W = .9S - .6S
.1W = .3S
W, or amount of water to add = , or 3S.
This means that the amount of water to be added is 3 times the amount of the original solution.
Now, the solution consists of 4S (S + 3S), of which 3S units of water were added to the original .6S units of water, which results in 3.6S units of water.
This makes the ratio: 3.6S:.4S, or
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