mixture problem

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memorial

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Please help me get started with this problem:

How much water must be added to 100 gallons of a mixture which is 80 per cent alcohol to reduce it to a 75 per cent mixture?

I don't want the answer because I want to work it out myself.

I've tried mainly by letting x = the amount of water to be added and using
.20(100+X)=.25(????). but I keep getting confused no matter how I approach this problem. I've been at this all day!

I would appreciate just a start (like what should I let x represent)and I think I can take it from there.

Thanks
 
There will always be 80 gallons of alcohol. So after you add x gallons, then:

80/(100+x) = 75%
 
memorial said:
Please help me get started with this problem:

How much water must be added to 100 gallons of a mixture which is 80 per cent alcohol to reduce it to a 75 per cent mixture?

I don't want the answer because I want to work it out myself.

I've tried mainly by letting x = the amount of water to be added and using
.20(100+X)=.25(????). but I keep getting confused no matter how I approach this problem. I've been at this all day!

I would appreciate just a start (like what should I let x represent)and I think I can take it from there.

Thanks
Click to expand...

You start with 100 gallons of something which contains 80% alcohol. The amount of alcohol in this solution is .80(100)

Ok...you add x gallons of water, which contains 0% alcohol.

You'll end up with 100 + x gallons of the mixture, which contains 75% alcohol, so the amount of alcohol at the end is .75(100 + x)

alcohol to begin with + alcohol added = alcohol at the end
.80(100) + 0(x) = .75(100 + x)

.80(100) = .75(100 + x)

Solve this for x...
 
0r 75(x) = 80(100) ; x = 106 2/3 (resulting total gallons)
 
Hello, memorial!

Your equation is close, but . . .

How much water must be added to 100 gallons of a mixture
which is 80% alcohol to reduce it to a 75% mixture?
Click to expand...

Since we're adding water (not alcohol), think in terms of water.

We start with 100 gallons which is 20% water \(\displaystyle \:\Rightarrow\;20\) gallons water.

We add \(\displaystyle x\) gallons of water. .The mixture contains: \(\displaystyle 20\,+\,x\) gallons of water.


But the mixture contains \(\displaystyle 100\,+\,x\) gallons which is 25% water.
. . That is, \(\displaystyle \,0.25(100\,+\,x)\) gallons of water


There is our equation! . . . \(\displaystyle \L20\,+\,x\:=\:0.25(100\,+\,x)\)

 
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