SOLUTION: a 90% antifreeze solution is to be mixed with a 75% solution to make 30 liters of an 80% solution. How many liters of the 90% and 75% solutions should be used?

Question 289107: a 90% antifreeze solution is to be mixed with a 75% solution to make 30 liters of an 80% solution. How many liters of the 90% and 75% solutions should be used?
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A 90% antifreeze solution is to be mixed with a 75% solution to make 30 liters of an 80% solution.
How many liters of the 90% and 75% solutions should be used?
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Equation:
alcohol + alcohol = alcohol
0.9x + 0.75(30-x) = 0.30(80)
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Multiply thru by 100 and solve for "x":
90x + 75*30 - 75x = 30*80
15x = 5*30
x = 6 liters (amt. of 90% antifreeze needed in the mixture)
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30-x = 24 liters (amt. of 75% antifreeze needed in the mixture)
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Cheers,
Stan H.
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Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
If you think in terms of how much antifreeze you are dealing with, I think that the problem becomes a little easier. For example, 10 liters of a 90% antifreeze solution contains (10*(0.9) = 9) liters of antifreeze, so...let x equal the required number of liters of the 90% antifreeeze solution.
Since you want to end up with 30 liters, you will also need (30-x) liters of the 75% antifreeze solution. In algebra this can be expressed as:
0.9x+0.75(30-x) = 0.8(30) Simplify and solve for x.
0.9x+22.5-0.75x = 24 Combine like-terms.
0.15x+22.5 = 24 Subtract 22.5 from both sides.
0.15x = 1.5 Finally, divide both sides by 0.15
x = 10
You will need to mix 10 liters of 90% antifreeze solution with (30-10 = 20) liters of 75% antifreeze solution to obtain 30 liters of 80% antifreeze solution.

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