Question 752567: how many gallons of 20% antifreeze should be mixed with 10 gallons of 83% antifreeze to obtain 55% antifreeze mixture? Found 2 solutions by stanbon, josgarithmetic:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! how many gallons of 20% antifreeze should be mixed with 10 gallons of 83% antifreeze to obtain 55% antifreeze mixture?
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Equation:
alcohol + alcohol = alcohol
0.20x + 0.83*10 = 0.55(x+10)
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20x + 83*10 = 55x + 55*10
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35x = 28*10
x = (2/7)28
x = 8 gallons (amt. of 20% solution needed)
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Cheers,
Stan H.
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You can put this solution on YOUR website! Would matter not if it were antifreeze or some other kind of mixture using percents concentration or other kind of strength. What if you had this, generally the same:
how much of L% mixture should be mixed with v units of H% mixture to obtain T% mixture?
Let u = the units of the L% mixture to use.
Equation to use is .
The only unknown variable is u.
Solve for u.
You would now have a formula for u and you can substitute the values you have to find the actual value of u for your particular problem.
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