SOLUTION: Suppose that one solution contains 50% alcohol and the other solution contains 80% alcohol. How many liters of each solution should be mixed to make 10.5 of a 70% solution?

Question 994150: Suppose that one solution contains 50% alcohol and the other solution contains 80% alcohol. How many liters of each solution should be mixed to make 10.5 of a 70% solution?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose that one solution contains 50% alcohol and the other solution contains 80% alcohol. How many liters of each solution should be mixed to make 10.5 of a 70% solution?
----------------
Equation::
alcohol + alcohol = alcohol
0.50x + 0.80(10.5-x) = 0.70*10.5
----------
50x + 80*10.5 - 80x = 70*10.5
----------
-30x = -10*10.5
----------
x = (1/3)10.5
-----
x = 3.5 liters (amt. of 50% solution needed)
10.5-x = 7 liters (amt. of 80% solution needed)
-----------
Cheers,
Stan H.
-------------
RELATED QUESTIONS
Suppose that one solution contains 40% alcohol and another solution contains 80% alcohol. (answered by Boreal)

Suppose that one solution contains 40% alcohol and another solution contains 80% alcohol. (answered by ikleyn,amalm06)

Suppose that one solution contains 30% alcohol and another solution contains 80%... (answered by greenestamps,Edwin McCravy)

Suppose that one solution contains 50% alcohol and another solution contains 90% alcohol. (answered by scott8148)

Suppose that one solution contains 20% alcohol and another solution contains 50% alcohol. (answered by ewatrrr)

Please help me solve the story problem : One solution contains 50% alcohol and another... (answered by nerdybill)

One solution contains 50% alcohol and another solution contains 80% alcohol. How many... (answered by John10,josmiceli)

One solution contains 50% alcohol and another solution contains 80% alcohol. How many... (answered by josmiceli)

One solution contains 50% alcohol and another solution contains 80% alcohol. How many... (answered by edjones)