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?
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> 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?
Log On
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.
-------------
(