Question

if a mixture of a 6% acid solution with an 11% acid solution is to be made, how much of each solution is needed to make 10 liters of an 8% acid solution

Answer 1

x - 6%
y - 11%
We need 10 liters of 8%.
A) x + y = 10
B) .06x + .11y = (10 * .08)
Multiplying equation A by -.06
A) -.06x + -.06y = -.6   Then adding this to B)
B) .06x + .11y = (10 * .08)
.05y = .2
y = 4 liters of 6%
x = 6 liters of 11%

Answer 2

To understand the mixture problems, it is recommended to use
the following table:

                       |   Amount                   Part                       Total
-------------------|-------------------------------------------------------------------
1st Item          |    x                               6%                         0.06x
-------------------|--------------------------------------------------------------------
2nd Item         |    y                              11%                       0.11y
-------------------|-------------------------------------------------------------------
TOTAL           | (x + y) =10                    8%            0.08(10 =0.8)
                             (a)                                                          (b)
a) x + y =10
b) 0.06x +0.11y = 0.8

Solving this equation gives x = 6  and y = 4, in other terms:
 x = 6  at 6% and y = 4 at 11%