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

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%

Answer 2

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%