Question

If the 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?
Please show work! Word problems have never been my forte, but this one...I cant even grasp what they are asking me to do! xD haha thanks for any help in advance! :D

Answer 1

x = liters of first acid solution
y = liters of second acid solution

Equation for total liters:
x + y = 10

Equation for percent of total liters:
.06x + .11y = .08(10)

Thus the system of equations to be solved is:

x + y = 10
.06x + .11y = .8

For convenience, multiply the second equation by 100 which results in the following system:

x + y = 10
6x + 11y = 80

Answer 2

Hopefully, you know how to solve systems of equations.

Answer 3

Sorry about the wait! My stupid computer kept shutting down and starting up.... -_-
But what I did next was multiply the first equation by -6
then added the two new equations together to get -6x+6x-6y+11y-60+80
(I think that is the system of equations...right??)
and I ended up with 5y=20 ; y=4
but now idk how to get x?

Answer 4

* -6x+6x-6y+11y=-60+80

Answer 5

use the first equation (both work, but the first equation is simpler)

Answer 6

oh okay I see now! lol that was easy :p

Answer 7

btw, the trick of these type questions is the 2nd equation

6% of one solution is acid
11% of the 2nd solution is acid
8% of 10 (combined) is acid

the idea is the sum of the acid in both solutions adds up to the acid in the result.

Answer 8

the amount of solution 1 is x , and the amount of acid is 6% of x or 0.06*x
ditto for solution 2: 11% of y or 0.11y is acid
the sum of the acid is 0.08*10 = 0.8

Answer 9

alright, now it makes a little more sense! haha thanks for the explanation!