Algebra question!?

Question

If you put together a 10% acid solution and a 30% acid solution and get a 500mL 25% acid solution, how much of the 30% solution did you use? The options are 125 mL, 150mL, 375 mL, and 450 mL.

THANKS! you guys rock. i just posted that question like a milisecond ago.

Answer 1

Let A be the amount of 10% solution used and B the amount of 30% solution.

A + B = 500 ml

( .10 ) A + ( .30 ) B = ( .25 ) ( 500 )

Substitute and solve...

( .10 ) ( 500 - B ) + ( .30 ) B = 125
50 - .10 B + .30 B = 125
.20 B = 75
B = 375

So you used 375 ml of B and 125 ml of A.

Checking:

125 ml of A contains 12.5 ml of acid
375 ml of B contains 112.5 ml of acid

Final solution contains 125 ml of acid, which is 25% of 500 ml.

Answer 2

% problems like this use the formula:

.10x + .30y = .25*500

x is the ml of 10% solution, y is the ml of 30% solution

we know that x and y have to add up to 500 (because the problem said that is how much we end up with) so:

x + y = 500

Change this to find y:

y = 500 - x

Now that we have a little formula for y, we can subsitute the little formula into the first formula:

.10x + .30(500-x) = .25*500

Solve it for x...
.10x + 150 - .30x = 125
.10x - .30x = 125 - 150
-.20x = -25
(divide by -.2) x = -25 / -.2 ==> x = 125

now, if x is 125, then y is 500 - 125, or 375

x is 125ml of the 10%
y is 375ml of the 30%

so, the answer is 375ml !

Answer 3

10x + (500 -- x)*30 = 500*25 => x = 500(30--25)/20 = 125
125mL 10% and 375mL 30% are required to get 500mL 25% acid.