1. ## Word Problem....ahhh

A chemist at a university needs to mix a 25% ethanol solution with a 40%ethanol solution to obtain 120mL of a 30% ethanol solution. How much of each kind is needed?

okay yeah, the two equations i got out of this was
.25a+.40b=120--->25a+40b=12000
a+b=120(.30)---->a+b=36
______________________________
yeah? no objections??? okay.......
______________________________
yeah so i times the bottom equation by (-25) and that cancelled out the A's so i was left with
15b=11100
so i divided that out and got B=740...
i plugged that into the original equation and got A=(-704)

(-704)mL of 25% ethanol solution and 740 mL of 40% ethanol solution...

THAT CANT BE RIGHT! A NEGATIVE AMOUNT OF SOLUTION?

2. Originally Posted by CONFUSED_ONE
A chemist at a university needs to mix a 25% ethanol solution with a 40%ethanol solution to obtain 120mL of a 30% ethanol solution. How much of each kind is needed?

okay yeah, the two equations i got out of this was
.25a+.40b=120--->25a+40b=12000
a+b=120(.30)---->a+b=36
______________________________
yeah? no objections??? okay.......
______________________________>3
yeah so i times the bottom equation by (-25) and that cancelled out the A's so i was left with
15b=11100
so i divided that out and got B=740...
i plugged that into the original equation and got A=(-704)

(-704)mL of 25% ethanol solution and 740 mL of 40% ethanol solution...

THAT CANT BE RIGHT! A NEGATIVE AMOUNT OF SOLUTION?
I would say the equations should be
$x+y=120$
$.25x+.4y=.3(120)=36$
Thus,
$D=\left|\begin{array}{cc}1&1\\.25&.4\end{array}\ri ght|=.4-.25=.15$
$D_x=\left|\begin{array}{cc}120&1\\36&.4\end{array} \right|=12$
$D_y=\left|\begin{array}{cc}1&120\\.25&36\end{array }\right|=6$
Thus,
$x=\frac{D_x}{D}=12/.15=80$
$y=\frac{D_y}{D}=6/.15=40$
Q.E.D.

Note, I used determinants because I was practing my LaTeX coding for arrays of numbers. But apparently you made a mistake.

3. thank you for the help. could you point out to me what mistake i made?

4. Originally Posted by CONFUSED_ONE
thank you for the help. could you point out to me what mistake i made?
.25a+.40b=120--->25a+40b=12000
The RHS of the first of these equations should be $0.3\times 120$ as you end with 120 ml of 30% solution.

RonL

5. [QUOTE=CONFUSED_ONE][center]A chemist at a university needs to mix a 25% ethanol solution with a 40%ethanol solution to obtain 120mL of a 30% ethanol solution. How much of each kind is needed?

The amount of 25% is x, so the amount of 40% is 120 -x
.25x + .40(120-x) = .30(120)
.25x + 48 -.40x = 36
-.15x = -12
x = 80

6. Thank You So Very Much!