# Mixture problem - end up with a negative answer???

• Apr 14th 2009, 01:27 PM
cdbmath
Mixture problem - end up with a negative answer???
Quote:

One alloy is one part copper to three parts tin. A second alloy is one part copper to four parts tin. How much of a second alloy shold be combined with 24 pounds of the first allow to obtain a new alloy that is two parts copper to seven parts tin?
I made this chart:

Code:

            Amount (lbs)  % Copper  Total Copper Alloy 1  24                .25          .25(24) Alloy 2  x                  .20          .20x Mix        24+x            2/7          2/7(24+x)
And then got:

$\displaystyle .25(24) + .20x = .29(24+x)$
$\displaystyle 6 + .20x = 6.96 + .29x$
$\displaystyle -.96 = .09x$
$\displaystyle -10.67 = x$

Where did I go wrong?
• Apr 15th 2009, 04:03 AM
earboth
Quote:

Originally Posted by cdbmath
I made this chart:

Code:

            Amount (lbs)  % Copper  Total Copper Alloy 1  24                .25          .25(24) Alloy 2  x                  .20          .20x Mix        24+x            2/7          2/7(24+x)
And then got:

$\displaystyle .25(24) + .20x = .29(24+x)$ <<<<<< here
$\displaystyle 6 + .20x = 6.96 + .29x$
$\displaystyle -.96 = .09x$
$\displaystyle -10.67 = x$

Where did I go wrong?

The first line must read like this:

$\displaystyle .25(24) + .20x = \dfrac29 \cdot (24+x)$

And $\displaystyle \dfrac29 \neq 0.29$

You should come out with x = 30