1. MIxture word problem

How would I go about setting this up?

How many liters of a 30% alcohol solution must be mixed with 80 liters 0f a 90% solution to get a 40% solution ?

Thanks

2. One more problem I'm having trouble with:

Walt made an extra $7000 last year from a part time job. He invested part of the money at 6% and the rest at 8%. He made a total of$500 in interest. How much was invested at 8%?

3. Originally Posted by Jerry99
How would I go about setting this up?

How many liters of a 30% alcohol solution must be mixed with 80 liters 0f a 90% solution to get a 40% solution ?

Thanks
30x + 90(80) = 40(x+80)

solve for x

4. Originally Posted by Jerry99
One more problem I'm having trouble with:

Walt made an extra $7000 last year from a part time job. He invested part of the money at 6% and the rest at 8%. He made a total of$500 in interest. How much was invested at 8%?
assuming simple interest ...

0.06(7000-x) + 0.08(x) = 500

solve for x

5. Thanks man...

6. Hello, Jerry99!

$\text{How many liters of a 30\% alcohol solution must be mixed}$
$\text{with 80 liters 0f a 90\% solution to get a 40\% solution?}$

Here is one way to reason it out . . .

We will use $\,x$ liters of the 30% solution.
. . It contains: . $0.30x$ liters of alcohol.

We have $80$ liters of the 90% solution.
. . It contains: . $0.90 \times 80 \:=\:72$ liters of alcohol.

The mixture will have: . $0.30x + 72$ liters of alcohol. .[1]

But we know that the mixture will be $x+80$ liters which is 40% alcohol.
. . The mixture will have: . $0.40(x+80)$ liters of alcohol. .[2]

We just described the final amount of alcohol in two ways.

There is our equation! . . . $0.30x + 72 \;=\;0.40(x + 80)$