Steps I have taken so far:

x+y=200

0.35x+.65y=200(.55)

x=200-y

plugged 200-y for x

0.35(200-y)+.65y=110

70-0.35y+.65y=110

-70+70-0.35y+.65y=110

0.3y=40

___ __

0.3 0.3

Lost from here... Are my steps taken so far correct? :shock:

Thank you

- Thread starter Stine
- Start date

Steps I have taken so far:

x+y=200

0.35x+.65y=200(.55)

x=200-y

plugged 200-y for x

0.35(200-y)+.65y=110

70-0.35y+.65y=110

-70+70-0.35y+.65y=110

0.3y=40

___ __

0.3 0.3

Lost from here... Are my steps taken so far correct? :shock:

Thank you

Stine said:Jane has a solution that is 65% slfuric acid and a solution that is 30% sulfric acid. How much of each should she mix to obtain 200ml of a 55% sulfuric acid solution?

Code:

```
a @ x (like 25ml of 40% solution)
b @ y
=======
a+b @ z
(ax + by) / (a + b) = z
```

let x = 65% solution; then 200 - x = 30% solution ; SO:

Code:

```
x @ 65
200-x @ 30
==========
200 @ 55
```

65x + 6000 - 30x = 11000

35x = 5000

x = 5000/35 = 1000/7 = 142 6/7

So we need 142 6/7 ml of 65% and 200 - 142 6/7 = 57 1/7 ml of 30%

Notice there is no need to use decimals like .65; 65 is easier...