word problem

Stine

New member
Joined
Nov 29, 2006
Messages
49
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?

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
 

skeeter

Senior Member
Joined
Dec 15, 2005
Messages
2,396
doing fine ... finish up

y = 400/3 mL

x = 200 - (400/3) = ?
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,445
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
Little "reminder how"; tatoo that on your wrist (under your watch) :idea:
 

Stine

New member
Joined
Nov 29, 2006
Messages
49
OK you say I am doing fine but I am lost. What is the question mark for am I not supposed to do that. I am still lost in dealing with this problem. What I am coming out with is __
6.66666 repeating. Can I get some more help?

:wink: Thank you...
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,445
To start with, you used 35% : that should be 30%.

let x = 65% solution; then 200 - x = 30% solution ; SO:
Code:
   x  @ 65
200-x @ 30
==========
 200  @ 55
65x + 30(200 - x) = 55(200)
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...
 
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