# Word problem, interested in the reasoning process

1. Dec 3, 2011

### mindauggas

1. The problem statement, all variables and given/known data

A 6-gallon radiator is filled with a 40% solution of antifreeze in water. How much of the solution must be drained and replaced with pure antifreeze to obtain a 65% solution?

Ans. 2.5 gallons

3. The attempt at a solution

Let x be the amount of added pure antifreeze (which is equal to the amount drained from the original solution (or so I presume)).

Thus I reason

40/100*(6) is the amount of antifreeze in the orig. solution
60/100*(6) water
40/100*(6-x) is the amount of antifreeze after the drainage
60/100*(6-x) water
65/100*(6) is the amount of antifreeze in the new solution
35/100*(6) water

I recon that 40/100*(6-x)+60/100*(6-x)+65/100*(6) should be equal to the new full tank, so 6. But after the calculation i get 3,9, not 2,5.

I would be interested in the reasoning here especially, if someone will bother.

2. Dec 3, 2011

### Ray Vickson

You add x gallons of anitfreeze to the drained solution.

RGV

3. Dec 3, 2011

### mindauggas

Could you expand on that? I actually do can't decode what you say...

4. Dec 3, 2011

### Ray Vickson

Look at the first sentence YOU wrote under heading 3.

RGV

5. Dec 3, 2011

### mindauggas

How should your insight reflect in the proposed solution? Does it invalidate the reasoning process?

6. Dec 3, 2011

### HallsofIvy

Staff Emeritus
And (40/100)(6- x) was the amount of antifreeze after the drainage so (40/100)(6- x)+ x= (65/100)(6)

$(40/100)(6- x)$ is the amount of anti-freeze after draining and (60/100)(6- x) is the amount of water so you should add x anti-freeze. You added the total amount of anti-freeze in the radiator so the amount left in after draining is counted twice.

7. Dec 3, 2011

### Ray Vickson

It means that you added the wrong quantity.

RGV

8. Dec 3, 2011

### mindauggas

the x value I get is 1,9 so I guess you were not proposing a solution here, but how come it;s not a solution if I have the antifreeze thats left and add only pure antifreeze (x) I should get the amount of antifreeze in the new admixture. But the answer is clearly incorrect.

Once again (40/100)(6- x)+(60/100)(6- x)+x=6 (if that's what you propose (i'm sorry if i misunderstood your intentions)) leads to incorrect answer: 1.2x = 0

9. Dec 4, 2011

### HallsofIvy

Staff Emeritus
No, (40/100)(6- 1)+ (60/100)(6- x)+ x= 6 leads to 6= 6! That equation does not yet use the new percentage.

10. Dec 4, 2011

### mindauggas

Can someone help a little more, I need this one sorted out.

Last edited: Dec 4, 2011
11. Dec 4, 2011

### eumyang

Double-check your work. I solved the equation in bold and I got x = 2.5.

12. Dec 5, 2011