Word problem, interested in the reasoning process

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mindauggas
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Homework Statement



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?

The answer is given:
Ans. 2.5 gallons

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.
 
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mindauggas said:

Homework Statement



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?

The answer is given:
Ans. 2.5 gallons

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.

You add x gallons of anitfreeze to the drained solution.

RGV
 
Could you expand on that? I actually do can't decode what you say...
 
How should your insight reflect in the proposed solution? Does it invalidate the reasoning process?
 
mindauggas said:

Homework Statement



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?

The answer is given:
Ans. 2.5 gallons

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

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.
[itex](40/100)(6- x)[/itex] 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.

I would be interested in the reasoning here especially, if someone will bother.
 
HallsofIvy said:
And (40/100)(6- x) was the amount of antifreeze after the drainage so (40/100)(6- x)+ x= (65/100)(6)

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 that's left and add only pure antifreeze (x) I should get the amount of antifreeze in the new admixture. But the answer is clearly incorrect.


HallsofIvy said:
[itex](40/100)(6- x)[/itex] 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.

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
 
mindauggas said:
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 that's 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
No, (40/100)(6- 1)+ (60/100)(6- x)+ x= 6 leads to 6= 6! That equation does not yet use the new percentage.
 
Can someone help a little more, I need this one sorted out.
 
Last edited:
HallsofIvy said:
And (40/100)(6- x) was the amount of antifreeze after the drainage so (40/100)(6- x)+ x= (65/100)(6)
mindauggas said:
the x value I get is 1,9 so I guess you were not proposing a solution here...
Double-check your work. I solved the equation in bold and I got x = 2.5.
 
eumyang said:
Double-check your work. I solved the equation in bold and I got x = 2.5.

Indeed I made a mistake.

Thank's to all for helping.