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Word problem, interested in the reasoning process

  1. Dec 3, 2011 #1
    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?

    The answer is given:
    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. jcsd
  3. Dec 3, 2011 #2

    Ray Vickson

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    You add x gallons of anitfreeze to the drained solution.

    RGV
     
  4. Dec 3, 2011 #3
    Could you expand on that? I actually do can't decode what you say...
     
  5. Dec 3, 2011 #4

    Ray Vickson

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    Look at the first sentence YOU wrote under heading 3.

    RGV
     
  6. Dec 3, 2011 #5
    How should your insight reflect in the proposed solution? Does it invalidate the reasoning process?
     
  7. Dec 3, 2011 #6

    HallsofIvy

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

    [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.

     
  8. Dec 3, 2011 #7

    Ray Vickson

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    It means that you added the wrong quantity.

    RGV
     
  9. Dec 3, 2011 #8
    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
     
  10. Dec 4, 2011 #9

    HallsofIvy

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    No, (40/100)(6- 1)+ (60/100)(6- x)+ x= 6 leads to 6= 6! That equation does not yet use the new percentage.
     
  11. Dec 4, 2011 #10
    Can someone help a little more, I need this one sorted out.
     
    Last edited: Dec 4, 2011
  12. Dec 4, 2011 #11

    eumyang

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    Double-check your work. I solved the equation in bold and I got x = 2.5.
     
  13. Dec 5, 2011 #12
    Indeed I made a mistake.

    Thank's to all for helping.
     
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