Results 1 to 11 of 11

Math Help - Problem on mixture. Please explain with steps.

  1. #1
    Member
    Joined
    Aug 2012
    From
    Utah
    Posts
    88

    Problem on mixture. Please explain with steps.

    A cask contains 3 parts wine and one part water. What part of the mixture must be drawn off and substituted by water so that the resulting mixture may be half wine and half water.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,569
    Thanks
    1409

    Re: Problem on mixture. Please explain with steps.

    Since there is no specific amount given, lets take the total liquid in the cask to be 1. Then the amount of water in the cask is 1/4 and the amount of wine is 3/4. Suppose you take out amount x (x< 1). The amount of water in that is x/4 and the amount of wine is 3x/4. That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water.

    Since we "substitute" that by water, we simply replace the amount taken out, so that we are adding x water, then we will have 1- x/4+ x= 1+ 3x/4 water and 1- 3x/4 wine. To have equal amounts of water and wine, those must be equal: 1+3x/4= 1- 3x/4. Solve that for x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member MaxJasper's Avatar
    Joined
    Aug 2012
    From
    Canada
    Posts
    482
    Thanks
    54

    Exclamation Re: Problem on mixture. Please explain with steps.

    Quote Originally Posted by HallsofIvy View Post
    To have equal amounts of water and wine, those must be equal: 1+3x/4= 1- 3x/4. Solve that for x.
    This results in x=0.

    I suspect we should have:

    3/4-3x/4 = 1/4-x/4+x

    which results in x = 1/3 (not sure yet)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2012
    From
    Utah
    Posts
    88

    Re: Problem on mixture. Please explain with steps.

    Solving that X gives 0.

    Your 3rd line "That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water."
    Will that not leave (1/4)-(x/4) water and (3/4)-(3x/4) wine and then adding x water gives water equation as (1/4)-(x/4)+x
    and leaving wine the same amount (3/4)-(3/4)x. Now solving (1/4)-(x/4)+x and (3/4)-(3/4)x yields x=1/3 which is the correct answer.

    I have another question. I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
    yields 4/3 as x.
    You started the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Problem on mixture. Please explain with steps.

    Quote Originally Posted by hisajesh View Post
    Solving that X gives 0.

    Your 3rd line "That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water."
    Will that not leave (1/4)-(x/4) water and (3/4)-(3x/4) wine and then adding x water gives water equation as (1/4)-(x/4)+x
    and leaving wine the same amount (3/4)-(3/4)x. Now solving (1/4)-(x/4)+x and (3/4)-(3/4)x yields x=1/3 which is the correct answer.

    I have another question. I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
    yields 4/3 as x.
    You started the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference.
    Here is my 2 cents. I visualized the problem as this

    At the begining the mixture has 75% wine and since it is well mixed what ever volume is removed with be 75% wine. At the end we want the percent to be 50%

    This gives

    .75-.75x=.5 \iff -.75x =-.25 \iff x =\frac{25}{75}=\frac{1}{3}

    Also where abouts are you from in Utah. I went to school there for a while.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629

    Re: Problem on mixture. Please explain with steps.

    Hello, hisajesh!

    A cask contains 3 parts wine and one part water.
    What part of the mixture must be drawn off and substituted by water
    so that the resulting mixture may be half wine and half water?

    Suppose the cask holds 60 liters of liquid.
    It contains 45 liters of wine and 15 liters of water.

    We remove x liters of the mixture.
    . . We remove \tfrac{3}{4}x liters of wine.
    . . We remove \tfrac{1}{4}x liters of water.
    The cask contains: \begin{Bmatrix}45-\tfrac{3}{4}x\text{ liters of wine}\; \\ 15-\tfrac{1}{4}x\text{ liters of water} \end{Bmatrix}

    We add x liters of water.
    The cask contains: . \begin{Bmatrix}45-\frac{3}{4}x\text{ liters of wine}\; \\ 15 + \frac{3}{4}x\text{ liters of water} \end{Bmatrix}

    But these two quantities are supposed to be equal.

    There is our equaton! . . . 45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}x
    Last edited by Soroban; October 16th 2012 at 04:29 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member MaxJasper's Avatar
    Joined
    Aug 2012
    From
    Canada
    Posts
    482
    Thanks
    54

    Lightbulb Re: Problem on mixture. Please explain with steps.

    Quote Originally Posted by Soroban View Post
    There is our equaton! . . . 45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}
    adding missing x in above equations we have:

    45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}x

    resulting x=20 which is 1/3 original volume.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Aug 2012
    From
    Utah
    Posts
    88

    Re: Problem on mixture. Please explain with steps.

    @ everyone: Thank you for your flurry of answers!!!!!
    Please answer my second question in my above reply:

    I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
    yields 4/3 as x. If I start the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member MaxJasper's Avatar
    Joined
    Aug 2012
    From
    Canada
    Posts
    482
    Thanks
    54

    Re: Problem on mixture. Please explain with steps.

    Whatever you assume the original volume to be you always end up with 1/3 of it. Start with vol v and get v/3.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Problem on mixture. Please explain with steps.

    Quote Originally Posted by hisajesh View Post
    @ everyone: Thank you for your flurry of answers!!!!!
    Please answer my second question in my above reply:

    I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
    yields 4/3 as x. If I start the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference
    You found the volume removed, but if you want the RATIO you need find

    \frac{\frac{4}{3}}{4}=\frac{1}{3}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629

    Re: Problem on mixture. Please explain with steps.

    Hello, hisajesh!

    A cask contains 3 parts wine and one part water.
    What part of the mixture must be drawn off and substituted by water
    so that the resulting mixture may be half wine and half water?

    Suppose the cask contains W units of the wine/water blend.
    . . It is 25% water; it contains \tfrac{1}{4}W units of water.

    We draw off x units of the mixture.
    . . We remove \tfrac{1}{4}x units of water.

    We add x units of pure water.
    . . This contains x units of water.

    The final amount of water is: . \tfrac{1}{4}W - \tfrac{1}{4}x + x \:=\:\tfrac{1}{4}W + \tfrac{3}{4}x units.

    But we know that the final mixture is W units which is 50% water.
    . . So it contains \tfrac{1}{2}W units of water.


    There is our equation! . . . . \tfrac{1}{4}W + \tfrac{3}{4}x \:=\:\tfrac{1}{2}W

    Therefore: . \tfrac{3}{4}x \:=\:\tfrac{1}{4}W \quad\Rightarrow\quad x \:=\:\tfrac{1}{3}W

    We should draw off one-third of the cask and replace it with water.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 26th 2010, 02:45 AM
  2. Replies: 1
    Last Post: February 8th 2010, 05:31 PM
  3. I want to explain the latest steps from this Q
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 25th 2009, 01:24 AM
  4. Replies: 1
    Last Post: September 19th 2009, 01:12 AM
  5. Explain why these steps were taken...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 14th 2009, 06:50 AM

Search Tags


/mathhelpforum @mathhelpforum