Results 1 to 8 of 8

Math Help - S.O.S from a chicken farm in Australia!

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    3

    S.O.S from a chicken farm in Australia!

    We have just received 25 litres of an antimicrobial solution that is to be added to the hens drinking water, but the solution that we have is 31% of the concentrate, whereaswe require it to be 28%.

    My question is this... If I have 1 litre of concentrate at 31%, how much water must I add to this in order to bring this down to the specified 28% concentration?

    This was an error on the manufacturers side, and as I am not mathematically equipped any help from you guys would be much appreciated!

    Kind regards,

    Adam H
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member arpitagarwal82's Avatar
    Joined
    Feb 2009
    Posts
    103
    1 liter of solution has .31 liter antimicrobial and .69 liter water.

    Now suppose we add x
    liter water to bring down the concentration

    .31/(.31 + .69 + x) = 28/100

    solve for x
    x = .107.

    So add .107 liter of water in 1 liter of solution to get cncentration of 28%
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    3

    Just one more thing...

    Thankyou so much, but just a little clarification...

    Does that mean 107 ml (mililitres) per litre of concentrate?

    You are a champion!...

    Kind regards

    Adam H
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member arpitagarwal82's Avatar
    Joined
    Feb 2009
    Posts
    103
    Yes its 107 ml (approx) per liter of solution.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Normally I wouldn't post an (essentially) identical working of a problem that was already solved, but since errors can have significant consequences in real-life applications like these, I thought you might like confirmation of arpitagarwal82's solution:

    In one liter of the solution, the volume of the concentrate is 0.31 L and the volume of the water is 0.69 L. Let V represent the amount of water that must be added. The volume of the solution will then become 1+V\text{ L.} To get the appropriate mixture, we need

    \frac{0.31}{1+V}=0.28.

    Solve for V\colon

    1+V=\frac{0.31}{0.28}=\frac{31}{28}

    \Rightarrow V=\frac{31}{28}-1=\frac3{28}

    \approx0.1071429\text{ L.}

    So that is about 107 ml, which does agree with the above answer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member arpitagarwal82's Avatar
    Joined
    Feb 2009
    Posts
    103
    Thanks Reckoner for verifying it. I really wanted someone to look into this.
    I thought all of the Adam's hens are on stake of my solution.

    Solving maths on paper and applying it in real life are kinda different feeling.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2009
    Posts
    3

    Thankyou all...

    Ok, all is well... Thanks for all who helped and for those that double checked - its a fairly dose dependant type solution and too much and the hens would likely fly the coop!

    Thanks again, and much appreciated...

    Adam
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Multiplying factors

    Yet another confirmation - perhaps this one is really for the budding mathematicians. Can I urge you to try to understand what we used to call (about 200 years ago, when I was taught arithmetic) 'multiplying factors'? These are basically fractions that increase or decrease things in a certain ratio. The numbers you are given to play with here are 28 and 31. And the multiplying factors you can make with these are simply \frac{28}{31} and \frac{31}{28}.

    To make something bigger in this ratio, then, you multiply by \frac{31}{28}, and to make something smaller you multiply by \frac{28}{31}.

    We have 1 litre of solution at 31%; if we increase this to 1 litre \times \frac{31}{28} = 1.107 litres the concentration will be 28%. It really is that easy!

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 13th 2008, 01:12 PM

Search Tags


/mathhelpforum @mathhelpforum