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A problem on quadratic equations .

  1. Aug 13, 2011 #1
    1. The problem statement, all variables and given/known data

    There are 2 beakers P and Q. P contains 50 litres of water and Q contains 40 litres of acid. x litres is taken from both the beakers, mixed and replaced. This process is repeated 2 times and after the second time, there is 71/8 litres of acid in P. What is the quantity of x ?

    2. Relevant equations

    Form the equation in the form of ax2 + bx + c = 0 where a is not equal to 0 and a,b and c are constants.

    3. The attempt at a solution
     
  2. jcsd
  3. Aug 13, 2011 #2

    SammyS

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    What does it mean by: "... x litres is taken from both the beakers, mixed and replaced."

    What is replaced? ... By what is it replaced?
     
  4. Aug 13, 2011 #3

    PeterO

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    Just checking, this reads as though the mixing is done three times in all - done once then repeated twice. Is that what is meant, or should it merely have stated "This process is done twice".

    I suggest you choose a couple of values for x and see what is happening , to get a feel for the situation.

    For example, what happens if the samples are 5 litres from each? 10 litres from each instead.

    eg: when x is 10 the first process will transfer 5 litres of acid to the water and 5 litres of water to the acid, 1st repeat will transfer less than 5 litres to each, and so on.
     
  5. Aug 13, 2011 #4

    PeterO

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    Sounds very much like half the mix is put back in each bucket, so that there is 50 and 40 litres of liquid in each bucket after each sample, mix and replacement
     
  6. Aug 14, 2011 #5
    DETAILED EXPLANATION OF QUADRATIC EQUATION PROBLEM


    1. Suppose there is a beaker P. It contains 50 litres of water .
    2. Also suppose there is beaker Q. It contains 40 litres of acid. Got it ?
    3. Now "x" litres of acid is taken from beaker Q .
    4. "x" litres of water is also taken from beaker P.
    5. This x litres of water from beaker P and also the x litres of acid from beaker Q are mixed with each other. Got this statement ?
    6. After they are mixed and form mixture , the same x litres are taken from mixed mixture and added to both the beakers P and Q.
    7. Now this process from step 3 to 6 is repeated again .
    8. Now the question asks what is the value of x ? How many litres ?


    I hope you get it . Please solve it .

    Thanks in advance .

    (I am 14 years , in class 10th . The above problem was from book "elementary algebra" by H.S. Hall and Knight . )
     
    Last edited: Aug 14, 2011
  7. Aug 14, 2011 #6
    OK, let's try one iteration of removal and replacement.

    If x litres is removed from the acid container, how much is placed into the water container?
     
  8. Aug 14, 2011 #7

    umm umm:confused:

    x litres removed from acid container ?? How much placed ?? :confused::confused:

    I don't know where to start . Please solve it step by step .
     
  9. Aug 14, 2011 #8
    Remove x litres from Q. Add it to x litres from P. Mix. Replace x litres into P. How much acid is there in P?
     
  10. Aug 14, 2011 #9
    x litres from Q mixed with x litres from P becomes 2x , it is diffused uniformly .
    Now x litres put back to Q , so
    40 l of acid + x/2 l of water - x/2 l of acid .


    In P :
    50 l of water - x/2 l of water + x/2 l of acid .

    Right ? :confused:



    Sorry my bad . I edited the post .
     
    Last edited: Aug 14, 2011
  11. Aug 14, 2011 #10
    I am already at step 6.
     
  12. Aug 14, 2011 #11
    So if there is 40- (x/2) litres of acid in Q, what is its concentration?
     
  13. Aug 14, 2011 #12
    No idea . I cant solve this problem . :confused:Maybe you can .

    In Q :
    40 l of acid + x/2 l of water - x/2 l of acid .


    In P :
    50 l of water - x/2 l of water + x/2 l of acid .
     
  14. Aug 14, 2011 #13

    SammyS

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    We're here to help you solve problems. According to the rules of this site, we're not supposed to solve them for you.
     
  15. Aug 14, 2011 #14
    See post #5 . Question corrected to extent .

    In Q :
    40 l of acid + x/2 l of water - x/2 l of acid .


    In P :
    50 l of water - x/2 l of water + x/2 l of acid .


    After this what ?
     
  16. Aug 14, 2011 #15
    Can you define concentration?
     
  17. Aug 14, 2011 #16

    ehild

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    So there is x/2 l of acid in P and 40-x/2 l acid in Q after the first step. You take out x litres from both bakers again. How much acid is in the x litres taken from P and how much acid is in the x litres taken from Q?

    ehild
     
  18. Aug 14, 2011 #17

    Umm :confused:


    we again take out x l from both . So in
    P:

    How will I form quadratic equation ? I don't know . Please just suggest only one step so that i can get it . :confused:
     
  19. Aug 15, 2011 #18

    ehild

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    The quadratic comes at the end. Answer my question. There is x/2 l of acid in 50 l mixture. You take x litres again. How much acid is in that x litres?

    If it is easier with number, let be 5 l acid in 50 l mixture. You take out 10 l. How much acid is in that 10 litres?

    ehild
     
  20. Aug 15, 2011 #19

    PeterO

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    Try some actual numbers [ and hopefully not happen upon the correct answer ] just to see how it works. To dive straight in to x is really quite advanced - especially if this is the first such problem you have done.

    For example let the sample be 40 litres from each - all of the smaller [acid] container. [ie x = 40]

    After mixing and replacing, you now have 20 litres of acid in each.
    The 40 litre container is 20 litres of water plus 20 litres of acid, the 50 litre beaker is 30 / 20.

    Now you again take 40 litres from each [1st repeat]

    You get 20 lites of water plus 20 litres of acid from the 40l beaker [everything again] and 40l from the 50 litre pot which is 60% water / 40% acid so 24 litres of water - 16 litres of acid.

    mix together and the 80 litres is 44 litres water - 36 litres acid

    So you now return 22l water - 18l acid to each beaker

    The 50 litre beaker already had the remaining 6 l water and 4 l acid to make a new mix of 28 l water and 22 l acid. The 40 litre container is just re-filled with the 22-18 mix.

    Now you do it again - you calculate this time.

    You will then end up with a certain amount of acid /water in each beaker.

    If you keep track of what has been happening to the original 40, you can then do those same things to the general sample size, x.

    For example, after the first sample - mix - replace cycle, there is x/2 litres of acid in the water and x/2 litres of water in the acid.

    Work through a couple of times with numbers if necessary, and you should see how this works. For example let x = 6,10,14 or 22 - each of those is 2 times a prime. We know we are going to halve the first time, so we may as well choose an even number to make that work. From there on you will be doing something to a prime number, so should be able to keep track of what is happening.

    [For example if we chose 8, it would become 4 after the first operation. If it became 2 after the second, we wouldn't know if it had been halved or "square rooted"]

    Peter
     
  21. Aug 15, 2011 #20

    Hmm a bit complicated though ! :surprised

    Ummmm :confused:

    50 l -> x/2 l of acid
    so x l -> x2/100 l of acid ?
    :confused:
     
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