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Math Help - challenging ratio problem

  1. #1
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    challenging ratio problem

    Dear Sir,
    Would appreciate very much if someone can help me to solve the below problem.
    thanks
    Kingsman

    Boxes A and B are each filled with some sand. If sand from Box B is poured into Box A until the sand in Box A reaches the brim, there will be 8 litres of sand left in Box B. If sand from Box A is poured into Box B until the sand in Box B reaches the brim, there will be 26 litres of sand left in Box A. The ratio of the volume of Box A to the volume of Box B is 5:3. How many more litres of sand are needed to fill box Boxes to their brim?
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  2. #2
    Junior Member lanierms's Avatar
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    I got it. I'll start posting it right now. Might take around 15 minutes.
    Last edited by lanierms; February 27th 2011 at 07:24 PM.
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  3. #3
    Junior Member lanierms's Avatar
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    Volume of A : Volume of B = 5:3

    So let's say that Volume of A = 5k, Volume of B = 3k.

    Right now there are x liters of sand in box A, y liters of sand in box B.

    So let's pour sand from from box B to box A. Let's say that we poured g liters of sand.

    x+g=5k, y-g=8.

    Now, pour from box A to box B. Let's say we poured l liters of sand.

    x-l=26, y+l=3k.

    Let's do some algebra now.

    x+g-(x-l)=5k-26

    g+l=5k-26.

    y+l-(y-g)=3k-8

    g+l=3k-8.

    \therefore 5k-26=3k-8.

    k=9.

    So volume of box A is 45, and volume of box B is 27.

    Let's rewrite the equations:

    x+g=45, y-g=8.

    x-l=26, y+l=27.

    Here we go again:
    x+g-(x-l)=19

    \therefore g+l=19.

    y-g+(y-l)=35

    2y-(g+l)=35

    2y-19=35

    y=27.

    Now, let's find x.

    y-g=8

    27-g=8

    \therefore g=19.

    x+g=45

    x=26.

    So, in box A there are 26 liters of sand, and in box B there are 27 liters of sand.

    So the amount needed to fill both boxes are:

    45+27-(26+27)=19.

    \therefore 19.

    Tell me if this is wrong. I might've done some calculations wrong since I did most of them in my head.
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  4. #4
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    Thanks very much for the solution but I wonder whether it can be done shorter by not using algebraic approach; instead by taking advantage of some special relationship or properties in the question.
    Thanks
    Kingsman
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  5. #5
    Junior Member lanierms's Avatar
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    We can't solve this by not using algebra at all; you would be a genius if you can just guess and check.

    It wouldn't be possible to set up any equations whatsoever if I didn't use any relationship or properties in the problem.

    There could be other solutions, but basic idea will probably be the same.
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  6. #6
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    Can be shortened; using Lanierms' variables :

    boxA= 5k: sand in boxA = x : sand required to fill boxA = 5k - x

    boxB= 3k: sand in boxB = y : sand required to fill boxB = 3k - y

    "If sand from Box A is poured into Box B until the sand in Box B reaches the brim,
    there will be 26 litres of sand left in Box A" :
    x - (3k - y) = 26 ; x - 3k + y = 26 ; x + y = 3k + 26 [1]

    "If sand from Box B is poured into Box A until the sand in Box A reaches the brim,
    there will be 8 litres of sand left in Box B" :
    y - (5k - x) = 8 ; y - 5k + x = 8 ; x + y = 5k + 8 [2]

    [1][2]: 5k + 8 = 3k + 26
    k = 9
    So boxA = 45, boxB = 27

    ......
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  7. #7
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    Hello, kingman!


    Boxes A and B are each filled with some sand.
    If sand from Box B is poured into Box A until the sand in Box A reaches the brim,
    there will be 8 litres of sand left in Box B.
    If sand from Box A is poured into Box B until the sand in Box B reaches the brim,
    there will be 26 litres of sand left in Box A.
    The ratio of the volume of Box A to the volume of Box B is 5:3.
    How many more litres of sand are needed to fill both boxes to their brim?

    Box A has volume \,5k and contains \,a liters of sand.
    Box B has volume \,3k and contains \,b liters of sand.


    Code:
             5k
          *      *
          |      |       3k
          * - - -*    *      *
          |::::::|    |      |
          |::::::|    * - - -*
          |:: a :|    |::::::|
          |::::::|    |:: b :|
          |::::::|    |::::::|
          *-------    *-------
              A           B


    Sand from box B is poured into box A, filling box A.
    The amount of sand transferred is: . 5k-a liters.

    Code:
             5k
          * - - -*
          |:5k-a:| ←     3k
          * - - -*    *      *
          |::::::|    |      |
          |::::::|    |      |
          |:: a :|    |      |
          |::::::|    * - - -*
          |::::::|    |:: 8 :|
          *-------    *------*
              A           B

    This leaves: b - (5k-a) \:=\:8 liters in box B.

    . . We have: . a + b - 5k \:=\:8 .[1]



    Sand from box A is poured into box B, filling box B.
    The amount of sand transferred is: . 3k-b liters.


    Code:
             5k
          *      *
          |      |       3k
          |      |    * - - -*
          |      |    |:3k-b :
          * - - -*    * - - -*
          |::::::|    |::::::|
          |::26::|    |:: b :|
          |::::::|    |::::::|
          *-------    *-------
              A           B

    This leaves a - (3k-b) \,=\,26 liters in box A.

    . . We have: . a + b - 3k \:=\:26 .[2]



    Subtract [2] - [1]: . 2k \,=\,18 \quad\Rightarrow\quad k \,=\,9

    Substitute into [1]: . a + b - 5(9) \:=\:8 \quad\Rightarrow\quad a + b \:=\:53

    . . The boxes originally contains 53 liters of sand.


    The total capacity of the two boxes is: . 5k + 3k \:=\:8k \:=\:8(9) \:=\:72 liters.


    Therefore: . 72 - 53 \:=\:19 liters of sand will fill both boxes.

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  8. #8
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    Hi Soroban,
    Thanks very much for the solution; and I really admire your great ability for good insight in seeing great gem in a question and most important providing very simple graphical solution for them thus rendering meaning and appreciation for readers.
    Bye
    Kingsman
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