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Math Help - Please solve it by easy and arithmetic method(if possible)

  1. #1
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    Question Please solve it by easy and arithmetic method(if possible)

    a jug weighed 1.8 KG wgen 2/7 of it was filled with water & weighs 2.85 KG when it was filled 5/7.
    Now,
    find
    (a)weight of jug when it was empty
    (b)what was mass of jug when it was completely filled
    DRAW IMAGE IF IT GIVES YOU NO DIFFICULTY!
    THANX
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  2. #2
    Super Member

    Joined
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    From
    Lexington, MA (USA)
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    Hello, Peacewanters007!

    I couldn't find an Arithmetic solution.
    I had to use Algebra . . . sorry.


    A jug weighed 1.8 kg when 2/7 of it was filled with water
    and weighed 2.85 kg when it was filled 5/7.

    Find:
    (a) Weight of jug when it was empty.
    (b) What was mass of jug when it was completely filled?

    Let J = weight of the empty jug.
    Let W = weight of water when jug is completely filled.


    Code:
          *       *
          |       |
          |       |
          |       |
          | - - - |   1.8 kg
          |:::::::|
          | 2/7 W |
          |:::::::|
          *-------*
              J
    When the jug is \tfrac{2}{7} full, the total weight is 1.8 kg.

    Equation #1: . J + \tfrac{2}{7}W \:=\:1.8\;\;{\color{blue}[1]}





    Code:
          *       *
          |       |
          | - - - |
          |:::::::|
          |:::::::|   2.85 kg
          | 5/7 W |
          |:::::::|
          |:::::::|
          *-------*
              J
    When the jug is \tfrac{5}{7} full, the total weight is 2.95 kg.

    Equation #2: . J + \tfrac{5}{7}W \:=\:2.85\;\;{\color{blue}[2]}



    Subtract [1] from [2]: . . \begin{array}{ccc}J + \frac{5}{7}W &=& 2.85 \\ \\[-3mm]J + \frac{2}{7}W &=& 1.80  \\ \hline \\[-4mm] \qquad \frac{3}{7}W &=& 1.05 \end{array}

    Hence: . \tfrac{3}{7}W \:=\:1.05 \quad\Rightarrow\quad \boxed{W \:=\:2.45}

    Substitute into [1]: . J + \tfrac{2}{7}(2.45) \:=\:1.8 \quad\Rightarrow\quad\boxed{ J \:=\:1.1}



    (a) The empty jug weighed: . 1.1\text{ kg.}

    (b) The filled jug weighed: . 2.45 + 1.1 \:=\:3.55\text{ kg.}

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  3. #3
    Junior Member Singular's Avatar
    Joined
    Dec 2006
    From
    Solo, Java
    Posts
    57
    <br />
\begin{array}{l}<br />
 \frac{2}{7}W + E = 1.8{\rm  }kg \\ <br />
 \frac{5}{7}W + E = 2.85{\rm  }kg \\ <br />
 \end{array}<br /> <br />

    W = weight of the water (full)
    E = weight of empty jug

    solve the equation ...

    <br />
\displaylines{<br />
  \frac{2}{7}W + E = 1.8{\rm  }kg........(1) \cr <br />
  \frac{5}{7}W + E = 2.85{\rm  }kg......(2) \cr <br />
   \cr <br />
  \frac{2}{7}W + E = 1.8{\rm  }kg \cr <br />
  E = 1.8{\rm  }kg - \frac{2}{7}W \cr}<br /> <br />

    substitue it to eq. (2)

    <br />
\displaylines{<br />
  \frac{5}{7}W + 1.8{\rm  }kg - \frac{2}{7}W = 2.85{\rm  }kg \cr <br />
  \frac{3}{7}W = 1.05{\rm  }kg \cr <br />
  W = 2.45{\rm  }kg \cr}<br /> <br />

    so ...

    <br />
\displaylines{<br />
  \frac{2}{7}W + E = 1.8{\rm  }kg \cr <br />
  \frac{2}{7}(2.45{\rm  }kg) + E = 1.8{\rm  }kg \cr <br />
  E = 1.1{\rm  }kg \cr}<br /> <br />

    the weight of empty jug is 1.1 kg
    the weight of full filled jug is 3.55 kg

    < wew Soroban solved it first ... >
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