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 
= weight of the empty jug.
Let 
= weight of water when jug is completely filled.
Code:
* *
| |
| |
| |
| - - - | 1.8 kg
|:::::::|
| 2/7 W |
|:::::::|
*-------*
J When the jug is 
full, the total weight is 1.8 kg.
Equation #1: . ![J + \tfrac{2}{7}W \:=\:1.8\;\;{\color{blue}[1]}](http://latex.codecogs.com/png.latex?J + \tfrac{2}{7}W \:=\:1.8\;\;{\color{blue}[1]})
Code:
* *
| |
| - - - |
|:::::::|
|:::::::| 2.85 kg
| 5/7 W |
|:::::::|
|:::::::|
*-------*
J When the jug is 
full, the total weight is 2.95 kg.
Equation #2: . ![J + \tfrac{5}{7}W \:=\:2.85\;\;{\color{blue}[2]}](http://latex.codecogs.com/png.latex?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}](http://latex.codecogs.com/png.latex?\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: . 
Substitute into [1]: .  \:=\:1.8 \quad\Rightarrow\quad\boxed{ J \:=\:1.1})
(a) The empty jug weighed: . 
(b) The filled jug weighed: . 
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Please solve it by easy and arithmetic method(if possible)
