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 $\displaystyle J$ = weight of the empty jug.
Let $\displaystyle W$ = weight of water when jug is completely filled.
Code:
* *
| |
| |
| |
| - - - | 1.8 kg
|:::::::|
| 2/7 W |
|:::::::|
*-------*
J
When the jug is $\displaystyle \tfrac{2}{7}$ full, the total weight is 1.8 kg.
Equation #1: .$\displaystyle J + \tfrac{2}{7}W \:=\:1.8\;\;{\color{blue}[1]}$
Code:
* *
| |
| - - - |
|:::::::|
|:::::::| 2.85 kg
| 5/7 W |
|:::::::|
|:::::::|
*-------*
J
When the jug is $\displaystyle \tfrac{5}{7}$ full, the total weight is 2.95 kg.
Equation #2: .$\displaystyle J + \tfrac{5}{7}W \:=\:2.85\;\;{\color{blue}[2]}$
Subtract [1] from [2]: . .$\displaystyle \begin{array}{ccc}J + \frac{5}{7}W &=& 2.85 \\ \\[-3mm]J + \frac{2}{7}W &=& 1.80 \\ \hline \\[-4mm] \qquad \frac{3}{7}W &=& 1.05 \end{array}$
Hence: .$\displaystyle \tfrac{3}{7}W \:=\:1.05 \quad\Rightarrow\quad \boxed{W \:=\:2.45}$
Substitute into [1]: .$\displaystyle J + \tfrac{2}{7}(2.45) \:=\:1.8 \quad\Rightarrow\quad\boxed{ J \:=\:1.1}$
(a) The empty jug weighed: .$\displaystyle 1.1\text{ kg.}$
(b) The filled jug weighed: .$\displaystyle 2.45 + 1.1 \:=\:3.55\text{ kg.}$