1. ## 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

2. 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: .$\displaystyle 2.45 + 1.1 \:=\:3.55\text{ kg.}$

3. $\displaystyle \begin{array}{l} \frac{2}{7}W + E = 1.8{\rm }kg \\ \frac{5}{7}W + E = 2.85{\rm }kg \\ \end{array}$

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

solve the equation ...

$\displaystyle \displaylines{ \frac{2}{7}W + E = 1.8{\rm }kg........(1) \cr \frac{5}{7}W + E = 2.85{\rm }kg......(2) \cr \cr \frac{2}{7}W + E = 1.8{\rm }kg \cr E = 1.8{\rm }kg - \frac{2}{7}W \cr}$

substitue it to eq. (2)

$\displaystyle \displaylines{ \frac{5}{7}W + 1.8{\rm }kg - \frac{2}{7}W = 2.85{\rm }kg \cr \frac{3}{7}W = 1.05{\rm }kg \cr W = 2.45{\rm }kg \cr}$

so ...

$\displaystyle \displaylines{ \frac{2}{7}W + E = 1.8{\rm }kg \cr \frac{2}{7}(2.45{\rm }kg) + E = 1.8{\rm }kg \cr E = 1.1{\rm }kg \cr}$

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

< wew Soroban solved it first ... >