Let  = the no. of kilos of gold in the bar 
(when weighed in air).
Let  = the no. of kilos of silver in the bar 
(when weighed in air).

>>...a bar...weighing 106 kilos in air...<<

Therefore one equation is



>>...19 kilos of gold loses 1 kilo and 10 kilos of 
silver loses 1 kilo when weighted in water...<<

Therefore

1 kilo of gold loses  kilo and 1 kilo of 
silver loses  kilo when weighted in water.

Therefore

 = the no. of kilos the gold in the 
bar weighs in water.
 = the no. of kilos the silver in the 
bar weighs in water.

>>...a bar...weighing...99 kilos in water?

Therefore



is the other equation.

Simplifying it:


Multiply through by the LCD of 190:





That can be divided through by 3 getting
only whole numbers:



So we have this system:



Solve that by substitution or elimination
and get:



Edwin