Question 160275
If 19 kilos of gold loses 1 kilo and 10 kilos of silver loses 1 kilo when weighted in water, find the weight of the gold in a bar of gold and silver weighing 106 kilos in air and 99 kilos in water?
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Let {{{g}}} = the no. of kilos of gold in the bar 
(when weighed in air).
Let {{{s}}} = the no. of kilos of silver in the bar 
(when weighed in air).

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

Therefore one equation is

{{{g + s = 106}}}

>>...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 {{{1/19}}} kilo and 1 kilo of 
silver loses {{{1/10}}} kilo when weighted in water.

Therefore

{{{g - g(1/19)}}} = the no. of kilos the gold in the 
bar weighs in water.
{{{s - s(1/10)}}} = the no. of kilos the silver in the 
bar weighs in water.

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

Therefore

{{{(g - g(1/19)) + (s - s(1/10)) = 99}}}

is the other equation.

Simplifying it:
{{{(g - g(1/19)) + (s - s(1/10)) = 99}}}
{{{g - g/19 + s - s/10 = 99}}}
Multiply through by the LCD of 190:

{{{190g - 10g + 190s - 19s = 18810}}}

{{{180g + 171s = 18810}}}

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

{{{60g + 57s = 6270}}}

So we have this system:

{{{system(g + s = 106, 60g + 57s = 6270)}}}

Solve that by substitution or elimination
and get:

{{{matrix(1,5,g=76,kilos,",",s=30,kilos)}}}

Edwin</pre>