SOLUTION: 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 9

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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?
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
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?

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+%2B+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%2F19 kilo and 1 kilo of 
silver loses 1%2F10 kilo when weighted in water.

Therefore

g+-+g%281%2F19%29 = the no. of kilos the gold in the 
bar weighs in water.
s+-+s%281%2F10%29 = the no. of kilos the silver in the 
bar weighs in water.

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

Therefore

%28g+-+g%281%2F19%29%29+%2B+%28s+-+s%281%2F10%29%29+=+99

is the other equation.

Simplifying it:
%28g+-+g%281%2F19%29%29+%2B+%28s+-+s%281%2F10%29%29+=+99
g+-+g%2F19+%2B+s+-+s%2F10+=+99
Multiply through by the LCD of 190:

190g+-+10g+%2B+190s+-+19s+=+18810

180g+%2B+171s+=+18810

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

60g+%2B+57s+=+6270

So we have this system:

system%28g+%2B+s+=+106%2C+60g+%2B+57s+=+6270%29

Solve that by substitution or elimination
and get:

matrix%281%2C5%2Cg=76%2Ckilos%2C%22%2C%22%2Cs=30%2Ckilos%29

Edwin