SOLUTION: a dishonest jeweler claims that the articles he sells are made of pure gold but actualy he mixes silver with gold. pure gold weights 19gm/cm^3 and silver 8 gm/cm^3. one cm^3 of th

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: a dishonest jeweler claims that the articles he sells are made of pure gold but actualy he mixes silver with gold. pure gold weights 19gm/cm^3 and silver 8 gm/cm^3. one cm^3 of th      Log On


   



Question 696353: a dishonest jeweler claims that the articles he sells are made of pure gold but actualy he mixes silver with gold. pure gold weights 19gm/cm^3
and silver 8 gm/cm^3. one cm^3 of the articles sold by the jeweler weighs 18gm/cm^3. the ratio of weight of gold an d silver he mixes is??

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = cm3 of gold in jewelers 1 cm^3
Let +b+ = cm3 of silver in jewelers 1 cm^3
(1) +19a+%2B+8b+=+18+
Note that in terms of units, this is
( gm / cm3 )x( cm3 ) + ( gm / cm3 ) = ( gm )
So I have ( gm ) = ( gm )
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Also,
(2) +a+%2B+b+=+1+
( cm3 ) + ( cm3 ) = ( cm3 )
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Multiply both sides of (2) by +8+
and subtract (2) from (1)
(1) +19a+%2B+8b+=+18+
(2) +-8a+-+8b+=+-8+
+11a+=+10+
+a+=+10%2F11+
and, since
(2) +a+%2B+b+=+1+
(2) +b+=+1%2F11+
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There is 10/11 cm3 of gold in jewelers 1 cm^3
There is 1/11 cm3 of silver in jewelers 1 cm^3
The weight of gold in 10/11 cm3 is
+19%2A%2810%2F11%29+=+190%2F11+
The weight of silver in 1/11 cm3 is
+8%2A%281%2F11%29+=+8%2F11+
( weight of gold ) / ( weight of silver ) = +%28+190+%2F+11+%29+%2F+%28+8+%2F+11+%29+
+%28+190+%2F+11+%29+%2F+%28+8+%2F+11+%29+=+190%2F8+
+23.75+
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( weight of gold ) / ( weight of silver ) = 23.75
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check answer:
+%28+19a+%29+%2F+%28+8b+%29+=+23.75+
+a+=+1+-+b+
+%28+19%2A%28+1+-+b+%29+%29+%2F+%28+8b+%29+=+23.75+
+%28+19+-+19b+%29+%2F+%28+8b+%29+=+23.75+
+19+-+19b+=+190b+
+209b+=+19+
+b+=+19%2F209+
+b+=+1%2F11+
OK