SOLUTION: The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourt
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Question 30587: The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
The amount (by weight) of gold, silver and lead in three alloys of these metals are in ratios:
4:3:2 - Alloy 1
3:5:1 - Alloy 2
2:2:5 - Alloy 3
It is desired to make a fourth alloy containing equal amounts of gold, silver and lead. How many grams each alloy should be used for every 10 grams of the new alloy?
LET X GMS OF ALLOY1 ,Y GMS OF ALLOY2 AND 10-X-Y GMS OF ALLOY3 BE USED TO GET
X+Y+10-X-Y=10 GMS OF ALLOY 4
SO.................GOLD..............SIVER............LEAD IN THE MIX IS GIVEN BY
X GMS A1...........4X/9..............3X/9..........2X/9
Y GMS A2...........3Y/9..............5Y/9...........Y/9....
10-X-Y GMS A3....(20-2X-2Y)/9 ......(20-2X-2Y)/9...(50-5X-5Y)/9
-------------------------------------------------------------------------------
10 GMS A4.......(20+2X+Y)/9........(20+X+3Y)/9.....(50-3X-4Y)/9
THESE ARE ALL EQUAL...HENCE
20+2X+Y = 20+X+3Y...OR......................X-2Y=0..............I
20+2X+Y = 50-3X-4Y...OR...5X+5Y=30....OR....X+Y=6......II
EQN.II - EQN I...GIVES
X+Y-X+2Y=6......OR 3Y=6.....Y=2
SO X=6-Y=6-2=4
Z=10-4-2=4...
HENCE 4 GMS OF A1,2 GMS OF A2 AND 4 GMS OF A3 ARE TO BE ADDED TO GET 10 GMS OF A4.
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