A and b are hvo alloys of gold and copper prepared by mixing metals in the ratio 7: 2 and 7:11, respectively . If equal quantities of the alloys are melted to from a third alloy c the ratio of gold and copper in c will be
Answers
Answer:
Alloy C contain the gold and copper in ratio = 7 : 12
Step-by-step explanation:
Since
Alloy A contain gold and copper in ratio = 7 : 2
Alloy B contain gold and copper in ratio = 7 : 11
Then
9 unit of Alloy A contain 7 units gold and 2 unit copper
That is
9 units of Alloy A = 7 units of gold + 2 units of copper
Now
1 units of Alloy A = 7/9 units of gold + 2/9 units of copper
Similarly
18 unit of Alloy B contain 7 units gold and 11 unit copper
That is
18 units of Alloy A = 7 units of gold + 11 units of copper
Now
1 units of Alloy B = 7/18 units of gold + 11/18 units of copper
NOW
If we create the new Alloy C by 1 units of Alloy A and 1 unit of Alloy B then
2 units of Alloy C = 1 unit of Alloy A + 1 unit of Alloy B
= (7/9 units of gold + 2/9 units of copper) + (7/18 units of gold + 11/18 units of copper)
= ( 7/9 + 7/18 ) units of gold + ( 2/9 + 11/18 ) units of copper
= 21/18 units of gold + 15/18 units of copper
So
2 units of new Alloy C = 21/18 units of gold + 15/18 units of copper
This implies that
1 unit of new Alloy C = 21/36 units of gold + 15/36 units of copper
Thus
Alloy C contain the gold and copper in ratio = 21/36 : 15/36 = 21 : 36
= 7 : 12
Hence
Alloy C contain the gold and copper in ratio = 7 : 12