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MAT 2005 Question-5

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Q.) A and B two 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 form a third alloy C , the ratio of gold and copper in C will be

1)5:7
2)7:5
3)8:9
4)9:5

Solution :

let the quantity of each alloy A and B be X kg. mixed to form alloy C.

gold in A = $\frac { 7X }{ (7+2) } =\frac { 7X }{ 9 } $

copper in A = $\frac { 2X }{ (7+2) } =\frac { 2X }{ 9 } $

gold in B =$\frac { 7X }{ (7+11) } =\frac { 7X }{ 18 } $

copper in B = $\frac { 11X }{ (7+11) } =\frac { 11X }{ 18 } $

Total gold in C = $\frac { 7X }{ 9 } +\frac { 7X }{ 18 } =\frac { 21X }{ 18 } $

Total copper in C = $\frac { 2X }{ 9 } +\frac { 11X }{ 18 } =\frac { 15X }{ 18 } $

Hence , the ratio of gold and copper in C = $\frac { 21X }{ 18 } :\frac { 15X }{ 18 } $ = 7:5

Hence Option(2)

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