SSC
Quantitative Aptitude
Mixture And Alligation
Question Bank
Mixture and Allegation (II)
question_answer
A and B are 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
A)5 : 7
B)5 : 9
C)7 : 5
D)9 : 5
Correct Answer:
C
Solution :
[c]
Gold
Copper
Alloy A
\[\frac{7}{9}\]
\[\frac{2}{9}\]
Alloy B
\[\frac{7}{18}\]
\[\frac{11}{18}\]
Since, alloys A and B are melted in the ratio 1 : 1 to make the alloy, therefore in the alloy C, the ratio of gold and copper. \[\left( \frac{7}{9}\times \frac{1}{2}+\frac{7}{18}\times \frac{1}{2} \right):\left( \frac{2}{9}\times \frac{1}{2}+\frac{11}{18}\times \frac{1}{2} \right)\] \[=\left( \frac{7}{9}+\frac{7}{18} \right):\left( \frac{2}{9}+\frac{11}{18} \right)\] \[=\frac{21}{18}:\frac{15}{18}=21:15=7:5\]