A) 10 : 3
B) 3 : 10
C) 3 : 7
D) 2 : 7
Correct Answer: B
Solution :
Let them be mixed in the ratio \[x:y\]. Then, in 1st alloy, zinc \[=\frac{x}{3}\] and copper \[=\frac{2x}{3}\] 2nd alloy, zinc \[=\frac{2y}{5}\] and copper \[=\frac{3y}{5}\] Now, we have, \[\frac{x}{3}+\frac{2y}{5}:\frac{2x}{3}+\frac{3y}{5}=5:8\] \[\Rightarrow \] \[\frac{5x+6y}{10x+9y}=\frac{5}{8}\] \[\Rightarrow \] \[40x+48y=50x+45y\] \[\Rightarrow \] \[10x=3y\] \[\Rightarrow \] \[\frac{x}{y}=\frac{3}{10}\] \[\therefore \] Required ratio = 3 : 10 Alternatively Let us consider the fractional value of zinc Therefore, they should be mixed in the ratio. \[\frac{1}{65}:\frac{2}{39}\] \[\Rightarrow \] \[\frac{1}{65}\times \frac{39}{2}=\frac{3}{10}=3:10\]You need to login to perform this action.
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