A) 49 : 221
B) 39 : 231
C) 94 : 181
D) None of these
Correct Answer: A
Solution :
Copper in \[4\,kg\,=\frac{4}{5}\,kg\] and zinc in \[4\,kg\,=4\times \frac{4}{5}=\frac{16}{5}\,kg\] Copper in 5 \[kg\,=\,5\times \frac{1}{6}=\frac{5}{6}\,kg\] and zinc in \[5\,kg\,=5\times \frac{5}{6}=\frac{25}{6}\,kg\] Therefore, copper m mixture \[=\frac{4}{5}+\frac{5}{6}=\frac{49}{30}\,kg\] and zmc in the mixture \[=\frac{16}{5}+\frac{25}{6}=\frac{221}{30}\,kg\] Therefore, the required ratio = 49: 221You need to login to perform this action.
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