In a club there are certain number of males and females. If 15 females are absent then number of males will be half of females. If 45 males are absent then female strength will be 5 times that of males. Number of males actually present is
Let us take x number of male and y number of female. \[\] \(\frac{1}{2}(y-15)=x\) \[\] \(5(x-45)=y\) \[\] \(2 x-y=-15\) \[\] \(5 x-y=225\) \[\] \(\Rightarrow 3 x=240\) or \(x=80\) \[\] Hence number of males will be 80 . \[\] Choice (B)\[\]
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