AP ICET 2015 Paper

Let m and w represent 1 day work of a man and a woman respectively.
We have,
$2\mathit{m}+3\mathit{w}=\frac{1}{10}\Rightarrow 20\mathit{m}+30\mathit{w}=1$.....(i)
$4\mathit{m}=\frac{1}{10}\Rightarrow 40\mathit{m}=1$...(ii)
Equating (i) and (ii) we get$40\mathit{m}=20\mathit{m},+30\mathit{w}$
$\Rightarrow 20\mathit{m}=30\mathit{w}\Rightarrow 3\mathit{w}=2\mathit{m}$
$\therefore 3\mathit{m}+3\mathit{w}=3\mathit{m}+2\mathit{m}=5\mathit{m}$
$4\mathit{m}\equiv \frac{1}{10}$
$\therefore 5\mathit{m}=\frac{5}{4\times 10}=\frac{1}{8}$
∴ Required no. of days = 8 days
Aliter.
4 men complete the work in 10 days.
∴ Part of work done by 2 men in 10 days $=\frac{1}{2}$
So, the remaining half work was done by 3 women in 10 days.
⇒ 3 women = 2men.
∴ 3 men + 3women = 5 men
∴ Required no. of days$=10\times \frac{4}{5}=8$days