A man can drink a cask of wine in 20 days, but if his wife drinks with him it will take only 14 days—how long would it take for the wife alone?
$$\dfrac{1}{20}+\dfrac{1}{x}=\dfrac{1}{14}$$
$$\dfrac{1}{x}=\dfrac{20-14}{14*20}=\dfrac{6}{14*20}$$
$$x=\dfrac{14*20}{6}=46.6666666667$$
for the wife alone 46.6666666667 days
P.S.
$$\begin{array}{lcr}
\frac{1\;V}{20\;days} & \mbox{rate for 1 day} & \mbox{man}\\\\
\frac{1\;V}{x\;days} & \mbox{rate for 1 day} & \mbox{woman}\\\\
\frac{1\;V}{20\;days} + \frac{1\;V}{x\;days} & \mbox{rate for 1 day} & \mbox{man and woman}\\\\
\left(\frac{1\;V}{20\;days} + \frac{1\;V}{x\;days}\right) & \times \mbox{ 14 days}
&=1\mbox{ V}
\end{array}\\\\\\
\Rightarrow \dfrac{1}{20}+\dfrac{1}{x}=\dfrac{1}{14}$$
$$\dfrac{1}{20}+\dfrac{1}{x}=\dfrac{1}{14}$$
$$\dfrac{1}{x}=\dfrac{20-14}{14*20}=\dfrac{6}{14*20}$$
$$x=\dfrac{14*20}{6}=46.6666666667$$
for the wife alone 46.6666666667 days
P.S.
$$\begin{array}{lcr}
\frac{1\;V}{20\;days} & \mbox{rate for 1 day} & \mbox{man}\\\\
\frac{1\;V}{x\;days} & \mbox{rate for 1 day} & \mbox{woman}\\\\
\frac{1\;V}{20\;days} + \frac{1\;V}{x\;days} & \mbox{rate for 1 day} & \mbox{man and woman}\\\\
\left(\frac{1\;V}{20\;days} + \frac{1\;V}{x\;days}\right) & \times \mbox{ 14 days}
&=1\mbox{ V}
\end{array}\\\\\\
\Rightarrow \dfrac{1}{20}+\dfrac{1}{x}=\dfrac{1}{14}$$
For those who don't understand why heureka manipulated reciprocal times (which is the correct thing to do) here's a more long-winded approach.
Here's the way I like to think about this sort of problem. Each day the man drinks 1/20 of the cask. When his wife helps him, they drink 1/14 of the cask each day. So 1/14 - 1/20 must be the portion that she drinks everyday.
1/14 - 1/20 = 6/280 ....... now, just take the reciprocal of the fraction on the right
280/6 = 46+2/3 days
And there you go.....