A is twice as good a workman as B and together they finish a piece of work in 18 days.in how many days will A alone finish this work?
Lets suppose that worker A works finishes a job in A days
and worker B finishes a job in B days
Then worker A's rate is $\displaystyle \frac{1}{A}$
Then worker B's rate is $\displaystyle \frac{1}{B}$
But we know that worker A is "twice" as good (2 times as fast) as worker B
so $\displaystyle \frac{1}{A}=2\cdot \frac{1}{B}$
We can now set the rates equal
$\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{1}{18}$
$\displaystyle 2\cdot \frac{1}{B}+\frac{1}{B}=\frac{1}{18}$
$\displaystyle \frac{3}{B}=\frac{1}{18} \iff 54=B$
Plugging in and solving for A gives A=27.
I hope this helps