Time and work

Here is a complete question to start with: A takes $4$ hours to finish a job, and B takes $6$ hours to finish the same job.
Working together, how long do they take? The idea is to combine how fast each one works, and there are two clean ways to do it.

Method 1 — count in whole units.
Pretend the whole job is a convenient number of units, say $4 \times 6 = 24$.
Then A, who takes $4$ hours, does $24 \div 4 = 6$ units each hour, and B does $24 \div 6 = 4$ units each hour; together that is $6 + 4 = 10$ units each hour, so the $24$-unit job takes $24 \div 10 = 2.4$ hours.

Method 2 — use fractions.
In one hour A does $\frac{1}{4}$ of the job and B does $\frac{1}{6}$, so together in one hour they do $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$ of the job.
Now use the unitary idea: if $1$ hour $\rightarrow$ $\frac{5}{12}$ of the job, and $x$ hours $\rightarrow$ $1$ whole job, then cross-multiplying gives $x = 1 \div \frac{5}{12} = \frac{12}{5} = 2.4$ hours.

The rate idea also covers someone undoing work.
Suppose a tap fills a tank in $3$ hours while an open drain empties the full tank in $6$ hours.
The tap adds $\frac{1}{3}$ per hour and the drain removes $\frac{1}{6}$, so the net rate is $\frac{1}{3} - \frac{1}{6} = \frac{1}{6}$ per hour, and the tank fills in $1 \div \frac{1}{6} = 6$ hours.

It also handles staggered starts, where one person begins alone.
Say A can finish a job in $5$ hours and B in $10$ hours, and A works alone for $2$ hours before B joins.
Take the job as $10$ units, so A does $2$ units per hour and B does $1$; in $2$ hours alone A does $4$ units, leaving $6$ units, and together they do $3$ units per hour, so the rest takes $6 \div 3 = 2$ more hours.

A finishes a job in $3$ hours and B finishes it in $6$ hours. How long do they take together?
In one hour they do $\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{1}{2}$ of the job.
So the whole job takes $1 \div \frac{1}{2} = 2$ hours.

A tap fills a tank in $5$ hours, and an open drain can empty the full tank in $10$ hours. If both are left open, how long does the tank take to fill?
The tap adds $\frac{1}{5}$ of the tank per hour while the drain removes $\frac{1}{10}$, so the net rate is $\frac{1}{5} - \frac{1}{10} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10}$ per hour.
So the tank fills in $1 \div \frac{1}{10} = 10$ hours.

A can finish a job in $6$ hours and B in $12$ hours. A works alone for $2$ hours before B joins. How much longer does the job take after B joins?
Take the whole job as $12$ units, so A does $2$ units per hour and B does $1$ unit per hour.
In $2$ hours alone, A completes $2 \times 2 = 4$ units, leaving $12 - 4 = 8$ units.
Once B joins, together they do $2 + 1 = 3$ units per hour, so the remaining $8$ units take $8 \div 3 = \frac{8}{3}$ hours, about $2.7$ hours.