Finding time needed for a task, given two person's distinct task/time rates John can dig constantly at $15$ inches per minute and Linda can dig constantly at $45$ inches per minute. 
A certain hole can be dug by John in $12$ hours. 
If hole is dug by John for half the time and both together for the rest of the time, how many minutes does it take to dig the hole?
 A: We need to use data about john to determine the depth of hole: $$\underbrace{\dfrac{15 \;\text{inches}}{\text{minute}}}_{\text{rate of john's digging}} \times \underbrace{\frac{12 \text{ hours}}{1}}_{\text{time needed for john to dig hole}} \times \dfrac{60 \;\text{ min}}{\text{hour}} = \underbrace{15 \times 12 \times 60 \text{ inches}}_{\text{depth of hole}}$$
Let $x$ be the total number of minutes to dig hole needed under the given restrictions. 
Then $$\left(\frac 12x\times 15\right) + \left(\dfrac 12 x(15 + 45)\right) = 15 \times 12\times 60$$
Solve for $x$.
$$7.5 x + 7.5 x + 22.5 x = 10800 \iff 37.5 x = 10800 $$ $$\iff x = \dfrac{10800}{37.5}  = 288\;\text{ minutes total time for john and Linda to dig hole}$$
A: Hint:  how many inches per minute do they dig together?  In one minute of John digging and one minute of both digging, how many inches are dug?
A: Let the total time taken be $2x $ minute.
Then we have $x\times15+x\times (15+45)=15\times 12\times 60$
$15x$ implies the depth Jhon has dug for first half at a rate of $15$ inches /minute.
$60x$ implies the depth Jhon & Linda has dug for next half at a rate of $15$ inches /minute and $45$ inches /minute.
$15\times 12\times 60$ implies the total depth of the hole which is $12\times 60$ minutes for jhon(at his rate of $15$ inches)
Now solve for $x$ .Remember that total time is 2 times $x$ and not just $x$.
