Question 413815
The tanker is to be emptied in 2 stages in 3 hrs.
Let {{{t}}} = hours that main pump alone is working
Amount pumped by main pump = 
(1) {{{(1/4)*t = t/4}}}
That is: (1 tank)/(4 hrs) x (t hrs)
The auxilliary pump + main pump will then pump out:
(2){{{(1/4)*(3 - t) + (1/7)*(3 - t)}}}
If I add (1) and (2), I should get {{{1}}} (1 tank pumped)
{{{t/4 + (3 - t)/4 + (3 - t)/7 = 1}}}
Multiply both sides by {{{4*7}}}
{{{7t + 7*(3 - t) + 4*(3 - t)  = 28}}}
{{{7t + (3 - t)*11 = 28}}}
{{{7t + 33 - 11t = 28}}}
{{{4t = 5}}}
{{{t = 5/4}}} hrs
So, the main pump runs alone for 1 hr 15 min. That means
the auxilliary pump comes on at 8:15
check answer:
{{{t/4 + (3 - t)/4 + (3 - t)/7 = 1}}}
{{{(5/4)/4 + (3 - (5/4))/4 + (3 - (5/4))/7 = 1}}}
{{{5/16 + 7/16 + 4/16 = 1}}}
{{{ 1 = 1}}}
OK