Question 897728
This solution will be in time quantities, and not points in time.


PIPE RATES FOR EMPTYING TANK UNITS of JOBS per HOUR.
Fast pipe, {{{1/4}}}
Slow pipe, {{{1/11}}}


Total time for the one job is to be 3 hours.  The fast pipe starts first and the slow pipe is
later started and works with the large pipe until all 3 hours pass; and the 1 job is done.


Fast pipe works alone for t hours; and both pipes work together for 3-t hours.


{{{highlight((1/4)t+(1/4+1/11)(3-t)=1)}}}
Solve this equation for {{{t}}}, the amount of time that the fast pipe works alone before
the slow pipe is opened.  (Lowest Common Denominator is {{{4*11}}}, so multiply both sides of
the equation by this.)


{{{11t+(11+4)(3-t)=44}}}
{{{11t+15(3-t)=44}}}
{{{11t+45-15t=44}}}
{{{-4t+45=44}}}
{{{-4t=-1}}}
{{{highlight(t=1/4)}}}-------FIFTEEN MINUTES, a fourth of an hour.