Question 136251
In all these type problems, the key usually lies in answering
the question "What is the same for both (planes, cars, whatever)?
In this case it is time. They both fly for 1/2 hr.
Let {{{d}}} = the distance covered by the San Francisco plane in 1/2 hr
Then {{{275 - d}}} will be the distance covered by the San Diago plane
{{{s}}} = the speed of the S.F. plane
{{{s-50}}} = the speed of the S.D. plane
For S.F. plane:
(1) {{{d = s*(.5)}}}
For S.D. plane:
(2){{{275 - d = (s-50)*(.5)}}}
Substitute {{{d}}} in (1) for {{{d}}} in (2)
{{{275 - .5s = .5*(s - 50)}}}
{{{275 - .5s = .5s - 25}}}
{{{s = 300}}} mi/hr
{{{s - 50 = 250}}}mi/hr
The speed of the San Francisco plane is 300 mi/hr
The speed of the San Diago plane is 250 mi/hr
Check answer:
For S.F. plane:
(1) {{{d = s*(.5)}}}
For S.D. plane:
(2){{{275 - d = (s-50)*(.5)}}}
{{{d = .5*300}}}
{{{d = 150}}}
{{{275 - 150 = .5*(300 - 50)}}}
{{{125 = 125}}}
OK