Question 785071
To deliver a message, an army officer will travel by car at 75mph from Camp A to Camp B, and then by plane to Camp C against a wind blowing 40 mph.
 The airplane can fly 280 mph in still air.
 If the messenger takes 3 2/3 hours in going from Camp A to Camp C and 3 1/6 hours for the return trip,
 what is the distance from Camp A to Camp C?
:
let x = dist from A to B
let y = dist from B to C
then
(x+y) = dist from A to C
:
We know the speed from A to B will be 75 mph both directions, however,
Speed from B to C will be 240 mph against the wind, and,
From C to B with the wind, it will be 320 mph
:
Change 2/3 to 4/6
:
Write a time equation for each trip
{{{x/75}}} + {{{y/240}}} = 3{{{4/6}}}
{{{x/75}}} + {{{y/320}}} = 3{{{1/6}}}
------------------------------------subtracting eliminates x, find y
{{{y/240}}} - {{{y/320}}} = {{{3/6}}}
{{{y/240}}} - {{{y/320}}} = {{{1/2}}}
the common denominator of 960
{{{4y/960}}} - {{{3y/960}}} = {{{480/960}}}
{{{y/960}}} = {{{480/960}}}
y = 480 miles from B to C
:
Find time from B to C
{{{480/240}}} = 2 hrs
:
Find time from A to B
3{{{4/6}}} - 2 = 1{{{4/6}}} hrs which is {{{10/6}}} hrs
:
Find the dist from A to B
{{{10/6}}} * 75 = 125 miles 
then
125 + 480 = 605 miles from A to C
:
:
:
See if that checks out; find the return trip time, use decimals this time
{{{125/75}}} + {{{480/320}}} = 
1.67 + 1.5 = 3.17 which is about 3{{{1/6}}} hrs