<PRE><B>Question 198883</B>
A man travels from Town x to town y at an average rate of 50mph and returns at
 an average rate of 40pmh. he takes 1/2 hour longer than he would take if he
 made the round trip at an average of 45 mph. What is the distance from town x to town y.
:
Let d = dist of town x to town y
:
Write a time equation: Time = dist/speed
:
To time + return time = 45 mph time + 1/2 hr
{{{d/50}}} + {{{d/40}}} = {{{(2d)/45}}} + {{{1/2}}}
:
Multiply equation by 1800:
1800*{{{d/50}}} + 1800*{{{d/40}}} = 1800*{{{(2d)/45}}} + 1800*{{{1/2}}}
Cancel the denominators, results:
36d + 45d = 40(2d) + 900
:
81d = 80d + 900
81d - 80d = 900
d = 900 mi from town x to town y
;
:
The solution in original time equation (round trip = 1800 mi)
900/50 + 900/40 = 1800/45 + 1/2
18 + 22.5 = 40 + .5</PRE>