Question

Solve the following word problem. A man travels from Town X to Town Y at an average rate of 50 mph and returns at an average rate of 40 mph. He takes a 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?

Answer 1

distance = rate*time

Let distance from X to Y be 'd'

From the equation we can solve for 'time'

time = distance/rate

The total time it takes to go 50 mph ,then 40 mph is:


`(d)/(50) + (d)/(40)`

The total time it takes to go 45 mph round trip is:


`(2d)/(45)`

We know that the round-trip at 45 mph takes half-hour shorter, so by adding '1/2' to its time it will be equal to the time to go 50 then 40.


`(2d)/(45) + (1)/(2) = (d)/(50) + (d)/(40) \\ \\ (4d+45)/(90) = (9d)/(200) \\ \\ 800d+9000 = 810d \\ \\ 10d = 9000 \\ \\ d = 900`

Answer: The distance from X to Y is 900 miles.

Answer 2

The distance from Town X to Town Y is 900 miles.

Given to us,

• Average speed from Town X to Town Y = 50 mph,

• Average speed from Town Y to Town X = 40 mph,

• Average speed for round trip = 45 mph,

• Time taken is
  `(1)/(2)` hour more if taken the round trip.

Assumption

Let the distance between town X and town Y be x.

Time Calculation

Time is taken for traveling from Town X to Town Y
`= (Distance)/(Speed)= (x)/(50)`

Time is taken for traveling from Town Y to Town X
`= (Distance)/(Speed)= (x)/(40)`

Total time = Time is taken for traveling from Town X to Town Y + Time is taken for traveling from Town Y to Town X

Total time =
`(x)/(50)+(x)/(40)`

Time is taken for a round trip at an average speed of 45 mph
`= (Distance)/(Speed)= (2x)/(45)`

Comparison of Time

`(x)/(50)+(x)/(40)= (1)/(2)+ (2x)/(45)`

`(4x)/(200)+(5x)/(200)= (45)/(90)+ (4x)/(90)\\\\(4x+5x)/(200)= (45+4x)/(90)\\\\(9x)/(200)= (45+4x)/(90)\\\\810x = 9000+800x\\810x-800x = 9000\\10x = 9000\\\\x=(9000)/(10)\\\\x=900`

Hence, the distance from Town X to Town Y is 900 miles.