MATH SOLVE

3 months ago

Q:
# 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? ______ miles

Accepted Solution

A:

distance = rate*time

Let distance from X to Y be 'd'Β

From the equation we can solve for 'time'

time = distance/rate

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

[tex]\frac{d}{50} + \frac{d}{40}[/tex]

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

[tex]\frac{2d}{45} [/tex]

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.

[tex]\frac{2d}{45} + \frac{1}{2} = \frac{d}{50} + \frac{d}{40} \\ \\ \frac{4d+45}{90} = \frac{9d}{200} \\ \\ 800d+9000 = 810d \\ \\ 10d = 9000 \\ \\ d = 900[/tex]

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

Let distance from X to Y be 'd'Β

From the equation we can solve for 'time'

time = distance/rate

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

[tex]\frac{d}{50} + \frac{d}{40}[/tex]

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

[tex]\frac{2d}{45} [/tex]

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.

[tex]\frac{2d}{45} + \frac{1}{2} = \frac{d}{50} + \frac{d}{40} \\ \\ \frac{4d+45}{90} = \frac{9d}{200} \\ \\ 800d+9000 = 810d \\ \\ 10d = 9000 \\ \\ d = 900[/tex]

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