lois rode her bike to visit a friend. She traveled at 10 miles an hour. While she was there it began to rain. Her friend drove her home at 25 miles per hour. Lois took 1.5 hours longer to go to a friends then to return home. How many hours did it take Lois to arrive at her friend's house?
distance = rate x time.
return trip = 25 mph x time
going = 10 mph x (time + 1.5)
distance is the same; therefore, set them equal to each other.
25t = (10(t+1.5)
solve for time
time + 1.5 = time it took Lois to go to her friends house. Post your work if you get stuck.
Call the two times t1 and t2. Call the distance between points D.
D = 10 t1
D = 25 t2
t1 = t2 + 1.5
t1 = 2.5 t2 would be a good substitution in the last equation. Then you will have one equation on one unknown (t2). Once you have that, you can compute t1.
To find the number of hours it took Lois to arrive at her friend's house, we can set up an equation based on the given information.
Let's assume the time taken by Lois to reach her friend's house is 't' hours.
Given that Lois traveled at a speed of 10 miles per hour, the distance covered by Lois to reach her friend's house is 10t miles.
When Lois returned home with her friend, they traveled at a speed of 25 miles per hour. Since the distance covered while returning is the same as the distance covered while going, Lois also traveled a distance of 10t miles during the return journey.
We are given that Lois took 1.5 hours longer to go to her friend's house than to return home. So, the time taken for the return journey is 't - 1.5' hours.
Using the equation for distance (distance = speed × time), we can write the equation as:
10t = 25 * (t - 1.5)
To solve this equation and find the value of 't', we can simplify it:
10t = 25t - 37.5 (distributing 25 to t and -1.5)
25t - 10t = 37.5 (subtracting 10t from both sides)
15t = 37.5 (combining like terms)
t = 37.5 / 15 (dividing both sides by 15)
t = 2.5 hours
Therefore, it took Lois 2.5 hours to arrive at her friend's house.