1. A Passenger train's speed is 60mi/h and a freight train's speed is 40mi/h. The passenger train travels the same distance in 1.5 h less time than the freight train. How long does each train take to make the trip?

2. Lois rode her bike to visit a friend. She traveled at 10 mi/h. While she was there it began to rain. Her friend drove her home in a car traveling at 20 mi/h. Lois took 1.5 h longer to go to her friend's than to return home. How many hours did it take Lois to ride to her friend's house?

distance = rate x time

passenger train rate = 60 mi/hr
freight train rate = 40 mi/hr.
distance traveled by P = 60*t
distance traveled by F = 40*t
time for F is 1.5 hours longer than P so
F dist = 40(t+1.5) = P dist = 60*t
set distance = to each other
60*t=40(t+1.5)
solve for t = time for passenger train.
Then t+1.5 = tiome for freight train.

#2 is done the same say as #1. Post your work if you get stuck.

2:30

40(t+1.5)=60x t

2/3(t+1.5)= T

To solve these problems, we can use the equation: Distance = Speed × Time.

1. Let's define the following variables:
- Let p be the time taken by the passenger train.
- Let f be the time taken by the freight train.

According to the problem, we know that the passenger train's speed is 60 mph, the freight train's speed is 40 mph, and the passenger train takes 1.5 hours less than the freight train to cover the same distance.

Using the formula Distance = Speed × Time, we can now set up two equations:
- For the passenger train: Distance = 60p
- For the freight train: Distance = 40f

Since the distance traveled by both trains is the same, we can set these two equations equal to each other:
60p = 40f

We also know that the passenger train takes 1.5 hours less than the freight train to complete the trip, so we can set up another equation:
p = f - 1.5

Now we have a system of equations:
60p = 40f
p = f - 1.5

To solve this system, we can substitute the second equation into the first equation:
60(f - 1.5) = 40f

Now we can solve for f:
60f - 90 = 40f
20f = 90
f = 4.5

Now we can substitute the value of f back into the equation p = f - 1.5 to find p:
p = 4.5 - 1.5
p = 3

So, the freight train takes 4.5 hours to make the trip, while the passenger train takes 3 hours.

2. Let's define the following variables:
- Let d be the distance Lois travels.
- Let t1 be the time it takes Lois to ride to her friend's house.
- Let t2 be the time it takes her friend to drive her home.

According to the problem, we know that Lois travels at a speed of 10 mph while riding her bike, and her friend travels at a speed of 20 mph while driving the car. Lois takes 1.5 hours longer to go to her friend's house than to return home.

Using the formula Distance = Speed × Time, we can now set up two equations:
- For Lois: d = 10t1
- For her friend: d = 20t2

We also know from the problem that Lois takes 1.5 hours longer to go to her friend's house than to return home, so we can set up another equation:
t1 = t2 + 1.5

To solve this problem, we need to find the value of t1.

We can substitute the equation d = 10t1 into the t2 equation:
10t1 = 20t2

Now we can substitute the equation t1 = t2 + 1.5 into the t2 equation:
10(t2 + 1.5) = 20t2

Now we can solve for t2:
10t2 + 15 = 20t2
10t2 = 15
t2 = 1.5

Now we can substitute the value of t2 back into the equation t1 = t2 + 1.5 to find t1:
t1 = 1.5 + 1.5
t1 = 3

So, it took Lois 3 hours to ride to her friend's house.