Question 1159562: Ms. Allen planned to meet her family at Redford Restaurant after work. Her work is 30 miles from the restaurant and she can drive an average speed of 60 mph. Her family lives closer to the restaurant. They live only 10 miles from the restaurant, but can only travel an average of 40 mph. If they both leave at 5:00 PM, who should arrive first? What time will Ms. Allen arrive?
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Ms. Allen has her work "30 miles from the restaurant and she can drive an average speed of 60 mph"
d = 30 miles is the distance
r = 60 mph is the rate or speed
t = unknown
distance = rate*time
d = r*t
30 = 60*t
30/60 = t
t = 30/60
t = 1/2
It takes her half an hour (30 minutes) to get to the restaurant.
If she leaves at 5:00 PM, then we expect her to arrive at around 5:30 PM
I'm using the phrasing "around" instead of "exactly" because it's possible that she might be a few minutes early or a few minutes late. Keep in mind that Ms. Allen, nor her family, is not traveling at the same exact speed the entire time. Instead, the car is slowing down and speeding up at various parts of the journey.
Now let's see how long it takes for Ms. Allen's family to get to the restaurant.
Her family lives only 10 miles away and they travel a speed of 40 mph
d = 10
r = 40
t = unknown
d = r*t
10 = 40*t
10/40 = t
t = 10/40
t = 1/4
The family only needs 1/4 of an hour, or 15 minutes, to get to the restaurant.
If they leave at 5:00 PM, then they should get to their destination at around 5:15 PM.
Therefore, the family arrives first before Ms. Allen.
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To summarize:
Question: Who arrives first?
Answer: Her family (family arrives at around 5:15 PM while Ms. Allen arrives at around 5:30 PM)
Question: What time will Ms. Allen arrive?
Answer: Around 5:30 PM
Answer by solver91311(24713)  (Show Source):
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