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 #8
avatar+26387 
+3
Jul 11, 2019
 #2
avatar+129845 
+2

For this problem....I'm assuming that the mother and daughter normally arrive at the train station at the same time

I'm also assuming that "between the station and their house" means halfway between the station and the house

 

Call the distance from the house to the train station , D

So....the distance that the mother would normally drive is 2D

And let her rate be   R

And let the time that she would normally drive be = T   (in hours)

So  we have that

2D / R   = T      (1)

 

But..on this day.....the mother only drives  (1/2)D + (1/2)D   = D

And....if she normaly returns at 5PM, but now only returns at 4:48PM....then her normal driving time must be 12 minutes less than normal  = 1/5  hr

So.....we have that

D / R  =  T - 1/5    multiply through by 2

2D / R  = 2(T - 1/5)      (2)

 

This implies that   (1)  = (2)...so....

T = 2(T - 1/5)

T = 2T - 2/5

2T - T  = 2/5

T  = 2/5.......this means that her normal round-trip driving time  is 2/5 hr  =  24 min

 

So....she must normally leave her house at   4:36 PM   (since 24 minutes after this = 5PM)

And she would normally reach the station at 12 minutes after 4:36 PM =  4:48 PM

But....this means that she reaches the halfway point at  6 mnutes after 4:36 PM  = 4:42 PM

 

Now.....let's look at this from the daughter's point of view..... she would normally meet her mother at the station at  4:48 PM.....but.....since she gets there an hour early......she must reach the station at 3:48 PM  and she immediately begins walking

 

So.....she walks from  3:48 PM   and meets her mother at the halfway point at  4:42 PM

 

So....she walks for 54 min

 

 

cool cool cool

Jul 11, 2019

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