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Betty and Carl live in the same apartment building, work at the same office, and set off for work in the morning at the same time. They must each travel at 42 km, and they arrive at work at the same time. Betty travels by car at 60 km/h, parks at the car park, and then walks at 4km/h the rest of the way to work. Carl takes the bus, which travels at 40 km/h, to the same car park as Betty uses, and then rides his bike the rest of the way at 12 km/h. If they both leave home at 6 a.m., at what time do they arrive at work?

 Oct 27, 2019
 #1
avatar
+1

Let the distance from their apartment to the car Park =D
It takes Betty: D / 60 hours. 
It takes Carl: D / 40  hours.
The distance from the car Park to the office =42 - D
The time it takes Betty to get to the office =42 - D / 4
The times it takes Carl to get to the office =42 - D/12
(42 - D) / 4 + D / 60 = (42 - D) / 12 + D/ 40, solve for D
D = 40 km. Sub this into Betty's time:
(42 - 40) / 4 + 40 / 60 = 7/6 hours, or 70 minutes to get to the office. 6 am + 70 minutes =7:10 am - when both arrive at the office.

 Oct 27, 2019
 #2
avatar+104932 
+1

Let  the  distance from the apartment bldg.  to the car park  =  x

Let the distance from the car park to the office  =  42-x

 

Since they take the same amount of time we have that

 

x / 60 + (42 - x) / 4  =  x/ 40 + (42 - x) / 12    simplify

 

[4x + 60(42 - x)]  / 240  = [12x + 40(42 - x) ] / 480

 

[4x + 2520 - 60x ] / 240  =  [12x + 1680 - 40x ] / 480

 

[ 2520 - 56x] / 240  =  [ 1680 - 28x] / 480

 

[2520 - 56x] / 240  =  [ 840 - 14x ] / 240

 

2520 - 56x  =  840 - 14x

 

1680  =  42x

 

x =  1680 / 42   = 40

 

So....the total time  (in hours)  =   

 

40/60  + (42 - 40) / 4  =

 

2/3 hrs   +  2/4  hrs  =

 

2/3 hrs  + 1/2  hrs  =

 

40 mi + 30 min  =

 

70 min  =   

 

7:10AM

 

 

cool cool cool

 Oct 27, 2019

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