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A family vacation trip took 12 hours. Of the total distance, 4/5 was in a car and 1/5 was in a boat, and the car traveled 5 times as fast as the boat. How much more time, in hours, was spent in the boat than in the car?

Guest Nov 19, 2017
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4/5D - Distance travelled by car
1/5D -Distance travelled by boat
S= Speed of boat
5S=Speed of car
Time=Distance/Speed
(4/5D)/(5S) + (1/5D) /S =12, solve for S
S =3D/100 Boat speed
15D/100 car speed
[(1/5D)/(3D/100)] =Boat time =20/3 =6 2/3 hours
[(4/5D)/(15D/100)] =Car time =16/3 =5 1/3 hours
20/3 - 16/3=4/3 =1 1/3 extra hours spent  on boat. 
Note: Regardless of the distance 
traveled or the speeds of the car and boat, there will always be a difference of 1 1/3 hours extra spent on the boat, given the conditions stipulated in the question.

Guest Nov 19, 2017
edited by Guest  Nov 19, 2017
 #2
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A family vacation trip took 12 hours. Of the total distance, 4/5 was in a car and 1/5 was in a boat, and the car traveled 5 times as fast as the boat. How much more time, in hours, was spent in the boat than in the car?

 

Call the total distance, D  ...call the boat's rate, R   and the car's rate 5R

 

And ........ D / Rate    = Time

 

So we have that

 

(4/5)D / [5R]  +  (1/5)D  / R  =  12

 

(4/25)D + (1/5)D = 12R

 

(9/25)D = 12R

 

D =  (100/3)R

 

So....the time in the boat was    (1/5)(100/3)R / R  =  100/15  =  20/ 3  hrs

 

And the time in the car was  (4/5)(100/3)R /[ 5R]  =   400 / 75  = 16/3  hrs

 

So....the time spent in the boat was   (20/3) - (16/3)  = 

4/3 hours more than in the car =

1 hr 20 min more

 

 

cool cool cool

CPhill  Nov 19, 2017

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