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?
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.
+130081 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
