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

#1**+1 **

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 **

Guest Nov 19, 2017

edited by
Guest
Nov 19, 2017

#2**+1 **

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

CPhill
Nov 19, 2017