## A family vacation trip took 12 hours.

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?

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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.

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

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## SOLUTION: A family vacation trip took 12 hours. Of the total, 4/5 was in car and 1/5 was in boat, and the car traveled 5 times as fast as the boat. How much more time, in hours, was spend in

## Distance Word Problems - Given the Total Time

These lessons, with videos, examples and step-by-step solutions, explain how to solve time-distance-rate problems.

Related Pages Rate, Time, Distance Solving Speed, Time, Distance Problems Using Algebra More Algebra Lessons Math Worksheets

Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of time.

The formula for distance problems is: distance = rate × time or d = r × t .

Things to watch out for: Make sure that you change the units when necessary. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately.

It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.

The following diagram shows how to set up a table to help solve time-distance problems. Scroll down the page for more examples and solutions on how to solve distance problems.

## Distance Problems: Given Total Time

Example: John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

Solution: Step 1: Set up a rtd table.

Step 2: Fill in the table with information given in the question.

John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

Let t = time to travel to town.

7 – t = time to return from town.

Step 3: Fill in the values for d using the formula d = rt

Step 4: Since the distances traveled in both cases are the same, we get the equation:

40 t = 120 The distance traveled by John to go back is also 120 So, the total distance traveled by John is 240

Distance, Rate, Time Word Problems Two examples of distance, rate, and time. One involves adding the distances in our chart, where as the other example involves setting the distances equal to each other.

- Two truck drivers leave a cafe at the same time, traveling in opposite directions. On truck goes 7 mph faster than the other one. After 4 hr, they are 404 miles apart. How fast is each truck going?
- Ryan left the science museum and drove south at a rate of 28 km/h. Jenna left three hours later driving 42 km/h faster in an effort to catch up to him. How long did Jenna have to travel to catch up with Ryan?

Distance-time word problem where the total time is given

Example: Gordon rode his bike at 15 mph to go get his car. He then drove back at 45 mph. If the entire trip took him 8 hours, how far was his car?

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Question 1149021: A

familyvacationtriptook12hours. 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? Answer by ikleyn(50911) (Show Source):A

familyvacationtriptook12hours. 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?Grammatically, the problem states that 4/5 and 1/5 are time. However, the question itself indicates that more time was spent in the boat which contradicts that premise. So, the problem contradicts itself. 4/5*

12=48/5 =9 3/5 hrs=9:36 hrs time in car. 1/5*12=12/5= 2 2/5 hrs=2:24 hrs time in boat.A

familyvacationtriptook12hrs. 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?The question asks to calculate the difference in time spent traveling by car and by boat during a family

vacationtrip. The total trip time is 12hours, with 4/5 in a car and 1/5 in a boat. The car travels 5 times as fast as the boat.Distance problems are word problems that involve the distance an object will

travelat a certain average rate for a given period of time. The formula for distance problems is: distance = rate × time or d = r × t. Things to watch out for: Make sure that you change the units when necessary.