Lesson Three kids get home from school using a tandem bike and walk

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Three kids get home from school using a tandem bike and walk

Problem 1

Peter, Reeta and Nikita have two ways for getting home from school: cycle on a tandem bike or walk.
The bike can carry either one or two riders at a time.
Regardless of the number of people pedaling, cycling speed is 5 times walking speed.
The kids leave school at the same time and use the same path between school and home,
whether walking or cycling. The school is 5 km from home and their walking speed is 4 kilometers per hour.

One day, Peter and Nikita ride the bike and Reeta walks. Peter drops Nikita off at a certain point
leaving him to walk home. Meanwhile he returns to pick up Reeta and they cycle home together.
If all three arrive home at the same time, how far from school are the drop-off and pick-up points?

Solution

So, the distance from the school to home is 5 km; walking speed is 4 km/h  and the cycling speed is 20 km/h.


Let x be the distance from the school to the drop-off point and 
let y be the distance from the school to the pick-up point.


The time for Peter to get the drop-off point is    hours.

The time for Peter to move from the drop-off point to pick-up point is    hours.

The time for Peter to get the drop-off point and then to return to the pick-up point is the sum

     = .   (1)


This time,   hours, is equal to the time for Reeta to get the pick-up pont from the school,  which is   hours.


It gives us this equation

     = .    (2)


Simplify it

    4(2x-y) = 20y,   or  2x-y = 5y,  or  2x = 5y + y,  or  2x = 6y,  or  x = 3y.    (3).


    +------------------------------------------------------------------------+
    |     At this point, we complete our consideration for moving Peter      |
    |  from the school to the drop-off point and then to the pick-up point.  |
    +------------------------------------------------------------------------+


Now we will consider the other part of moving.


The time for Nikita to get the home from the drop-off point walking is    hours.


During this time, Peter is cycling from point x to point y and then from point y to the home.

It takes for Peter   +  =   hours.


The time   for Nikita is the same as the time    for Peter.

It gives us this equation

     = .    (4)


Simplify it

    20(5-x) = 4(5+x- 2y),  or  100-20x = 20+4x-8y,  or  100-20 = 4x + 20x-8y,  or  24x-8y = 80,  or  3x - y = 10.    (5)


Thus, we have now two equations

    x = 3y  and  3x - y = 10.


Substitute first equation,  x = 3y, into the second equation

    3*(3y) - y = 10,  or  9y - y = 10,  8y = 10,  y =  =  = 1.25.


Then  x = 3y = 3*1.25 = 3.75.


Thus  the drop-off point is 3.75 km from the school, and the pick-up point is 1.25 km from the school.   ANSWER

My other additional lessons on Travel and Distance problems (section 3) in this site are
    - Traveling on moving walkway
    - Two ships in parallel courses in ocean at restricted visibility
    - OVERVIEW of additional lessons on Travel and Distance - section 3

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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