# Need help understanding this concept

#### Isiltir

Two people are going from place A to B, which is 5km away. One of them has bike, so they agree on following: Person X will start on the bike, Y goes by foot. When X reaches an agreed-on distance between them (him and Y), he will lay down the bike and continue by foot, leaving bike for Y. Y will ride the bike until he sees X and then leaves him the bike. They will do N such cycles until they get to B. Person on bike travels at the speed of 15 km/h, person going by foot travels at 5 km/h

a) How much time will they save when doing cycles? Does this depend on agreed-on distance d?
b) Determine conditions under which both X and Y will get to B at the same time and calculate number of cycles needed to accomplish this.

I need help to understand the concept and solution.

#### oz93666

Two people are going from place A to B, which is 5km away. One of them has bike, so they agree on following: Person X will start on the bike, Y goes by foot. When X reaches an agreed-on distance between them (him and Y), he will lay down the bike and continue by foot, leaving bike for Y.
So far so good ....

Y will ride the bike until he sees X and then leaves him the bike.
We must know how good Y's eyesight is ! ... At what distance does Y see X??? and is X's eyesight as good as Y's ...all this must be known ..

Have you given the question as written ???

#### Woody

I think that "sees" can be interpreted as meets or catches up with.

Note that if this were tried for real in most places
Person Z would come along, find a bike abandoned at the side of the road, and take it!

#### Isiltir

Yes, "sees" is interpreted as "meets". Maybe I should have written that instead .

#### oz93666

Reading the question again it's not obvious that "sees" does mean they are at the same position , what is the exact wording in the question ???

X and Y leave at the same time . When X is an "agreed distance(d) " from Y he leaves the bike and continues on foot ... now both are walking ... Y finally reaches the bike and rides to meet or perhaps overtake X to the "agreed distance"??

#### Isiltir

It is a translation mistake.

Your idea Is correct, they will be both walking for a while until Y reaches the bike and catches up with X. X then rides the bike again to the distance (d). Such Is the cycle.

#### oz93666

It is a translation mistake.

Your idea Is correct, they will be both walking for a while until Y reaches the bike and catches up with X. X then rides the bike again to the distance (d). Such Is the cycle.

Let's just focus on the bike ...

From start ... Relative speed between X and Y is 10 Km/Hr , distance between them is d when bike is laid down , so time is d/10 Hrs

Then the bike has zero speed for d/5 Hrs .. ( the walker travels d at 5 km/hr to reach the bike)

Then the bike travels the distance between X and Y in d/10 Hrs and X and Y are together again ....

This is the cycle , the bike travels at 15Km /Hr for two periods of d/10 = d/5Hrs ..... and is still for d/5Hrs

So the average speed of the bike is 7.5Km/Hr ... and is not dependent on d , so this should answer part (a) of the question ....

b) Determine conditions under which both X and Y will get to B at the same time and calculate number of cycles needed to accomplish this....

For X and Y to reach B at the same time a complete number of cycles must have occurred ...

How much distance is one cycle ??? there should be enough information in this post to find this distance ...

If the cycle distance divides into 5 (distance from A to B) and gives a whole number then X and Y will meet at B

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