Who Will Win the Race? You or I ?

Here’s a long corridor with a moving walkway. Let’s race to the far end and back. We’ll both run at the same speed, but you run on the floor and I’ll run on the walkway, going “downstream” to the far end and “upstream” back to this point. 

Who will win?

This person will be the winner of the race!

"You Will Be Winner of the Race!"

How race was conducted?

Let's assume that we have to run 60 units forward & 60 units backward i.e. total 120 units of distance.

Let 10 units be the speed of the moving walkway. Then I have to run faster than 10 while coming back "upstream" to reach at the source again.

So let 20 units be the our speed of running.

Speed = Distance/Time

Time = Distance/Speed

Time that you need to complete the race = 120/20 = 6 unit.

Time that I need to go forward = 60/(20+10) = 2 units.

Time that I need to come back = 60/(20-10) = 6 units.

Hence, 

Time that I need to complete the race = 4 + 12 = 8 units.

I will require more time to complete the race, that's why you will be the winner of the race!  

How Far Did I Run?

I leave my front door, run on a level road for some distance, then run to the top of a hill and return home by the same route. I run 8 mph on level ground, 6 mph uphill, and 12 mph downhill. 

If my total trip took 2 hours, how far did I run?

I ran the distance.....Click here for more!

Calculation of Avarage Speed is Tricky!

First read what was the question!

Let's first find my average speed when I was running uphill & downhill.

Assume 'x' be the distance that I have to run to reach at the top of the hill in time 'y'.

So x/y = 6 mph.

While running downhill, I cover same distance 'x' in time 'y/2' as I ran at double speed of 12 mph.

Average speed = Total Distance / Total Time

Average speed = (x + x) / (y + y/2)

Average speed = (4/3)(x/y)

Average speed = (4/3) x 6

Average speed = 8 mph.

That means my average speed on hill is equal to the my speed on level ground and that is 8 mph.

Since I ran for 2 hours in my trip the distance I ran is 8 x 2 = 16 miles.

"Share The Walk; Share The Ride!"

You and I have to travel from Startville to Endville, but we have only one bicycle between us. So we decide to leapfrog: We’ll leave Startville at the same time, you walking and I riding. I’ll ride for 1 mile, and then I’ll leave the bicycle at the side of the road and continue on foot. When you reach the bicycle you’ll ride it for 1 mile, passing me at some point, then leave the bicycle and continue walking. And so on — we’ll continue in this way until we’ve both reached the destination.

Will this save any time? 

You say yes: Each of us is riding for part of the distance, and riding is faster than walking, so using the bike must increase our average speed.

I say no: One or the other of us is always walking; ultimately every inch of the distance between Startville and Endville is traversed by someone on foot. So the total time is unchanged — leapfrogging with the bike is no better than walking the whole distance on foot.

Who’s right?

Look who is right in the case! 

"Okay, I'm Wrong in the Case!"

Where I went wrong?

That's going to save time for sure.

Let's assume that the distance between Startville and Endville is 2 miles. And suppose we walk at the same speed of 4 mph and ride bicycle at the speed of 12 mph.

Then I will travel for first 1 mile in 5 minutes leave the bicycle and start walking thereafter. You take 15 minutes to reach at the point to pick up bicycle and ride next mile. For next mile, I need 15 minutes as I am walking & you need only 5 minutes ride on bicycle. So exactly after 20 minutes we will reach at Endville.

And what if we had walked entire 2 miles distance? It would have taken 30 minutes for us to reach at the destination.

One thing you must have noticed, each of us walked for 1 mile only and ride on bicycle for other mile which saved 10 minutes of our journey. Imagine it as if we had 2 bicycles where we ride 1 mile in 5 minutes, leave bicycles and walk next mile in another 15 minutes.

So my argument in the case is totally wrong. It would have been correct if I had waited for you after finishing 1 mile ride on bicycle and then started to walk next mile. 

In that case, you will reach at the destination in 20 minutes but I need 30 minutes as I wasted 10 minutes in middle. 

Conclusion: 

My argument - 

"One or the other of us is always walking; ultimately every inch of the distance between Startville and Endville is traversed by someone on foot."

tells only half story.

Yes, ultimately every inch of the distance between Startville and Endville is traversed by someone on foot but the distance that each of us walk is equal though different parts of journey. And for the rest of distance we ride on bicycle where total time required for journey is saved.

Finding The Average Speed

A man drives his car to the office at 20miles/hr. After reaching the office, he realizes that it's a new year holiday so he went back home at a speed of 30miles/hr.

Discounting the time spent in the stoppage what was his average speed of his journey?

Right way to calculate the average speed! 

Source 

Right Way To Find The Average Speed

What was the given data?

If you are finding average speed of 2 given speeds as (Speed 1 + Speed 2)/2 then you are getting tricked by questioner. The speed itself is a distance covered per unit i.e. Speed = Distance/Time. Calculating average speed like that means,you are doing like,

Average Speed = (Speed1 + Speed2)/2
                     
                      = (Distance1/Time1 + Distance 2/Time2)/2              
            
So you need to find the total distance traveled & the total time taken to complete the entire journey.It should be like,


Average Speed = Total Distance/Total Time

                       = (Distance1 + Distance2)/(Time1 + Time2)

In the given problem, let D be the distance traveled by car to the office. Let T1 be the time required to go to the office & T2 be the time to return back.

Since, Speed = Distance / Time, Time = Distance / Speed.

Hence, 

T1 = D / 20

T2 = D / 30

Now,

Total distance traveled =  D + D = 2D, Total Time taken = T1 + T2

Hence,

Average Speed = 2D / (T1 + T2)

Average Speed = 2D / (D/20 + D/30)

Average Speed = 2D / (50D/600)

Average Speed = 2D / (D / 12)

Average Speed = 24 miles/hr.  

Therefore, the average speed of the journey is 24 miles/hr not 25 miles/hr [(20 + 30)/2].