"Go The Distance"

There are 50 bikes with a tank that has the capacity to go 100 km. Using these 50 bikes, what is the maximum distance that you can go? 

Here is the maximum distance calculation!

Maximizing The Distance!

What was the challenge?
 
Remember, there are 50 bikes, each with a tank that has the capacity to go 100 kms. 

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SOLUTION 1 : 

Any body can think that these 50 bikes together can travel 50 x 100 = 5000 km. But this is not true in the case as all bikes will be starting from the same point. And we need to find how far we can we go from that point. 

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SOLUTION 2 : 

Just launch all 50 bikes altogether from some starting point and go the distance of only 100 km with tanks of all bikes empty in the end.

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SOLUTION 3 : 

1. Take all 50 bikes to 50 km so that tank of each is at half.

2. Pour fuels of 25 bikes (half filled) into other 25 bikes so that their tanks are full.

3. Now, move these 25 bikes to another 50 km so that again their tanks are at half.

4. Pour fuel of 12 bikes into other 12 so that we have 12 bikes with full fuel tank. Leave 1 bike with half filled fuel tank and repeat above.

So for every 50 km distance, half of bikes are eliminated as - 

50 ---> 25 ---> 12 ---> 6 ---> 3 ---> 1

The last bike left with it's tank full can go 100 km. So. the total distance that can be traveled in the case is 5 x 50 + 100 = 350 km. 

However, we have wasted 1/2 fuel each whenever odd number of bikes are left i.e. at 25 and at 3. 

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SOLUTION 4 :

Let's optimize little further so that the 1/2 fuel is not wasted whenever odd bikes are left.


1. Take all 50 bikes to 50 km so that tank of each is at half.

2. Pour fuels of 25 bikes (half filled) into other 25 bikes so that their tanks are full.

3. Now take these 25 bikes to another 20 km using 1/5th (20/100) fuel of each. 

4. Make 5 groups of 5 bikes each. From each group, use 4/5th fuel of 1 bike to fill tank 1/5th emptied tanks of other 4 bikes.

5. Leave bike with empty tank and take 20 bikes to next 50 km. And again after 50 km, pour fuel of 10 bikes into other 10 to eliminate 10.

6. After moving 10 bike for another 50 km, again pour fuel of 5 bikes into another 5.

7. Now take these 5 bikes to another 20 km using 1/5th (20/100) fuel of each.

8. Use 4/5th fuel of 1 bike to fill tank 1/5th emptied tanks of other 4 bikes. 

9. Now these 4 bikes again taken to another 50 km where 2 more are eliminated by taking half of their fuel to fill tanks of other 2.

10. After taking those 2 bikes for another 50 km distance, 1 can be eliminated by taking away it's half fuel to fill up the tank of other bike.

11. The last bike can now go another 100 km distance as it's tanks is full.

To summarize,

50 ---50km---> 25 ---20km---> 20 ---50km---> 10 ---50km--- > 5 ---20km--- > 4 ---50km ---> 

--->2 ---50km---> 1 ---100km ---||

Total distance that can be traveled = 5 x 50 + 2 x 20 + 100 = 390 km.  

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SOLUTION 5 : 

Now we have got the idea from SOLUTION 4 how to maximize the distance further.

Instead of waiting for tanks to be at half or 4/5th we should empty the tank of 1 bike into others at the point where that bike has sufficient fuel for this process.

For example, to have 49/50th fuel in tank of 1 bike at some point, all bikes need to be taken so that 1/50th of each is used up. Since the bike goes 100 km with full tank, with 1/50th fuel it can go 100 x 1/50 = 2km distance.

In short, after 2km distance 49/50th fuel of 1 bike can be used to fill 1/50th empty tanks of other 49 bikes. Now, that 1 bike with empty tank can be left there.

For next phase, we have, 49 bikes. Now, after using up another 1/49th fuel for another distance of (1/49) x 100 = 100/49 km, the 48/49th fuel left in any one bike can fill up the tanks of other 48 bikes (each with 1/49th part is empty). Then, these 48 bikes can be taken for the next phase.

Now, again after consuming 1/48 fuel for another distance of 100/48km, 47/48th of fuel from 1 bike can be used to fill tanks of other 47 bikes (each bike with 1/48th tank empty after traveling 100/48km). So, now 47 bikes can be taken for the next phase.

This way, we are making sure that at each phase 1 bike uses it's all fuel to make tanks of other full.

Repeating this process, till 1 bike left which can go further 100km with full tank.

So the total distance that can be covered is - 

100/50 + 100/49 + 100/48 +.................100/1 = 449.92 km.

And this is the maximum distance that we can go with 50 bikes.


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Effect of Average Speed on Time

If a car had increased its average speed for a 210 mile journey by 5 mph, the journey would have been completed in one hour less. What was the original speed of the car for the journey?

Here is the calculation of averages speed!

Calculation of the Original Speed!

What was the question?

Let S1 be the original speed and S2 be the modified speed and T1 be the time taken with speed S1 and T2 be the time taken with speed S2.

As per given data,

T1 - T2 = 1 hr.

D/S1 - D/S2 = 1 hr.

Here, D = 210 miles, S2 = S1 + 5,

210/S1 - 210/(S1+5) = 1

210(S1+5) - 210s = 1S1(S1+5)

S1^2 + 5S1 - 1050 = 0

(S1-30)(S1+35) = 0 

S1 = 30 or S1 = -35.

Since speed can't be negative, S1 = 30 mph.

Hence, the original speed is 30 mph and average speed is 30 + 5 = 35 mph.

 
With the original speed it would have taken 210/30 = 7 hours but with average speed it took only 210/35 = 6 hours saving 1 hour of time. 

"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.