How long was he walking?

Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. 

One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? 

Distances are unspecified. Speeds are unspecified, but constant.

Give a number which represents the answer in minutes.

He must be walking for....minutes. Click to know! 

Jack's Walking Duration in Journey!

What was the question?

It's important to think from wife's point of view in the case.

Ideally, had Jack somehow informed earlier his wife about his 1 hour early arrival then his wife's (and his as well) total 60 minutes in round trip would have been saved. 

30 minutes of her round trip are saved which means only 15 minutes of each leg of her trip must have been saved. That is she meets her husband only 15 minutes earlier on the day instead of 60 minutes earlier (if Jack had informed her earlier). 

Hence, Jack must be walking for 45 minutes.

Let's understand this with example.

Suppose wife needs exactly one hour to reach the station every day. She leaves home at 4 PM everyday and reach station at 5 PM & drive Jack home at 6 PM

On one day, Jack arrived at 4 PM and started walking whereas wife leaves home at the same time as usual. They reach home at 5:30 PM.

30 minutes of wife saved indicates that she took 45 minutes to meet husband (instead of 1 hour) at 4:45 PM (instead of 5PM, only 15 minutes earlier) and took him to home at 5:30 PM (instead of 6PM) in next 45 minutes (instead of 1 hour) thereby saving 15 + 15 = 30 minutes only.

Since, Jack started walking at 4 PM and meet her wife at 4:45 PM, he must be walking for 45 minutes.   

The Tunnel Trouble!

A man needs to go through a train tunnel to reach the other side. He starts running through the tunnel in an effort to reach his destination as soon as possible. When he is 1/4th of the way through the tunnel, he hears the train whistle behind him. 

Assuming the tunnel is not big enough for him and the train, he has to get out of the tunnel in order to survive. We know that the following conditions are true - 

1. If he runs back, he will make it out of the tunnel by a whisker.

2. If he continues running forward, he will still make it out through the other end by a whisker.

What is the speed of the train compared to that of the man?

The train must be traveling at THIS speed!

Escape From The Tunnel Trouble!

What was the question?

LOGICAL APPROACH : 

As per condition, if the man runs back he will make it out of the tunnel by a whisker. That means while he runs back 1/4 th tunnel distance, the train travels from it's position to the start of the tunnel. 

In other words, the time taken by man to get back covering 1/4th to the start of the tunnel and the time taken by train to reach at the start of tunnel is same.

So if the man decides to go forward then by time the train reaches at the start of tunnel, man covers another 1/4th tunnel distance i.e. he will be halfway of the tunnel.

At this point of time, the man needs to cover another 1/2th tunnel distance while train has to cover entire tunnel distance. Since, man just manages to escape from accident with train at the exit of tunnel, the train speed has to be double than man's speed as it has to travel distance double of that man travels.

MATHEMATICAL APPROACH : 

Let us suppose - 

M - Speed of Man

T - Speed of Train

D - Tunnel Distance/length

S - Distance between train and the start of tunnel.

As per condition 1, 

Time needed for man to get back at the start of tunnel = Time needed for train to cover distance F to arrive at the start of tunnel

(D/4)/M = S/T  

D/4M =  S/T  .....(1)

As per condition 2,

Time needed for man to move forward at the end of tunnel = Time needed for train to cover distance S + time needed to cover tunnel distance.

(3D/4)/M = S/T + D/T 

Putting S/T = D/4M,

3D/4M - D/4M = D/T

2D/4M = D/T

T/M = 2

T = 2M.

That is speed of the train needs to be double of the speed of the man.

Interestingly, from (1),

D/S = 4M/T

D/S = 2 

D = 2S

S = D/2

That is train is 1/2th tunnel distance away from the start of tunnel. 

"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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"Save Your House From Me!"

Okay, I’ll ask three questions, and if you miss one I get your house. Fair enough? Here we go:

1.A clock strikes six in 5 seconds. How long does it take to strike twelve?

2.A bottle and its cork together cost $1.10. The bottle costs a dollar more than the cork. How much does the bottle cost?

3.A train leaves New York for Chicago at 90 mph. At the same time, a bus leaves Chicago for New York at 50 mph. Which is farther from New York when they meet?

You need little common sense in answering above!

Presence of Common Sense in Answers!

What where questions?

1. A clock strikes six in 5 seconds. How long does it take to strike twelve?

A: Not 10 seconds, it takes 11 seconds.

Here, interval between 2 strikes is 1 second i.e. if counter started at first strike, it will count 1 second after second strike, 2 seconds after third strike & so on. 

Hence, 11 seconds needed for strike 12.

2.A bottle and its cork together cost $1.10. The bottle costs a dollar more than the cork. How much does the bottle cost?

A: Not 1 dollar, it would cost $1.05.

If x is cost of cork,
x + (x +1) = 1.10
2x = 0.10
x = 0.05

Hence cost of bottle is $1.05 and cost of cork is $0.05



3.A train leaves New York for Chicago at 90 mph. At the same time, a bus leaves Chicago for New York at 50 mph. Which is farther from New York when they meet?

A: Obviously, when they meet at some point then that point must be at the some distance from New York. Hence, they are at the same distance from the city.

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.

The Camel and Banana Puzzle

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels.

What is the most bananas you can bring over to your destination?

As many as 'these' numbers of bananas can be saved!

The Camel and Banana Puzzle : Solution

What was the puzzle? 

 Let A be the starting point and B be the destination in this transportation. If the camel is taken with 1000 bananas at start, to reach the point B which is 1000 km away from A, it needs 1000 bananas. So there will be no bananas left to return back to point A.

That's why we need to break down the journey into 3 parts.

Part 1 :

For every 1 km the camel needs to -

1. Move ahead 1 km with 1000 bananas but eat 1 banana in a way.

2. Leave 998 bananas at the point and take 1 banana to return back to previous point.

3. Pick up another 1000 bananas and move forward while eating 1 banana.

4. Drop 998 bananas at the same point. Return back to previous point by consuming 1 banana.

5. Pick left over 1000 bananas and move 1km forward while consuming 1 banana to same point where 998 + 998 bananas are dropped. Now, the camel doesn't need to  return back to previous point. So, 998 + 998 + 999 are carried to the point.

That is for every 1km, the camel needs 5 bananas.

After 200 km from point A, the camel eats of 200x5 = 1000 bananas and at this point the part 1 ends.

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PART 2 :

1. Move ahead 1 km with 1000 bananas but eat 1 banana in a way.

2. Leave 998 bananas at the point and take 1 banana to return back to previous point.

3. Pick up another 1000 bananas and move forward to the point where 998 bananas left while eating 1 banana.

Now, the camel needs only 3 bananas per km.

So for next 333 km, the camel eats up 333x3 = 999 bananas.

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PART 3 :

So far, the camel has travelled 200 + 333 = 533 km from point A and needs to cover 1000 - 533 = 467 km more to reach at B.

Number of bananas left are 3000 - 1000 - 999 = 1001.

Now, instead of wasting another 3 bananas for next 1 km here, better drop 1 banana at the point P2 and move ahead to B with 1000 bananas. This time the camel doesn't need to go back at previous points & can move ahead straightaway.

For the remaining distance of 467 km, the camel eats up another 467 bananas and in the end 1000 - 467 = 533 bananas will be left.

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Sharing The Driving Time!

John and Mary drive from Westville to Eastville. John drives the first 40 miles, and Mary drives the rest of the way. That afternoon they return by the same route, with John driving the first leg and Mary driving the last 50 miles.

Who drives the farthest, and by what distance?

'This' is the person who drives the farthest!


Source

Uneven Sharing of Driving Time!

But actual how it was shared?

Let's assume for a moment, the distance between Westville and Eastville is 50 miles.

In this case, John drives only 40 miles while Mary drives 10 + 50 = 60 miles. For any distance beyond 50 miles, Mary drives east equal distance as John drives west.

So in any case the difference will remain same of 20 miles. So Mary drives the farthest by 20 miles.

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.

Journey In Parts

Someone drove from Aardvark to Beeville.

On the first, day they traveled 1/3 of the distance.

On day two, they traveled 1/2 of the remaining distance.

On day three, they traveled 2/3 of the remaining distance.

On day four, after covering 3/4 of the remaining distance, they were still 5 miles away from Beeville.

How many miles had they covered so far?

Know the total distance traveled!
 

Total Distance In The Journey

Click for the question! 

We need to start in reverse.

In last part after covering 3/4 still 5 miles left which accounts for 1/4 of remaining distance. Hence, 20 miles were left at the start of DAY 4.

On DAY 3, 2/3rd covered leaving 20 miles for DAY 4. That means 20 miles distance is remaining 1/3rd. Hence, at the start of DAY 3, 60 miles were left.

On DAY 2, 1/2 of covered leaving 60 miles for DAY 2. So that means 60 miles distance is remaining 1/2. So at the start of DAY 2, 120 miles yet to be covered.

On DAY 1, 1/3 of covered leaving 120 miles for DAY 2. Meaning 120 miles distance is remaining 2/3. Hence, 180 miles yet to be covered at the start of DAY 1.

Out of 180 miles, 175 covered in 4 days still 5 miles left.