A World Class Swimmer's Puzzle

A world-class swimmer can swim at twice the speed of the prevailing tide.

She swims out to a buoy and back again, taking four minutes to make the round-trip.

How long would it take her to make the identical swim in still water?

Click here for the answer!

Solution of A World Class Swimmer's Puzzle

What was the puzzle?

 Let C be the speed of water current then the speed of a world class swimmer will be 2C.

When she swims out to a buoy located at a distance say D and back again, the time needed is - 

D/(C+2C) + D/(2C-C) = 4 minutes

D/3C + D/C = 4

Multiplying both the sides by 3, 

D/C + 3D/C = 12

4D/C = 12

D/C = 3  

Now when she swims out to a buoy and back to shore again in still water, she needs only time -

D/2C + D/2C 

Since D/C = 3, then D/2C = 3/2

Hence, 
 
D/2C + D/2C  = 3/2 + 3/2 = 3.

That is, she needs only 3 minutes to swim out to a buoy and back to shore again in still water.

 
 

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. 

A Man Walking on Railroad Bridge

A man is walking across a railroad bridge that goes from point A to point B. He starts at point A, and when he is 3/8 of the way across the bridge, he hears a train approaching. The train's speed is 60 mph (miles per hour). The man can run fast enough so that if he turns and runs back toward point A, he will meet the train at A, and if he runs forward toward point B, the train will overtake him at B.

How fast can the man run?


He must be running at 'this' speed! 

Speed Needed For Run on Bridge

Why to calculate the speed?

If the man turns back and runs towards A for 3/8 of distance while train reaches at the point A. That means the train train reaches at point A when man runs for 3/8 distance.

So if man continues to run towards point B then, while he covers 3/8 distance the train reaches at point A. Now, man is 3/8 + 3/8 = 6/8 = 3/4 distance away from the point A where B is 1/4 away from him now.

Again, we know, the train will overtake the man at point B covering total distance between A and B. Till then man can run only 1/4 th distance between A and B to reach B.That mean the train must be 4 times faster than the speed of man.

Since, train is traveling at 60 mph, the speed of man = (1/4)x60 = 15 mph.

A Letter Delivery to the Leader

There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time.

Assuming that he ran the whole distance with uniform speed. 

Skip to the answer!

Distance Covered by Letter Delivery Person

What was the puzzle?

Let's suppose the first person of the platoon move X meters ahead by the time the last person reaches to him.

To reach at him, the last person has to cover distance of 50 + x.

Assume M be the speed of last person and P be the speed of platoon i.e. of first person.

If T1 is time needed for the last person to get to at the first person,

T1 = (50 + X)/M

T1 = X/P

(50 + X)/M = X/P

M/P = (50 + X)/X  ..........(1)

As per given data, by the time the last person gets back to it's original position (i.e. end of platoon), the platoon moves 50m ahead from it's position that was when the last person started his journey towards first person.

That is end of platoon is now at position at which the start of platoon was initially.

Since, the first person has already moved X meters ahead, he has to move only 50 - X meter to lead the platoon 50m ahead of it's original position.

And, the last person has to move only X meters to get back to  original position i.e. the end of platoon.

If T2 is the time taken by last person to get back to original position (i.e. time taken by first person to move ahead 50 - x) then,

T2 = X/M

T2 = (50 - X)/P

X/M = (50 - X)/P

M/P = X/(50 - X)  ..........(2)

Equating (1) and (2),

(50 + X)/X = X/(50 - X)

X^2 = (50 + X)(50 - X)

X^2 = 2500 - X^2

X^2 = 1250

X = 35.355 meters.

So,

the distance traveled by the last person = (50 + X) + X 
                                                          
                                                          = (50 +35.355) + 35.355 

the distance traveled by the last person = 120.71 meters.

The ratio of their speeds = M/P = (50 + X)/X = (50 + 35.355)/35.355 = 2.41

M = 2.41 P

That is the speed of last man is 2.41 times the speed of first man or the speed of platoon itself.

And that's the speed the last man needed to reach at the first person of the platoon.

A Visit To Grandmother's Home!

A father wants to take his two sons to visit their grandmother, who lives 33 kilometers away. His motorcycle will cover 25 kilometers per hour if he rides alone, but the speed drops to 20 kph if he carries one passenger, and he cannot carry two. Each brother walks at 5 kph. 

Can the three of them reach grandmother’s house in 3 hours?

Do you think it's impossible? Click here!

Planning Journey Towards Grandmother's Home

What was the challenge in the journey?

Yes, all three can reach at grandmother's home within 3 hours. Here is how.

Let M be the speed of motorcycle when father is alone, D be the speed of motorcycle when father is with son and S is speed of sons.  Let A and B are name of the sons.

As per data, M = 25 kph, D = 20 kph, S = 5 kph.

1. Father leaves with his first son A while asking second son B to walk. Father and A drives for 24 km in 24/20 = 6/5 hours. Meanwhile, son B walks (6/5) x 5 = 6 km.

2. Now father leaves down son A for walking and drives back to son B. The distance between them is 24 -6 = 18 km.

3. To get back to son B, father needs 18/(M+S) = 18/(25+5) = 18/30 = 3/5 hours & in that time son B walks for another (3/5) x 5 = 3 km. Now, son B is 6 + 3 = 9 km away from source where he meets his father. While son A walks another (3/5) x 5 = 3 km towards grandmother's home.

4. Right now father and B are 24 km while A is 6 km away from grandmother's home. So in another 24/20 = 6/5 hours father and B drive to grandmother's home. And son B walks further (6/5) x 5 = 6 km reaching grandmother's home at the same time as father & brother B.

In this way, all three reach at grandmother's home in (6/5) + (3/5) + (6/5) = 3 hours.

In this journey, both sons walks for 9 km spending 9/5 hours and drives 24 km with father taking (6/5) hours. Whereas, father drives forward for 48 km (24 km + 24 km) in (6/5) + (6/5) hours and 15 km backward in 3/5 hours. 
  

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.

The Race Between 2 Brothers

Zachary challenges his brother Alexander to a 100-meter race. Alexander crosses the finish line when Zachary has covered only 97 meters.

The two agree to a second race, and this time Alexander starts 3 meters behind the starting line.

 
If both brothers run at the same speed as in the first race, who will win?

He will win the second race for sure! 

Source 

Winner of The Race Between 2 Brothers

What was the race of?

Let's assume that, Alexander completes the first race in time 't'. That means, he reaches at the finish line after running 100m after time 't' since start of the race. In the same time interval, Zachary could reach only 97m.

Now, in second race too, Alexander covers 100m once again in time interval 't' & Zachary runs 97m distance. Since, Alexander started 3m ahead of start line, at this point of time both are at the same point with 3m left to complete the race. 

 
Since, Alexander had won first race with faster speed & speed of both are unchanged in second race, it's clear that Alexander will take less time to cover leftover 3m distance. Hence, Alexander will be winner of the second race. 

MATHEMATICAL APPROACH:

Let's suppose Alexander took 10s to complete the first race. Then, his speed is 100/10 = 10m/s.

In 10s, the Zachary could run only 97m. So, his speed is 97/10 = 9.7m/s.

In the second race, their respective speeds are unchanged but Alexander has to run 103m to reach at the finish line compared to 100m of Zachary.

Hence, time taken by Alexander to reach at the finish line = 103/10 = 10.3s and that taken by Zachary = 100/9.7 = 10.30929s.

It's clear that Zachary needs more time to finish this race too. Hence, Alexander will be the winner in this race as well. He beats Zachary by 100 - (10.3x9.7) = 0.09m.

Cars Around Interesting Circular Track

Around a circular race track are n race cars, each at a different location. At a signal, each car chooses a direction and begins to drive at a constant speed that will take it around the course in 1 hour. When two cars meet, both reverse direction without loss of speed. Show that at some future moment all the cars will be at their original positions.

Here is that proof!


Source 

Analysing Interesting Circular Race Track

What was the interesting fact about?

Just imagine that each car carries a flag on it and on meeting pass on that flag to the next car. Obviously, this flag will move at the constant speed around the track as cars carrying it are also moving at the constant speed. So, the flag will be back at the original position after 1 hour.

Let's assume there are only 2 cars on the track at diagonally opposite points as shown below. 

After 15 minutes, on meeting with Car 2, Car 1 will pass on the flag & both will reverse their own direction. 30 minutes later (i.e. 45 minutes after start) both cars again meet each other and Car 2 will pass on flag back to Car 1 & directions are reversed again. Again in another 15 minutes (i.e. after 1 hour from start), both cars are back at the original positions. 

Now, let's suppose that there are 4 cars on the track positioned as below.

The above image shows how cars will be positioned after different points of time & how they reverse direction after meeting.

Again, all are back to the original position after 1 hour including the flag position. One more thing to note that the orders in which cars are never changes. It remains as 1-2-3-4 clockwise. 

To conclude, for 'n' number of cars, at some point of time all the cars will be in original positions in future.