Maze Challenge For a Rat?

A rat is placed at the beginning of a maze and must make it to the end. There are four paths at the start that he has an equal chance of taking: path A takes 5 minutes and leads to the end, path B takes 8 minutes and leads to the start, path C takes 3 minutes and leads to the end, and path D takes 2 minutes and leads to the start.

What is the expected amount of time it will take for the rat to finish the maze?

This could be the average time that rat needed!
 

A Rat Finishing off The Maze!

The challenge ahead of rat?

For rat, there are 2 paths viz A (5 minutes) and C (3 minutes) leading to the end while paths B (8 minutes) and D (2 minutes) lead to the start again.

Since, there are 4 paths & each having equal chance of being chosen by rat, there is 1/4 th chance for each path for to be chosen by rat.

Let's assume T be the time needed for rat to finish the maze. 

But if rat selects path B or D then rat need T more time again as these paths lead to the start of the maze again.

Hence,

T = (1/4) x A + (1/4) x B + (1/4) x C + (1/4) x D

T =  (1/4) x 5 + (1/4) x (8 + T) + (1/4) x 3 + (1/4) x (2 + T)

T = (5/4) + (2) + (T/4) + (3/4) + (1/2) + (T/4)

T =  (9/2) + (T/2)

T/2 = 9/2

T = 9

That is rat needs 9 minutes to finish the maze. 

 

Puzzle : Set Timer without Clock?

You are a cook in a remote area with no clocks or other way of keeping time other than a four-minute and a seven-minute hourglass. On the stove is a pot of boiling water. Jill asks you to cook a nine-minute egg in exactly 9 minutes, and you know she is a perfectionist and can tell if you undercook or overcook the egg by even a few seconds. 

How can you cook the egg for exactly 9 minutes?

You should follow THIS process! 

Solution : Setting up Timer without Clock!

What was the challenge?

We can set up a timer of 9 minutes using 4 and 7 minutes hourglass. Below is the process-

1. Flip both the hourglasses and drop egg into the water. 

2. After 4 minutes, 4-minutes hourglass will run out. Flip and reset it. ( 4 minutes counted and 3 minutes countdown left in 7-minute hourglass).

3. After 3 minutes, the 7-minutes hourglass will run out while 1 minute countdown will be left in 4-minutes hourglass. ( So far 4 + 3 = 7 minutes counted ). 

Flip the 7-minutes hourglass thereby resetting it's timer.

4. After 1 minute, the 4-minutes hourglass will run out. ( 4 + 3 + 1 = 8 minutes counted). 

5. At this point of time there will be sand for 6 minutes countdown left in 7-minutes hourglass. Just flip it so that it count exactly 1 more minute.
   

Now, that's how 4 + 3 + 1 + 1 = 9 minutes are counted.

 

Who is older, Joe or Smoe?

Two friends, Joe and Smoe, were born in May, one in 1932, the other a year later. Each had an antique grandfather clock of which he was extremely proud. Both of the clocks worked fairly well considering their age, but one clock gained ten seconds per hour while the other one lost ten seconds per hour. 

On a day in January, the two friends set both clocks correctly at 12:00 noon. "Do you realize," asked Joe, "that the next time both of our clocks will show exactly the same time will be on your 47th birthday?" Smoe agreed. 

Who is older, Joe or Smoe?

Know who is older in the case! 

"Smoe is older than Joe"

What was the puzzle?

Since one of the clock looses and other gains 10 seconds per hour, that means one looses 240 seconds (4 minutes) & other gains 240 seconds (4 minutes) in a day.

Both the clocks are set at 12:00 PM correctly. One has to gain 6 hours (360 minutes) and other has to loose 6 hours (360 minutes) to show the same time again. At the speed of 4 minutes per day the would need 360/4 = 90 days to show the same time again. 

On 90th day, they will come together to show 6:00. Exactly at 12 noon on 90th day one clock must be showing 6:00 PM and other must be showing 6:00 AM, if they have feature of showing AM/PM.

Now as per Joe it would be 47th birthday of Smoe on the day on which the clocks will show the same time. That means, the clocks are set correctly on the noon of 90 days prior to Smoe's birthday which is 1 May for sure but year yet to be known. 

If the year is leap year then 90th day before 1st May will be on 1st February and if it's not a leap year then it would be on January 31. Since, they have set their clocks correctly at 12:00 on some day in January, the year must not be a leap year. 

But if Smoe had been born in 1933, his 47th birthday would have been on May 1, 1980 which is leap year. Hence, Smoe must have born in 1932 and Joe in 1933.

Therefore, Smoe is older than Joe.

The story must be of 1979!

The Toaster Twist!

Jasmine has a toaster with two slots that toasts one side of each piece of bread at a time, and it takes one minute to do so.

If she wants to make 3 pieces of toast, what is the least amount of time she needs to toast them on both sides?

She needs only....minutes. Click to know. 

Smarter Use of Toaster!

What was the task given?

Jasmine has 3 pieces of bread and she want to toast total 6 sides of toasts. Since, the toaster can toast 2 sides (in two slots) at a time in 1  minute, she needs only 3 minutes to toast all her 3 pieces. All that she need to do use the toaster smartly.

Let A-B, C-D and E-F be the names of sides of toast that Jasmine has.

1. She puts 2 pieces - A-B and C-D in the two slots of toaster so that 2 sides say A and C are toasted but B and D remains untoasted. 1 minute used in the process.

2. For next minute, she takes out 1 piece say C-D and puts E-F in toaster while she flips A-B so that B is toasted. With that, B and E are toasted in this minute.

3. Now, only F and D are remain untoasted. In third minute, using 2 slots of toaster she toasts these 2 sides. 

Confusing Ride on the Ferris Wheel

There are 10 two-seater cars attached to a Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute. 

The wheel began operation at 10 in the morning and shut down 30 minutes later. 

What's the maximum number of people that could have taken a ride on the wheel in that time period?

Here is count of people enjoying the ride!

Peoples Enjoying Ferris Wheel Ride

What was the puzzle?

Let's simplify the situation by naming 10 cars as A, B, C, D, E, F, G, H, I, J.

Suppose the Car A is at the exit platform at 10:00 AM. Obviously it can be 'loaded' with 2 peoples say A1-A2.

At 10:01 AM, the Car B will be at the exit platform which can be 'loaded' with 2 more peoples say B1-B2.

So continuing this way, at 10:09 AM the car J will be loaded with peoples J1-J2.

At 10:10 AM, the Car A will be again at the exit platform and now A1-A2 can get off the board to allow 2 more new peoples A3-A4 to get on the board.

At 10:11 AM, the Car B will be at the exit platform where B3-B4 will replace B1-B2. 

Continuing this way, at 10:19 AM the J1-J2 in the Car J will be replaced by J3-J4.

So far, 10 x 4 = 40 different peoples would have enjoyed the ride. 

It's clear that every car takes 10 minutes to be at the exit platform after once it goes through it. That's how Car A is at the exit platform at 10:10 AM, 10:20 AM and it can be at 10:30 AM as well when the wheel is supposed to be shut down.

Now, since wheel needs to be shut down 10:30 AM, emptying all the cars must be started except Car A for the reason stated above.

Therefore, at 10:20, peoples A5-A6 replace A3-A4. Thereafter, every car should be emptied. So, far 40 + 2 = 42 different people have boarded on the cars of the wheel.

So, at 10:21 AM, the Car B is emptied, at 10:22 AM, the Car C should be emptied and so on.

At 10:29 AM, J3-J4 of Car J get out of the car and finally at 10:30 AM, A5-A6 get out of the Car A. 

Now, the ferris wheel can be shut down with no one stuck at any of cars.

To conclude, 42 different peoples can enjoy the ride in given time frame.

The Mistimed Clock!

Andrea’s only timepiece is a clock that’s fixed to the wall. One day she forgets to wind it and it stops.

She travels across town to have dinner with a friend whose own clock is always correct. When she returns home, she makes a simple calculation and sets her own clock accurately.

 
How does she manage this without knowing the travel time between her house and her friend’s?


That's how she manages to set it accurately!
 

Correcting The Mistimed Clock!

 How it was mistimed?

Andrea winds her clock & sets it to the arbitrary time. Then, she leaves her house and when she reaches her friend's house, she note down the correct time accurately. Now, after having dinner, she notes down the correct time once again before leaving her friend's house.

After returning to home, she finds her own clock acted as 'timer' for her entire trip. It has counted time that she needed to reach her friend's house + time that she spent at her friend's house + time she needed to return back to home.

Since, Andrea had noted timings at which she reached & left her friend's house, she can calculate the time she spent at her friend's house. After subtracting this time duration from her unique timer count she gets the time she needed to reach to & return from her friend's house.

She must have taken the same time to travel from her home to her friend's home and her friend's home to her home. So dividing the count after subtracting 'stay time' she can get how much time she needed to return back to home.

Since, she had noted correct time when she left her friend's home, now by adding time that she needed to return back to home to that, she sets her own clock accurately with correct time.

Let's try to understand it with example.

Suppose she sets her own clock at 12:00 o' clock and leave her house. Suppose she reaches her friend's home and note down the correct time as 3:00 PM. After having dinner she leaves friend's home at 4:00 PM.

After returning back to home she finds her own clock showing say 2:00 PM. That means, she spent 120 minutes outside her home with includes time of travel to and from friends home along with time for which she spent with her friend. If time of stay at her friend is subtracted from above count, then it's clear that she needed 60 minutes to travel to & return back from friend's home.

That is, she needed 30 minutes for travel the distance between 2 homes. Since, she had noted correct time as 3:00 PM when she left friend's home, she can set her own clock accurately at 3:30 PM.

"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.

"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.

The Plane Landing Ahead of Schedule

A motorcyclist was sent by the post office to meet a plane at the airport.

The plane landed ahead of schedule, and its mail was taken toward the post office by horse. After half an hour the horseman met the motorcyclist on the road and gave him the mail.

The motorcyclist returned to the post office 20 minutes earlier than he was expected.

How many minutes early did the plane land?

Here is calculation of scheduled arrival time. 

Finding Scheduled Arrival Time of The Plane

What is the data given for calculation?

Since, motorcyclist returned to the post office 20 minutes earlier than he was expected, it mean, the horse had saved his 20 minutes of journey. That is, after meeting with horse at some point, the motorcyclist would have needed 20 minutes to go to & come back from airport to the same point. That's how the horse managed to save 20 minutes of motorcyclist. 

Let 'T' be the time at which horse met with motorcyclist.

It also means that, the motorcyclist would have taken another 10 minutes to reach at the airport exactly when plane was scheduled for landing. So the scheduled time of arrival of plane is T+10.

However, plane arrived at time T-30 where horse left airport with mail & met motorcyclist exactly half hour later at time T.

In short, plane landed at time T-30 instead of scheduled T+10 shows that plane landed 40 minutes ahead of schedule.
 

Lighting Up The Candles

In a group of 200 people, everybody has a non burning candle. On person has a match at lights at some moment his candle. With this candle he walks to somebody else and lights a new candle. Then everybody with a burning candle will look for somebody without a burning candle, and if found they will light it. This will continue until all candles are lit. Suppose that from the moment a candle is lit it takes exactly 30 seconds to find a person with a non burning candle and light that candle.

From the moment the first candle is lit, how long does it take before all candles are lit?

ESCAPE to answer! 

Time Calculation For Lighting Up The Candles

What is the exact situation?

From a moment from first candle is lit, 30 seconds later there would be total 2 candles lit. In next 30 seconds, each of these 2 candle holders will find 1 candle to lit. So there are now 4 candles lit after 60 seconds. In next 30 seconds, these 4 candles would lit another 4 candles making total 8 candles lit. 

In short, for every 30 seconds, the number of candles lit are doubled. So after 7 sets of 30 seconds, 2^7 = 128 candles would be lit. At 8th set of 30 seconds, 256 candles can be lit. But we have only 200 candles. Still 72 of 128 candles would lit another 72 in 8th set of 30 seconds. 

To conclude, 8 X 30 = 240 seconds = 4 minutes required to lit all 200 candles.