Puzzle : Which one is the car thief?

A car thief, who had managed to evade the authorities in the past, unknowingly took the automobile that belonged to Inspector Detweiler. 

The sleuth wasted no time and spared no effort in discovering and carefully examining the available clues. He was able to identify four suspects with certainty that one of them was the culprit.

The four make the statements below. In total, six statements are true and six false.

Suspect A:

1. C and I have met many times before today.
2. B is guilty.
3. The car thief did not know it was the Inspector's car.

Suspect B:

1. D did not do it.
2. D's third statement is false.
3. I am innocent.

Suspect C:
 
1. I have never met A before today.
2. B is not guilty.
3. D knows how to drive.

Suspect D:
 
1. B's first statement is false.
2. I do not know how to drive.
3. A did it.

Which one is the car thief?

Know here who is that car thief? 

Solution : The Unlucky Car Thief

What was the puzzle?

Take a look at the statements made by suspects.

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Suspect A:

1. C and I have met many times before today.
2. B is guilty.
3. The car thief did not know it was the Inspector's car.

Suspect B:

1. D did not do it.
2. D's third statement is false.
3. I am innocent.

Suspect C:

1. I have never met A before today.
2. B is not guilty.
3. D knows how to drive.

Suspect D:

1. B's first statement is false.
2. I do not know how to drive.
3. A did it.

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After investigation, it is found that,  in total, six statements are true and six false.

We will name statements as A1 for first statement of A, A2 for his second statement, B1 for B's first statement B2 for his second statement and so on.

1. Since, a car thief, who had managed to evade the authorities in the past, unknowingly took the automobile that belonged to Inspector Detweiler, we assume statement A3 is TRUE.

2. A1 and C1, C3 and D2 are contradicting statements. These statements are having least relevant in the process as they are not pointing to anybody else. Two of them must be TRUE and 2 must be FALSE. There are 4 TRUE and 4 FALSE statements from rest of statements.

We have, 2 FALSE statements among A1, C1, C3 and D2 for sure.

3. Assume A is a car thief. Then only A2, B2 and D3 turns out to be FALSE from rest giving in total of 5 FALSE statements only.

4. Assume C is a car thief. Then only A2, D1 and D3 are FALSE, hence total of 5 FALSE statement among all statements.

5. Assume D is a car thief. Then again only A2, B1 and D3 are FALSE, once again total 5 out of 12 given statements are FALSE in the case.

6. Assume B is a car thief. In this case, B3, C2, D1 and D3 turns out to be FALSE. Hence, total 6 out of 12 given statements are FALSE. 

This is exactly as per fact found in the investigation which suggests that exactly 6 out of 12 statements are FALSE. 

Hence, B must be a car thief. 

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.   

A Long Journey of 27000 Miles

The MacDonalds are planning a long car journey of 27,000 miles. If they use tires that last 12,000 miles each, how many tires will they need, and how can they make the best use of them?

This is how usage of tires can be optimized!

Optimizing The Use of Tires in Long Journey!

How long the actual journey is?

Since each tire would be traveling 27,000 miles, when car travels 27000 miles; tire -miles are equal to 27000 x 4 = 108000.

But the each tire lasts for 12000 then tires required = 108000/ 12000 = 9.

Now managing use of these 9 tires need some planning.

For first 12000 miles of 27000 miles they can use 4 out of 9 tires. For the rest of 15000 miles use of remaining 5 tires need to be planned.

Best way is to change 1 tire after every 3000 miles like below.

First 3000 miles: Tires 1, 2, 3, 4

Second 3000 miles: Tires 2, 3, 4, 5

Third 3000 miles: Tires 3, 4, 5, 1

Fourth 3000 miles: Tires 4, 5, 1, 2

Fifth 3000 miles: Tires 5, 1, 2, 3

This way, it's made sure that each tire is used for only 12000 miles. And this is how the journey can be completed using 9 tires only.

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. 

The Monty Hall Problem

You’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?”

Is it to your advantage to switch your choice?

Will you switch or stay with your door?

Note : Monty Hall was the host of game show called 'Let's Make a Deal' on which above puzzle is based. (Source-Wikipedia)



This is what you should do! 

Winning The Monty Hall Game Show

What was the game show?

Suppose you always choose DOOR 1. Then host will open DOOR 2 or DOOR 3 behind which car is not there.

If the car is behind the DOOR 1, then host will open the DOOR 2 or DOOR 3. And if you switch to remaining DOOR 3 or DOOR 2, you will find goat behind it & you will loose.

And if the car is behind the DOOR 2, then host will be forced to open DOOR 3. Now, if you switch your choice to DOOR 2 then you will win the car behind that door.

Again, if the car is behind the DOOR 3, then host has to open the DOOR 2 behind which goat is there. And now if you switch your selection from DOOR 1 to DOOR 3, then you will be winning the car.

So out of 3 possibilities, in 2 you will be winning this game show if you switch your choice. The probability of winning the game show is 2/3.

And if you stay with your first choice, then probability of having car behind selected door is 1/3.

To conclude, it's better to switch the choice as it increases the probability of winning the game show from 1/3 to 2/3. 
 

A Car on a Fragile Bridge

A car is crossing a 20km long bridge. The bridge can support at most 1500kg of weight over it. If somehow, the weight on the bridge becomes more than that, it will break.
Now, the weight of the car is exactly 1500kg. At the midway, a bird comes and sits on the roof of the car. This bird weighs exactly 200 gram. 

 
Can you tell if the bridge breaks at this point or not? 

Read what will happen next? 

Source 

 

Bridge Under Load

What was the situation?

At first look, the first impression would be that the bridge will break certainly. But if you wait for a while before concluding anything you will get the right answer.

The bridge will not break in the case! It's 20 km long bridge & now it's in the middle of bridge after traveling 10 km. By now, it must have used half of the fuel that was in the tank initially at the start. This amount of fuel must be weighing more than 200 gm. Hence even a bird sits on the car there are hardly any chances the total weight on the bridge goes beyond 1500 kg! Hence, no chance of breaking of it.

 

Cars Across the Desert

A military car carrying an important letter must cross a desert. 

There is no petrol station on the desert and the car has space only for petrol that lasts to the middle of the desert.

There are also other cars that can transfer their petrol into one another.

How can the letter be delivered?

This is how letter can be delivered!

Source 

Delivering Letter Across The Desert

What was the task?

We need 4 such cars to deliver the letter across the desert successfully.

Let's divide the entire route into 6 parts. That means the distance that car can travel (half the total path in desert) is divided into 3 parts. To travel each part car requires 1/3rd of it's petrol in the tank.

1. At first 1/6th of total path, all cars are 2/3rd full. Now 2/3rd of the petrol from 1 car can be used to fill 1/3rd of tanks in other 2 cars (1/3 + 1/3 = 2/3). This way, we would have 2 cars full while 1 car 2/3rd full. We are leaving behind the empty car, taking 3 cars forward.

Stage 1

2. At next 1/6th of the distance, 2 full cars will use 1/3rd of their petrol hence would be 2/3rd full. And the car that was 2/3rd at previous stage would be not 1/3rd full. At this stage, the petrol from car that is 1/3rd full can be used to fill tank of 1 car completely. So we are leaving behind one another empty car here & taking fully filled car & 2/3rd filled car for next stage.

Stage 2

3. For next 1/6th of the total distance, the car that was fully filled would have 2/3rd petrol. And the car which was 2/3rd at previous stage would be now 1/3rd filled. The petrol of this car can be used to fill the tank of the first car. Now we have 1 car fully filled while other one is empty. So we can leave behind the empty car & use fully filled car for the rest half of the journey. Remember, a car which tank is full can travel half the total path.

Stage 3

 

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].

  
 

The 500 Miles Race

Mike, Jimmy, Nader, Kevin and Larry were the top five finishers in the regional 500-mile race. They drove yellow, orange, green, red and blue cars but not necessarily in that order.

1. Neither Kevin nor Larry drove the green car.

2. Kevin finished faster than Mike and Larry.

3. The blue car finished earlier than Larry’s and Nader’s car.

4. The yellow car finished faster than the green car and the orange car.

5. Mike’s and Larry’s car finished ahead of the orange car.

6. Jimmy’s car finished before the blue and the yellow car.

Who drove what color car and what place did each driver finish?

To know that click here! 

Source 

Stats of The 500 Miles Race

Here is the scenario! 

Let's make a bit simple by putting clues into table. Rows representing owner while columns representing their cars. Now we need to use logical deduction by using all the clues & fill the table one by one. Finally, we get this.

Stats of The 500 Miles Race

 Filled circle represents rank of the particular car or participant. e.g. 1 circle means it's first, 2 means second & so on.