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Showing posts from July, 2019

### Escape Story of 3 Robbers!

Three robbers, Babylas, Hilary, and Sosthenes, are stealing a treasure chest from the top of an old tower. Unfortunately, they’ve had to destroy their ladder to avoid pursuit, so they’ll have to descend using a crude tackle — a single pulley and a long rope with a basket at each end.

Babylas weighs 170 pounds, Hilary 100 pounds, Sosthenes 80 pounds, and the treasure 60 pounds. If the difference in weight between the two baskets is greater than 20 pounds then the heavier basket will descend too quickly and injure its occupant (though the treasure chest can withstand this).

How can the three of them safely escape the tower with the treasure?

That's how they escape!

### Robbers' Plan To Escape The Tower!

What was the challenge in escaping the tower?

Note that difference between weights in 2 baskets can't exceed 20 pounds. However, chest can withstand the shock even if the difference exceeds that limit

1. They put 60 pounds chest in one of empty basket & sends it down. (60 pounds vs 0 Pounds)

2. Sosthenes of 80 pounds climbs in other empty basket and rides down to the ground while bringing up the chest. (60 pounds vs 80 pounds)

3. Now chest is taken out and Hilary is sent down while bringing up Sosthenes.(80 pounds vs 100 pounds).

4. Hilary gets out of the basket on ground and Sosthenes get out of the basket at top of the tower.

5. They again send chest of 60 pound to the ground putting it in 1 basket. This brings other empty basket up at the top of the tower.

6. Now, Hilary (who is on ground) climbs in the basket with chest (total weight = 60 + 100 = 160 pounds) while Babylas (170 pounds) enters in the basket at the top of tower. This allows Babylas to ride down to the ground. (160 pounds vs 170 pounds).

Right now, Babylas is on ground and Hilary, Sosthenes are on the top of the tower with chest.

7. Here, Babylas jumps on the ground & Hilary gets out of the basket at the top of tower.

8. Again, because of chest which weighs 60 pounds the basket descends down to the ground bringing up other basket up. (60 pounds vs 0 Pounds).

9. Now, Sosthenes (at top) rides down to the ground bringing up the chest. (60 pounds vs 80 Pounds).

10. Hilary replaces himself with the chest at the top and rides down while bringing Sosthenes up. (100 pounds vs 80 Pounds).

At the moment, Sosthenes is at the top with chest while Hilary and Babylas are on the ground.

11. Now, Sosthenes sends back chest on the ground & then rides to the ground while entering into other empty basket. (60 pounds vs 80 Pounds).

12. And when Sosthenes gets out of the basket on the ground the basket with chest descends on it's own to the ground.

13. Finally, all Sosthenes, Babylas and Hilary take chest out of the basket and safely escape the tower.

### 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?

### 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.

### Find Correct Digits For Correct Letters

If each letter represents a different nonzero digit, what must Z be?

### Correct Digits For Correct Letters

The equation in question was....

First of all X + Y + Z  = Z < = 10 in not possible since in that case, X +  Y + Z - Z =0 i.e.
X + Y = 0.

Hence, there must be carry 1 forwarded to digit's place. So,

X + Y + Z = 10 + Z

X + Y = 10.          ........(1)

Therefore, possible pairs for (X,Y) or for (Y,X) are (1,9), (2,8), (3,7), (4,6) and (5,5) out of which (5,5) is invalid as repeating digits are not allowed.

Now, since carry is forwarded to ten's place addition, it looks like,

1 + X + Y + Z = XY

Putting (1) in above,

1 + 10 + Z = XY

11 + Z = XY        ........(2)

The maximum value of Z can be 9 & in that case as per above equation XY = 20. But since 0 is not allowed Z must be less than or equal to 8. Then XY <= 11 + 8 = 19.

So X which seems to be carry to hundred place must be 1 (can't be 0 or 2 as proved above). Hence, X = 1.

If X = 1, then from (1), Y = 9.

And if XY = 19 then from (2) Z = 8.

To conclude, X = 1, Y = 9 and Z = 8.

Final equation, looks like,

11 + 99 + 88 = 198.

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Another Method :

The given equation is nothing but

10X + X + 10Y + Y + 10Z + Z = 100X + 10Y + Z

11(X + Y + Z) = 100X + 10Y + Z

89X = Y + 10Z = 10Z + Y

Since, only whole digits are allowed, maximum value of 10Z + Y can be 98 with Z = 9 & Y = 8. So if X = 2 then 89 x 2 = 198 = 10Z + Y is impossible case. Hence, X = 1.

And only Z = 8 and Y = 9 satisfies 89 x 1 = 10Z + Y = 10(8) + 9 = 89.

In short, X = 1, Y = 9 and Z = 8.

### Flip The Orientation of the Group

Assume these matchsticks as fishes. Can you flip the positions of the fishes horizontally i.e. 1 at the leftmost position and 4 at the rightmost position. Condition is that you can move only 3 matchsticks.

That's how it can be done!

### Flipping The Orientation of The Group

What was the task given?

All we need to do is just move these sticks as shown below.

### The Wedding Anniversary Puzzle

Recently I attended the twelfth wedding anniversary celebrations of my good friends Mohini and Jayant. Beaming with pride Jayant looked at his wife and commented, ‘At the time we were married Mohini was 3/4 of my age, but now she is only 5/6 th.
We began to wonder how old the couple must have been each at the time of their marriage!

Can you figure it out?

### The Wedding Anniversary Puzzle's Solution

What was the puzzle?

Let 'x' be the age of the Jayant & 'y' be the age of Mohini 12 years ago.

So 12 years ago,

y = (3/4)          .........(1)

And now 12 years later, the proportion is -

y + 12 = (5/6) (x+12)

Putting (1) in above,

(3/4) x + 12 = (5/6) x + 10

(1/12) x = 2

x = 24

Again putting this value in (1),

y = (3/4) 24 = 18

So, the age of Jayant was 24 and that of Mohini was 18 at the time of marriage. And now after 12 years, they are 36 and 30 years old respectively.

### Help The Policeman in Finding The Culprit

Late one evening, a car ran over a pedestrian in a narrow bystreet and drove away without stopping. A policeman who saw the vehicle leave the scene of the accident reported it moving at very high speed. The accident itself was witnessed by six bystanders. They provided the following conflicting accounts of what had happened:
• It was a blue car, driven by a man;
• The car was moving at high speed, its headlights were turned off;
• The car did have license plates, it wasn’t going very fast;
• It was a Toyota, it’s headlights were turned off;
• The car didn’t have license plates, the driver was a woman;
• It was a gray Ford.
When the the car and its driver were finally apprehended, it turned out that only one of the six eyewitnesses gave a fully correct description. Each of the other five provided one true and one false piece of information.

Keeping that in mind, can you determine the following:

— What was the car’s brand?
— What color was the car?
— Was the car going fast or slow?
— Did it have license plates?
— Were its headlights turned on?
— Was the driver a man or a woman?

### To Help The Policeman in Finding The Culprit

Let's recollect all the statements made by all 6 bystanders.

1.It was a blue car, driven by a man.

2.The car was moving at high speed, its headlights were turned off.

3.The car did have license plates, it wasn’t going very fast.

4.It was a Toyota, it’s headlights were turned off.

5.The car didn’t have license plates, the driver was a woman.

6 It was a gray Ford (It was gray car; it was Ford).

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If we believe in report made by Policeman where he stated that the car was moving at very high speed; then the part of the Statement 3 made by third bystander where he says car wasn't going fast turns out to be false. Hence, other part of his statement must be true. So the car must have license plates.

If the car has license plates; then 1st part of the Statement 5 will be false & other part must be true. Hence, the driver must be a woman.

If the driver was a woman, then 2nd part of the Statement 1 turns false making part 1 to be true. Hence, the color of the car must be blue.

If the car was at high speed then the entire Statement 2 must be true or it's 2nd part must be false.

Let's assume 2nd part of the statement 2 be false. Then 2nd part of statement 4 also must be false leaving 1st part to be true. That means the car was Toyota. But this makes statement 6 entirely false (as we already know color of car is blue). This contradicts the crucial data given - Each of the other five provided one true and one false piece of information. In the case, there will be no eyewitness giving full correct description.

So the entire Statement 2 must be true. Hence, the car was with it's headlight off.

If headlights were turned off then 2nd part of the Statement 4 must be true and 1st part false. That means, car wasn't Toyota.

And if car wasn't Toyota, as per Statement 6, it must be Ford but not of gray color.
This matches our early conclusion where we concludes color of the car was blue.

Conclusions:

1.What was the car’s brand?
- Ford
2.What color was the car?
- Blue
3.Was the car going fast or slow?
- Fast
4.Did it have license plates?
- Yes, it had.

5.Were its headlights turned on?
- No, those were off.
6.Was the driver a man or a woman?
- A woman.

### 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?”

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)

### 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.

### Aeroplane Probability Puzzle

People are waiting in line to board a 100-seat airplane. Steve is the first person in the line. He gets on the plane but suddenly can’t remember what his seat number is, so he picks a seat at random. After that, each person who gets on the plane sits in their assigned seat if it’s available, otherwise they will choose an open seat at random to sit in.

The flight is full and you are last in line. What is the probability that you get to sit in your assigned seat?

### Aeroplane Probability Puzzle - Solution

What was the puzzle?

The probability is really 1/2.

Let's assume that my name is Steve who was the first person who gets on the plane. And you are the person who is last in line. Also, let Seat No.1 was my assigned seat & Seat No.100 is your assigned seat.

Now if I choose seat no.1 itself, rest of all will get their assigned seat including you.

But what if I choose random seat say seat no.15? All the passengers from 2 to 14 will get their assigned seats.

Now, passenger no.15 can choose seat assigned to me i.e. seat no.1 or seat no.100 i.e. your seat. If he sits on my seat i.e. seat no.1 then rest of all including you will get the respective assigned seats. And if he chooses your seat i.e. seat no.100 then all passengers from 16-99 will get their respective assigned seats but you will be left with only option of seat left vacant by me i.e. seat no.1.

The passenger no.15 can choose any random seat as well. In that case, he leaves your 'result' to the passenger whose assigned seat he chooses. Suppose he chooses seat no.76. Again all passengers from 16 to 75 will be getting their respective seats. Now, passenger can choose seat no.1 or seat no.100 or any seat from 77-99.

Again, if he chooses my seat no.1 then rest of all i.e. 77 to 100 which includes you will be getting assigned seats. And if he selects your seat no.100 then again 77-99 will be getting their respective seats but my seat no.1 will be empty for you. However, he can leave your 'result' to any body from 77-99 whose seat he is going to occupy.

In short, your 'result' is in hand of the passenger whose assigned seat is already occupied. He can choose my seat to give you your assigned seat or he can select your seat leaving my seat for you as rest of all passenger will be getting their respective assigned seats. And he can leave your 'result' in hand of other passenger whose seat he randomly occupies.

Suppose, the passenger no.76 above selects seat no.99 in random. Now, all from 77-98 will have no problem in getting assigned seats but passenger 99 will have only 2 options either seat no.1 or seat no.99.

In the end, whenever the passenger ahead of you gets on the plane then either he will get his assigned seat if somebody occupied my seat hence you will get your assigned seat. And if my seat is vacant, then probability that he chooses my seat or your seat is 1/2.

So you will find either my seat or your seat empty when you get on the plane. That is it could be assigned seat or not assigned seat. The probability that the seat you have chosen is assigned must be 1/2.

### Prove The Mathematical Fact

Show that the numbers from 1 to 15 can’t be divided into a group A of 13 numbers and a group B of 2 numbers so that the sum of the numbers in A equals the product of the numbers in B.

Here is the proof!

### Proof of The Mathematical Fact!

What was that fact?

For a moment, let's assume that such group of 2 numbers exists whose product is equal to sum of rest 13 numbers taken out of 15 numbers.

Let x and y be those numbers in group B. Now x and y can be any number from 1 to 15.

As per condition,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 - x - y = xy

120 = xy + x + y

Adding 1 to both sides,

121 = xy + x + y + 1

121 = x( y + 1 ) + 1( y + 1 )

121 = ( x + 1 ) ( y + 1 )

Since x & y are the numbers in between 1 to 15, possible value of x & y satisfying the above equation is 10. But x & y are must be 2 different number. Hence, our assumption goes wrong here!

So, the numbers from 1 to 15 can’t be divided into a group A of 13 numbers and a group B of 2 numbers so that the sum of the numbers in A equals the product of the numbers in B.

### Count The Number of People From Handshakes

At a party, everyone shook hands with everybody else. There were 66 handshakes.
How many people were at the party?

### Getting Count of Number of Peoples From Handshakes

Let's suppose that there are 'n' number of people in the party.

The first person will shake hand with (n-1) people, the second person will shake hand with (n-2) people, the third will shake hand with (n-3) people.

In this way, (n-1) th person will shake hand with n-(n-1) = 1 person i.e. last person.

Adding all the number of handshakes,

(n-1) + (n-2) + (n-3) + ..... + 3 + 2 + 1 = n[(n-1)/2]

But total handshakes given are - 66

n[(n-1)/2] = 66

n(n-1) = 132

n^2 - n - 132 = 0

(n-12)(n+11) = 0

n = 12 or n = -11

Since number of people can't be negative, n = 12.

Hence there are 12 people in the party.

### Simple Logical Mathamatical Problem

There are 5 people who can build 5 houses in just 5 days.

How long would it take 100 people to build 100 houses?

100? Really? Check back or click here!

### Answer Of Simple Logical Mathematical Problem

Did you answer 100?

"There are 5 people who can build 5 houses in just 5 days."

The above statement suggests that 1 man takes 5 days to build 1 house & not 1 man take 1 day to construct 1 house.

So 100 people will make 100 houses in only 5 days as each one completing the task in 5 days!