What's The Right Answer?

On a multiple-choice test, one of the questions is illegible, but the choice of answers is listed clearly below. 

What’s the right answer?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.

'THIS' should be the right answer! 

Source
 

Well, 'THIS' is the Right Answer!

What was the question?

Let's recall the question once again.


What’s the right answer?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.


If (a) is the right answer then as per (a), all below including (b) and (f) are right answers. But (b) & (f) are contradicting each other. So, (a) can't be right.

Now if (c) is right, then as per (c), (a) also must be right. But as concluded above, (a) can't be right. So, (c) also can't be right.

Next, if (b) is correct, then (d) is also correct making 2 options right. Hence, (b) is also false.

If all (a), (b), (c) are false, then (d) has to be false.

If (f) is correct, then what (e) states also is correct making both (e) & (f) correct. Hence, (f) must be false.

That makes (e) is correct answer. 

Puzzle : Ants Walk on a Stick

Twenty-five ants are placed randomly on a meter stick. Each faces east or west. At a signal they all start to march at 1 centimeter per second. Whenever two ants collide they reverse directions. How long must we wait to be sure that all the ants have left the stick?

This sounds immensely complicated, but with a simple insight the answer is immediately clear. What is it?

You need to wait for.....seconds only!
 

Analysing Ants Walk on a Stick

Read the question associated with the walk.

For a moment, let's assume that there are only 2 ants 20 cm away from either end of the stick. Now, after 30 seconds they both will collide with each other & will reverse the direction.

At 50th seconds they will be at the end of the sticks falling off the stick.

So after 80 seconds they will fall off the stick. Now, imagine if ants avoid collision & pass through (or above) each other. Still, both ants would need 80 seconds to leave the stick.

In short, 2 ants' collision & reversal in direction is equivalent to their passing through each other. The other ant continues the journey on the behalf of the first ant & vice versa.

And in case, if they were 100 cm apart, they would need 100 seconds to get off the stick. Again, after collision at halfway mark here, the other ant travels the rest of distance that other ant was supposed to travel.

On the similar note, we can say that even if there are 25 ants on the stick then each ant will cover some distance on the behalf of some other ant. And we need to wait for maximum 100 seconds if 1 of 25 ants is at the edge of the stick. 

All 25 ants together completes the journey of each others in 100 seconds. The ant which is at the edge of stick might complete journey of some other ant which might be only 10 seconds long. But the 100 seconds journey of that ant will be shared by rest of ants. 

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 River Crossing Challenge!

There are 3 men, two Chimps, and one Gorilla on one side of a river :

• They have a boat but only the men and the Gorilla can row the boat across, so there must always be a human and/or Gorilla on the boat.

• The boat can only carry two people/monkeys.

• If monkeys and humans are together on one side of the river there must be as many or more people than monkeys for the men's safety. 

How can all men and monkeys make it to the other side ? 



Here is the PROCESS by which it can be done! 

Responding to The River Crossing Challenge!

What was the challenge ahead?

Recalling the conditions those need to be followed. 

--------------------------------------------------------------

• They have a boat but only the men and the Gorilla can row the boat across, so there must always be a human and/or Gorilla on the boat.

• The boat can only carry two people/monkeys.

• If monkeys and humans are together on one side of the river there must be as many or more people than monkeys for the men's safety.

--------------------------------------------------------------
 
Here, we go step by step process. (M - Men, G - Gorilla, C - Chimps)

1. The gorilla takes 1 chimp across the river and comes back. 

    (M - 3, G - 1, C - 1 | M - 0, G - 0, C - 1) 

2. Again, gorilla takes 1 man across the river and comes back. 

    (M - 2, G - 1, C - 1 | M - 1, G - 0, C - 1)

Now, here gorilla can't take chimp across the river as that will violate condition 3 on that side. Neither gorilla can take 1 man on other side and return back since number of monkeys on returning side will be more than people again violating condition 3.

3. Next, one man drops gorilla at the other side and bring back chimp.

    (M - 2, G - 0, C - 2 | M - 1, G - 1, C - 0) 

4. Now, 2 men has to cross the river and send back gorilla for the rest of work.

   (M - 0, G - 1, C - 2 | M - 3, G - 0, C - 0) 

5. Finally, gorilla takes 2 chimps across the river in 2 round trips.

   (M - 0, G - 0, C - 0 | M - 3, G - 1, C - 2) 

 

Find The Heaviest Ball

Given 27 table tennis balls, one is heavier than the others.

What is the minimum number of weighings (using a two-pan balance scale) needed to guarantee identifying the heavy one? The other 26 balls weigh the same.

Here is how to identify the heavier ball.

Identifying The Heavier Ball

 What was the task given?

We need to use the scale only 3 times.

1. Divide the 27 balls into 3 groups of 9 balls each. 

2. Use the scale to weigh 2 groups. This will tell us which group has the heavier ball. 

    If the scale is balanced then third group must be having heavier ball

3. Now, divide the 9 balls into 3 groups of 3 balls each and weigh 2 groups, and identify the group which has the heavier ball. Again, if the scale is balanced then the third group must be having the heavier ball.

4. From the heavier group, weigh 2 balls to determine the heavier ball. And, if the scale is balanced then the third ball must be heavier.

A Brillian Deception

A witch owns a field containing many gold mines. She hires one man at a time to mine this gold for her. She promises 10% of what a man mines in a day, and he gives her the rest. Because she is blind, she has three magic bags who can talk. They report how much gold they held each day, and this is how she finds out if men are cheating her. 

Upon getting the job, each man agrees that if he isn't honest, then he will be turned into stone. So around the witch's mines, many statues lay! 

Now comes an honest man named Garry. He accepts the job gladly. 

The witch, who didn't trust him said, "If I wrongly accuse you of cheating me, then I'll be turned into stone."  

That night, Garry, having honestly done his first day's job, overheard the bags talking to the witch. He then formulated a plan... 

The next night, he submitted his gold, and kept 1.6 pounds of gold. Later, the witch talked with her bags.

The first bag said it held 16 pounds that day. The second one said it held 5 pounds. The third one said it held 2 pounds. 

Beaming, the witch confronted Garry. "You scoundrel, you think you could fool me. Now you shall turn into stone!" the witch cried. One second later, the witch was hard as a rock, and very grey-looking.

How did Garry brilliantly deceive the witch? 

Here is the Garry's Master Plan!
 

Mathematics Behind The Brilliant Deception!

What was the deception?

As per the honest man, he must have mined 16 Pounds of gold since he kept 1.6 Pounds (10% of 16) gold for himself.

And as per magical bags, since Bag no.1 said it had 16 Pounds, Bag no.2 said 5 Pounds and Bag no.3 said 2 Pounds the total gold mined 16 + 5 + 2 = 23 Pounds.

We know, honest man Garry would never do any fraud and neither of magical bag would lie. 

So, it's clear that some of pounds are counted multiple times by magical bags. There are 23 - 16 = 7 Pounds gold extra found by those bags.

Now, Bag No.3 must have 2 Pounds in real.

And to make count of Bag No.2 as 5, Garry must had put 3 Pound + Bag No.3 itself in Bag No.2. Hence, the Bag No.2 informed witch that it had total 5 Pound of gold.

Finally, to force Bag No.1 to tell it's count as 16, Garry must had put 
11 Pounds + Bag No.2 (5 Pounds = 3 + Bag No.1) = 11 + 5 = 16 Pounds.

In short, Garry put 2 pounds of gold in Bag No.3 and put that Bag No.3 in Bag No.2 where he had already added 3 Pounds of gold. After that, he put this Bag No.2 in Bag No.3 where he had already added 11 Pounds of gold (somewhat like below picture).

 

This way, 2 Pounds of gold in Bag No.3 are counted 2 extra times and 3 Pounds gold of Bag No.2 are counted 1 more extra time. That is 2 + 2 + 3 = 7 Pounds of extra gold found by bags.

PUZZLE : The Case of Honest Suspects

Handel has been killed and Beethoven is on the case. 

He has interviewed the four suspects and their statements are shown below. Each suspect has said two sentences. One sentence of each suspect is a lie and one sentence is the truth. 

Help Beethoven figure out who the killer is.

Joplin: I did not kill Handel. Either Grieg is the killer or none of us is.

Grieg: I did not kill Handel. Gershwin is the killer.

Strauss: I did not kill Handel. Grieg is lying when he says Gershwin is the killer.

Gershwin: I did not kill Handel. If Joplin did not kill him, then Grieg did.

Click here to know who is the killer! 

DETECTION : Killer in The Case of Honest Suspects

What was the case?

Let's first see what are the statements made by 4 suspects.

Joplin: I did not kill Handel. Either Grieg is the killer or none of us is.

Grieg: I did not kill Handel. Gershwin is the killer.

Strauss: I did not kill Handel. Grieg is lying when he says Gershwin is the killer.

Gershwin: I did not kill Handel. If Joplin did not kill him, then Grieg did. 

ANALYSIS :

1] If Joplin is the killer then his first statement would be lie & other statement must be true. But his second statement contradicts assumption that he is killer. So Joplin can't be the killer.

2] Let's assume Grieg is the killer. His first statement must be lie and second must be true. His second statement points to Gershwin as a killer. But there is one killer among 4 so Grieg and Gershwin can't be killers together. Therefore, Grieg can't be the killer.

3] Suppose Gershwin is the killer. Again, his first statement is lie and second is true. His true second statement suggests that Grieg is the killer as Joplin is assumed to be innocent in the case. Again, both Grieg and Gershwin can't be killers together since there is only one killer. Hence, Gershwin is not the killer.

4] Therefore, Strauss must be the killer. His first statement is lie and second statement is true. And that's how Grieg is lying when he is saying Gershwin is the killer.

CONCLUSION : 

STRAUSS is the one who committed the crime and his first statement is lie and second is true.

The first statements of rest of all suspects are true and second statements of each are lie.  

 

The Challenge Ahead of New Manager

In a wood-cutting factory, five large sawing machines stand in a windowless room. Each machine has an on/off switch attached, there being no doubt as to which switch controls which machine.

Outside the door to the room are five back-up on/off switches, one for each machine inside. The power for each machine must first pass through the back-up switch, and then the machine switch before reaching the saw.

The problem is, the new manager cannot decide how these back-up switches match with the machines inside the room. One day, the manager's brother visits. The manager takes him inside the sawing room where all five machines are at work and explains the problem. The brother announces that he intends to leave the room and that when he returns he will be able to match correctly the five switches outside the room to the five machines inside. The brother works alone, cannot see the machines from outside the room and solves the problem purely by operating switches. 

How is it possible?

Read how brother managed to do it! 

Genius Moves by Brother of Manager!

What was the challenge?

Suppose you are the brother of that manager & have accepted the challenge. 

For sake of convenience, we will name switches inside the room operating machines as an 'operating switches' & those which are outside the room as a 'back up switches'.

Though windowless, we assume that operating sound of machine(s) can be heard from outside room but can't see which machine(s) is (are) in operating condition. 

Here is what you should do!

1. Let's label the machines which we are going to keep ON as ACTIVE & others as INACTIVE. 

2. Turn off operating switches of 2 machines so that there are three machine ACTIVE and two are INACTIVE inside the room.

3. Go outside and now your task is to find the 3 back up switches controlling 3 active machines inside the room.

4. You have to switch off three switches in all possible combinations. There are 10 such possible combinations (5C3) of 3 controlling switches out of 5.

5. If 0 represents OFF position and 1 represents ON position then you should try all possible below combination. 

11000
10100
10010
10001
01100
01010
01001
00110
00101
00011

6. There will be exactly one combination in which all three ACTIVE machines  will be OFF & there will be no sound coming out of the room. The three switches having value of 0 are controlling ACTIVE machines while other 2 must be controlling INACTIVE machines.

7. Still we don't know the exact switch operating the each machine. Label 3 switches controlling ACTIVE machines as A, B and C & those controlling INACTIVE machines as D & E. 

Allow some time to cool down all those ACTIVE machines.

8. Now, turn on 2 switches A & B and keep switch E ON. Machines connected to A and B will start working.

9. After a while, turn off the switch B and go inside the room. 

10. The machine which is still operating must be controlled by back up switch A.

11. Touch other 2 which were labelled as ACTIVE & check which has got warmer.

12. The machine which is warmer must be controlled by back up switch B.

13. And since we didn't turn on the switch C (after giving time all to cool down), the machine having normal temperature must be controlled by the switch C.

14. Remember, we had turned off operating switches of INACTIVE machines initially. And before entering into room again we have turned on switch E. 

15. Now, turn on operating switch of one of the INACTIVE labelled machine. 

If the machine starts working it must be connected to the switch E & other to the switch D. And if the machine doesn't start working it must be connected to D & other to E. 

This way, you will find every back up switch located outside the room controlling operating switch of each machine inside the room.

 

What day of the week is it?

A group of campers have been on vacation so long, that they've forgotten the day of the week. 

The following conversation ensues. 

Darryl: "What's the day? I don't think it is Thursday, Friday or Saturday." 

Tracy: "Well that doesn't narrow it down much. Yesterday was Sunday." 

Melissa: "Yesterday wasn't Sunday, tomorrow is Sunday." 

Ben: "The day after tomorrow is Saturday." 

Adrienne: "The day before yesterday was Thursday." 

Susie: "Tomorrow is Saturday." 

David: "I know that the day after tomorrow is not Friday." 

If only one person's statement is true, what day of the week is it?

So it was THIS day of the week!

The Forgotten Day of Week is Wednesday!

What was the puzzle?

Let's see once again the conversation that campers had - 

--------------------------------------------------------


Darryl: "What's the day? I don't think it is Thursday, Friday or Saturday." 

Tracy: "Well that doesn't narrow it down much. Yesterday was Sunday." 

Melissa: "Yesterday wasn't Sunday, tomorrow is Sunday." 

Ben: "The day after tomorrow is Saturday." 

Adrienne: "The day before yesterday was Thursday." 

Susie: "Tomorrow is Saturday." 

David: "I know that the day after tomorrow is not Friday."

-----------------------------------------------------

Let's see what day statement of each is suggesting -

Darryl - Sunday, Monday, Tuesday, Wednesday.

Tracy - Monday.

Melissa - Saturday.

Ben - Thursday.

Adrienne - Saturday.

Susie - Friday.

David - Sunday, Monday, Tuesday, Thursday, Friday or Saturday.  

If we assume David's statement is TRUE then one of statements of Darryl (Sunday, Monday, Tuesday are common) or Tracy (Monday is common) or Melissa & Adrienne (Saturday is common) or Susie (Friday is common) or Ben (Thursday is common) has to be also TRUE. But this is against the given data that only 1 of the statement is TRUE.

Hence, David's statement must be FALSE and the only day that isn't pointed by David is Wednesday.

So the day must be Wednesday as suggested correctly by Darryl and thereby making statements of every other camper including David FALSE. 

Crack The 10-Digits Password

In an attempt to further protect his secret chicken recipe, Colonel Sanders has locked it away in a safe with a 10-digit password.  Unable to resist his crispy fried chicken, you'd like to crack the code. 

Try your luck using the following information:

All digits from 0 to 9 are used exactly once.

No digit is the same as the position which it occupies in the sequence.

The sum of the 5th and 10th digits is a square number other than 9.

The sum of the 9th and 10th digits is a cube.

The 2nd digit is an even number.

Zero is in the 3rd position.

The sum of the 4th, 6th and 8th digits is a single digit number.

The sum of the 2nd and 5th digits is a triangle number.

Number 8 is 2 spaces away from the number 9 (one in between).

The number 3 is next to the 9 but not the zero.

Click here to know how to CRACK it! 

Cracking The 10-Digits Password

What was the challenge? 

Let's suppose the 10-digits password is ABCDEFGHIJ.

---------------------------------

Take a look at the given clues to crack down the password.

1. All digits from 0 to 9 are used exactly once.

2. No digit is the same as the position which it occupies in the sequence.

3. The sum of the 5th and 10th digits is a square number other than 9.
 
4. The sum of the 9th and 10th digits is a cube.
 
5. The 2nd digit is an even number.
 
6. Zero is in the 3rd position.
 
7. The sum of the 4th, 6th and 8th digits is a single digit number.
 
8. The sum of the 2nd and 5th digits is a triangle number.
 
9. Number 8 is 2 spaces away from the number 9 (one in between).
 
10. The number 3 is next to the 9 but not the zero. 

--------------------------------------

STEPS : 

Respective Clues are in ().

1] Since, every digit is used only once (1), the sum of 2 digits can't exceed 
9 + 8 = 17 and minimum sum of 2 digits can be only 0 + 1 = 1.

2] So, the sum of 5th & 10th digits which is square (3) can be - 1, 4, 16.
The sum of 9th & 10th digits which is cube (4) can be - 1, 8.
The sum of 2nd & 5th digits which is triangle number (8) can be - 1, 3, 6, 10, 15.

3] As per (6), 0 is already at third position i.e. C = 0. Hence, 2nd, 5th, 9th or 10th digits can't be 0. Therefore, the sum of 5th & 10th or 9th & 10th or 2nd & 5th can't be 1. Revising list of possible values of those in step 2.

The sum of 5th & 10th digits which is square (3) can be -  4, 16.
The sum of 9th & 10th digits which is cube (4) can be -  8.
The sum of 2nd & 5th digits which is triangle number (8) can be -  3, 6, 10, 15.

4] Possible pairs of 9th & 10th digits - (1,7), (7,1), (2,6), (6,2), (3,5), (5,3)

If IJ = 71, then 5th digit E must be 4 - 1 = 3 & possible values of 2nd digit B are 0,3,7.
But as per (5), B must be even & C = 0 already.

IJ = 26 or 62 or 35 doesn't leave any valid value for 5th digit E.

If IJ = 53, then 5th digit E must be 1 & possible values of 2nd digit B are 2,5,9
That is B = 2 (5). But as per (2), no digit is the same as the position which it occupies in the sequence. So, the number 2 can't be at 2nd position where B is there.

Hence, IJ must be 17 i.e. I = 1, J = 7.

5] This leaves only 9 as valid value for 5th digit E so that (3) is true. 

So, E = 9 and hence B = 6 with sum of 2nd digit B and 5th digit E as 15.

6] So far, we have, B = 6, C = 0, E = 9, I = 1 and J = 7. 

As per (10), the number 3 is next to the 9 but not the zero. Hence, the 6th digit F must be 3. F = 3.

7] As per (9), the 8 is 2 spaces away from 9 i.e. here 5th digit E=9. With 3rd position already occupied by C = 0, the digit 8 must be at 7th position. 
So, G = 8. 

8] As per (7), D + F + H must be single digit. We have, F = 3 already, 
so possible values of D + H are 0, 1, 2, 3, 4, 5, 6.

D + H can't be 0 (0+0), 1 (1+0), 2(1+1/2+0), 3(1+2/3+0), 4(2+2/3+1/4+0), 5(2+3/4+1/5+0) since digits are repeated with C=0, I=1, F=3 already.

Hence, D + H = 6. 

9] Now, D = 5 and H = 1 is not possible. Also, D/H can't be 0 or 6. Both D and H can't be 3 at the same time. D can't be 4 as D is at 4th position. 

Hence, D = 2, H = 4.

10] With B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7, the only digit left for the first position A is 5. So A = 5.

CONCLUSION : 

A = 5, B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7.

So, the 10-digits password ABCDEFGHIJ is 5602938417.