Dating

Puzzle:

Tanya wants to go on a date and prefers her date to be tall, dark and handsome.

• Of the preferred traits - tall, dark and handsome - no two of Adam, Bond, Cruz and Dumbo have the same number.
• Only Adam or Dumbo is tall and fair.
• Only Bond or Cruz is short and handsome.
• Adam and Cruz are either both tall or both short.
• Bond and Dumbo are either both dark or both fair.
  

Who is Tanya's date?

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Solution:

Cruz is Tanya's date.

As no two of them have the same number of preferred traits - from (1), exactly one of them has none of the preferred traits and exactly one of them has all the preferred traits.

From (4) and (5), there are only two possibilities:
* Adam & Cruz both are tall and Bond & Dumbo both are fair.
* Adam & Cruz both are short and Bond & Dumbo both are dark.

But from (2), second possibility is impossible. So the first one is the correct possibility i.e. Adam & Cruz both are tall and Bond & Dumbo both are fair.

Then from (3), Bond is short and handsome.

Also, from (1) and (2), Adam is tall and fair. Also, Dumbo is the person without any preferred traits. Cruz is Dark. Adam and Cruz are handsome. Thus, following are the individual preferred traits:

Cruz - Tall, Dark and Handsome
Adam - Tall and Handsome
Bond - Handsome
Dumbo - None :-(

Hence, Cruz is Tanya's date.

The Party

Puzzle:

Yesterday in a party, I asked Mr. Shah his birthday. With a mischievous glint in his eyes he replied. "The day before yesterday I was 83 years old and next year I will be 86." Can you figure out what is the Date of Birth of Mr. Shah? Assume that the current year is 2000.

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Solution:

Mr. Shah's date of birth is 31 December, 1915

Today is 1 January, 2000. The day before yesterday was 30 December, 1999 and Mr. Shah was 83 on that day. Today i.e. 1 January, 2000 - he is 84. On 31 December 2000, he will be 85 and next year i.e. 31 December, 2001 - he will be 86. Hence, the date of birth is 31 December, 1915. Many people do think of Leap year and date of birth as 29th February as 2000 is the Leap year and there is difference of 3 years in Mr. Shah's age. But that is not the answer.

Females

Puzzle:

• A is the father of two children - B and D who are of different sexes.
• C is B's spouse.
• E is the same sex as D.
• B and C have the two children - F who is the same sex as B and G who is the same sex as C.
• E's mother, H who is married to L, is the sister of D's mother, M.
• E and E's spouse, I have two children - J and K who are the same sex as I.
  

Note that no persons have married more than once. Also, there are more number of females than males. Can you tell how many females are there?

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Solution:

There are 7 females and 6 males.

Assume that there are four sexes - male, female, X and Y.

L(m) H(f) M(f) A(m) E(x) I(y) D(x) B(y) C(x) J(y) K(y) F(y) G(x)

It is clear that there are altogether 13 persons - 2 males, 2 females, 4 Xs and 5 Ys. It is given that there are more number of females than male. Hence, all Y must represent female. Thus, there are 7 females and 6 males.

Order Of Visits

Puzzle:

Annie, Bunnie, Candy and Dina visited Edy on 14th February.

• The time of each visit was as follows:
• Annie at 8:00
• Bunnie at 9:00
• Candy at 10:00
• Dina at 11:00
• Each time mentioned above may be either AM or PM.
• Candy did not visit Edy between Bunnie and Dina.
• At least one female visited Edy between Annie and Bunnie.
• Annie did not visit Edy before both Candy and Dina.
  

Can you tell at what time did they individually visit Edy?

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Solution:

Bunnie (9:00AM) - Dina (11:00AM) - Annie (8:00PM) - Candy (10:00PM)

From the given data, it is clear that at least one female visited Edy in the morning and at least one female visited Edy in the evening. Also, from (4), Annie did not visit Edy first. It means that Annie visited Edy at 8:00 PM. From (3), Bunnie must have visited Edy at 9:00 AM. Also, either Candy or Dina or both visited Edy in the morning. But from (2), only Dina must have visited Edy in the morning at 11:00 AM and hence, Candy visited Edy at 10:00 PM.

Two Juices

Puzzle:

An orange colored glass has Orange juice and white colored glass has Apple juice both of equal volumes. 50ml of the orange juice is taken and poured into the white glass. After that similarly, 50ml from the white glass is poured into the orange glass. Of the two quantities, the amount of apple juice in the orange glass and the amount of orange juice in the white glass, which one is greater and by how much?

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Solution:

The two quantities are equal.

Solve it by taking example. Let's assume that both glasses contain 450 ml of juice each. Now, 50ml of the orange juice is taken and poured into the White glass. Hence, orange colored glass contains 400 ml of Orange juice and white glass contains 450 ml of Apple juice and 50 ml of Orange juice i.e. total of 500 ml from white glass contains 450 ml of Apple juice and 50 ml of Orange juice. It means that every 50 ml from white glass contains 45 ml of Apple juice and 5 ml of Orange juice. Similarly, 50 ml of juice from white glass is poured into orange glass. Now this 50 ml is not a pure apple juice. It contains 45 ml of Apple juice and 5 ml of Orange juice. Hence, Orange glass contains 405 ml of Orange juice and 45 ml of Apple juice. Similarly, white glass contains 405 ml of Apple juice and 45 ml of Orange juice.

Four Men

Puzzle:

Four men - Abraham, Bobby, Clinton and Denial - are standing in a straight line.

• One man is fair, handsome and unscarred.
• Two men who are not fair, are each standing next to Abraham.
• Bobby is the only man standing next to exactly one handsome man.
• Clinton is the only man not standing next to exactly one scarred man.
  

Who is fair, handsome and unscarred?

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Solution:

Clinton is fair, handsome and unscarred.

From (2), both the men standing next to Abraham are not fair. Also, exactly one man is fair, handsom and unscarred. Hence, there are two cases:

Case 1 :: ? (N, ?, ?) : Abraham (Y, Y, N) : ? (N, ?, ?) : ? (?, ?, ?)
Case 2 :: ? (N, ?, ?) : Abraham (?, ?, ?) : ? (N, ?, ?) : ? (Y, Y, N)

Note the representation - Name (Fair, Handsome, Scarred). "Y" stands for Yes and "N" stabds for No. Abraham (Y, Y, N) means Abraham is Fair, Handsome and Unscarred.

It is clear that either Abraham or the man at the extreme right is fair, handsome and unscarred.

From (4), it is deduced that Clinton is standing next to unscarred man and each of the other men standing next to exactly one scarred man.

Case 1 :: Clinton (N, ?, N) : Abraham (Y, Y, N) : ? (N, ?, Y) : ? (?, ?, Y)
Case 2 :: ? (N, ?, Y) : Abraham (?, ?, Y) : ? (N, ?, N) : Clinton (Y, Y, N)

From (3), Bobby is the only man standing next to exactly one handsome man. But in Case 1, Clinton is standing next to exactly one handsome man. Hence, Case 1 is not possible and Case 2 is the correct one.

Case 2 :: ? (N, ?, Y) : Abraham (?, ?, Y) : ? (N, ?, N) : Clinton (Y, Y, N)

Again from (3) and (4), there are 2 possibilities as shown below.

Case 2a :: Denial (N, N, Y) : Abraham (?, N, Y) : Bobby (N, N, N) : Clinton (Y, Y, N)
Case 2b :: Bobby (N, N, Y) : Abraham (?, Y, Y) : Denial (N, N, N) : Clinton (Y, Y, N)

Thus, Clinton is fair, handsome and unscarred. Also, Abraham may be either fair or not fair.

The Secret Agent

Puzzle:

The secret agent X emailed a code word to his head office. They are "AIM DUE OAT TIE MOD". But four of these five words are fake and only one contains the information. The agent X also mailed a sentence as a clue - if I tell you any one character of the code word, you would be able to tell the number of vowels in the code word. Can you tell which is the code word?

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Solution:

The code word is TIE.

If you were told any one character of MOD, then you would not be able to determine whether the number of vowels are one or two. e.g. if you were told M, there are two words with M - AIM with 2 vowels and MOD with 1 vowel. So you would not be able to say the number of vowels. Same arguments can be given for characters O and D. Hence, the word with any one of M, O or D is not a code word i.e. AIM, DUE, OAT and MOD are not the code word. Thus, TIE is the code word.
T : two words - TIE and OAT, both with 2 vowels
I : two words - TIE and AIM, both with 2 vowels
E : two words - TIE and DUE, both with 2 vowels.

Boys and Girls

Puzzle:

Eleven boys and girls wait to take their seats in the same row in a movie theater. There are exactly 11 seats in the row. They decided that after the first person sits down, the next person has to sit next to the first. The third sits next to one of the first two and so on until all eleven are seated. In other words, no person can take a seat that separates him/her from at least one other person. How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.

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Solution:

There are 1024 different ways.

This is the type of Brain Teaser that can be solved using the method of induction. If there is just a one person and one seat, that person has only one option. If there are two persons and two seats, it can be accomplished in 2 different ways. If there are three persons and three seats, it can be accomplished in 4 different ways. Remember that no person can take a seat that separates him/her from at least one other person. Similarly, four persons and four seats produce 8 different ways. And five persons with five seats produce 16 different ways. It can be seen that with each additional person and seat, the different ways increase by the power of two. For six persons with six seats, there are 32 different ways. For any number N, the different possible ways are 2(N-1)

Thus, for 11 persons and 11 seats, total different ways are 210 i.e. 1024

Five Students

Puzzle:

Five students - Akash, Chintan, Jignesh, Mukund and Venky - appeared for an exam. There were total five questions - two multiple choice (a, b or c) and three true/false questions. They answered five questions each and answered as follow.

	I	II	III	IV	V
Chintan	c	b	True	True	False
Akash	c	c	True	True	True
Jignesh	a	c	False	True	True
Mukund	b	a	True	True	False
Venky	b	b	True	False	Tru

Also, no two students got the same number of correct answers. Can you tell which are the correct answers? What are their individual score?

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Solution:

The correct answers are b, a, True, False and False. Also, the scores are Jignesh (0), Akash (1), Chintan (2), Venky (3) and Mukund (4).

As no two students got the same number of correct answers, the total number of correct answers must be either 15 (1+2+3+4+5) or 10 (0+1+2+3+4).

Let's find out the maximum number of correct answers possible from the answers given by them.
For Question I = 2 (b or c)
For Question II = 2 (b or c)
For Question III = 4 (True)
For Question IV = 4 (True)
For Question V = 3 (True)

Thus, the maximum number of correct answers possible are 15 (2+2+4+4+3) which means that Akash would have given all correct answers as only he answered True for questions III, IV and V. But then Chintan and Jignesh would have exactly 3 correct answers. And also, Mukund and Venky would have 2 correct answers. So no one got all five correct. One can also arrive at this conclusion by trial-and-error, but that would be bit lengthy.

Now, it is clear that total number of correct answers are 10 (0+1+2+3+4). Questions III and IV both can not be False. If so, total number of correct answers would not be 10. So the student who got all wrong can not be Chintan, Akash and Mukund.

If Venky got all wrong, then Chintan, Jignesh and Mukund each would have atleast 2 correct answers. It means that Akash would have to be the student with only one correct answer and the correct answers for questions I and II would be a and a respectively. But then the total number of correct answers would be 1 (a) + 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus, Jignesh is the student with all wrong answers. The correct answers are b, a, True, False and False. Also, the scores are Jignesh (0), Akash (1), Chintan (2), Venky (3) and Mukund (4).

Counter

Puzzle:

How long would it take you to count 1 billion orally if you could count 200 every minute and were given a day off every four years? Assume that you start counting on 1 January 2001.

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Solution:

9 Years, 187 Days, 5 Hours, 20 minutes

As you can count 200 per minute, to count 1 billion you require
= 1,000,000,000/200 minutes
= 5,000,000 minutes
= 83,333.3333 hours
= 3,472.2222 days
= 9.512937 years
= 9 Years, 187 Days, 5 Hours, 20 minutes

The Multiplication

Puzzle:

In the following multiplication, certain digits have been replaced with asterisks (*). Replace all the asterisks such that the problem holds the result.

.......* * 7
X.....3 * *
.....---------
......* 0 * 3
......* 1 *
...* 5 *
...------------
...* 7 * * 3

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Solution:

.......1 1 7
X.....3 1 9
.....---------
......1 0 5 3
......1 1 7
...3 5 1
...------------
...3 7 3 2 3

The Flowers

Puzzle:

In a small town, there are three temples in a row and a well in front of each temple. A pilgrim came to the town with certain number of flowers.

Before entering the first temple, he washed all the flowers he had with the water of well. To his surprise, flowers doubled. He offered few flowers to the God in the first temple and moved to the second temple. Here also, before entering the temple he washed the remaining flowers with the water of well. And again his flowers doubled. He offered few flowers to the God in second temple and moved to the third temple. Here also, his flowers doubled after washing them with water. He offered few flowers to the God in third temple.

There were no flowers left when pilgrim came out of third temple and he offered same number of flowers to the God in all three temples. What is the minimum number of flowers the pilgrim had initially? How many flower did he offer to each God?

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Solution:

The pilgrim had 7 flowers, initially and he offered 8 flowers to each God.

Assume that the pilgrim had X flowers initially and he offered Y flowers to each God. From the data in puzzle, there are (8X - 7Y) flowers when the pilgrim came out of the third temple. But it is given that there were no flowers left when he came out of third temple. It means that

(8X - 7Y) = 0 or 8X = 7Y

The minimum values of X and Y are 7 and 8 respectively to satisfy above equation. Hence, the pilgrim had 7 flowers and he offered 8 flowers to each God.

In general, the pilgrim had 7N flowers initially and he offered 8N flowers to each God, where N = 1, 2, 3, 4,

The cigars

Puzzle:

Because cigars cannot be entirely smoked, a Bobo who collects cigar butts can make a cigar to smoke out of every 3 butts that he finds. Today, he has collected 27 cigar butts. How many cigars will he be able to smoke?

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Solution:

He will be able to smoke 13 cigars.

He makes 9 originals from the 27 butts he found, and after he smokes them he has 9 butts left for another 3 cigars. And then he has 3 butts for another cigar. So 9+3+1=13

Construction

Puzzle:

A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives. He paid five rupees per day to each man, four rupees to each woman and one rupee to each child. He gave out 200 rupees every evening. How many men, women and children were working with the constructor?

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Solution:

There were 16 men, 12 women and 72 children working with the constructor.

Let's assume that there were X men, Y women and Z children working with the constructor. Hence,
X + Y + Z = 100
5X + 4Y + Z = 200

Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z

As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get

if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428

if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727

But Z must be an integer, hence Z=72. Also, X=16 and Y=12

100 Bulbs

Puzzle:

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, …). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, …). This continues until all 100 people have passed through the room.

What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

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Solution:

Light Bulb 64 is on. The total number of bulbs which are on including #64 is 10.

First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This will result in the bulb being back at it's original state.

So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers.

There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.

100 Factorial

Puzzle:

How many consecutive zeros are there at the end of 100! (100 factorial) ?

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Solution:

There are 24 zeros on the end of 100!

One solution would be to work this out and count the zeros, well I've done this so you don't have to. The answer is...

9332621544394415268169923885626670049071596826438162146859296389521759999322991560894146397615651828625369792082722375825118520916864000000000000000000000000

Well now you can clearly just count the zeros but actually working out the number is not practical so we need another plan.

The clever bit here is thinking what numbers when multiplied together will end in a zero.
So the product of what numbers when multiplied ends in a zero:
1. When one of the things being multiplied ends in zero itself
2. A number ending in 5 multiplied by an even number

3. 25, 50 and 75 when multiplied by some of the small numbers available eg (4, 2 and 6) generate an extra zero

Below is tabulated the origin of all the zeros:

Number	Zeros	Number	Zeros
100	2	95	1
90	1	85	1
80	1	75	2
70	1	65	1
60	1	55	1
50	2	45	1
40	1	35	1
30	1	25	2
20	1	15	1
10	1	5	1
Total	12	Total	12

So that's it then there are 24 zeros at the end of 100!

The Apple Vendor

Puzzle:

An apple vendor has 1000 apples and 10 empty boxes. He asks his son to place all the 1000 apples in all the 10 boxes in such a manner that if he asks for any number of apples from 1 to 1000, his son should be able to pick them in terms of boxes. How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box.

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Solution:

1, 2, 4, 8, 16, 32, 64, 128, 256, 489

Let's start from scratch.
• The apple vendor can ask for only 1 apple, so one box must contain 1 apple.
• He can ask for 2 apples, so one box must contain 2 apples.
He can ask for 3 apples, in that case box one and box two will add up to 3.
• He can ask for 4 apples, so one box i.e. third box must contain 4 apples.
• Now using box number one, two and three containing 1, 2 and 4 apples respectively, his son can give upto 7 apples. Hence, forth box must contain 8 apples.
• Similarly, using first four boxes containing 1, 2, 4 and 8 apples, his son can give upto 15 apples. Hence fifth box must contain 16 apples.

You must have noticed one thing till now that each box till now contains power of 2 apples. Hence the answer is 1, 2, 4, 8, 16, 32, 64, 128, 256, 489. This is true for any number of apples, here in our case only upto 1000.

Mathematicians

Puzzle:

There are 4 mathematicians - Brahma, Sachin, Prashant and Nakul - having lunch in a hotel. Suddenly, Brahma thinks of 2 integer numbers greater than 1 and says, "The sum of the numbers is..." and he whispers the sum to Sachin. Then he says, "The product of the numbers is..." and he whispers the product to Prashant. After that following conversation takes place :

Sachin : Prashant, I don't think that we know the numbers.
Prashant : Aha!, now I know the numbers.
Sachin : Oh, now I also know the numbers.
Nakul : Now, I also know the numbers.

What are the numbers?

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Solution:

The numbers are 4 and 13.

As Sachin is initially confident that they (i.e. he and Prashant) don't know the numbers, we can conclude that -
1) The sum must not be expressible as sum of two primes, otherwise Sachin could not have been sure in advance that Prashant did not know the numbers.
2) The product cannot be less than 12, otherwise there would only be one choice and Prashant would have figured that out also.

Such possible sum are - 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, 101, 107, 113, 117, 119, 121, 123, 125, 127, 131, 135, 137, 143, 145, 147, 149, 155, 157, 161, 163, 167, 171, 173, 177, 179, 185, 187, 189, 191, 197, ....

Let's examine them one by one.

If the sum of two numbers is 11, Sachin will think that the numbers would be (2,9), (3,8), (4,7) or (5,6).

Sachin : "As 11 is not expressible as sum of two primes, Prashant can't know the numbers."

Here, the product would be 18(2*9), 24(3*8), 28(4*7) or 30(5*6). In all the cases except for product 30, Prashant would know the numbers.

- if product of two numbers is 18:
Prashant : "Since the product is 18, the sum could be either 11(2,9) or 9(3,6). But if the sum was 9, Sachin would have deduced that I might know the numbers as (2,7) is the possible prime numbers pair. Hence, the numbers must be 2 and 9." (OR in otherwords, 9 is not in the Possible Sum List)

- if product of two numbers is 24:
Prashant : "Since the product is 24, the sum could be either 14(2,12), 11(3,8) or 10(4,6). But 14 and 10 are not in the Possible Sum List. Hence, the numbers must be 3 and 8."

- if product of two numbers is 28:
Prashant : "Since the product is 28, the sum could be either 16(2,14) or 11(4,7). But 16 is not in the Possible Sum List. Hence, the numbers must be 4 and 7."

- if product of two numbers is 30:
Prashant : "Since the product is 30, the sum could be either 17(2,15), 13(3,10) or 11(5,6). But 13 is not in the Possible Sum List. Hence, the numbers must be either (2,15) or (5,6)." Here, Prashant won't be sure of the numbers.

Hence, Prashant will be sure of the numbers if product is either 18, 24 or 28.

Sachin : "Since Prashant knows the numbers, they must be either (3,8), (4,7) or (5,6)." But he won't be sure. Hence, the sum is not 11.

The Fly

Puzzle:

A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine, reverses itself immediately, and flies back to the other engine, repeating the process each time. The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?

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Solution:

Initially, the trains were 240 km apart.

The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started. It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of 60 km/h. So, each train traveled 120 km before they met. Hence, the trains were 240 km apart initially.

The color of the horse

Puzzle:

Pinto says, "The horse is not Black."
Sandy says, "The horse is either Brown or Grey."
Andy says, "The horse is Brown."

At least one is telling truth and at least one is lying.

Can you tell the color of the horse?

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Solution:

The color of the horse can be any color other than Black and Brown.

If the color of the horse is Black - all are lying. If the color of the horse is Brown - all are telling truth. Thus, the horse is neither Black nor Brown. If the color of the horse is Grey - Pinto and Sandy are telling truth whereas Andy is lying. If the color of the horse is other than Black, Brown and Grey - Pinto is telling truth whereas Sandy and Andy are lying. You must have noticed that for the given conditions, Pinto is always telling truth whereas Andy is always lying.

Five bales

Puzzle:

Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121.

How much does each bale weigh?

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Solution:

They weigh 54, 56, 58, 59, 62 pounds.

Let's assume that the weight of five bales are B1, B2, B3, B4 and B5 pounds respectively. Also, B1 <= B2 <= B3 <= B4 <= B5

It is given that five bales of hay are weighed two at a time in all possible ways. It means that each of the bale is weighted four times.
Thus,
4*(B1 + B2 + B3 + B4 + B5) = (110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121)
4*(B1 + B2 + B3 + B4 + B5) = 1156
(B1 + B2 + B3 + B4 + B5) = 289 pounds

Now, B1 and B2 must add to 110 as they are the lightest one.
B1 + B2 = 110

Similarly, B4 and B5 must add to 121 as they are the heaviest one.
B4 + B5 = 121

From above three equation, we get B3 = 58 pounds

Also, it is obvious that B1 and B3 will add to 112 - the next possible higher value. Similarly, B3 and B5 will add to 120 - the next possible lower value.
B1 + B3 = 112
B3 + B5 = 120

Substituting B3 = 58, we get B1 = 54 and B5 = 62
From 2 & 3 equations, we get B2 = 56 and B4 = 59

Hence, the weight of five bales are 54, 56, 58, 59 and 62 pounds.

Replace Each Letter

Puzzle:

Replace each letter by a digit. Each letter must be represented by the same digit and no beginning letter of a word can be 0.

O N E
O N E
O N E
+ O N E
-------
T E N

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Solution:

Use trial and error. 0 =1, N = 8 ,E = 2, T = 7

1 8 2
1 8 2
1 8 2
+ 1 8 2
------
7 2 8

Find Next Three Numbers

Puzzle:

Find next three numbers in the given series...

1 2 3 2 1 2 3 4 2 1 2 3 4 3 2 ? ? ?

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Solution:

The next three numbers are 3, 4 and 5.

The pattern is - the number of letters in the Roman numeral representation of the numbers i.e. number of letters in I, II, III, IV, V, VI, VII, VIII, IX, X, XI, XII, XIII, XIV, XV, ..... Hence, the next numbers in the given series are 3(XVI), 4(XVII), 5(XVIII), 3(XIX), 2(XX), 3(XXI), 4(XXII), 5(XXIII), 4(XXIV), 3(XXV), etc...

Seating Arrangement

Puzzle:

Mrs. F has invited several wives of delegates to the United Nations for an informal luncheon. She plans to seat her 9 guests ina row such that each lady will be able to converse with the person directly to her left and right. She has prepared the following list.

Mrs. F speaks English only.
Mrs. G speaks English and French.
Mrs. H speaks English and Russian.
Mrs. J speaks Russian only.
Mrs. K speaks English only.
Mrs. L speaks French only.
Mrs. M speaks French and German.
Mrs. N speaks English and German.
Mrs. O speaks English only.

How many distinct seating arrangements are possible? Give all possible seating arrangements.

Note that ABCD and DCBA are the same.

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Solution:

126 distinct seating arrangements are possible.

Mrs. J and Mrs. H must be together and Mrs. J must be at the end as Mrs. J speaks only Russian and Mrs. H is the only other Russian speaker.

Mrs. L speaks only French and there are two others - Mrs. G and Mrs. M - who speak French. Here there are 2 cases.
• CASE A : Mrs. L is at the other end
If Mrs. L is at the other end, either Mrs. G or Mrs. M must seat next to her.
o CASE AA : Mrs. G seats next to Mrs. L
Then, Mrs. M must seat next to Mrs. G and Mrs. N must seat next to Mrs. M. This is because Mrs. M speaks French and German, and Mrs. N is the only other German speaker. Thus, the possible seating arrangement is JHxxxNMGL, where x is the English speakers. Mrs. F, Mrs. K and Mrs. O can be arranged in remaining 3 positions in 3! different ways i.e. 6 ways.
o CASE AB : Mrs. M seats next to Mrs. L
If so, then either Mrs. N or Mrs. G must seat next to Mrs. M
? CASE ABA : Mrs. N seats next to Mrs. M
Thus, the possible seating arrangement is JHxxxxNML, where x is the English speakers. Mrs. F, Mrs. G, Mrs. K and Mrs. O can be arranged in remaining 4 positions in 4! different ways i.e. 24 ways.
? CASE ABB : Mrs. G seats next to Mrs. M
Thus, the possible seating arrangement is JHxxxxGML, where x is the English speakers. Mrs. F, Mrs. K, Mrs. N and Mrs. O can be arranged in remaining 4 positions in 4! different ways i.e. 24 ways.

• CASE B : Mrs. L does not seat at the end
It means that Mrs. G, Mrs. L and Mrs. M must seat together. Also, Mrs. L must seat between Mrs. G and Mrs. M.

o CASE BA : Mrs. G seats left and Mrs. M seats right to Mrs. L i.e. GLM

? CASE BAA : GLM is at the other end
Thus, the possible seating arrangement is JHxxxxGLM, where x is the English speakers. Mrs. F, Mrs. K, Mrs. N and Mrs. O can be arranged in remaining 4 positions in 4! different ways i.e. 24 ways.
? CASE BAB : GLM is not at the other end
Then Mrs. N must seat next to Mrs. M. Now, we have a group of four GLMN where Mrs. G and Mrs. N speak English. Thus, the possible seating arrangement is JHxxxX, where x is the individual English speakers and X is the group of four females with English speakers at the both ends. Thus, there are 4! different ways i.e. 24 ways.

o CASE BB : Mrs. M seats left and Mrs. G seats right to Mrs. L i.e. MLG
Then, Mrs. N must seat next to Mrs. M. Now, we have a group of four NMLG where Mrs. G and Mrs. N speak English. Thus, the possible seating arrangement is JHxxxX, where x is the individual English speakers and X is the group of four females with English speakers at the both ends. Thus, there are 4! different ways i.e. 24 ways.
Thus, total different possible seating arrangements are :
= 6 (case AA) + 24 (case ABA) + 24 (case ABB) + 24 (case BAA) + 24 (case BAB) + 24 (case BB)
= 126 seating arrangements

Thus, 126 distinct seating arrangements are poosible.

N secret agents

Puzzle:

There are N secret agents each know a different piece of secret information. They can telephone each other and exchange all the information they know. After the telephone call, they both know anything that either of them knew before the call. What are the minimum number of telephone calls needed so that all of the them know everything?

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Solution:

(2N - 3) telephone calls, for N = 2,3
(2N - 4) telephone calls, for N > 3

Divide the N secret agents into two groups. If N is odd, one group will contain one extra agent.

Consider first group: agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. Similarly in second group, agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. After (N - 2) calls, two agents in each the group will know anything that anyone knew in his group, say they are Y1 & Y2 from group 1 and Z1 & Z2 from group 2.

Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in next two calls total of 4 agents will know everything.

Now (N - 4) telephone calls are required for remaining (N - 4) secret agents.

Total telephone calls require are
= (N - 2) + 2 + (N - 4)
= 2N - 4

Travel

Puzzle:

A person travels on a cycle from home to church on a straight road with wind against him. He took 4 hours to reach there. On the way back to the home, he took 3 hours to reach as wind was in the same direction. If there is no wind, how much time does he take to travel from home to church?

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Solution:

3 hours 25 minutes 42 seconds

Let distance between home and church is D. A person took 4 hours to reach church. So speed while travelling towards church is D/4. Similarly, he took 3 hours to reach home. So speed while coming back is D/3. There is a speed difference of 7*D/12, which is the wind helping person in 1 direction, & slowing him in the other direction. Average the 2 speeds, & you have the speed that person can travel in no wind, which is 7*D/24. Hence, person will take D / (7*D/24) hours to travel distance D which is 24/7 hours.

Age

Puzzle:

Sam and Mala have a conversation.

• Sam says I am certainly not over 40
• Mala says I am 38 and you are atleast 5 years older than me
• Now Sam says you are atleast 39

All the statements by the two are false. How old are they really?

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Solution:

Sam is 41 and Mala is 37.

Let's invert the teaser and read it like this :
• Sam says I am certainly over 40
• Mala says I am not 38 and you are atmost 4 years older than me
• Now Sam says you are atmost 38
From first statement it is clear that Sam is over 40. Also, from next 2 statements it is clear that Mala is less then 38. Hence the possibilities are :
Sam = 41, 42, 43, 44, 45, ......
Mala = 37, 36, 35, 34, 33, ......

It also says that the difference between their age is maximum 4 years. Hence, there is only one possible pair i.e. 41 and 37, all other combination have differences more then 4.

Hence the answer - Sam is 41 and Mala is 37.

Robbery

Puzzle:

After a local shop robbery, four suspects were being interviewed. Below is a summary of their statements. Police know that each of them told the truth in one of the statements and lied in the other. From this information can you tell who committed the crime?

Anil said:
It wasn't Deepak
It wasn't Bhaskar
Bhaskar said:
It wasn't Charan
It was Deepak
Charan said:
It was Anil
It wasn't Depak
Deepak said:
It was Charan
It wasn't Anil

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Solution:

Bhaskar robbed the store.

The original number

Puzzle:

I bought a car with a peculiar 5 digit numbered licence plate which on reversing could still be read. On reversing, its value is increased by 78633. Whats the original number if all digits are different?

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Solution:

Only 0 1 6 8 and 9 can be read upside down. So on rearranging these digits we get the answer as 10968.

A Farmer

Puzzle:

A farmer needs 8 gallons of water. He has only three unmared buckets, two 6 gallon and one 11 gallon bucket. How can he collect 8 gallons of water using three unmarked buckets? Provide solution with minimal water wastage.

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Solution:

Here is the solution with 10 gallon water wastage.
OPERATIONS 6 6 11
Fill 6 gallon bucket with water 6 0 0
Empty 6 gallon bucket into 11 gallon bucket 0 0 6
Fill 6 gallon bucket with water 6 0 6
Fill 11 gallon bucket to full using filled 6 gallon bucket. This will leave 1 gallon water in 6 gallon bucket 1 0 11
Empty 11 gallon bucket into second 6 gallon bucket. 1 6 5
Empty 11 gallon bucket - wastage of 5 gallon water 1 6 0
Empty second 6 gallon bucket into 11 gallon bucket 1 0 6
Fill seccond 6 gallon bucket with water 1 6 6
Fill 11 gallon bucket to full using filled second 6 gallon bucket. This will leave 1 gallon water in second 6 gallon bucket 1 1 11
Fill first 6 gallon bucket with 1 gallon water which is in second 6 gallon bucket 2 0 11
Empty 11 gallon bucket into second 6 gallon bucket. 2 6 5
Empty 11 gallon bucket - wastage of 5 gallon water 2 6 0
Fill 11 gallon bucket with water in both the 6 gallon buckets 0 0 11

Movie

Puzzle:

Last Saturday Milan went for the late night show and came late. In the morning family members asked him which movie did he see. He gave different answers to everyone.

• He told to his father that he had gone to see MONEY.
• According to his mom, he saw either JOHNY or BABLU.
• His elder brother came to know that he saw BHABI.
• To his sister, he told ROBOT.
• And his grandpa heard that he saw BUNNY.
  

Thus, Milan gave six movie names, all five letter words. But he saw some other movie with five letter word. Moreover, each of the six movie names mentioned above has exactly two letters common with the movie he saw. (with the same positions)

Can you tell which movie did Milan see?

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Solution:

Milan saw BOBBY.

The six movie names are - MONEY, JOHNY, BABLU, BHABI, ROBOT and BUNNY.

Compare MONEY and JOHNY. They have O common at the second place and Y common at the fifth place. Also, they can't have two different letters each, common with the required movie as the letters in remaining three places are all different. Thus, the required movie must have either O at the second place or Y at the fifth place or both.

Similarly, comparing JOHNY and BUNNY - the required movie must have either N at the fourth place or Y at the fifth place or both. Also, comparing MONEY and BUNNY - the required movie must have either N at the third place or Y at the fifth place or both.

From the above 3 deduction, either Y is at fifth place or O is at the second place and N is at the third & fourth place. The later combination is not possible as BABLU, BHABI & ROBOT will need at least 3 other letters which makes the required movie 6 letter long. Hence, the required movie must have Y at the fifth place.

Now Y is not there in BABLU and BHABI at the fifth place and they have only B common at the first place. Hence, B must be the first letter.

As B is at the first place and Y is at the fifth place and every movie has exactly 2 letters common with the required movie. From BUNNY, the required movie do not have U at the second place and N at the third and fourth place. Now looking at JOHNY and MONEY, they must have O common at the second place.

Using the same kind of arguments for BABLU, BHABI and ROBOT, we can conclude that Milan saw BOBBY.

Doctor & Engineer

Puzzle:

One of Mr. Bajaj, his wife, their son and Mr. Bajaj’s mother is an Engineer and another is a Doctor.

• If the Doctor is a male, then the Engineer is a male.
• If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.
• If the Engineer is a female, then she and the Doctor are blood relatives.
  

Can you tell who is the Doctor and the Engineer?

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Solution:

Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.

Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.

Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.

Now from 2, Mr. Bajaj's mother cannot be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.

Number Cycles

Puzzle:

Find a cycle of five 4-digit numbers such that the last 2 digits of each number are equal to the first 2 digits of the next number in the cycle. Each of the 5 numbers must be exactly one of the following types (with each of the 5 types being represented exactly once): Square, Cube, Triangular, Prime, Fibonacci. The solution is unique.

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Solution:

8117, 1728, 2850, 5041, 4181 (Prime, Cube, Triangular, Square, Fibonacci)

Six 9's make 100

Puzzle:

Can you just use only +, -, *, / to make six 9’s equal 100? Do you think this is too easy. There are at least 6 solutions.

9 ..9 ..9 ..9 ..9 ..9 = 100

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Solution:

9 * 9 + 9 + 9 + 9/9 = 100
99 + 99 / 99 = 100
(99 + 9/9) * 9 / 9 = 100
(99 + 9/9) + 9 - 9 = 100
(9 + 9/9) * (9 + 9/9) = 100
(999 - 99) / 9 = 100
99 / 9 * 9 + 9 / 9 = 100
(99 + 9/9) / (9 / 9) = 100

No Intersection

Puzzle:

Within the rectangle, draw a line to connect from circle 1 to the other circle 1, then draw a line from circle 2 to the other circle 2, and so on. There should not be any intersections.

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Solution:

Draw a line (line A) from 3 to 3 and draw another line (line B) from 4 to 4 first. Since there is no intersection for line A and B, there must be a gap between lines. Draw a line from 1 to 1 to go through the gap. Draw another line from 2 to 2 to go through the gap again.

Mixed Nuts

Puzzle:

A mixed packet of nuts contains 1 pound of walnuts and 2 pounds of Brazil nuts. It costs exactly $2. A packet containing 4 pounds of filberts and 1 pound of walnuts costs $3. And for only $1.50, you can buy a mixed packet of 3 pounds of almonds, 1 pound of walnuts, and 1 pound of filberts. How much should you pay for a mixture of 1 pound of each of the four kinds?

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Solution:

A mixture of 1 pound of each of the four kinds costs $2

It is not possible to work out the cost per pound of each of the different nuts. However, the cost of various combinations can be worked out to get one pound of each of the four kind of nuts.

5 Digit Number

Puzzle:

Can you find a five digit number which has no zeros no digit is repeated, where:

• The first digit is a prime number.
• The second digit is the fifth digit minus the first digit.
• The third digit is twice the first digit.
• The fourth digit is the third digit plus three.
• The fifth digit is the difference between the first digit and the fourth digit.
  

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Solution:

The five digit number is 23475.

Pharmacy

Puzzle:

The pharmacy has an old scale which has only two measuring weights 30 grams and 5 grams. How can you divide 300 grams of powder medicine into 3 groups, one 150 grams, one 100 and the other 50 grams. You can only weigh 3 times.

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Solution:

First, we use 300 grams of medicine and 30 gram weight to split 300 medicine into 165 and 135 grams.

Then we use 165 grams of medicine and 30 + 5 gram weights to separate 165 grams into 100 and 65 grams.

Then we further separate the 65 grams into 50 grams and 15 grams.

Finally, Combine 135 and 15 grams to 150 grams.

Plus & Minus

Puzzle:

Below is an equation that isn't correct yet. By adding a number of plus signs and minus signs between the ciphers on the left side (without changes the order of the ciphers), the equation can be made correct.

123456789 = 100

How many different ways can you find that makes the equation correct?

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Solution:

There are 12 different ways:

123+45-67+8-9=100
123+4-5+67-89=100
123-45-67+89=100
123-4-5-6-7+8-9=100
12+3+4+5-6-7+89=100
12+3-4+5+67+8+9=100
12-3-4+5-6+7+89=100
1+23-4+56+7+8+9=100
1+23-4+5+6+78-9=100
1+2+34-5+67-8+9=100
1+2+3-4+5+6+78+9=100
-1+2-3+4+5+6+78+9=100

The Cucumber

Puzzle:

On a sunny morning, a greengrocer places 200 kilograms of cucumbers in cases in front of his shop. At that moment, the cucumbers are 99% water. In the afternoon, it turns out that it is the hottest day of the year, and as a result, the cucumbers dry out a little bit. At the end of the day, the greengrocer has not sold a single cucumber, and the cucumbers are only 98% water. How many kilograms of cucumbers has the greengrocer left at the end of the day?

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Solution:

The greengrocer has 100 kilograms of cucumbers left at the end of the day.

In the morning, the 200 kilograms of cucumbers are 99% water. So the non-water part of the cucumbers has a mass of 2 kilograms. At the end of the day, the cucumbers are 98% water. The remaining 2% is still the 2 kilograms of non-water material (which does not change when the water evaporates). If 2% equals 2 kilograms, then 100% equals 100 kilograms. So, the greengrocer has 100 kilograms of cucumbers left at the end of the day.

Sweets

Puzzle:

A kind old gentleman decided to give 12 sweets to each of the girls in his town and 8 sweets to each of the boys. Of the 612 children in his town, only half the girls and three quarters of the boys were allowed to take the sweets. How many sweets did the kind old gentleman have to buy?

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Solution:

3,672 Sweets

Escalator

Math Puzzle:

You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator. How many steps do you need if the escalator stands still?

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Solution:

100 Steps.

Let v be the speed of the escalator, in steps per second. Let L be the number of steps that you need to take when the escalator stands still. Upwards (along with the escalator), you walk 1 step per second. You need 50 steps, so that takes 50 seconds. This gives: L - 50 × v = 50. Downwards (against the direction of the escalator), you walk 5 steps per second. You need 125 steps, so that takes 25 seconds. This gives: L + 25 × v = 125. From the two equations follows: L = 100, v = 1. When the escalator stands still, you need 100 steps.