Solve One Puzzle A Day

Dating

2007-11-22T00:54:00.000+05:30

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

2007-11-21T00:15:00.000+05:30

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

2007-11-20T00:46:00.000+05:30

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

2007-11-19T01:52:00.000+05:30

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

2007-11-18T00:40:00.000+05:30

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

2007-11-17T00:14:00.000+05:30

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

2007-11-16T00:59:00.000+05:30

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

2007-11-15T00:40:00.000+05:30

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

2007-11-14T00:10:00.000+05:30

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

2007-11-13T01:02:00.000+05:30

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

2007-11-12T00:29:00.000+05:30

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

2007-11-11T00:13:00.000+05:30

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

2007-11-10T00:09:00.001+05:30

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

2007-11-09T01:36:00.000+05:30

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

2007-11-08T00:25:00.000+05:30

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

2007-11-07T00:21:00.000+05:30

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

2007-11-06T00:09:00.000+05:30

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

2007-11-05T00:41:00.000+05:30

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

2007-11-04T00:10:00.000+05:30

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

2007-11-03T00:55:00.000+05:30

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

2007-11-02T00:20:00.000+05:30

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

2007-11-01T00:19:00.000+05:30

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

2007-10-31T00:10:00.000+05:30

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

2007-10-30T00:42:00.000+05:30

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

2007-10-29T01:37:00.000+05:30

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

2007-10-28T00:34:00.000+05:30

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

2007-10-27T00:44:00.000+05:30

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

2007-10-26T00:23:00.000+05:30

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

2007-10-25T00:26:00.000+05:30

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

2007-10-24T00:24:00.000+05:30

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

2007-10-23T00:08:00.000+05:30

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

2007-10-22T00:07:00.000+05:30

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

2007-10-21T02:14:00.000+05:30

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

2007-10-20T00:29:00.000+05:30

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

2007-10-19T00:01:00.000+05:30

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

2007-10-18T07:29:00.000+05:30

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

2007-10-17T02:09:00.000+05:30

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

2007-10-16T00:23:00.000+05:30

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

2007-10-15T00:33:00.000+05:30

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

2007-10-14T00:27:00.000+05:30

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

2007-10-13T00:10:00.000+05:30

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

2007-10-12T00:18:00.000+05:30

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.

Secret Code

2007-10-11T00:46:00.000+05:30

Puzzle:

The only way to get a free drink in this bar, is to prove to the barman that you are a friend of his boss. Here's how its done. You go to the barman, he tells you a number and you tell him the "countersign" number. If your answer is correct you get your drink. There's this mathematician who thought that there can't be a standard list of numbers and countersign numbers. There must be an algorithm that provides the answer from the number given by the barman. So he joins a party of three men who ARE friends of the boss and walks with them to the bar hopping to understand the algorithm from the three answers. HERE'S WHAT HE HEARD: The barman says "6" to the first man, he says "3" and got his drink. The barman said "12" to the second man, who replied "6" and got his drink. The barman said "14" to the third man and he answered "8" (!) and got his drink. Now its the mathematicians turn. "20", said the barman. What is the counter sign?

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Solution:

Six

You count the number of letters in the word. Twenty has 6 letters so the mathematician should say "six".

Sack of Apples

2007-10-10T00:14:00.000+05:30

Puzzle:

A man had to pack a sack of apples into packets but as each packet had to have exactly the same number of apples he was having difficulty. If he packed 10 apples per packet, one packet only had 9. If he packed 9 apples per packet, one packet only had 8. If he packed 8 apples per packet, one packet only had 7. If he packed 7 apples per packet, one packet only had 6. And so on down to 2 apples. How many apples did he start with?

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Solution:

2519 apples.

The solution for the answer is the LCM (Lowest Common Multiple) of 10,9,8,7,6,5,4,3,2,1 minus 1. LCM would give the least number which is divisible by all of these number and subtracting one would give us the number of apples which were initially there.

Fruit Stall

2007-10-09T00:21:00.000+05:30

Puzzle:

At a fruit stall in Puzzle land, an Apple costs Rs.17, an Orange costs Rs.21, and a Pear costs Rs.14. What is the cost of a mango?

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Solution:

Cost of Mango is Rs.17

Each vowel cost Rs.4 and each consonants cost Rs.3

Positive Integer

2007-10-08T07:34:00.000+05:30

Puzzle:

Find a positive integer (base 10) which has the following four properties:

Each of the 10 digits (0 to 9) appears exactly once in the number.
For each pair of digits whose sum is 9, the number of other digits positioned strictly between the pair is equal to the smaller digit of the pair.
The sum of each pair of digits positioned at the same distance from opposite ends of the number is a prime number.
The difference between any 2 adjacent digits is greater than 1.

The solution is unique. Find the Number.

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Solution:

The number is 5831746290

Time

2007-10-07T01:44:00.000+05:30

Puzzle:

I was having trouble sleeping last night and I tossed and turned well into the night. Our local town hall has a clock which strikes on the hour and also strikes just once on the half hour. During one of my more awake moments I heard the clock strike once, but I could not tell what time it was. Half an hour later it struck once again, but I still could not tell what time it was. Finally, half an hour later it struck once again and I knew what the time was. What time was it?

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Solution:

1.30 in the morning. The initial single strike was at 12.30.

Find Next Number

2007-10-05T23:59:00.002+05:30

Puzzle:

Find next number in the series...

1/1...3/2 ...7/5...17/12...41/29...?

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Solution:

Next number in the series is 99/70

The successive term for a/b is formed as (a + 2b) / (a + b).

Birds and Trees

2007-10-05T00:41:00.000+05:30

Puzzle:

There is a forest with trees. On each and every tree in the forest, exactly the same number of birds are sitting on it (ex. "There are 5 trees and each tree has 5 birds on it"). The birds in the forest is a given number between 200 and 300. How many trees are in our forest(such that there is no other combination of trees and birds to acquire the same product)?

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Solution:

There are 289 birds and 17 trees in the forest (with 17 birds per tree).

We are looking for a number whose square is between 200 and 300. The candidates are 15, 16 and 17. It can't be 15, because 15*15 = 45*5 = 5*45, so again NO unique answer on the number of trees. Similarly it can't be 16, because 16*16 = 8*32 = 32*8. But 17*17 = 289 and there isn't any other pair of numbers whose product is 289. So, for 289 birds there can be only 17 trees in the forest (with 17 birds per tree) and this is the only unique answer available.

5 Consecutive Integers

2007-10-04T00:16:00.000+05:30

Puzzle:

In the equation shown below, A, B, C, D, and E are 5 consecutive positive integers. What are they?

A*A + B*B + C*C = D*D + E*E

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Solution:

A=10, B=11, C=12, D=13, E=14.
10x10 + 11x11 + 12x12 = 13x13 + 14x14

Gold in a Box

2007-10-03T00:25:00.000+05:30

Puzzle:

We have two closed boxes with a label on each one. Label A on box A: "The label on box B is true and the gold is in box A" Label B on box B: "The label on box A is false and the gold is in box A" We don't know if the labels (one or both of them) are true or false. Knowing FOR SURE that the gold is in box A or in box B, where's the gold?

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Solution:

The Gold is in box B.

Lets assume that label A is true. This means that label B is also true and the gold is in box A. In this case, the label A is false. This CAN'T be, because we assumed that the label A is true. Thus, our first assumption is invalid. So now we know that label A is lying. This leads us to two cases: (1) The label B is false. (2) The gold is in box B (now that we know that it cant be in box A). If (1) was true, then the gold is in box B or label A is true - which can't be because we know that label A is lying. So in BOTH CASES the gold IS IN BOX B

MEATS

2007-10-02T00:29:00.000+05:30

Puzzle:

The MEATS is the result of multiplication. Each letter stands for a distinct digit. What is the 'MEATS'?

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Solution:

MEATS, the result of multiplication is 39784

4973 x 8 = 39784

Bathtub

2007-10-01T01:13:00.000+05:30

Puzzle:

You want to fill a bathtub by running water from both the hot and cold taps. The hot tap takes 6 minutes to fill the tub alone and the cold tap takes 4 minutes to fill alone. So you turn them both on at the same time but you are unaware that you left the plug out. The plug empties the tub in 12 minutes. How long will it take to fill the tub?

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Solution:

3 minutes

Solve this by:
(1/6 + 1/4 - 1/12) ^ (-1)
= (1/3)^(-1) = 3

$10K Will

2007-09-30T02:24:00.000+05:30

Puzzle:

A man leaves $10,000 in his will, to which there are 6 beneficiaries - his three sons and their wives. The 3 wives receive $3,960 of which Jennifer gets $100 more than Rachel and Holly gets $100 more than Jennifer. Of the sons, James gets twice as much as his wife, Eric gets the same as his wife, and Sam gets 50% more than his wife. Who is married to whom?

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Solution:

Jennifer is married to Sam.
Rachel is married to Eric.
James is married to Holly.

Three Kids

2007-09-29T00:24:00.000+05:30

Puzzle:

There are three kids and each one have some pencils. The first kid gave the second and third as many pencils as they each already had. Then, the second gave the first and third as many as they each had. Still later, the third gave the first and second as many as they each already had. Each kid now have 24 pencils. How many pencils did each kid have originally?

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Solution:

You can solve the problem easily backward

First+Second+Third

24 + 24 + 24 = 72

12 + 12 + 48 = 72

6 + 42 + 24 = 72

39 + 21 + 12 = 72

The first had 39 pencils, the second had 21 and the third had 12.

Clock in the Mirror

2007-09-28T00:21:00.000+05:30

Puzzle:

Today, in the morning, while going to college, I looked at the clock in the mirror. The clock has no number indication and I made a mistake in interpreting the time (mirror-image). Assuming that the clock must be out of order, I went to college, where I arrive after twenty minutes. At that moment the clock at college shows a time that is two and a half hours later than the time that I saw on the clock (mirror-image) at home. At what time did I reach college?

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Solution:

I reached college at twenty-five minutes past seven.

The difference between the real time and the time of the mirror image is two hours and ten minutes (two and a half hours, minus the twenty minutes to reach college). Therefore, the original time on the clock at home that morning could only have been five minutes past seven (4:55 in the mirror). I reaches college at five minutes past seven plus twenty minutes, which is twenty-five minutes past seven!...

Guessing Contest

2007-09-27T00:10:00.000+05:30

Puzzle:

In a contest, five fruits (an apple, a banana, an orange, a strawberry and a pear) have been placed in five closed boxes (one fruit per box). People may guess which fruit is in which box. 107 people participate in the contest. When the boxes are opened, it turns out that 40 people have guessed none of the fruits correctly, 30 people have guessed one fruit correctly, 20 people have guessed two fruits correctly and 10 people have guessed three fruit correctly. How many people have guessed four fruits correctly and how many people have guessed all the fruits correctly?

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Solution:

It is not possible to guess only four fruits correctly: the fifth fruit is then correct too! So nobody has guessed four fruits correctly and 107-40-30-20-10 = 7 people have guessed five fruits correctly.

Hourglasses

2007-09-26T00:46:00.000+05:30

Puzzle:

There are two hourglasses. The small one can measure 5 minutes and the large one can measure 7 minutes. How can we measure 16 minutes with 2 hourglasses running together? (If one of the hourglasses finishes leaking, then it must be flipped over immediately to keep it running. )

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Solution:

Start both hourglasses at the same time. You flip over the small hourglass when it finishes the first 5 minutes. Then you flip over the large one when it finishes its first 7 minutes. When the small hourglass runs to 10 minutes, flip over the small and large hourglass at the same time. It means the large hourglass will run 3 minutes again. When the large hourglass run to 13 minutes (7 + 3 + 3), flip both the small and large hourglass over again to run another 3 minutes.

Digits 1 to 6

2007-09-25T00:05:00.000+05:30

Puzzle:

Take a look at the following multiplication problem. Replace all "?" with digits 1, 2, 3, 4, 5 and 6 (use one digit only once) to make it true.

? ?

x ?

___

? ? ?

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Solution:

54
x 3
-----
162
-----

Spider, Dragonfly & Cicada

2007-09-24T00:23:00.001+05:30

Puzzle:

A spider has 8 legs. A dragonfly has 6 legs and 2 pairs of wings. A cicada has 6 legs and one pair of wings. Now we have all 3 kinds and a total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of wings. How many insects do we have of each kind?

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Solution:

There are Five Spiders, Seven Dragonflies & Six Cicada.

Let X be the number of spiders, Y be number of Dragonflies and Z be number of cicada. Now from the puzzle,

X+Y+Z = 18
8X+6Y+6Z = 118
2Y+Z = 20
Solving this, X=5, Y=7, Z=6

A Barrel

2007-09-23T01:45:00.000+05:30

Puzzle:

There is a barrel with no lid and some wine in it.
"This barrel of wine is more than half full," said Kiran.
"No it's not," says Rahul. "It's less than half full."

Without any measuring implements and without removing any wine from the barrel, how can they easily determine who is correct?

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Solution:

Tilt the barrel until the wine barely touches the lip of the barrel. If the bottom of the barrel is visible then it is less than half full.If the barrel bottom is still completly coverd by the wine, then it is more than half full.

Marbles

2007-09-22T01:15:00.000+05:30

Puzzle:

In a box you have 13 white marbles and 15 black marbles. You also have 28 black marbles outside the box. Remove two marbles, randomly, from the box. If they are of different colors, put the white one back in the box. If they are the same color, take them out and put a black marble back in the box. Continue this until only one marble remains in the box. What color is the last marble?

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Solution:

The last marble will be WHITE.

Since marbles can only be taken out in pairs and you started off with an odd number of whites there is always going to be one white left over that you'll keep putting back in the box until it's left on it's own.

5 Digit Number

2007-09-21T01:38:00.000+05:30

Puzzle:

There is a 5 digit number. With a 1 after it, it is three times as large as it would be with a one before it. What is the 5 digit number?

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Solution:

The 5 digit number is: 42857

The number with a one after it is 10x + 1. With a 1 before it, it is 100,000 + x. Then 10x +1 = 3(100,000 + x), and x = 42,857. Therefor the 5 digit number is: 42857

Guests

2007-09-20T01:11:00.000+05:30

Puzzle:

A cook wanted to find out how many guests she cooked for. An official told her that for every dish of rice, there were 2 people. And every three people had a dish of broth. There was 1 dish of meat for every four people. There were 65 dishes in all. The cook then figured out how many guests there were. How many guests did the cook cook for?

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Solution:

60 guests

Let x be the number of guests. According to the puzzle, the equation for total number of dishes is x/2 + x/3 + x/4 = 65. Solving this, we get x=60.

Dog, Cow and Goat

2007-09-19T00:09:00.000+05:30

Puzzle:

A Dog, a Cow and a Goat were stolen. Three suspects got caught: Raghu, Kiran and Sam. Each person stole one animal, but we do not know who stole which. Here are the investigation statements.

Raghu said: Sam stole the Dog.
Kiran said: Sam stole the Cow.
Sam said: They both were lying. I did not steal the Dog or the Cow.

Later on, the police found out the man who stole the goat told a lie. The man who stole the dog told the truth. Can you find out who stole which?

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Solution:

Sam stole the Cow
Raghu stole the Goat
Kiran stole the Dog

The man who stole the dog told the truth. All we need to do is to find out who was that.
If Raghu had stolen the dog he would not said: Sam stole the dog. He should said: I stole the dog or Sam did not steal the dog. If Sam had stolen the dog, he would not said: I did steal the dog.

So Kiran stole the dog. What he said was true: Sam stole the cow. Then Raghu stole the goat and told a lie.

A to I are 1 to 9

2007-09-18T00:38:00.000+05:30

Puzzle:

Look at the calculation below. Every alphabet represents a different digit from 1 to 9 to make the calculation. What are they?

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Solution:

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

Two Numbers

2007-09-17T00:12:00.000+05:30

Puzzle:

There are two numbers. One larger and the other is smaller. If we subtract half of the smaller number from both numbers then the resultant larger number is 3 times larger as large as the resultant with the orignianl smaller and its half. How many times is the original larger number bigger than the original smaller number?

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Solution:

The original larger number is two times bigger than the original smaller number.

Make a Square

2007-09-16T00:11:00.000+05:30

Puzzle:

Put together all the five pieces shown below to form a square. (Print the picture to solve it easily).

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Solution:

Walking Home

2007-09-15T00:34:00.000+05:30

Puzzle:

Every day, a wife picks up her husband at train station at 5. One day he arrives early at 4 and he begins to walk home along the road which his wife would be traveling. She meets him and takes him the rest of the way. If he would have waited at the train station, she would have arrived there at 5. As it turned out, they reached home 20 minutes early. What time was it when she picked him up?

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Solution:

It was 4:50

If they made it home 20 minutes early, then the trip was 20 minutes shorter. Then it was 10 minutes shorter going, and 10 minutes shorter coming back home. Since it was 10 minutes shorter going, then she picked him up at 10 'til 5:00, or 4:50.

Handshaking

2007-09-14T00:34:00.000+05:30

Puzzle:

Jack and his wife went to a party where four other married couples were present. Every person shook hands with everyone he or she was not acquainted with. When the handshaking was over, Jack asked everyone, including his own wife, how many hands they shook. To his surprise, Jack got nine different answers. How many hands did Jack's wife shake ?

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Solution:

Jack's wife shook 4 hands.

Because, obviously, no person shook hands with his or her partner, nobody shook hands with more than eight other people. And since nine people shook hands with different numbers of people, these numbers must be 0, 1, 2, 3, 4, 5, 6, 7, and 8. The person who shook 8 hands only did not shake hands with his or her partner, and must therefore be married to the person who shook 0 hands. The person who shook 7 hands, shook hands with all people who also shook hands with the person who shook 8 hands (so in total at least 2 handshakes per person), except for his or her partner. So this person must be married to the person who shook 1 hand. The person who shook 6 hands, shook hands with all people who also shook hands with the persons who shook 8 and 7 hands (so in total at least 3 handshakes per person), except for his or her partner. So this person must be married to the person who shook 2 hands. The person who shook 5 hands, shook hands with all people who also shook hands with the persons who shook 8, 7, and 6 hands (so in total at least 4 handshakes per person), except for his or her partner. So this person must be married to the person who shook 3 hands. The only person left, is the one who shook 4 hands, and which must be Jack's wife. The answer is: Jack's wife shook 4 hands.

The Circus

2007-09-13T00:12:00.000+05:30

Puzzle:

A circus performance is witnessed by 120 people who have paid a total of $120. The men pay $5, women pay $2, and children pay 10 cents each. How many of each went to the circus?

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Solution:

17 men, 13 women, 90 children went to the circus.

Snail in the well

2007-09-12T00:10:00.000+05:30

Puzzle:

A snail is at the bottom of a well 30 feet deep. It crawls up 3 feet each day, but at night, it slips down 2 feet. How long does it take for the snail to crawl out the well?

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Solution:

28 days. Once it reaches the top, it does not slip down 2 feet.

On a Train

2007-09-11T00:07:00.000+05:30

Puzzles:

On a train, Smith, Robinson, and Jones are the fireman, brakeman, and the engineer, but not respectively. Also aboard the train are three businessmen who have the same names: a Mr. Smith, a Mr. Robinson, and a Mr. Jones.

Mr. Robinson Lives in Detroit.
The brakeman lives exactly halfway between Chicago and Detroit
Mr. Jones earns exactly $20,000 per year.
The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
Smith beats the fireman in billiards.
The passenger whose name is the same as the brakeman's lives in Chicago.

Who is the Engineer?

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Solution:

Smith is the engineer.

Love Story

2007-09-10T00:21:00.000+05:30

Puzzle:

Four men and four women are shipwrecked on a desert island. Eventually each one falls in love with one another, and is himself loved by one person. John falls in love with a girl who is in love with Jim. Arthur loves a girl who loves the man who loves Ellen. Mary is loved by the man who is loved by the girl who is loved by Bruce. Gloria hates Bruce and is hated by the man whom Hazel loves. Who loves Arthur?

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Solution:

Gloria loves Arthur.

Find Next Number

2007-09-09T00:16:00.000+05:30

Puzzle:

Do you know numbers and letters? Now find the next number in the series.

5.. 15 ..5.. 18.. 5.. 24 ..14.. 20.. ?

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Solution:

Next number is 5.

Write the numbers in words. onE, twO, threE, fouR......
Look at the last letter in each word and convert it into position of letter in the alphabet (E=5, O=15, E=5, R=18.....). Following the pattern, the next word is ninE and position of E is 5 in alphabet.

Count Triangles

2007-09-08T00:29:00.000+05:30

Puzzle:

Take a look at the figure below. Can you count number of different triangles in this hexagonal pattern?

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Solution:

There are 76 triangles.

Box and Locks

2007-09-07T00:21:00.000+05:30

Puzzle:

You want to send a valuable object to a friend. You have a box to contain the object. The box has a locking ring which is more than large enough to have a lock attached. You and your friend have several locks with keys. But your friend does not have the key to any lock that you have, and vice versa. How do you do it? Note that you cannot send a key in an unlocked box, since it might be copied.

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Solution:

Attach a lock to the ring. Send it to her. She attaches her own lock and sends it back. You remove your lock and send it back to her. She removes her lock.

A Number

2007-09-06T01:29:00.000+05:30

Puzzle:

A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.

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Solution:

The number is 58.

LCM of 3, 4, 5 and 6 is 60. So these numbers when divides 60 gives remainder 0. Now to get the remainder 1, 2, 3 and 4 respectively, the number should be 2 (3-1 ; 4-2 ; 5-3 ; 6-4 ) short of 60.

Odd One Out

2007-09-05T00:30:00.000+05:30

Puzzle:

Which of the following formula is different from others?

100/11
100/22
100/33
100/44
100/55
100/66
100/77
100/88
100/99

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Solution:

The answer is 100/77

100 / 11 = 9.09090909090909090909090909090909
100 / 22 = 4.54545454545454545454545454545455
100 / 33 = 3.03030303030303030303030303030303
100 / 44 = 2.27272727272727272727272727272727
100 / 55 = 1.818181818181818181818181818181818
100 / 66 = 1.51515151515151515151515151515152
100 / 77 = 1.2987012987012987012987012987013
100 / 88 = 1.13636363636363636363636363636364
100 / 99 = 1.01010101010101010101010101010101

The result of 100/77 is odd to all other result.

Paddy Field

2007-09-04T00:19:00.000+05:30

Puzzle:

A farmer was getting old and so he decided to divide his paddy field between himself and his four sons in proportion to their work rates. He knew that Rahul, Anil, Sunil and Vineel together could plant a field in five hours whereas Anil, Sunil and Vineel and himself together could manage the same task in six hours. So the farmer divided his field into a two-digit square number of parts and kept just one part for himself. Now, Anil, Sunil and Vineel were identical triplets so each received the same whole number of parts. How many parts did each son get?

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Solution:

Ravi received 9 parts and the triplets each got 13 parts. Let the work rates of the triplets combined, Ravi and farmer be x, y, z, respectively. Then

5(x + y) = 6(x + z) and so y = (x + 6z)/5

Since z has one part, we can let x = nz where n is a whole number. We then have y = (6 + n)z/5. The ratio of the work rates of x, y, z is now

n : (6 + n)/5 : 1

Let the total of parts be m, a two-digit square number. We then have n + (6 + n)/5 + 1 =m so that n = (5m - 11)/6. The only solutions are n=19, m=25 or n=39, m=49. Since n is the total number of parts the triplets receive, only the value divisible by three applies (n=39).

Make "+" Sign

2007-09-03T00:15:00.000+05:30

Puzzle:

Look at the six pieces below. Combine these pieces such that they form a symmetrical plus sign.

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Solution:

Departmental Store

2007-09-02T01:00:00.000+05:30

Puzzle:

I went to the departmental store and spent one half of the money that was in my purse. When I came out, I found that I had as many paise as I had rupees and half as many rupees as I had paise when I went in. Find the money in my purse when I entered the store.

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Solution:

I have Rs. 99.98 in my purse when I entered the store.

Ten True Sentences

2007-09-01T00:13:00.000+05:30

Puzzle:

Take a look at the following sentences:

The number of times the digit 0 appears in this puzzle is _____.
The number of times the digit 1 appears in this puzzle is _____.
The number of times the digit 2 appears in this puzzle is _____.
The number of times the digit 3 appears in this puzzle is _____.
The number of times the digit 4 appears in this puzzle is _____.
The number of times the digit 5 appears in this puzzle is _____.
The number of times the digit 6 appears in this puzzle is _____.
The number of times the digit 7 appears in this puzzle is _____.
The number of times the digit 8 appears in this puzzle is _____.
The number of times the digit 9 appears in this puzzle is _____.

Fill these sentences with digits such that all the sentences holds true.

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Solution:

The number of times the digit 0 appears in this puzzle is 1.
The number of times the digit 1 appears in this puzzle is 7.
The number of times the digit 2 appears in this puzzle is 3.
The number of times the digit 3 appears in this puzzle is 2.
The number of times the digit 4 appears in this puzzle is 1.
The number of times the digit 5 appears in this puzzle is 1.
The number of times the digit 6 appears in this puzzle is 1.
The number of times the digit 7 appears in this puzzle is 2.
The number of times the digit 8 appears in this puzzle is 1.
The number of times the digit 9 appears in this puzzle is 1.

Alternative Solution:
The number of times the digit 0 appears in this puzzle is 1.
The number of times the digit 1 appears in this puzzle is 11.
The number of times the digit 2 appears in this puzzle is 2.
The number of times the digit 3 appears in this puzzle is 1.
The number of times the digit 4 appears in this puzzle is 1.
The number of times the digit 5 appears in this puzzle is 1.
The number of times the digit 6 appears in this puzzle is 1.
The number of times the digit 7 appears in this puzzle is 1.
The number of times the digit 8 appears in this puzzle is 1.
The number of times the digit 9 appears in this puzzle is 1.

Find Next Number

2007-08-31T00:17:00.000+05:30

Puzzle:

Find the next number in the series.

1211
122111
312211
12221131
?

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Solution:

Answer is 1113213211.

Just reverse numbers 2nd and 4th (alternative) and look at that series.

1211
111221
312211
13112221

So after starting the sequence with 1211, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit. For example, first term has "one 1", "one 2" and "two 1s". So the second numbers becomes 111221. Now the 5th number is 1113213111. Since we reverse only even terms, the answer is 1113213211.

Grandfather's Age

2007-08-30T00:18:00.000+05:30

Puzzle:

My grandson is about as many days as my son in weeks, and my grandson is as many months as I am in years. My grandson, my son and I together are 120 years. Can you tell me my age in years?

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Solution:

I am 72 years old.

Let m be my age in years. If s is my son's age in years, then my son is 52s weeks old. If g is my grandson's age in years, then my grandson is 365g days old. Thus,

365g = 52s.
Since my grandson is 12g months old,
12g = m.
Since my grandson, my son and I together are 120 years,
g + s + m = 120.
The above system of 3 equations in 3 unknowns (g, s and m) can be solved as follows.
m / 12 + 365 m / (52 x 12) + m = 120 or
52 m + 365 m + 624 m = 624 x 120 or
m = 624 x 120 / 1041 = 72.
So, I am 72 years old.

Hourglasses

2007-08-29T00:49:00.000+05:30

Puzzle:

You have two hourglasses: a 7 minute one and an 11 minute one. Using just these hourglasses, how can you accurately measure time 15 minutes?

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Solution:

1. Start both the 7 minute hour glass & 11 minute hour glass.
2. Wait till the 7 minute hour glass times out. Time is 7 minute!

3. Restart the 7 minute hour glass. At this time 11 minute hour glass will have 4 minutes left to time out.

4. As soon as 11 minute glass times out invert the 7 minute hour glass. Total time now is 11 minutes.

5. After inverting 7 minute hour glass, it will now have 4 minutes left for time out.

6. After these 4 minutes times out, the total time is 15 minutes.

Self Referencing Number

2007-08-28T00:17:00.000+05:30

Puzzle:

ABCDEFGHI is a nine digit number. A is the count of number of 0's in the number, B is the number of 1's, C is number of 2's and so on. Find the number.

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Solution:

The self referencing nine digit number is 521001000.

The Six Doors

2007-08-27T00:22:00.000+05:30

Puzzle:

You are in a room that has six doors. You must pass through all six doors, each door once only, in the correct order. Three are entrances only and three are exits only. Door A must be followed by door B or E; D by A or F; B by C or E; C by D or F; E by B or D; and F by C or D. Also A, F are entrance and B is exit. In what order you must pass through the doors?

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Solution:

The order is CFDABE.

Doors A, F are entrance and B is exit. We must alternate exits and entrances. Let the unknown entrance be X. With a "_" denoting an exit door the possible orders are:
(a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A

Neither (a) nor (c) are possible since A is followed by B or E and neither leads to F. Since A can only be preceded by D, (e) and (f) are possible as DA_X_F and _F_XDA, respectively. With D used, door F can only be preceded by C to give DA_XCF and CF_XDA, respectively. Door B must occupy the remaining space in each possibility (third door is exit) so X=E but C cannot follow E in the former and for the latter, B cannot follow F. This eliminates (e) and (f). Since only D precedes A, (b) is possible as _FDA_X and (d) as _X_FDA. For (d), the exit B must occupy a "_" but cannot be followed by F, and if it is the first door, neither the C or E that must follow as X can be followed by the remaining E or C, respectively. This only leaves (b) and only C (of the possible B,C,E) can precede F as CFDA_X and B must occupy the final "_" as an exit leaving E=X. Notice that the condition that E is followed by B or D cannot be satisfied.

Find the Number

2007-08-26T00:57:00.000+05:30

Puzzle:

There is a five digit number ABCDE (A, B, C, D & E are 5 different single digit numbers). If you multiply this number with 4, you get reverse of the number. Find A, B, C, D & E respectively.

ABCDE * 4 = EDCBA

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Solution:

21978 * 4 = 87912

Look at the A in ABCDE. A can be only 1 or 2 because E can be 9 at the most. On the other hand, we look at the A in EDCBA. We found that A must be an even number, which should be 2. Since E x 4 produces a number within the one's place as 2, we know E must be 8. Since there is no carry from B x 4 to E's place, we know B must be 1. Since B = 1, we know D x 4 will produce a number within one's place as 8 so that 8 + 3 (carry) = 11. The only choice for D is 7. C can only be 8 or 9 to make the product of BC x 4 be close to DC. Since E is 8 already, C must be 9.

So ABCDE=21978

25 August 2007

2007-08-25T00:47:00.000+05:30

Puzzle:

At a fruit stall in Puzzle land, an Apple costs Rs.20, an Orange costs Rs.24, and a Pear costs Rs.16.

What is the cost of a mango?

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Solution:

Cost of mango is Rs.20.

1. Each letter in the word costs Rs.4 Since the word mango has 5 words, its cost is Rs20.

2. Assign A = 1; B = 2; C = 3.... and so on upto Z = 26. Now find the score of the fruits.
APPLE - 1+16+16+12+5= 50
PEAR - 16+5+1+18 =40
ORANGE - 15+18+1+14+7+5 = 60
Multiply the score with Rs.0.4 to get the cost of the fruit.
So mango - 13+1+14+7+15 = 50, 50*0.4=Rs.20

24 August 2007

2007-08-24T00:44:00.000+05:30

Puzzles:

There are five competitor A, B, C, D, and E and they enter a running race that awards gold, silver, and bronze medals. Each of the following compound statements about the race is false, although one of the two clauses in each may be true.

A didn't win the gold, and B didn't win the silver.
D didn't win the silver, and E didn't win the bronze.
C won a medal, and D didn't.
A won a medal, and C didn't.
D and E both won medals.

Who won each of the medals?

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Solution:

A won the Gold medal.
D won the Silver medal.
C won the Bronze medal.

23 August 2007

2007-08-23T00:35:00.000+05:30

Puzzle:

You have three boxes of fruit. First box contains just Apples, second contain just Oranges and the third box contains a mixture of both. Each box is labeled -- one says "Apples", one says "Oranges", and one says "Apples and Oranges." However, it is known that none of the boxes are labeled correctly. How can you label the boxes correctly if you are only allowed to take and look at just one piece of fruit from just one of the boxes?

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Solution:

Take a piece of fruit from the box marked "apples and oranges." Suppose the fruit you take is an apple. Then that box must be the box containing just apples. Therefore, the box marked "oranges" can't be the box containing just apples, and it can't be the box containing just oranges either -- so it must be the box containing apples and oranges. The remaining box is therefore the box containing just oranges.

If the fruit you take out is an orange, the solution is derived in a similar fashion: the box marked "apples and oranges" is the box containing just oranges; the box marked "apples" is the box containing both apples and oranges; and the box marked "oranges" is the one containing just apples.

22 August 2007

2007-08-22T01:34:00.000+05:30

Puzzle:

Find next number in the series.

6 ..25 .. 64 ..81 ..32 ..?

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Solution:

The answer is 1.
Well, you can write every number as a power of another number:
6 equals 6^1,
25 equals 5^2,
64 equals 4^3,
81 equals 3^4,
32 equals 2^5.

So the next number in the series is 1^6 that equals 1.

21 August 2007

2007-08-21T00:30:00.000+05:30

Puzzle:

Find the next letter in this sequence.

F S S E N T T T T T T T T T T T T T T T T T T T T ?

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Solution:

Answer is F. The sequence is the first letter of numbers starting at fifteen, as far as thirty-nine. The next letter would be F for forty.

20 August 2007

2007-08-20T00:36:00.000+05:30

Puzzle:

You have a 12 liter jug, an 8 liter jug, and a 5 liter jug. The 12 liter jug is full, and the other two are empty. How can you divide the 12 liters of water equally (i.e., so two of the jugs have exactly 6 liters of water in them, and the third is empty)?

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Solution:

Fill the 8 liter jug with the 12 liter jug, leaving 4 liters remaining. Fill the 5 liter jug with the 8 liter jug, leaving 3 liters remaining. Empty the 5 liter jug into the 12 liter jug. Now there are 9 liters in the 12 liter jug and 3 liters in the 8 liter jug. Pour the 3 liters from the 8 liter jug into the 5 liter jug. Now fill the 8 liter jug with water from the 12 liter jug, leaving 1 liter in the 12 liter jug. Fill the 5 liter jug (which already has 3 liters in it) from the 8 liter jug, leaving 6 liters in the 8 liter jug. Empty the 5 liter jug into the 12 liter jug. Now there are 6 liters in the 12 liter jug, 6 liters in the 8 liter jug, and the 5 liter jug is empty.

19 August 2007

2007-08-19T00:48:00.000+05:30

Puzzle:

Take six 6s and place any mathematical operator in the equation to make it equal to 100.

6 6 6 6 6 6 = 100

There are multiple ways of doing this. One of them is given below.

(666-66)/6 = 100

Find other ways.

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Solution:

1. ((6 + 6) / 6)^6 + 6 * 6 = 100

2. 6! / (6 * 6) * (6 - (6 / 6)) = 100

3. ((6!-(6!/6))/6)*(6/6) = 100

18 August 2007

2007-08-18T00:55:00.000+05:30

Puzzle:

1. Ravi is as old as Ram will be when Ravi is twice as old as Ram was when Ravi's age was half the sum of their present ages.

2. Ram is as old as Ravi was when Ram was half the age he will be 10 years from now.

How old are Ravi and Ram?

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Solution:

Ravi is 40 years old and Ram is 30 years old.

Let Ravi's age be a and let Ram's age be b.
The first sentence tells us that "Ravi is as old as Ram will be when Ravi is twice as old as Ram was when Ravi's age was half the sum of their present ages".
"Half the sum of their present ages" is: 0.5*(a+b)
So "Ravi's age was half the sum of their present ages", the following number of years ago: a - (0.5*(a+b)) = 0.5*a-0.5*b
At that moment, "Ram's age" was: b - (0.5*a-0.5*b) = 1.5*b-0.5*a.
So "twice that age" is: 3*b-a
"Ravi's age will be twice that age", in the following number of years: (3*b-a) - a = 3*b-2*a
At that moment, Ram will be: b + (3*b-2*a) = 4*b-2*a
And we are told that "that's Ravi's age", so: 3*a = 4*b.

The second sentence tells us that "Ram is as old as Ravi was when Ram was half the age he will be 10 years from now.".
"10 years from now", Ram will be: b+10
"Half that age" is: 0.5*b+5
"Ram was that age", the following number of years ago: b - (0.5*b+5) = 0.5*b-5
At that time, "Ravi was", the following number of years old: a - (0.5*b-5) = a-0.5*b+5
And that "is Ram's current age", so: b = a-0.5*b+5 or a = 1.5*b-5.

Now we have two equations:
3*a = 4*b and a = 1.5*b-5.
Solve these equations. So Ravi's age is 40, and Ram's age is 30.

17 August 2007

2007-08-17T00:37:00.000+05:30

Puzzle:

Take a look at the clock below. Break the clock into exactly five pieces such that sums of all the numbers on each piece are 8, 10, 12, 14 and 16.

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Solution:

1+2+...+11+12 = 78 and 8+10+12+14+16 = 60. So you probably should try to decrease the total on the clock. How could you decrease the total without throwing away something? Have a good look at the numbers 10, 11 and 12. If you break the clock between the 1 and 2 of twelve, you loose 12 in the total but you gain 1 and 2. So overall the total would decrease by 9. Breaking between the 1 and 1 of 11 and between the 1 and 0 of 10 also decreases the total by 9. So if you break between two of the numbers 10, 11 and 12, the total will become 78-9-9 = 60.

16 August 2007

2007-08-16T00:24:00.000+05:30

Puzzle:

You have 5 boxes and each box has 5 marbles in it. Marbles in 4 bags weight 10g each and in one bag weight 9g each. You don't know which bag has 9g marbles and you are given a chance to use a weight scale for one time. How can you find the bag with 9g marbles?

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Solution:

Number the bags from 1 through 5. Take 1 marble from bag 1, 2 marbles from bag 2, 3 marbles from bag 3, 4 marbles from bag 4 and 5 marbles from bag 5. Now put them on the weightscale. Now there are 5 possible measurements. If all marbles would weigh 10g, the total would be 150g. But one or more marbles weigh 9g. So if you took one 9g marble, the total would be 149g. If you took two 9g marbles, the total would be 148g. And so on.

Therefore, 150-(Measured Weight) gives the bag number which contain 9g marbles.
(Ex: If measured weight is 147g, 150-147=3 is the bag number as you have taken three 9g marbles from bag number 3)

15 August 2007

2007-08-15T00:32:00.000+05:30

I Wish To All Indians
A
Happy Independence Day

Puzzle:

Crack the code!
A message is hidden inside the code. Can you decode it?

L4A1Q10P1Q6.. P3M2A3Q3P1P8Z6L7Q3M2Z3P8.. L7A1P5

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Solution:

The message is "HAPPY INDEPENDENCE DAY".

The message can be easily decoded using a QWERTY keyboard. Each two characters (except Q10 for P and P10 for Q, which are three characters) represent a letter. The first character of a letter is always Q, A, Z, P, L or M, specifying which row on a QWERTY keyboard the letter is on. The number preceding the letter is the position of the letter on a QWERTY keyboard from left to right (in case of Q, A and Z) or from right to left (in case of P, L and M) along the row. For example, F can be represented as A4 or L6, E can be represented as Q3 or P8.