Infosys Puzzles

Brain Teaser No : 00765
Three men, including Gianni and three woman, including Sachi are in line at the BrentWood post office. Each has two different pieces of business to conduct.
1. The first person is a woman.
2. Carlos wants to send an overnight package.
3. Lau is just ahead of Pimentelli who is the same sex as Lau.
4. Gianni is two places ahead of the person who wants to buy stamps.
5. Knutson - who is the opposite sex than Rendler - isn't the person who wanted to complain about a mail carrier.
6. The six people, not necessarily in the same order are - Anthony, Donna, the person who wants to fill out a change-of-address form, the one who wants to buy a money order, the one who wants to send Airmail to Tibet and the second person in the line.
7. The four tasks of the last two people in line, not necessarily in the same order are - sending books fourth class, buying a money order, picking up a package and complaining about a mail carrier.
8. The person who wants to send books fourth class is just behind a person of the same sex.
9. Mary is just behind a person who wants to send an insured package.
10. The person who wants to send Airmail to Tibet is either two places ahead of or two places behind the one who wants to add postage to his or her meter.
11. Anthony isn't two places behind the who wants to pickup a registered letter.
12. Toriseza is two places ahead of the person who wants to pick up a package.
13. Knutson isn't just ahead of the person who wants to send an item parcel post.
Can you figure out where each customer is in the line, his or her full name (one surname is Loti) and the two things he or she wants to accomplish? Provide your answer is POSITION - FIRST NAME - LAST NAME - BUSINESS format.
A very TOUGH puzzle !!!
POS FIRST NAME LAST NAME BUSINESS
1 Sachi Loti • Fill Out a Change-of-Address Form
• Add Postage to Meter
2 Gianni Lau • Pick Up a Registered Letter
• Send an Item Parcel Post
3 Carlos Pimentelli • Overnight Package
• Send Airmail to Tibet
4 Donna Toriseza • Buy Stamps
• Send an Insured Package
5 Mary Knutson • Buy a Money Order
• Send Books fourth Class
6 Anthony Rendler • Complain About a Mail Carrier
• Pick Up a Package
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Brain Teaser No : 00250
There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.
In the mean time the whole platoon has moved ahead by 50m.
The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
Submitted by : manojaba
Answer
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.
Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.
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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 ruppes 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?
Answer
16 men, 12 women and 72 children were 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
There were 16 men, 12 women and 72 children working with the constructor.
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Brain Teaser No : 00438
Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of sheep that they owned.
It was found that Guran had ten more sheep than Lakha.
If Arjan gave one-third to Bhuvan, and Bhuvan gave a quarter of what he then held to Guran, who then passed on a fifth of his holding to Lakha, they would all have an equal number of sheep.
How many sheep did each of them possess? Give the minimal possible answer.
Answer
Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.
Assume that Arjan, Bhuvan, Guran and Lakha had A, B, G and L sheep respectively. As it is given that at the end each would have an equal number of sheep, comparing the final numbers from the above table.
Arjan's sheep = Bhuvan's sheep
2A/3 = A/4 + 3B/4
8A = 3A + 9B
5A = 9B
Arjan's sheep = Guran's sheep
2A/3 = A/15 + B/5 + 4G/5
2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)
30A = 3A + 5A + 36G
22A = 36G
11A = 18G
Arjan's sheep = Lakha's sheep
2A/3 = A/60 + B/20 + G/5 + L
2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18)
2A/3 = A/6 + L
A/2 = L
A = 2L
Also, it is given that Guran had ten more sheep than Lakha.
G = L + 10
11A/18 = A/2 + 10
A/9 = 10
A = 90 sheep
Thus, Arjan had 90 sheep, Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55 sheep and Lakha had A/2 i.e. 45 sheep.
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Brain Teaser No : 00590
You are locked inside a room with 6 doors - A, B, C, D, E, F. Out of which 3 are Entrances only and 3 are Exits only.
One person came in through door F and two minutes later second person came in through door A. He said, "You will be set free, if you pass through all 6 doors, each door once only and in correct order. Also, door A must be followed by door B or E, door B by C or E, door C by D or F, door D by A or F, door E by B or D and door F by C or D."
After saying that they both left through door B and unlocked all doors. In which order must you pass through the doors?
Answer
The correct order is CFDABE
It is given that one person came in through door F and second person came in through door A. It means that door A and door F are Entrances. Also, they both left through door B. Hence, door B is Exit.
As Exit and Entrance should alter each other and we know two Entrances, let's assume that the third Entrance is W. Thus, there are 6 possibilities with "_" indicating Exit.
(1) _W_A_F (2) _W_F_A (3) _F_W_A (4) _F_A_W (5) _A_W_F (6) _A_F_W
As door A must be followed by door B or E and none of them lead to the door F, (1) and (6) are not possible.
Also, door D must be the Exit as only door D leads to the door A and door A is the Entrance.
(2) _W_FDA (3) _F_WDA (4) _FDA_W (5) DA_W_F
Only door D and door C lead to the door F. But door D is used. Hence, door C must be the Exit and precede door F. Also, the third Exit is B and the W must be door E.
(2) BECFDA (3) CFBEDA (4) CFDABE (5) DACEBF
But only door B leads to the door C and both are Exits. Hence, (2) and (5) are not possible. Also, door F does not lead to door B - discard (3). Hence, the possible order is (4) i.e. CFDABE.
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Brain Teaser No : 00187
There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11.
Find the number.
Answer
65292
As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)
It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)
Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.
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Brain Teaser No : 00012
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?
Answer
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.
Answer is 3 hours 25 minutes 42 seconds
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Brain Teaser No : 00306
If a bear eats 65 pounds in fish every day EXCEPT every 6th day which it only eats 45 pounds of fish.
If the bear continues this, how many pounds of fish will it eat in 200 days?
Submitted by : David
Answer
The bear will eat 12,340 pounds of fish in 200 days.
It is given that on every 6th day beareats 45 pounds of fish i.e. on day number 6, 12, 18, 24, .... 192, 198 the bear eats 45 pounds of fish.
Total number of 6th days = 200/6 = 33 (the bear eats 45 pounds)
Hence, the normal days are = 200 - 33 = 167 (the bear eats 65 pounds)
Thus, in 200 days, the bear will eat
= (167) * (65) + (33) * (45)
= 10855 + 1485
= 12,340 pounds
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Brain Teaser No : 00546
Ankit and Tejas divided a bag of Apples between them.
Tejas said, "It's not fair! You have 3 times as many Apples I have." Ankit said, "OK, I will give you one Apple for each year of your age." Tejas replied, "Still not fair. Now, you have twice as many Apples as I have." "Dear, that's fair enough as I am twice older than you.", said Ankit.
Ankit went to Kitchen to drink water. While Ankit was in Kitchen, Tejas took apples from Ankit's pile equal to Ankit's age.
Who have more apples now?
Answer
At the end, Ankit and Tejas, both have the same number of apples.
Let's assume that initially Tejas got N apples and his age is T years. Hence, initially Ankit got 3N apples and his age is 2T years.
Operation Ankit's Apples Tejas's Apples
Initially 3N N
Ankit gave T apples to Tejas
(equals age of Tejas) 3N - T N + T
Tejas took 2T apples from Ankit's pile
(equals age of Ankit) 3N - 3T N + 3T
It is given that after Ankit gave T apples to Tejas, Ankit had twice as many apples as Tejas had.
3N - T = 2*(N + T)
3N - T = 2N + 2T
N = 3T
From the table, at the end Ankit have (3N - 3T) apples and Tejas have (N + 3T) apples. Substituting N = 3T, we get
Ankit's apples = 3N - 3T = 9T - 3T = 6T
Tejas's apples = N + 3T = 3T + 3T = 6T
Thus, at the end Ankit and Tejas, both have the same number of apples.
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Brain Teaser No : 00326
The sum of their (father, mother and son) ages is 70. The father is 6 times as old as the son.
When the sum of their ages is twice 70, the father will be twice as old as the son.
How old is the mother?
Submitted by : Tim Sanders
Answer
The mother is 29 years and 2 months old.
Let's assume that son is X years old. Hence, father is 6X years old and mother is (70-7X) years old.
It is given that the sum of their ages is 70, which will total 140 after 70/3 years.
After 70/3 years, son will be (X + 70/3) years old and father will be (6X + 70/3) years old. Also, it is given that after 70/3 years, the father will be twice as old as the son. Thus,
(6X + 70/3) = 2 * (X + 70/3)
6X + 70/3 = 2X + 140/3
4X = 70/3
X = 35/6
Hence, their ages are
Son = X = 35/6 = 5 years and 10 months
Father = 6X = 6(35/6) = 35 years
Mother = (70 - 7X) = 70 - 7(35/6) = 29 years and 2 months
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Brain Teaser No : 00040
Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the circumference of the earth was placed along the equator, and drawn to a tight fit.
Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.
By how many cm. will the thread be separated from the earth's surface?
Answer
The cicumference of the earth is
= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm
where r = radius of the earth, PI = 3.141592654
Hence, the length of the thread is = 1280000000 * PI cm
Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm
This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is nothing but (1280000000 * PI) + 12 cm
Assume that radius of the outer circle is R cm
Therefore,
2 * PI * R = (1280000000 * PI) + 12 cm
Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm
Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm
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Brain Teaser No : 00319
Substitute digits for the letters to make the following relation true.
S T I L L
+ W I T H I N
--------------------
L I M I T S
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all other S in the puzzle must be 3.
Answer
The value of L must be one more than W i.e. L=W+1 and there must be one carry from S+I=I. Also, the value of S must be 9 as S+I=I with one carry from T+T=M, which means that the value of T must be greater than 4.
From I+H=I, the value of H must be 0 as the value of S is 9.
Now, applying all those constraints and using trial-n-error, we get two possible answers.
9 7 1 6 6 9 8 5 3 3
+ 5 1 7 0 1 3 + 2 5 8 0 5 6
--------------- ---------------
6 1 4 1 7 9 3 5 6 5 8 9
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Brain Teaser No : 00271
Substitute digits for the letters to make the following Division true
O U T
-------------
S T E M | D E M I S E
| D M O C
-------------
T U I S
S T E M
----------
Z Z Z E
Z U M M
--------
I S T
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.
Submitted by : Calon
Answer
C=0, U=1, S=2, T=3, O=4, M=5, I=6, Z=7, E=8, D=9
It is obvious that U=1 (as U*STEM=STEM) and C=0 (as I-C=I).
S*O is a single digit and also S*T is a single digit. Hence, their values (O, S, T) must be 2, 3 or 4 (as they can not be 0 or 1 or greater than 4).
Consider, STEM*O=DMOC, where C=0. It means that M must be 5. Now, its simple. O=4, S=2, T=3, E=8, Z=7, I=6 and D=9.
O U T 4 1 3
------------- -------------
S T E M | D E M I S E 2 3 8 5 | 9 8 5 6 2 8
| D M O C | 9 5 4 0
------------- -------------
T U I S 3 1 6 2
S T E M 2 3 8 5
---------- ----------
Z Z Z E 7 7 7 8
Z U M M 7 1 5 5
-------- --------
I S T 6 2 3
Also, when arranged from 0 to 9, it spells CUSTOMIZED.
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Brain Teaser No : 00107
If you look at a clock and the time is 3:15.
What is the angle between the hour and the minute hands? ( The answer to this is not zero!)
Answer
7.5 degrees
At 3:15 minute hand will be perfactly horizontal pointing towards 3. Whereas hour hand will be towards 4. Also, hour hand must have covered 1/4 of angle between 3 and 4.
The angle between two adjacent digits is 360/12 = 30 degrees.
Hence 1/4 of it is 7.5 degrees.
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Brain Teaser No : 00170
A series comprising of alphabets contains 13 letters. The first seven letters in the given series are A, E, F, H, I, L, M
Can you find the next two letters?
Answer
The next letters in the series are N, O, R, S, U, X.
The pattern is - letters whose English names (Phonetic Pronunciations) start with vowels.
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Brain Teaser No : 00180
Which number in the series does not fit in the given series:
1 4 3 16 6 36 7 64 9 100
Answer
This is a series with odd positions containing position number whereas even positions containing square of the position.i.e. even position numbers are 4 16 36 64 100 and odd position numbers are 1 3 5 7 9
Hence, 6 does not fit in the series. It should be 5.
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Brain Teaser No : 00434
What are the next three numbers in the given series?
1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 ?
Answer
The next three numbers in the series are 4, 3, 4.
The pattern is - the number of 1's in the binary expansion of the positive integers starting from 1.
Number Binary Equivalent # of 1's
1 1 1
2 10 1
3 11 2
4 100 1
5 101 2
6 110 2
7 111 3
8 1000 1
9 1001 2
10 1010 2
11 1011 3
12 1100 2
13 1101 3
14 1110 3
15 1111 4
16 10000 1
17 10001 2
18 10010 2
19 10011 3
20 10100 2
21 10101 3
22 10110 3
23 10111 4
24 11000 2
25 11001 3
26 11010 3
27 11011 4
28 11100 3
29 11101 4
The other way of looking at it is - break up the series into lines as follow:
1
1 2
1 2 2 3
1 2 2 3 2 3 3 4
1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5
A new line can be created by writing previous line followed by the previous line with 1 added to each number.
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Brain Teaser No : 00552
A cricket team of 11 players lined up in a straight line to have their photograph. The captain was asked to stand in the center of the line-up.
1) Bharat and Bhavin stood to the right of the captain
2) Two players stood between Bhagat and Bhairav
3) Seven players stood between Bhadrik and Bhanu
4) Bhavesh stood to the right of Bhuvan
5) Bhola and Bhumit stood either side of Bhagat
6) Bhavik and Bhumit stood to the left of the captain
7) Six players stood between Bhavin and Bhagat
8) Two players stood between Bhagat and Bhavik
Who is the captain? Can you tell the positions of all the palyers?
Answer
Players from left to right : Bhavik, (Bhadrik/Bhanu), (Bhola/Bhumit), Bhagat, (Bhola/Bhumit), BHUVAN, Bhairav, (Bharat/Bhavesh), (Bharat/Bhavesh), (Bhadrik/Bhanu), Bhavin
Let's number the positions 1 to 11 from left to right. Hence, the captain is at position 6. Now, looking at the clues 7, 5, 2 and 8 together:
Poistion 1 - Bhavik or Bhairav
Position 3 - Bhumit or Bhola
Position 4 - Bhagat
Position 5 - Bhumit or Bhola
Poistion 7 - Bhavik or Bhairav
Position 11 - Bhavin
From clue (3), the only possible positions for Bhadrik and Bhanu are Position 2 and Position 10.
Now there are 3 positions remaining - 6, 8 and 9 and remaining 3 players are Bhuvan, Bharat and Bhavesh. But from clue (1), Bharat stood to the right of the captain i.e. Bharat must be on position 8 or 9 as position 6 is for the captain. So either Bhuvan or Bhavesh is the captain.
From (4), Bhavesh stood to the right of Bhuvan. Hence, Bhuvan is the captain.
Players from left to right are : Bhavik, (Bhadrik/Bhanu), (Bhola/Bhumit), Bhagat, (Bhola/Bhumit), BHUVAN, Bhairav, (Bharat/Bhavesh), (Bharat/Bhavesh), (Bhadrik/Bhanu), Bhavin.
Thus,
* Bhavik(1), Bhagat(4), Bhuvan(6), Bhairav(7) and Bhavin(11) are the players whose positions are fixed.
* Bhadrik and Bhanu are at position 2 or 10.
* Bhola and Bhumit are at position 3 or 5.
* Bharat and Bhavesh are at position 8 or 9.
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Brain Teaser No : 00854
In the middle of the confounded desert, there is the lost city of "Ash". To reach it, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for five days and the farthest we can travel in one day is 30 miles. Also, the city is 120 miles from the starting point.
What I am trying to figure out is the fewest number of persons, including myself, that I will need in our Group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies.
How many persons (including myself) will I need to accomplish this mission?
Answer
Total 4 persons (including you) required.
It is given that each person can only carry enough rations for five days. And there are 4 persons. Hence, total of 20 days rations is available.
1. First Day : 4 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 15 days.
2. Second Day : The remaining three people use up 3 days of rations. One person goes back using 2 days of rations for the return trip. The rations remaining for the further trek is for 10 days.
3. Third Day : The remaining two people use up 2 days of rations. One person goes back using 3 days of rations for the return trip. The rations remaining for the further trek is for 5 days.
4. Fourth Day : The remaining person uses up one day of rations. He stays overnight. The next day he returns to the coast using 4 days of rations.
Thus, total 4 persons, including you are required.
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Brain Teaser No : 00348
There are 10 statements written on a piece of paper:
1. At least one of statements 9 and 10 is true.
2. This either is the first true or the first false statement.
3. There are three consecutive statements, which are false.
4. The difference between the numbers of the last true and the first true statement divides the number, that is to be found.
5. The sum of the numbers of the true statements is the number, that is to be found.
6. This is not the last true statement.
7. The number of each true statement divides the number, that is to be found.
8. The number that is to be found is the percentage of true statements.
9. The number of divisors of the number, that is to be found, (apart from 1 and itself) is greater than the sum of the numbers of the true statements.
10. There are no three consecutive true statements.
Find the minimal possible number?
Submitted by : Milind Gadagkar
Answer
The numebr is 420.
If statement 6 is false, it creates a paradox. Hence, Statement 6 must be true.
Consider Statement 2:
* If it is true, it must be the first true statement. Otherwise, it creates a paradox.
* If it is false, it must be the second false statement. Otherwise, it creates a paradox.
In both the cases, Statement 1 is false.
As Statement 1 is false, Statement 9 and Statement 10 both are false i.e. there are three consecutive true statements.
1 2 3 4 5 6 7 8 9 10
False - - - - True - - False False
Let\'s assume that Statement 3 is false i.e. there are no three consecutive false statements. It means that Statement 2 and Statement 8 must be true, else there will be three consecutive false statements.
1 2 3 4 5 6 7 8 9 10
False True False - - True - True False False
Also, atleast two of Statements 4, 5 and 7 must be true as there are three consecutive true statements.
According to Statement 8, the number that is to be found is the percentage of true statements. Hence, number is either 50 or 60. Now if Statement 7 is true, then the number of each true statement divides the number, that is to be found. But 7 and 8 do not divide either 50 or 60. Hence, Statement 7 is false which means that Statement 4 and 5 are true. But Statement 5 contradicts the Statement 8. Hence, our assumption that Statement 3 is false is wrong and Statement 3 is true i.e. there are 3 consecutive false statements which means that Statement 8 is false as there is no other possibilities of 3 consecutive false statements.
Also, Statement 7 is true as Statement 6 is not the last true statement.
1 2 3 4 5 6 7 8 9 10
False - True - - True True False False False
According to Statement 7, the number of each true statement divides the number, that is to be found. And according to Statement 5, the sum of the numbers of the true statements is the number, that is to be found. For all possible combinations Statement 5 is false.
There 3 consecutive true statements. Hence, Statement 2 and Statement 4 are true.
1 2 3 4 5 6 7 8 9 10
False True True True False True True False False False
Now, the conditions for the number to be found are:
1. The numebr is divisible by 5 (Statement 4)
2. The numebr is divisible by 2, 3, 4, 6, 7 (Statement 7)
3. The number of divisors of the number, that is to be found, (apart from 1 and itself) is not greater than the sum of the numbers of the true statements. (Statement 9)
The minimum possible number is 420.
The divisors of 420, apart from 1 and itself are 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210. There are total of 22 divisors. Also, the sum of the numbers of the true statements is 22 (2+3+4+6+7=22), which satisfies the third condition.
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Brain Teaser No : 00186
500 men are arranged in an array of 10 rows and 50 columns according to their heights.
Tallest among each row of all are asked to come out. And the shortest among them is A.
Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B.
Now who is taller A or B ?
Answer
No one is taller, both are same as A and B are the same person.
As it is mentioned that 500 men are arranged in an array of 10 rows and 50 columns according to their heights. Let's assume that position numbers represent their heights. Hence, the shortest among the 50, 100, 150, ... 450, 500 is person with height 50 i.e. A. Similarly the tallest among 1, 2, 3, 4, 5, ..... 48, 48, 50 is person with height 50 i.e. B
Now, both A and B are the person with height 50. Hence both are same.
===================================================
Brain Teaser No : 00021
A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager.
According to his will: First give one coin to manager. 1/5th of the remaining to the elder son. Now give one coin to the manager and 1/5th of the remaining to second son and so on..... After giving coins to 5th son, divided the remaining coins among five daughters equally.
All should get full coins. Find the minimum number of coins he has?
Answer
We tried to find out some simple mathematical method and finally we wrote small C program to find out the answer. The answer is 3121 coins.
Here is the breakup:
First son = 624 coins
Second son = 499 coins
Third son = 399 coins
Forth son = 319 coins
Fifth son = 255 coins
Daughters = 204 each
Manager = 5 coins
=====================================
Brain Teaser No : 00302
There are 20 people in your applicant pool, including 5 pairs of identical twins.
If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins? (Needless to say, this could cause trouble ;))
Submitted by : Tim Sanders
Answer
The probability to hire 5 people with at least 1 pair of identical twins is 25.28%
5 people from the 20 people can be hired in 20C5 = 15504 ways.
Now, divide 20 people into two groups of 10 people each :
G1 - with all twins
G2 - with all people other than twins
Let's find out all possible ways to hire 5 people without a single pair of indentical twins.
People from G1 People from G2 No of ways to hire G1 without a single pair of indentical twins No of ways to hire G2 Total ways
0 5 10C0 10C5 252
1 4 10C1 10C4 2100
2 3 10C2 * 8/9 10C3 4800
3 2 10C3 * 8/9 * 6/8 10C2 3600
4 1 10C4 * 8/9 * 6/8 * 4/7 10C1 800
5 0 10C5 * 8/9 * 6/8 * 4/7 * 2/6 10C0 32
Total 11584
Thus, total possible ways to hire 5 people without a single pair of indentical twins = 11584 ways
So, total possible ways to hire 5 people with at least a single pair of indentical twins = 15504 - 11584 = 3920 ways
Hence, the probability to hire 5 people with at least a single pair of indentical twins
= 3920/15504
= 245/969
= 0.2528
= 25.28%
============================================
Brain Teaser No : 00408
A woman took a certain number of eggs to the market and sold some of them.
The next day, through the industry of her hens, the number left over had been doubled, and she sold the same number as the previous day.
On the third day the new remainder was tripled, and she sold the same number as before.
On the fourth day the remainder was quadrupled, and her sales the same as before.
On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stock.
What is the smallest number of eggs she could have taken to market the first day, and how many did she sell daily? Note that the answer is not zero.
Submitted by : Dan Allen
Answer
She took 103 eggs to market on the first day and sold 60 eggs everyday.
Let's assume that she had N eggs on the first day and she sold X eggs everyday. Putting down the given information in the table as follow.
Days Eggs at the start of the day Eggs Sold Eggs Remaining
Day 1 N X N-X
Day 2 2N-2X X 2N-3X
Day 3 6N-9X X 6N-10X
Day 4 24N-40X X 24N-41X
Day 5 120N-205X X 120N-206X
It is given that she disposed of her entire stock on the fifth day. But from the table above, the number of eggs remaining are (120N-206X). Hence,
120N - 206X = 0
120N = 206X
60N = 103X
The smallest value of N and X must be 103 and 60 respectively. Hence, she took 103 eggs to market on the first day and sold 60 eggs everyday.
=======================================
Brain Teaser No : 00483
Four couples are going to the movie. Each row holds eight seats. Betty and Jim don't want to sit next to Alice and Tom. Alice and Tom don't want to sit next to Gertrude and Bill. On the otherhand, Sally and Bob don't want to sit next to Betty and Jim.
How can the couples arrange themselves in a row so that they all sit where they would like?
Submitted by : Tara Smith
Answer
From the given data, it can be inferred that:
(Sally & Bob) NOT (Betty & Jim) NOT (Alice & Tom) NOT (Gertrude & Bill)
(A) NOT (B) means A and B can not seat next to each other.
Now, it is obvious that (Betty & Jim) and (Alice & Tom) will occupy the corner seats as both of them can have only one neighbour. Therefore,
(Gertrude & Bill) will seat next to (Betty & Jim)
(Sally & Bob) will seat next to (Gertrude & Bill)
(Alice & Tom) will seat next to (Sally & Bob)
Thus, there are two possible arrangements - a mirror images of each other.
1. (Betty & Jim) - (Gertrude & Bill) - (Sally & Bob) - (Alice & Tom)
2. (Alice & Tom) - (Sally & Bob) - (Gertrude & Bill) - (Betty & Jim)
=========================================================
Brain Teaser No : 00528
Substitute digits for the letters to make the following relation true.
N E V E R
L E A V E
+ M E
-----------------
A L O N E
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.
Answer
A tough one!!!
Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to
solve it. Now use trial-n-error method.
N E V E R 2 1 4 1 9
L E A V E 3 1 5 4 1
+ M E + 6 1
----------------- -----------------
A L O N E 5 3 0 2 1
===================================================
Brain Teaser No : 00118
Shahrukh speaks truth only in the morning and lies in the afternoon, whereas Salman speaks truth only in the afternoon and lies in the morning.
A says that B is Shahrukh.
Is it morning or afternoon and who is A - Shahrukh or Salman?
Answer
It is Afternoon and A can be Salman or Shahrukh. If A is Salman, he is speaking truth. If A is Shahrukh, he is lying.
Want to confirm it? Consider following 4 possible answers and check for its truthness individually.
1. It is Morning and A is Shahrukh
2. It is Morning and A is Salman
3. It is Afternoon and A is Shahrukh
4. It is Afternoon and A is Salman
====================================
Brain Teaser No : 00161
Mr. D'souza has bought four cars - Merc, Honda, Ford, Zen - as presents for his sons' birthdays, all of which are next week. Given the following information, what will each son get?
Alan will not get the Honda unless Barry gets the Merc and Denzil gets the Ford. Barry will not get the Ford unless Carl gets the Zen and Alan gets the Merc. Denzil will not get the Zen unless Alan gets the Honda and Barry gets the Merc. Alan will not get the Merc unless Carl gets the Zen and Denzil gets the Ford. Barry will not get the Merc unless Alan gets the Zen and Denzil gets the Ford. Alan will not get the Zen unless Barry gets the Honda and Carl gets the Merc. Carl will not get the Zen unless Barry gets the Honda and Alan gets the Ford. Alan will not get the Ford unless Barry gets the Zen and Denzil gets the Honda. Carl will not get the Merc unless Denzil gets the Honda.
Answer
Let's put given 9 information in a table. The person in Bold Font will not get the corresponding car unless the persons in Normal Font get the corresponding cars. Also, the person will Italics will get the remaining car.
Merc Honda Ford Zen
1 Barry Alan Denzil Carl
2 Alan Denzil Barry Carl
3 Barry Alan Carl Denzil
4 Alan Barry Denzil Carl
5 Barry Carl Denzil Alan
6 Carl Barry Denzil Alan
7 Denzil Barry Alan Carl
8 Carl Denzil Alan Barry
9 Carl Denzil ? ?
Now, let's assume that Alan gets the Merc. Then from (4), Barry gets the Honda, Denzil gets the Ford and Carl gets the Zen. But from (7), Carl will not get the Zen unless Barry gets the Honda and Alan gets the Ford. Thus, it contradicts the original assumption. Hence, Alan will not get the Merc.
Let's assume that Alan gets the Honda. Then from (1), Barry gets the Merc, Denzil gets the Ford and Carl gets the Zen. But from (5) or from (7), it contradicts the original assumption. Hence, Alan will not get the Honda.
Let's assume that Alan gets the Ford. Then from (8), Carl gets the Merc, Denzil gets the Ford and Barry gets the Zen - which does not contradict any of the statement.
Similaly, you can assume that Alan gets the Zen. (which is contradictory to (9))
Hence, Alan gets the Ford, Barry gets the Zen, Carl gets the Merc and Denzil gets the Honda.
=====================================================
Brain Teaser No : 00110
In a certain game, if 2 wixsomes are worth 3 changs, and 4 changs are worth 1 plut, then 6 plutes are worth how many wixsomes?
Answer
It is given that
2 wixsomes = 3 changs
8 wixsomes = 12 changs ----- (I)
Also, given that
4 changs = 1 plut
12 changs = 3 plutes
8 wixsomes = 3 plutes ----- From (I)
Therefore,
6 plutes = 16 wixsomes
========================================
Brain Teaser No : 00267
A stick of length 1 is divided randomly into 3 parts.
What is the probability that a triangle can be made with those three parts?
Answer
The probability, that a triangle can be made by randomly dividing a stick of length 1 into 3 parts, is 25%
A triangle can be made, if and only if, sum of two sides is greater than the third side. Thus,
X1 < x3 =" 1" is =" (1/2)" is =" (1/2)" is =" (1/2)" made =" (1/4)" made =" 1" is =" 1" 9="11" 6="9" 12="14" 8="11" 6="10" 12="14" 8="11" 6="10" 15="17" 10="13" 6="11" 15="17" 10=" 13" 6="11" 21="23" 14="17" 7="13" 26="28" 13="17" 30="32" 20="23" 15="19" 12="17" 10="16" 33="35" 22="25" 11="17" 35="37" 14="19" 10="17" 36="38" 24="27" 18="22" 12="18" 9="17" 40 =" 20" 40 =" 10" 40 =" 5" 40 =" 4" 40 =" 1" to =" 0.5" 025 =" 1" are =" (" 22770 =" 22935" is =" (Total" 9="18:" 8="9" n =" any" n="369)" d =" the" n="369" d="3)" p =" is" n="369" d="3" n="369" d="3," p="1001001" p="369369369" n="111." d="3" p="1001001" p="111111111" n="111111111." d="9" p="1000000001000000001" p="111111111..." n="1111111..." n="1111111..." number =" 29" number =" 2" 9 =" 11" number =" (11)" 1 =" 13" number =" (13)" 3 =" 17" number =" (17)" 7 =" 25" number =" (25)" 5 =" 32" number =" (32)" 2 =" 37" number =" (37)" 7 =" 47" number =" (47)" 7 =" 58" m="1." s="9" m="1." o="0," m="1," y="2," e="5," n="6," d="7," r="8," s="9" 2 ="=" are =" 11" 21 =" 11" 21 =" 192" directions =" 1/2" r =" (-1/4)" directions =" 1" r =" (-1/4)" 0 ="=" 2 =" C" 15 =" C" f =" 16" c =" 47." son ="=" 64p2 =" 64*63" bishop =" 64*14" rook =" 28*7" 13 =" 560" is =" (896" 4032 =" 13/36" 3611 ="=" a="1," car="CAR" f="0," a="F" r="F" i="2," yyau="T," r="3," ilyi="YYI," l="4," ilyi="YYI" i="2" y="5," 4="24Y2" c="6," 4="2452" e="9," c="R" u="7" t="8" 8 ="=" b="C," c="A" b="C," a =" 1," b =" 2," c =" 3," d =" 4" b =" C" 2 =" 3" c =" A" 3 =" 1" b =" C" 2 =" 3" f =" J" 6 =" 10" 12480 =" 26N2" 12480 =" 0" n="24." kg ="=" 100 =" 19%" is =" PI" 7 =" 3.14159265" 7 =" 21.99115" y =" NLY" h="0" h =" L" p =" 2N" n =" N" y =" P" l =" p" n="1," p="2," y="5," l="7" e="9." l="8" e="10" y="5," n="1," p="2," l="7," e="9," h="0" ay="PNH," f="6" a="3." 0 ="=" number =" 0!" number =" 1!" number =" 2!" number =" 3!" number =" 4!" number =" 5!" 120 =" 20" number =" 6!" 720 =" 20" number =" 7!" 5040 =" 40" number =" 8!" 40320 =" 20" number =" 9!" 362880 =" 80" number =" 10!" 3628800 =" 00" number =" 11!" 39916800 =" 00" 11131221133112132113212221 ="=" x =" 120" x =" 240" x =" 240" x =" 21.8182" x =" 21" x =" 5*X" 6 =" 0" 6 ="=" boys =" 12" girls =" 5" m =" 60" 14 =" 60" 6m =" 46" m2 =" 60" 15m =" 60" 8m =" 60" m2 =" 65" 14m =" 65" 8m =" 65" m2 =" 72" 14m =" 72" 8m =" 72" m =" 9" 36 =" 38" 18 =" 21" 12 =" 16" 9 =" 14" 6 =" 13" 9 =" 13" 6 =" 11" 4 =" 10" d="12," 10="25," 19="25" b="3," 28 =" 25," sam =" 41," mala =" 37," 3 =" one" 9 =" one" 27 =" one" 2 =" one" 0 =" zEro" 4 =" Four" 8 =" eiGht" 3 =" tHree" 5 =" fIve" hand =" 180" 11x =" 30" x =" 30/11" are =" 2" 11 =" 60/11" is =" 4" seconds ="=" burnt =" XL/6" burnt =" XL/4" remaining =" L" remaining =" L" 6 =" (4" x =" 12" 2x =" 6" x =" 3" in =" (" 50 =" 18250." oc2 =" OD2" oc2 =" 202" oc =" 28.28427" cx =" 20" ox =" OC" ox =" 28.28427" ox =" 8.28427" op2 =" OQ2" op2 =" (3.5)2" op =" 4.94974" py =" 3.5" oy =" OP" oy =" 4.94974" oy =" 8.44974"> OX i.e. smaller sphere requires more space than the space available. Hence, smaller sphere of 7 cms diameter can not pass through the space between the big sphere, the wall and the floor.
The puzzle can be solved by another method.
Draw a line tangent to the big sphere at the point X such that X is the closest point to the origin O on sphere. The tanget will cut X and Y axes at A and B respectively such that OA=OB. [See Fig III] From above, OX=8.28427 cms.
From the right angle triangle OAB, we can deduct that
OA = OB = 11.71572 cms
AB = 16.56854 cms
Now, the diameter of the inscribed circle of right angle triangle is given by d = a + b - c where a <= b < is =" OA" ab =" 11.71572" 56854 =" 6.86291" first =" 5" second =" 6" third =" 5" fourth =" 6" fifth =" 5" sixth =" 5" seventh =" 7" eighth =" 6" ninth =" 5" tenth =" 5" number =" 2" one =" 3" number =" 6" one =" 7" number =" 62" number =" 63" a =" 1/51" b =" 2/53" 1 =" (1/51)" a =" 50/51" b =" 1/53" 2 =" (50/51)" is =" 2" 2703 =" 0.0192378" 16p16 =" 16!" 8p8 =" 8!" 2p2 =" 2!" is =" (16!)" 800 ="=" mony =" 500-X" tv =" 800-X" tv =" 1000-X" plus =" 1500" x =" 200" mony =" 2000" x =" 500" tv =" 2500" x =" 700" 3700 =" 4000" x =" 300" s="1," i="2," l="3," a="4," n="5," e="6" d="9" 2 =" 2^1" 11 =" 2^1" 75 =" 2^1" 700 =" 2^1" 8476 =" 2^1" 126125 =" 2^1" 2223277 =" 2^1" 45269998 =" 2^1" 8 ="=" p =" PI" every =" (12" minutes =" 65.45" minutes =" 65" 38 =" 65" a="1," b="2," c="3," d="4" number =" 15(O)" number =" 20(T)" number =" 20(T)" number =" 6(F)" number =" 6(F)" number =" 19(S)" number =" 19(S)" number =" 5(E)" number =" 14(N)" 42 ="=" plane =" 10" plane =" 180" started =" 20" point =" 180" 20 =" 200" miles ="=" y ="=" 1000 =" 1" 1000 =" 1001" 1000 =" 1" 1000 =" 1000" x="56" y="72" number =" 1" 2 =" 2" number =" 2" 3 =" 6" number =" 3" 4 =" 12" number =" 4" 5 =" 20" number =" 5" 6 =" 30" number =" 6" 7 =" 42" number =" 7" 8 =" 56" number =" 8" 9 =" 72" number =" 2" number =" 2" 4 =" 6" number =" 6" 6 =" 12" number =" 12" 8 =" 20" number =" 20" 10 =" 30" number =" 30" 12 =" 42" number =" 42" 14 =" 56" number =" 56" 16 =" 72" g="0," h="1," y="2," t="3," e="4," i="5," w="6," f="7," n="8," o="9" e =" T" 2y =" Y" 2 ="=" is =" 41664" favourables =" 2" 8c3 =" 2" 56 =" 112" is =" 112" 41664 =" 1" 372 =" 0.002688" t =" N(N-1)" 46 =" (N+X)(N+X-1)" 46 =" (N+X)(N+X-1)" 46 =" N2" 46 =" 2NX" 46 =" (2N" 46 =" 0" 1 =" 45" 2n =" 46" n =" 23" x =" 1" 1 =" 21" 2n =" 22" n =" 11" x =" 2" n =" 16," 16 =" 10" 15 =" 11" 14 =" 12" 13 =" 13" 12 =" 14" 11 =" 15" 10 =" 16" 9 =" 17" 8 =" 20" 7 =" 22" 6 =" 24" 5 =" 31" 4 =" 100" 3 =" 121" 2 =" 10000" 10000 ="=" glory ="=" 2m =" 3B" m =" B" m =" B" 2m =" 3B" 2 =" 3B" k =" B" b =" 9" b =" 9" b =" 3" kapil="2," madanlal="4," binni="3," runout="1" kapil="1," madanlal="3," binni="2," runout="1" kapil="0," madanlal="2," binni="1," runout="1" school =" B" school =" 2*B" college =" (3/4)(2*B)" graduation =" [(7/3)B]" 9 ="=" a="2," b="1," c="9," d="4," e="3," f="8," g="6," h="5," i="7" ghi =" 9" def =" 6" abc =" 3" a="2," b="1," c="9," d="4," e="3," f="8," g="6," h="5," i="7" 9 ="=" 4 =" 12" 6 =" 32" 8 =" 66" 10 =" ??" 80 ="=" 9x =" X" 8y =" 10X" 4y =" 5X" x="4," y="5" 1 =" 720" x =" (1/6)" remaining ="=" 120 =" 72" 120 =" 80" 48 =" 32" 32 =" 40" bangles ="=" 2 =" 2^1" 12 =" 2^2" 360 =" 2^3" 75600 =" 2^4" 174636000 =" 2^5" 5244319080000 =" 2^6" x =" N2" x2 =" N2" 2x2 =" N2" x2 =" (N2" x2 =" N(N+1)/2" n =" 8," x =" 6" n =" 49," x =" 35" n =" 288," x =" 204" n =" 1681," x =" 1189" n =" 9800," x =" 6930" truth =""> today is Thursday
Assume that Prof. Ahmad is lying => today is Monday
Similarly, Assume Prof. Joshi is telling truth => today is Sunday
Assume that Prof. Joshi is lying => today is Thrusday.
Hence, today is Thrusday, Prof. Ahmad is telling truth and Prof. Joshi is lying.
Teaser 2 :
Assume that First Prof. is telling truth => Thursday, Friday, Saturday or Sunday
Assume that First Prof. is lying => Thursday, Friday or Saturday
Similarly, Assume Second Prof. is telling truth => Monday, Tuesday, Wednesday or Sunday
Assume that Second Prof. is lying => Monday, Tuesday, Wednesday
The only possibility is Sunday and both are telling truth.
Teaser 3 :
A simple one. First Prof. says - "I lie on Sunday" which is false as both the Prof. tell truth on sunday. It means the first statement made by the First Prof. is also false. It means the First Prof. tells truth on Saturday. Hence First Prof. is Prof. Ahmad and he is lying. It means that today is either Monday, Tuesday or Wednesday.
It is clear that Second Prof. is Prof. Joshi.
Assume that he is telling truth => today is Wednesday
Assume that he is lying => today is Saturday.
Hence, today is Wednesday !!!
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Brain Teaser No : 00088
Find the values of each of the alphabets.
N O O N
S O O N
+ M O O N
----------
J U N E
Answer
Using trial and error. There are 2 solutions to it and may be more.
2 4 4 2
1 4 4 2
+ 5 4 4 2
----------
9 3 2 6
4 1 1 4
5 1 1 4
+ 0 1 1 4
----------
9 3 4 2
=================================
Brain Teaser No : 00252
A man received a cheque. The rupees has been transposed for paise and vice versa. After spending 5 rupees 42 paise, he discovered that he now had exactly six times the value of the correct cheque amount.
What amount should he have received?
Answer
He received 6 rupees and 44 paise.
Let's assume that he received a cheque of X rupees and Y paise i.e. (100X + Y)
The amount received by him = 100Y + X
After spending 5 rupees 42 paise, the remaining amount is (100Y + X - 542) which is 6 times the original amount. Thus,
(100Y + X - 542) = 6*(100X + Y)
100Y + X - 542 = 600X + 6Y
94Y = 599X + 542
Using trial-n-error, we get X=6 and Y=44
Hence, he should have received 6 rupees and 44 paise.
====================================
Brain Teaser No : 00024
A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that.
After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.
Find X and Y. (1 Rupee = 100 Paise)
Answer
As given, the person wanted to withdraw 100X + Y paise.
But he got 100Y + X paise.
After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is
2 * (100X + Y) = 100Y + X - 20
200X + 2Y = 100Y +X - 20
199X - 98Y = -20
98Y - 199X = 20
Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1
Case I : Y=2X
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 0
We get X = - 20/3 & Y = - 40/2
Case II : Y=2X+1
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 1
We get X = 26 & Y = 53
Now, its obvious that he wanted to withdraw Rs. 26.53
======================================================
Brain Teaser No : 00158
Consider the sum: ABC + DEF + GHI = JJJ
If different letters represent different digits, and there are no leading zeros, what does J represent?
Answer
The value of J must be 9.
Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC + DEF + GHI = 14? + 25? + 36? = 7??)
Now, the remainder left after dividing any number by 9 is the same as the remainder left after dividing the sum of the digits of that number by 9. Also, note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9 and JJJ has a remainder of 0, 3, or 6.
The number 9 is the only number from 7, 8 and 9 that leaves a remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9. Hence, it follows that J must be 9.
===================================================
Brain Teaser No : 00323
Find next two numbers in the series :
100, 365, 24, 60, ?, ?
Answer
The next two numbers are 60 and 1000.
The pattern is breaking down the time from Century to miliseconds.
There are 100 years in a century.
There are 365 days in a year.
There are 24 hours in a day.
There are 60 minutes in an hour.
There are 60 seconds in a minute.
There are 1000 miliseconds in a second.
Hence, the sequence is : 100, 365, 24, 60, 60, 1000
======================================================
Brain Teaser No : 00031
Find sum of digits of D.
Let
A= 19991999
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C
(HINT : A = B = C = D (mod 9))
Answer
The sum of the digits od D is 1.
Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
Consider,
A = 19991999
< 20002000 =" 22000" 10002000 =" 1024200" 106000 =" 106800" 9 =" 61200" 9 =" 45" 9 =" 18" 1999 =" 1" 19991999 =" 1" e="1." 3 =" 4" 6 =" 8" 9 =" 12" boys =" 90" girls =" 81" class =" 84" g =" 84*(B+G)" 81g =" 84B" 6b =" 3G" 2b =" G"> Amar-6, Akbar-4, Anthony-9
2> Amar-6, Akbar-8, Anthony-5
In both the cases, we know that Amar have 6 children and hence Amar is the speaker.
============================================
Brain Teaser No : 00089
There are 20 poles with a constant distance between each pole. A car takes 24 second to reach the 12th pole.
How much will it take to reach the last pole.
Answer
The car will take 41.45 seconds to reach the last pole.
Let the distance between two poles is X.
The car takes 24 seconds to reach the 12th pole. It means it travels distance of 11X in 24 seconds.
To reach the 20th pole car has to travel 19X. So time taken to reach there is
= (19X * 24) / 11X
= (19 * 24) / 11
= 41.45 seconds
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Brain Teaser No : 00032
Find the smallest number N which has the following properties:
1. its decimal representation has 6 as the last digit.
2. If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as great as the original number N.
Answer
The smallest such number is 153846.
Assume that the number N is
N = BnBn-1Bn-2 ... B3B26
as its given that 6 is the last digit.
Now after erasing 6 and putting it in front of the remaining digits, we get
Nnew = 6BnBn-1Bn-2 ... B3B2
Also given that Nnew is 4 times the N. Also note that the last digit Nnew is second last digit of N and so on. The required result is
BnBn-1Bn-2 ... B3B26
X 4
--------------------
6BnBn-1Bn-2 ... B3B2
So start multiplying and put nth digit of Nnew to (n + 1)th digit of N and you will get result as
1 5 3 8 4 6
X 4
---------------
6 1 5 3 8 4
Hence, the number is 153846
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Brain Teaser No : 00167
What are the chances that at least two out of a group of fifty people share the same birthday?
Submitted by : Erin
Answer
The probability of atleast two out of a group of 50 people share the same birthday is 97%
Probability of atleast two share the same birthday = 1 - probability of all 50 have different birthdays
Probability of all 50 have different birthday
= 365/365 * 364/365 * 363/365 * ... * 317/365 * 316/365
= (365 * 364 * 363 * 362 * ... * 317 * 316)/36550
= 0.0296264
Probability of atleast two share the same birthday
= 1 - 0.0296264
= 0.9703735
= 97% approx.
Thus, the probability of atleast two out of a group of 50 people share the same birthday is 97%
This explains why in a school/college with classrooms of 50 students, there are at least two students with a birthday on the same day of the year. Also, if there are 23 people in the room, then there are 50% chances that atleast two of them have a birthday on the same day of the year!!!
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Brain Teaser No : 00168
Chintu put some Black marbles and some White marbles into a jar. He then asked his brother Pintu to take out a marble. Pintu drew out a Black marble. Chintu asked Pintu to draw out another marble, and again he drew out a Black marble.
Pintu thought there must be more Black marbles than White marbles in the jar and asked Chintu, "I wonder what is the probability of me drawing a Black marble on a third try?"
Chintu replied, "Exactly 9/10 of what it was of drawing a Black marble on your first draw."
Can you help Pintu to determine how many marbles of each colour had been in the jar in the beginning? Give the minimal possible answer. Also, Pintu knew that there were at least seven marbles in the jar in the beginning.
Answer
There were 8 Black marbles and 4 White marbles in the jar.
Let's assume that initially there are total N marbles. Also, B and W are the number of Black and White marbles respectively. Thus, N = B + W
The probability of drawing a Black marble on the first draw = B/(B+W)
Similarly, after drawing 2 Black marbles, the probability of drawing a Black marble on the third draw = (B-2)/(B+W-2)
Chintu said that the probability of drawing a Black marble on third draw (after drawing Black marbles on first two draws) is exactly 9/10 of what it was of drawing a Black marble on first draw.
(B - 2) / (B + W - 2) = (9 / 10) * B / (B + W)
10 * (B + W) * (B - 2) = 9 * B * (B + W - 2)
10B2 - 20B + 10BW - 20W = 9B2 + 9BW - 18B
B2 - 2B + BW - 20W = 0
We know that N = B + W, hence substitute W = N - B
B2 - 2B + B(N - B) - 20(N - B) = 0
B2 - 2B + BN - B2 - 20N + 20B = 0
18B + BN - 20N = 0
B(18 + N) = 20N
B = 20N / (18 + N)
Now, we know that the value of N is at least 7. Hence, using trial-n-error on the equation, the minimal value of N must be 12 so that B=8 and W=4.
Hence, initially there were 8 Black marbles and 4 White marbles in the jar.
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Brain Teaser No : 00669
Anna, her brother - Andre, her daughter - Ami, and her son, Adam are tennis players. As a game of doubles
1. Anna's brother is directly across the net from Ami.
2. Adam is diagonally across the net from the worst player's sibling.
3. The best player and the worst player are on the same side of the
net.
Who is the best player?
Answer
The best player is Andre.
Find out all possible arrangement such that no condition is contradicted.
From (1), there are two possible arrangements. Discarding two arrangements which are identical to following two.
Andre Anna Andre Adam
------------------ AND ------------------
Ami Adam Ami Anna
(I) (II)
From (2), Anna is the worst player in Case I and Adam is the worst player in Case II.
From (3), Andre is the best player in both the cases.
Hence, the best player is Andre.
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Brain Teaser No : 00675
Elizabeth is engaged.
* Her fiancé is either Arthur, Barry, Colin or Derek.
* Each of the four men and Elizabeth either always tells the truth or always lies.
* Arthur says: "Exactly one of us four men always tells the truth."
* Barry says: "Exactly one of us four men always tells the lies."
* Colin says: "Arthur or Barry is Elizabeth's fiancé."
* Elizabeth says: "My fiancé and I either both always tell truth or both always lie."
Who is Elizabeth's fiancé?
Answer
Barry is Elizabeth's fiancé.
Analyse the statements made by Colin and Elizabeth first.
From (6), if Elizabeth always tells the truth, then her fiancé always tells the truth and if Elizabeth always lies, then her fiancé always tells the truth. Hence, Elizabeth's fiancé always tells the truth.
Let's assume that statement made by Colin (5) is false. Then Colin lied and Derek is the truth-telling fiancé. But then statement made by Arthur can not be true (if true, then there are 2 truth-tellers which contradicts itself). And it can not be false either (if false, then there are at least 2 truth-tellers, which makes Barry's statement false and
hence there is just a one truth-teller i.e. Derek, again contradiction!!!). Hence statement made by Arthur contradicts itself. So the statement made by Colin is true i.e. Arthur or Barry is Elizabeth's fiancé.
As statement made by Colin is true, the statement made by Arthur is false. Then because Elizabeth's fiancé always tells the truth, Barry is Elizabeth's fiancé. Then the statement made by Barry is true and thus Derek always tells the truth.
Thus, Barry is Elizabeth's fiancé.
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Brain Teaser No : 00680
A group of fewer than 10 girls found a number of gold-coins which they were able to divide equally among them.
After this division had been done, Lalita - one of the girls, suggested that it would be more equitable to divide the gold-coins by families rather than by individuals. Among the them, there were two groups with two sisters, of course Lalita was not in either group. The rest of the girls were unrelated to each other. A re-division by families would have meant that the gold-coins per family were 5 more than the gold-coins per girl.
The girls argued among themselves over this way of dividing the gold-coins. Before a final decision is made, Ash - one of the girls, decided that she did not want any gold-coins. Her share was equally divided (without breaking/cutting any gold-coin) among the other girls.
Finally, Lalita decided to withdraw her suggestion of dividing the gold-coins by families.
How many girls were there and how many gold-coins did each girl end up with?
Answer
There were total 6 girls. Each end up with 12 gold-coins.
The number of gold-coins is evenly divisible by the number of girls as well as the number of families.
Let's assume that N is the number of gold-coins each girl received initially and G is the total number of girls.
Then, total number of gold-coins = NG
If the gold-coins had been divided by families rather than by individuals, the number of recipients would be (G - 2) and each share would be (N + 5).
Again, total number of gold-coins = (N + 5)(G - 2)
But, the total number of the gold-coins is the same.
NG = (N + 5)(G - 2)
NG = NG -2N +5G -10
2N = 5G -10
N = (5/2)G - 5
Now, N and G are the positive integers and also total number of gold-coins must be divisible by G, (G-1) and (G-2). This is because initially there were G girls; then since it was divided family wise, the total number of units, the coins were to be divided would be (G-2) (as two groups had two sisters, so two girls got combined as one group, one per group); and after Ash backed out, there were (G-1) girls.
Now trying different EVEN values for G, starting with 2; there were total 6 girls and 60 gold-coins. The gold-coins are divided among 5 girls and hence each girl ends up with 12 gold-coins.
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Brain Teaser No : 00150
What are the next two letters in this sequence?
A, B, H, F, M, C, I, G, T, D, O, ?, ?
Answer
The next letters in the sequence are J, U, E, X etc.
A is the first letter of the alphabet with symmetry about the vertical axis. B is the first letter of the alphabet with symmetry about the horizontal axis. H is the first letter of the alphabet with symmetry about both, the vertical axis and the horizontal axis. F is the first letter of the alphabet with no symmetry.
The sequence then repeats with the second letter of the alphabet with symmetry about the vertical axis, the horizontal axis, both the axes and no axes i.e. M, C, I and G
The third letters of the alphabet are T, D, O and J.
The fourth letters of the alphabet are U, E, X and L.
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Brain Teaser No : 00160
A family photo contained:
one grandfather, one grandmother;
two fathers, two mothers;
six children, four grandchildren;
two brothers, two sisters;
three sons, three daughters;
one father-in-law, one mother-in-law, one daughter-in-law, one son-in-law
30 people you may think, but no, what is the least number of people here?
Answer
There are total 8 people.
Four children - 2 boys and 2 girls, their mother and father, and one set of grandparents - EITHER mother's mother and father's father OR mother's father and father's mother.
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Brain Teaser No : 00263
There are 4 novels - Thriller, Mystery, Romance and Science Fiction - writen by Ambrose, Richards, Hooper and Walsh, and published by Alpha, Beta, Gamma and Theta not in necessarily in the same order.
* The book by Ambrose is published by Theta.
* The Thriller is published by Alpha.
* The Science Fiction novel is by Hooper and is not published by Gamma.
* The Romance novel is by Walsh.
Who are the Author and the Publisher of each novels?
Answer
(Thriller-Richards-Alpha), (Mystery-Ambrose-Theta), (Romance-Walsh-Gamma), (Science Fiction-Hooper-Beta)
From (1), (2) and (3) the Science Fiction novel is by Hooper and published by Beta.
From (4) the Romance novel is by Walsh and published by Gamma.
Now it is obvious that the Mystery is published by Theta and written by Ambrose. And the Thriller is published by Alpha and written by Richards.
Novel Type Author Publisher
Thriller Richards Alpha
Mystery Ambrose Theta
Romance Walsh Gamma
Science Fiction Hooper Beta
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Brain Teaser No : 00103
Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.
Answer
18
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X - 4) bullets each.
But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X
Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
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Brain Teaser No : 00278
What are the next two in the series?
st, nd, rd, th, th, ..., ...
Submitted by : Christopher Colon
Answer
The answer is th and th
The pattern is the suffix of each number when counted i.e. 1st, 2nd, 3rd, 4th, 5th, 6th, 7th and so on
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Brain Teaser No : 00288
Five friends with surname Batliwala, Pocketwala, Talawala, Chunawala and Natakwala have their first name and middle name as follow.
1. Four of them have a first and middle name of Paresh.
2. Three of them have a first and middle name of Kamlesh.
3. Two of them have a first and middle name of Naresh.
4. One of them have a first and middle name of Elesh.
5. Pocketwala and Talawala, either both are named Kamlesh or neither is named Kamlesh.
6. Either Batliwala and Pocketwala both are named Naresh or Talawala and Chunawala both are named Naresh.
7. Chunawala and Natakwala are not both named Paresh.
Who is named Elesh?
Answer
Pocketwala is named Elesh.
From (1) and (7), it is clear that Batliwala, Pocketwala and Talawala are named Paresh.
From (6) and (5), if Pocketwala or Talawala both are named Kamlesh, then either of them will have three names i.e. Paresh, Kamlesh and Naresh. Hence, Pocketwala and Talawala both are not named Kamlesh. It means that Batliwala, Chunawala and Natakwala are named Kamlesh.
Now it is clear that Talawala and Chunawala are named Naresh. Also, Pocketwala is named Elesh.
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Brain Teaser No : 00414
Sarika multiplied 414 by certain number and obtained 69958 as the answer. But she found that there is some error in the answer - both the 9s in the answer are wrong and all the other digits are correct.
Can you find the correct answer?
Answer
The correct answer is 60858.
If you divide 69958 by 414, you will get 168.98. Hence, assume some three digit number and multiply it by 414 and use 6**58 as the answer.
Assume three digit number such that
* * *
4 1 4
-------------
* * *
* * * 0
* * * 0 0
-------------
6 * * 5 8
It is obvious that the last digit of the assumed number must be 7.
* * 7
4 1 4
-------------
* * 8
* * 7 0
* * 8 0 0
-------------
6 * * 5 8
Now, the second last digit of the assumed number must be 4 or 9. Also, the first digit of the assumed number must be 1 as the first digit of the answer is 6. Using trial and error for above two conditions, the answer is
1 4 7
4 1 4
-------------
5 8 8
1 4 7 0
5 8 8 0 0
-------------
6 0 8 5 8
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Brain Teaser No : 00422
What are the next three numbers in the given series?
0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 ? ? ?
Answer
The next three numbers are 4, 2 and 2
The pattern is - the number of factors in prime factorization of positive integers. Note that all prime number positions are 1 in the given series.
Number Prime Factorization Factors in Prime Factorization
1 1 0
2 2 1
3 3 1
4 2*2 2
5 5 1
6 2*3 2
7 7 1
8 2*2*2 3
9 3*3 2
10 2*5 2
Thus, the next few numbers in the given series are:
4 2 2 3 3 1 3 1 5 2 2 2 4 1 2 2 4 1 3 1 3 3 2 1 5 2 3 2 3 1 4 2 4 2 2 1 ...
Note that 1 is neither a Prime number nor a Composite number.
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Brain Teaser No : 00464
Find the next two numbers in the series:
1, 8, 11, 18, 80, 81, _, _
Answer
The next two numbers are 82 and 83.
The pattern is : the list of positive integers beginning with a VOWEL.
1(One), 8(Eight), 11(Eleven), 18(Eighteen), 80(Eighty), 81(Eighty-one)
Hence, the next numbers are 82(Eighty-two), 83(Eighty-three), 84(Eighty-four), .....
It is not a series comprise of 0, 1 and 8. If so, 10 must be there.
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Brain Teaser No : 00154
There are 4 mugs placed upturned on the table. Each mug have the same number of marbles and a statement about the number of marbles in it. The statements are: Two or Three, One or Four, Three or One, One or Two.
Only one of the statement is correct. How many marbles are there under each mug?
Answer
A simple one.
As it is given that only one of the four statement is correct, the correct number can not appear in more than one statement. If it appears in more than one statement, then more than one statement will be correct.
Hence, there are 4 marbles under each mug.
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Brain Teaser No : 00254
One of the four people - Mr. Clinton, his wife Monika, their son Mandy and their daughter Cindy - is a singer and another is a dancer. Mr. Clinton is older than his wife and Mady is older than his sister.
1. If the singer and the dancer are the same sex, then the dancer is older than the singer.
2. If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer.
3. If the singer is a man, then the singer and the dancer are the same age.
4. If the singer and the dancer are of opposite sex then the man is older than the woman.
5. If the dancer is a woman, then the dancer is older than the singer.
Whose occupation do you know? And what is his/her occupation?
Answer
Cindy is the Singer. Mr. Clinton or Monika is the Dancer.
From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.
CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.
CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.
In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.
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Brain Teaser No : 00272
What are the next two numbers in this series?
1, 4, 1, 5, 9, 2, _, _
Submitted by : Alex Crosse
Answer
Next two numbers are 6 and 5.
The patten is the each digit in the value of PI after the decimal point. The value of mathematical constant PI is 3.1415926536.
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Brain Teaser No : 00275
There are two ropes. Each one can burn in exactly one hour. They are not necessarily of the same length or width as each other. They also are not of uniform width (may be thiner/wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.
By burning the ropes, how do you measure exactly 45 minutes worth of time?
Answer
If you light both ends of one rope simultaneously, it will burn in exactly a 1/2 hour. Thus, burn one rope from both ends and the other rope from only one end. Once the one rope (which is burning from both ends) finally burns out (and you know a 1/2 hour has elapsed), you also know that the other rope (which is burning from only one end) has exactly 1/2 hour left to burn. Since you only want 45 minutes, light the second end of the rope. This remaining piece will burn in 15 minutes. Thus, totaling 45 minutes.
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Brain Teaser No : 00393
Which of the following day(s) can't be the last day of a century?
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
Justify your answer.
Answer
The last day of a century can not be Tuesday, Thursday or Saturday.
A normal year has 365 days whereas a leap year has 366 days. Every year which is divisible by 4 is called a leap year. Also, every 4th century is a leap year but no other century is a leap year.
1 normal year = 365 days = 52 weeks + 1 day
1 leap year = 366 days = 52 weeks + 2 day
Thus, a normal year has 1 odd day whereas a leap year has 2 odd days.
100 years
= 76 normal years + 24 leap years
= 76*[52 weeks + 1 day] + 24*[52 weeks + 2 day]
= (76*52) weeks + 76 days + (24*52) weeks + 48 days
= 5200 weeks + 124 days
= 5217 weeks + 5 days
i.e. 100 years contain 5 odd days
Similarly,
200 years contain 10 odd days i.e. 3 odd days.
300 years contain 15 odd days i.e. 1 odd days.
400 years contain (20+1) odd days i.e. 0 odd days.
Note that 400 years contain one more leap year.
Also, we have Sunday for 0 odd day, Monday for 1 odd day, Tuesday for 2 odd days, and so on...
Thus, last day of first century is Friday. (5 odd days)
Last day of second century is Wednesday. (3 odd days)
Last day of third century is Monday. (1 odd days)
Last day of forth century is Sunday. (0 odd days)
Since the order is repeating in successive cycles, the last day of a century can not be Tuesday, Thursday or Saturday.
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Brain Teaser No : 00474
Given any whole number take the sum of the digits, and the product of the digits, and multiply these together to get a new whole number.
For example, starting with 6712, the sum of the digits is (6+7+1+2) = 16, and the product of the digits is (6*7*1*2) = 84. The answer in this case is then 84 x 16 = 1344.
If we do this again starting from 1344, we get (1+3+4+4) * (1*3*4*4) = 576
And yet again (5+7+6) * (5*7*6) = 3780
At this stage we know what the next answer will be (without working it out) because, as one digit is 0, the product of the digits will be 0, and hence the answer will also be 0.
Can you find any numbers to which when we apply the above mentioned rule repeatedly, we never end up at 0?
Answer
Three such numbers are 1, 135 and 144.
It seems that most numbers will eventually end up at 0 when we apply the rule repeatedly. But there are a few numbers that have the property that when we apply the rule repeatedly, we never end up at 0.
Start with 332, then we get (3+3+2) * (3*3*2) = 144
And then (1+4+4) * (1*4*4) = 144
Thus if we reach 144, we stay there however many times we apply this rule. We say that 144 is fixed by this rule. Now try 233 or 98 or 332 or 1224. They all fall into the same group i.e. we reach 144.
There is another number that is fixed by this rule; it is 1 (because the sum of the digits of 1 is 1, and the product of the digits is 1 so, starting with 1, the answer is 1 * 1 = 1).
And the third one is 135.
If you know some other numbers, do let us know.
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Brain Teaser No : 00510
Substitute digits for the letters to make the following addition problem true.
W H O S E
T E E T H
A R E
+ A S
-------------------
S W O R D S
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter H, no other letter can be 3 and all other H in the puzzle must be 3.
Answer
It is obvious that S=1 and T=9.
Also, (H + E) should be greater than 10 and hence, (E + H + E) must 20. Thus, there are 3 possible values for (E, H) pair: (6, 8) or (7, 6) or (8, 4). Use trial-n-error and everything will fit-in.
W H O S E 2 8 5 1 6
T E E T H 9 6 6 9 8
A R E 4 7 6
+ A S + 4 1
------------------- -------------------
S W O R D S 1 2 5 7 3 1
==============================================
Brain Teaser No : 00595
Substitute digits for the letters to make the following addition problem true.
I
A G R E E
+ I T S
-------------------
T O U G H
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all other S in the puzzle must be 3.
Answer
I=7, A=2, G=9, R=4, E=5, T=3, S=6, O=0, G=9, H=8
It is obvious that T=A+1. Also, G=9, O=0 and R+I>10. Hence, T>1
There must be a carry from the units. Hence E+T=8. So (E, T) can be (6, 2), (5, 3), (3, 5), (2, 6) or (1, 7).
Now, use trial-n-error and solve it.
I 7
A G R E E 2 9 4 5 5
+ I T S + 7 3 6
------------------- -------------------
T O U G H 3 0 1 9 8
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Brain Teaser No : 00010
In a contest of intelligence, three problems A, B and C were posed.
* Among the contestants there were 25 who solved at least one problem each.
* Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.
* The number of participants who solved only problem A was one more than the number who solved problem A and at least one other problem.
* Of all students who solved just one problem, half did not solve problem A.
How many students solved only problem B?
Answer
6 students solved only problem B
X => Students who solved only problem A
Y => Students who solved only problem B
Z => Students who solved only problem C
P => Students who solved both problem B and problem C
From 4 :
Students who solved only problem A = Students who solved only problem B + Students who solved only problem C
X = Y + Z
From 3 :
Students who solved problem A and at least one other = X - 1
From 2 :
(Y + P) = 2 * (Z + P)
Y + P = 2 * Z + 2 * P
Z = (Y - P) / 2
From 1 and Figure:
X + X - 1 + Y + Z + P = 25
2*X + Y + Z + P = 26
2*(Y + Z) + Y + Z + P = 26 (from 4)
3*Y + 3*Z + P = 26
3*Y + 3* (Y - P) / 2 + P = 26 (from 2)
6*Y + 3*Y - 3*P + 2*P = 52
9*Y - P = 52
Y = (52 + P) / 9
Now, it is obvious that all values are integer. Hence, P must be 2 and Y must be 6.
So 6 students solved only problem B.
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Brain Teaser No : 00355
Montu, Bantu, Chantu and Pintu have pets.
Montu says, "If Pintu and I each have a dog, then exactly one of Bantu and Chantu has a dog."
Bantu says, "If Chantu and I each have a cat, then exactly one of Montu and Pintu has a dog."
Chantu says, "If Montu and I each have a dog, then exactly one of Bantu and Pintu has a cat."
Pintu says, "If Bantu and I each have a cat, then exactly one of Bantu and I has a dog."
Only one of the four is telling the truth. Who is telling the truth?
Answer
Bantu is telling the truth.
For a IF-THEN statement to be false, IF part has to be true and THEN part has to be false.
Since only one statement is true and remaining three are false, IF part of three statements are true & THEN part of one statement is true. Let's put the given information in table. The pet-name in the normal text represents the IF part and the pet-name in round brackets represents the THEN part.
Montu Bantu Chantu Pintu
Montu says Dog (Dog) (Dog) Dog
Bantu says (Dog) Cat Cat (Dog)
Chantu says Dog (Cat) Dog (Cat)
Pintu says Cat
(Dog) Cat
(Dog)
It is clear that the IF part of the statements made by Montu, Chantu and Pintu are true as they do not contradict each other. And the IF part of the statement made by Bantu is false.
Thus, Bantu is telling the truth.
Montu have a Dog and may or may not have a Cat.
Bantu have a Cat.
Chantu have a Dog.
Pintu have a Dog and a Cat.
=================================
Brain Teaser No : 00496
Karan bought a little box of midget matches, each one inch in length. He found that he could arrange them all in the form of a triangle whose area was just as many square inches as there were matches.
He then used up six of the matches, and found that with the remainder he could again construct another triangle whose area was just as many square inches as there were matches.
And using another six matches he could again do precisely the same.
How many matches were there in the box originally?
Note that the match-box can hold maximum of 50 matches.
Answer
Initially, there were 42 or 36 matches in the match-box.
There are 42 matches in the box with which he could form a triangle 20, 15, 7, with an area of 42 square inches. After 6 matches had been used, the remaining 36 matches would form a triangle 17, 10, 9, with an area of 36 square inches. After using another 6 matches, the remaining 30 matches would form a triangle 13, 12, 5, with an area of 30 square inches. After using another 6, the 24 remaining would form a triangle 10, 8, 6, with an area of 24 square inches.
Thus, there are two possible answers. There were either 42 or 36 matches in the match-box.
Also it is interesting to know that there are just 5 such triangles for which the perimeter and the area is the same (assuming all sides are integers) and they are :
1. 24 (10, 8, 6)
2. 30 (13, 12, 5)
3. 36 (17, 10, 9)
4. 42 (20, 15, 7)
5. 60 (29, 25, 6)
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Brain Teaser No : 00524
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?
Answer
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.
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Brain Teaser No : 00499
Six cabins numbered 1-6 consecutively, are arranged in a row and are separated by thin dividers. These cabins must be assigned to six staff members based on following facts.
1. Miss Shalaka's work requires her to speak on the phone frequently throughout the day.
2. Miss Shudha prefers cabin number 5 as 5 is her lucky number.
3. Mr. Shaan and Mr. Sharma often talk to each other during their work and prefers to have adjacent cabins.
4. Mr. Sinha, Mr. Shaan and Mr. Solanki all smoke. Miss Shudha is allergic to smoke and must have non-smokers adjacent to her.
5. Mr. Solanki needs silence during work.
Can you tell the cabin numbers of each of them?
Answer
The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan, Mr. Sharma, Miss Shudha and Miss Shalaka.
From (2), cabin number 5 is assigned to Miss Shudha.
As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan & Mr. Solanki all smoke, they must be in cabin numbers 1, 2 and 3 not necessarily in the same order. Also, Miss Shalaka and Mr. Sharma must be in cabin 4 and 6.
From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4. Thus, Miss Shalaka is in cabin 6.
As Mr. Solanki needs silence during work and Mr. Shaan is in cabin 3 who often talks to Mr. Sharma during work, Mr. Solanki must be in cabin 1. Hence, Mr. Sinha is in cabin 2.
Thus, the cabins numbers are
1# Mr. Solanki,
2# Mr. Sinha,
3# Mr. Shaan,
4# Mr. Sharma,
5# Miss Shudha,
6# Miss Shalaka
===============================
Brain Teaser No : 00541
A CAR has a value of 1 and a TRACTOR has a value of 2.
What is the value of an AIRPLANE? Note that AIRPLANE costs more than CAR or TRACTOR.
Submitted by : Lynda
Answer
AIRPLANE has a value of 4.
The value of each vehicle equals the total number of vowels in the word.
CAR has 1 vowel i.e. A
TRACTOR has 2 vowels i.e. A O
AIRPLANE has 4 vowels i.e. A I A E
There is one more possible answer: The value of each vehicle equals the total number of DISTINCT vowels in the word. So the value of AIRPLANE will be 3.
========================================
Brain Teaser No : 00235
Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5.
How many such 3-digit numbers are there?
Answer
There are 45 different 3-digit numbers.
The last digit can not be 0.
If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-digit number)
If the last digit is 2, the possible numbers are 202 and 112.
If the last digit is 3, the possible numbers are 303, 213 and 123.
If the last digit is 4, the possible numbers are 404, 314, 224 and 134.
If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.
Note the pattern here - If the last digit is 1, there is only one number. If the last digit is 2, there are two numbers. If the last digit is 3, there are three numbers. If the last digit is 4, there are four numbers. If the last digit is 5, there are five numbers. And so on.....
Thus, total numbers are
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Altogether then, there are 45 different 3-digit numbers, where last digit is the sum of first two digits.
==============================
Brain Teaser No : 00284
While sitting in a club where all single men tell the truth and all married men lie, a woman is approached by three men.
She asks the first guy if he is married, but the music is so loud that she can not hear his answer.
So she turns to the second guy, who tells her, "The first guy said, 'I am married', but he is really single."
Then she turns to the third guy, who says, "The second guy is single."
Can you determine the marital status of each of the three men?
Submitted by : Marie
Answer
All three are married.
A single man would always say that he is single. And a married man would also always say that he is single. It means that no man in the club would say that he is married. Thus, the second man is lying, so the second man is married.
Since the second man (always lies) says the first man is single, the first man is married.
Similarly, the third man incorrectly says that the second man is single. Hence, the third man is lying and he is married.
Thus, all three are lying and hence, married.
=============================
Brain Teaser No : 00435
Substitute digits for the letters to make the following Division true
G E T
-------------
N U T | G I N G E R
| N U T
-------------
E I N E
G U A O
---------
T T O R
E O G D
-------
U A O
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter U, no other letter can be 3 and all other U in the puzzle must be 3.
Submitted by : Calon
Answer
I=0, G=1, E=2, T=3, A=4, R=5, O=6, U=7, N=8, D=9
At a first glance, it is obvious that G=1, E=2 and T=3. Everything else is pretty simple now.
1 2 3
-------------
8 7 3 | 1 0 8 1 2 5
| 8 7 3
-------------
2 0 8 2
1 7 4 6
---------
3 3 6 5
2 6 1 9
-------
7 4 6
=======================================
Brain Teaser No : 00023
There is a grid of 20 squares by 10 squares. How many different rectangles are possible?
Note that square is a rectangle.
Answer
11550
The Generic solution to this is:
Total number of rectangles = (Summation of row numbers) * (Summation of column numbers)
Here there are 20 rows and 10 columns or vice versa. Hence, total possible rectangles
= ( 20 + 19 + 18 + 17 + 16 + .... + 3 + 2 + 1 ) * ( 10 + 9 +8 + 7 + .... + 3 + 2 + 1)
= ( 210 ) * (55)
= 11550
Hence, total 11,550 different rectangles are possible.
If you don't believe it, try formula on some smaller grids like 4x2, 3x2, 3x3 etc...
===================================
Brain Teaser No : 00115
Four prisoners escape from a prison.
The prisoners, Mr. East, Mr. West, Mr. South, Mr. North head towards different directions after escaping.
The following information of their escape was supplied:
* The escape routes were North Road, South Road, East Road and West Road
* None of the prisoners took the road which was their namesake
* Mr. East did not take the South Road
* Mr.West did not the South Road
* The West Road was not taken by Mr. East
What road did each of the prisoners take to make their escape?
Answer
Put all the given information into the table structure as follow:
North Road South Road East Road West Road
Mr. North No
Mr. South No
Mr. East No No No
Mr. West No No
Now from table, two things are obvious and they are:
* Mr.North took the South Road
* Mr.East took the North Road
Put this information into the table, Also keep in mind that the prisoners head towards different directions after escaping.
North Road South Road East Road West Road
Mr. North No YES No No
Mr. South No No
Mr. East YES No No No
Mr. West No No No
Now from the table:
* Mr.West took the East Road
* Mr.South took the West Road
So the answer is:
* Mr.North took the South Road
* Mr.South took the West Road
* Mr.East took the North Road
* Mr.West took the East Road
======================================
Brain Teaser No : 00216
When Socrates was imprisoned for being a disturbing influence, he was held in high esteem by his guards. All four of them hoped that something would occur that would facilitate his escape. One evening, the guard who was on duty intentionally left the cell door open so that Socrates could leave for distant parts.
Socrates did not attempt to escape, as it was his philosophy that if you accept society's rules, you must also accept it's punishments. However, the open door was considered by the authorities to be a serious matter. It was not clear which guard was on that evening. The four guards make the following statements in their defense:
Aaron:
A) I did not leave the door open.
B) Clement was the one who did it.
Bob:
A) I was not the one who was on duty that evening.
B) Aaron was on duty.
Clement:
A) Bob was the one on duty that evening.
B) I hoped Socrates would escape.
David:
A) I did not leave the door open.
B) I was not surprised that Socrates did not attempt to escape.
Considering that, in total, three statements are true, and five statements are false, which guard is guilty?
Answer
David is the guilty.
Note that "All four of them hoped that something would occur that would facilitate his escape". It makes Clement's statement B True and David's statement B False.
Now consider each of them as a guilty, one at a time.
Aaron Bob Clement David True
Stmts
A B A B A B A B
If Aaron is guilty False False True True False True True False 4
If Bob is guilty True False False False True True True False 4
If Clement is guilty True True True False False True True False 5
If David is guilty True False True False False True False False 3
Since in total, three statements are true and five statements are false. It is clear from the above table that David is the guity.
==================================
Brain Teaser No : 00296
Substitute digits for the letters to make the following subtraction problem true.
S A N T A
- C L A U S
-----------------
X M A S
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.
Answer
One of the simplest brain teaser as there are total 26 possible answers.
It is obvious that S=C+1. Since A-S=S, it is clear that A=2*S or 2*s-10. Also, L and X are interchangeable.
SANTA - CLAUS = XMAS
24034 - 16492 = 7542
24034 - 17492 = 6542
24074 - 15432 = 8642
24074 - 18432 = 5642
24534 - 16492 = 8042
24534 - 18492 = 6042
24794 - 16452 = 8342
24794 - 18452 = 6342
24804 - 15462 = 9342
24804 - 19462 = 5342
24974 - 16432 = 8542
24974 - 18432 = 6542
36806 - 27643 = 9163
36806 - 29643 = 7163
36156 - 27693 = 8463
36156 - 28693 = 7463
62132 - 54206 = 7926
62132 - 57206 = 4926
62172 - 53246 = 8926
62172 - 58246 = 3926
62402 - 53276 = 9126
62402 - 59276 = 3126
62712 - 53286 = 9426
62712 - 59286 = 3426
62932 - 58206 = 4726
62932 - 54206 = 8726
========================
Brain Teaser No : 00844
Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.
Six months later, there were 1000Z rabbits. At the beginning of the 3rd year, there were roughly 2828Z rabbits, which was 4 times what the scientists placed in there at the beginning of the 1st year.
If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?
Submitted by : David Johnson
Answer
At the beginning of the 11th year, there would be 1,024,000 rabbits.
At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e. 1.414 (the square root of 2)
Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z
Also, it is given that at the end of 6 months, there were 1000Z rabbits.
It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(Z^(2N)) i.e. 1000*(2^N)
Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 rabbits.
=====================================
Brain Teaser No : 00036
A number of 9 digits has the following properties:
* The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9. * Each digit in the number is different i.e. no digits are repeated. * The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order. Find the number. Answer The answer is 381654729 One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur. The other way to solve this problem is by writing a computer program that systematically tries all possibilities. ================================= Brain Teaser No : 00127 Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass. How many cows are needed to eat the grass in 96 days? Answer 20 cows g - grass at the beginning r - rate at which grass grows, per day y - rate at which one cow eats grass, per day n - no of cows to eat the grass in 96 days From given data, g + 24*r = 70 * 24 * y ---------- A g + 60*r = 30 * 60 * y ---------- B g + 96*r = n * 96 * y ---------- C Solving for (B-A), (60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y) 36 * r = 120 * y ---------- D Solving for (C-B), (96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y) 36 * r = (n * 96 - 30 * 60) * y 120 * y = (n * 96 - 30 * 60) * y [From D] 120 = (n * 96 - 1800) n = 20 Hence, 20 cows are needed to eat the grass in 96 days. ============================================= Brain Teaser No : 00164 Substitute digits for the letters to make the following relation true. W O R L D + T R A D E ------------- C E N T E R Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter W, no other letter can be 3 and all other W in the puzzle must be 3. Answer A tough one. It is obvious that C=1. Also, the maximum possible value of E is 7. Now, start putting possible values of D, E and R as they occure frequently and use trial-n-error. W O R L D 5 3 6 8 4 + T R A D E + 7 6 0 4 2 ------------ ------------ C E N T E R 1 2 9 7 2 6 ============================================= Brain Teaser No : 00391 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 Answer A simple one. 1 1 7 X 3 1 9 ---------- 1 0 5 3 1 1 7 3 5 1 ------------- 3 7 3 2 3 ================================= Brain Teaser No : 00468 A blindfolded man is asked to sit in the front of a carrom board. The holes of the board are shut with lids in random order, i.e. any number of all the four holes can be shut or open. Now the man is supposed to touch any two holes at a time and can do the following. * Open the closed hole. * Close the open hole. * Let the hole be as it is. After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of what are the holes which are open or closed. How many minimum number of turns does the blindfolded man require to either open all the holes or close all the holes? Note that whenever all the holes are either open or close, there will be an alarm so that the blindfolded man will know that he has won. Submitted by : Vikrant Ramteke Answer The blindfolded man requires 5 turns. 1. Open two adjacent holes. 2. Open two diagonal holes. Now atleast 3 holes are open. If 4th hole is also open, then you are done. If not, the 4th hole is close. 3. Check two diagonal holes. * If one is close, open it and all the holes are open. * If both are close, open any one hole. Now, two holes are open and two are close. The diagonal holes are in the opposite status i.e. in both the diagonals, one hole is open and one is close. 4. Check any two adjacent holes. * If both are open, close both of them. Now, all holes are close. * If both are close, open both of them. Now, all holes are open. * If one is open and one is close, invert them i.e. close the open hole and open the close hole. Now, the diagonal holes are in the same status i.e. two holes in one diagonal are open and in other are close. 5. Check any two diagonal holes. * If both are open, close both of them. Now, all holes are close. * If both are close, open both of them. Now, all holes are open. ======================================== Brain Teaser No : 00807 How many possible combinations are there in a 3x3x3 rubics cube? In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)? How many for a 4x4x4 cube? Submitted by : Sarah Flod Answer There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics. Let's consider 3x3x3 Rubics first. There are 8 corner cubes, which can be arranged in 8! ways. Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7) Similarly, 12 edge cubes can be arranged in 12! ways. Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11) Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2. Total different possible combinations are = [(8!) * (3^7)] * [(12!) * (2^11)] / 2 = (8!) * (3^7) * (12!) * (2^10) = 4.3252 * 10^19 Similarly, for 4x4x4 Rubics total different possible combinations are = [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24 = 7.4011968 * 10^45 Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24. ====================================== Brain Teaser No : 00093 There is a number that is 5 times the sum of its digits. What is this number? Answer is not 0. Answer The number is 45, simply because 45 = 5 * (4 + 5) How does one find this number? Let T be the digit in the tens place and U be the digit in the units place. Then, the number is 10*T + U, and the sum of its digits is T + U. The following equation can be readily written: 10*T + U = 5*(T + U) or 10*T + U = 5*T + 5*U or 5*T = 4*U Thus, T / U = 4 / 5 Since T and U are digits, T must be 4 and U must be 5. ==================================== Brain Teaser No : 00273 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. Answer 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 ======================== Brain Teaser No : 00340 Find the next in the set: AZFR, LARU, AMAS, SBNS, KICI, ???? Submitted by : Milind Gadagkar Answer The next word in the series is AAEA. If you collect the corresponding letter from each word, they are names of the places. Consider the following five names : A L A S K A Z A M B I A F R A N C E R U S S I A First letter of each word : A Z F R Second letter of each word : L A R U Third letter of each word : A M A A Forth letter of each word : S B N S Fifth letter of each word : K I C I Sixth letter of each word : A A E A Hence, the answer is AAEA. ============================ Brain Teaser No : 00477 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? Submitted by : Travis Lara Answer 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. ================================ Brain Teaser No : 00461 Buffalo produces 10 pounds of milk daily, a cow produces 1 pound of milk daily, and sheep produce 1/4 pound of milk daily. I need exactly 100 pounds of milk from exactly 100 different animals daily. Also, I should use all 3 types of animals. How do I get it? Submitted by : JK Answer There are 7 possible answers to it. Assume that there are B Buffaloes, C Cows and S Sheep. It is given that a Buffalo produces 10 pounds of milk, a Cow produces 1 pound of mile and a Sheep produces 0.25 pound of milk. Also, we need 100 pound of milk from exactly 100 animals, we get 2 equations. 10*B + C + 0.25*S = 100 --------- (I) B + C + S = 100 --------- (II) Subtracting equation (II) from equation (I) 9*B - 0.75*S = 0 36*B - 3*S = 0 (multiplied by 4) 12*B - S = 0 (division by 3) 12*B = S Now, for different values of B, we will get multiple solutions. * 1 Buffalo, 12 Sheep, 87 Cows * 2 Buffaloes, 24 Sheep, 74 Cows * 3 Buffaloes, 36 Sheep, 61 Cows * 4 Buffaloes, 48 Sheep, 48 Cows * 5 Buffaloes, 60 Sheep, 35 Cows * 6 Buffaloes, 72 Sheep, 22 Cows * 7 Buffaloes, 84 Sheep, 9 Cows ======================================= Brain Teaser No : 00526 What are the next two number in the series? 1, 2, 4, 7, 28, 33, 198, ?, ? Answer The next two numbers are 205 and 1640. The pattern is : 1 + 1 * 2 + 3 * 4 + 5 * 6 + 7 * 8 + 9 * 10 or to the current number add and multiply alternatively by the number's position to get the next number. First number = 1 Second number = 1 + 1 = 2 Third number = 2 * 2 = 4 Fourth number = 4 + 3 = 7 Fifth number = 7 * 4 = 28 Sixth number = 28 + 5 = 33 Seventh number = 33 * 6 = 198 Eight number = 198 + 7 = 205 Ninth number = 205 * 8 = 1640 Tenth number = 1640 + 9 = 1649 ================================== Brain Teaser No : 00537 There is a 4-character code, with 2 of them being letters and the other 2 being numbers. How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive. Submitted by : Destructo_girl Answer The maximum number of attempts required are 16,22,400 There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers). Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@ Hence, the required answer is = 52*52*10*10*6 = 16,22,400 attempts = 1.6 million approx. Thanks to Tim Sanders for opening BrainVista's brain !!! ============================= Brain Teaser No : 00643 A, B and C are three points on a straight line, not necessarily equidistant with B being between A and C. Three semicircles are drawn on the same side of the line with AB, BC and AC as the diameters. BD is perpendicular to the line ABC, and D lies on the semicircle AC. If the funny shaped diagram between the three semicircles has an area of 1000 square cms, find the length of BD. Answer The length of BD is 35.68 cms There are 3 right-angled triangles - ABD, CBD and ADC. From ABD, AB^2 + BD^2 = AD^2 ------ I From CBD, CB^2 + BD^2 = CD^2 ------ II From ADC, AD^2 + CD^2 = AC^2 ------ III Adding I and II, AB^2 + BC^2 + 2*BD^2 = AD^2 + CD^2 ------ IV FROM III and IV AB^2 + BC^2 + 2*BD^2 = AC^2 AB^2 + BC^2 + 2*BD^2 = (AB+CB)^2 2*BD^2 = 2*AB*CB BD^2 = AB*CB BD = SQRT(AB*CB) Given that funny shaped diagram beween three semicircles has an area of 1000 square cms. [PI/2 * (AC/2)^2] - [PI/2 * (AB/2)^2] - [PI/2 * (BC/2)^2] = 1000 PI/8 * [AC^2 - AB^2 - BC^2] = 1000 PI * [(AB+BC)^2 - AB^2 - BC^2] = 8000 PI * [2*AB*BC] = 8000 AB * BC = 4000/PI Hence BD = SQRT(4000/PI) = 35.68 cms where PI = 3.141592654 Hence, the length of BD is 35.68 cms. ================================== Brain Teaser No : 00142 There are four groups of Mangoes, Apples and Bananas as follows: Group I : 1 Mango, 1 Apples and 1 Banana Group II : 1 Mango, 5 Apples and 7 Bananas Group III : 1 Mango, 7 Apples and 10 Bananas Group IV : 9 Mango, 23 Apples and 30 Bananas Group II costs Rs 300 and Group III costs Rs 390. Can you tell how much does Group I and Group IV cost? Answer Group I costs Rs 120 and Group IV costs Rs 1710 Assume that the values of one mango, one apple and one banana are M, A and B respectively. From Group II : M + 5A + 7B = 300 From Group III : M + 7A + 10B = 390 Subtracting above to equations : 2A + 3B = 90 For Group I : = M + A + B = (M + 5A + 7B) - (4A + 6B) = (M + 5A + 7B) - 2(2A + 3B) = 300 - 2(90) = 300 - 180 = 120 Similarly, for Group IV : = 9M + 23A + 30B = 9(M + 5A + 7B) - (22A + 33B) = 9(M + 5A + 7B) - 11(2A + 3B) = 9(300) - 11(90) = 2700 - 990 = 1710 Thus, Group I costs Rs 120 and Group IV costs Rs 1710. ============================= Brain Teaser No : 00580 In Column-I below, are given some words. These have been translated into a code language. The code equivalents of the words in Column-I are given in Column-II, not necessarily opposite to the corresponding words. Also, the codes for the different letters in each word have also not been given the same order as these letter occur in the original word. COLUMN-I COLUMN-II ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ TAPE moi COP lhhpok TIE nls ROTATE nhpk SAY nkpl TREAT msr YEAR khlph SIP hrp TYRE pmlh Can you decode the individual letter codes? Answer We first find the exact codes of the each given words. ROTATE is a 6-letter word. So its code is lhhpok. And h is for T. TREAT is a 5-letter word. So its code is khlph. The 4-letter words are TAPE, YEAR, TYRE and codes are nhpk, nkpl, pmlh. YEAR and TYRE have 3 letters in common (Y, E, R). They must be either nhpk or nkpl. Hence, the code for TAPE is pmlh and m is for P. Also the code for TYRE is nhpk (as h is for T) and the code for YEAR is nkpl. The 3-letter words are COP, TIE, SAY, SIP and codes are moi, nls, msr, hrp. The code for TIE is hrp. The code for SIP is msr. The code for COP is moi. And the code for SAY is nls. Thus, the words and their codes are: ROTATE lhhpok TREAT khlph TAPE pmlh TYRE nhpk YEAR nkpl TIE hrp COP moi SIP msr SAY nls So far we know that h is for T and m is for P. In SAY and SIP, the common letter is S stands for s. In TYRE and SAY, the common letter Y stands for n. Thus, in SAY, the remaining letter A stands for l. In TIE and SIP, the common letter I stands for r. Thus, in TIE, the remaining letter E stands for p. In ROTATE and COP, the common letter O stands for o. Thus, in ROTATE, the remaining letter R stands for k. Also, in COP, the remaining letter C stands for i. Summerizing h-T, i-C, k-R, l-A, m-P, n-Y, o-O, p-E, r-I, s-S ======================================== Brain Teaser No : 00266 What are the next two letters in the series? E, O, E, R, E, X, N, T, E, _, _ Submitted by : Katie Cinquegrana Answer The next two letters in the series are N and N The pattern is - Last letter of the numbers starting from 1 when written in English i.e. onE, twO, threE, fouR, fivE, siX, seveN, eighT, ninE, teN, eleveN, twelvE, etc... ===================================== Brain Teaser No : 00274 Find all sets of consecutive integers that add up to 1000. Submitted by : James Barberousse Answer There are total 8 such series: 1. Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000. (-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000 2. Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202. (-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000 3. Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70. (-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000 4. Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000 5. Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52. 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000 6. Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70. 55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000 7. Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202. 198 + 199 + 200 +201 + 202 = 1000 8. Sum of 1 number starting from 1000. 1000 = 1000 ============================== Brain Teaser No : 00437 One side of the bottom layer of a triangular pyramid has 12 balls. How many are there in the whole pyramid? Note that the pyramid is equilateral and solid. Submitted by : Angela Parr Answer There are total 364 balls. As there are 12 balls along one side, it means that there are 12 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55, 66 and 78 balls in the remaining layers. Hence, the total number of balls are = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 = 364 balls ================================ Brain Teaser No : 00146 Sachin, Dravid and Ganguly played in a Cricket match between India and England. * None of them scored more than 99 runs. * If you add the digits of the runs scored by Sachin to his own score, you will get the runs scored by Dravid. * If you reverse the digits of the runs scored by Dravid, you will get the runs scored by Ganguly. * The total runs scored by them is 240. Can you figure out their individual scores? Answer Sachin, Dravid and Ganguly scored 75, 87 and 78 respectively. Sachin's score must be less than 86, otherwise Dravid's score would be more than 99. Also, he must have scored atleast 42 - incase Dravid and Ganguly scored 99 each. Also, as none of them scored more than 99 and the total runs scored by them is 240; their individual scores must be around 80. Now, use trial-n-error method to solve the teaser. =========================== Brain Teaser No : 00196 Veeru says to Jay, "Can you figure out how many Eggs I have in my bucket?" He gives 3 clues to Jay: If the number of Eggs I have 1. is a multiple of 5, it is a number between 1 and 19 2. is not a multiple of 8, it is a number between 20 and 29 3. is not a multiple of 10, it is a number between 30 and 39 How many Eggs does Veeru have in his bucket? Answer 32 eggs Let's apply all 3 condition separately and put all possible numbers together. First condition says that if multiple of 5, then the number is between 1 and 19. Hence, the possible numbers are (5, 10, 15, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39) Second condition says that if not a multiple of 8, then the number is between 20 and 29. Hence, the possible numbers are (8, 16, 20, 21, 22, 23, 25, 26, 27, 28, 29, 32) Third condition says that if not a multiple of 10, then the number is between 30 and 39. Hence, the possible numbers are (10, 20, 31, 32, 33, 34, 35, 36, 37, 38, 39) Only number 32 is there in all 3 result sets. That means that only number 32 satisfies all three conditions. Hence, Veeru have 32 eggs in his bucket. =================================== Brain Teaser No : 00222 Consider a chessboard with a single Rook. A Rook can move any number of square sideways/forward, but not diagonally. What is the minimum number of moves the Rook needs to make, in order to pass over all the squares on the chessboard and return to the original position? Note: Take any square as a starting position for the Rook. Answer 16 moves As a Rook can move any number of square sideways/forward, but not diagonally and there are 8 rows and 8 columns on the chessboard; the Rook needs minimum 16 moves to pass over all the squares and return to the original position. ================================= Brain Teaser No : 00295 For a TV talk show on Bollywood, the producer must choose a group of two Directors and two Musicians. At least one of them must be an Actor and at least one a Singer. 1. Available Directors are Mahesh Bhatt, Karan Johar, Subhash Ghai, Aditya Chopra and Ashutosh. 2. Available Musicians are A R Rehman, Annu Malik, Sandeep Chowta and Aadesh Srivastava. 3. Shubash Ghai, A R Rehman and Annu Malik are Actors. 4. Aditya Chopra and Aadesh Srivastava are Singers. 5. A R Rehman will not seat in the same room with Subhash Ghai, and will take part only if Mahesh Bhatt is there. 6. Aditya Chopra refuses to take part with Annu Malik. 7. Ashutosh refuses to take part with Aadesh Srivastava. How many acceptable groups can the producer put together? Answer The producer can put 9 acceptable groups together. Your basic solution step is to make a table of the possible persons and qualifications on the Musicians side, with the possible combinations on the Directors side. Note that A R Rehman insists on Mahesh Bhatt being present, but the reverse is not true. Also, remember that there must be at least one actor and at least one singer. The valid combinations are: Musician Director ---------------------------------------------------------- (A R Rehman, Sandeep Chowta) (Mahesh Bhatt, Aditya Chopra) (A R Rehman, (Mahesh Bhatt, Karan Johar) Aadesh Srivastava) (Mahesh Bhatt, Aditya Chopra) (Annu Malik, (Mahesh Bhatt, Karan Johar) Aadesh Srivastava) (Mahesh Bhatt, Subhash Ghai) (Karan Johar, Subhash Ghai) (Sandeep Chowta, (Mahesh Bhatt, Subhash Ghai) Aadesh Srivastava) (Karan Johar, Subhash Ghai) (Aditya Chopra, Subhash Ghai) Thus, there are total 9 acceptable groups. ======================================== Brain Teaser No : 00570 * 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? Answer There are 7 females and 6 males. Assume that there are four sexes - male, female, X and Y. Prepare the following tree based on the data given : sister 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. =========================================== Brain Teaser No : 00614 Place the numbers from 1 to 15 in the blocks below, without repeating any of them, such a way that the sum of the numbers in any two adjacent blocks is always a perfect square. [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] A whole number is said to be a perfect square if it is equal to some whole number multiplied by itself. e.g. 36 (6*6), 121 (11*11) Answer [9] [7] [2] [14] [11] [5] [4] [12] [13] [3] [6] [10] [15] [1] [8] Find out all the possible pairs of the numbers from 1 to 15 that sum up to a perfect square. 1: (1,3), (1,8), (1,15) 2: (2,7), (2,14) 3: (3,6), (3,13) 4: (4,5), (4,12) 5: (5,4), (5,11) 6: (6,3), (6,10) 7: (7,2), (7,9) 8: (8,1) 9: (9,7) 10: (10,6), (10,15) 11: (11,5), (11,14) 12: (12,4), (12,13) 13: (13,3), (13,12) 14: (14,2), (14,11) 15: (15,1), (15,10) Note that there are 2 numbers that with just a one pair i.e. (8,1) and (9,7). Thus, 8 and 9 must be at the end. [9] [7] [] [] [] [] [] [] [] [] [] [] [] [1] [8] Now, 7 have only one other pair i.e. (7,2) [9] [7] [2] [] [] [] [] [] [] [] [] [] [] [1] [8] Similarly, there is only one other pair of 2 i.e. (2,14) [9] [7] [2] [14] [] [] [] [] [] [] [] [] [] [1] [8] Following the same procedure, the final solution is [9] [7] [2] [14] [11] [5] [4] [12] [13] [3] [6] [10] [15] [1] [8] =================================== Brain Teaser No : 00311 Replace the letters with the correct numbers. T W O X T W O --------- T H R E E Submitted by : Timmy Chan Answer T=1, W=3, O=8, H=9, R=2, E=4 1 3 8 x 1 3 8 ------------ 1 9 0 4 4 You can reduce the number of trials. T must be 1 as there is multiplication of T with T in hundred's position. Also, O can not be 0 or 1. Now, you have to find three digit number whose square satisfies above conditions and square of that has same last two digits. Hence, it must be between 102 and 139. ======================================= Brain Teaser No : 00319 Substitute digits for the letters to make the following relation true. S T I L L + W I T H I N -------------------- L I M I T S Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all other S in the puzzle must be 3. Answer The value of L must be one more than W i.e. L=W+1 and there must be one carry from S+I=I. Also, the value of S must be 9 as S+I=I with one carry from T+T=M, which means that the value of T must be greater than 4. From I+H=I, the value of H must be 0 as the value of S is 9. Now, applying all those constraints and using trial-n-error, we get two possible answers. 9 7 1 6 6 9 8 5 3 3 + 5 1 7 0 1 3 + 2 5 8 0 5 6 --------------- --------------- 6 1 4 1 7 9 3 5 6 5 8 9 ========================================= Brain Teaser No : 00444 Silu and Meenu were walking on the road. Silu said, "I weigh 51 Kgs. How much do you weigh?" Meenu replied that she wouldn't reveal her weight directly as she is overweight. But she said, "I weigh 29 Kgs plus half of my weight." How much does Meenu weigh? Answer Meenu weighs 58 Kgs. It is given that Meenu weighs 29 Kgs plus half of her own weight. It means that 29 Kgs is the other half. So she weighs 58 Kgs. Solving mathematically, let's assume that her weight is X Kgs. X = 29 + X/2 2*X = 58 + X X = 58 Kgs ================================== Brain Teaser No : 00665 The three of us made some bets. First, Waldo won from Molly as much as Waldo had originally. Next, Molly won from Spike as much as Molly then had left. Finally, Spike won from Waldo as much as Spike then had left. We ended up having equal amounts of money. I started with 50 cents. Who am I? Answer I am Molly. Let Waldo start with W, Molly with M, and Spike with S. Following the above account we get the following progression of money: W = 2*W, M = M-W M = 2*(M-W), S = S-(M-W) = S+W-M S = 2*(S+W-M), W = 2*W-(S+W-M) = W+M-S And so Waldo finished with W+M-S, Molly with 2*(M-W), and Spike with 2*(S+W-M). Since these must all be equal, we have three equations and three unknowns, so solve; this gives us that 4*M = 5*S and 3*S = 4*W. Now, if S = 1/2 this implies that W = 3/8 = 37.5 cents, an impossible amount of money to start with. If W = 1/2 this implies that S = 2/3, again an impossible amount to start with. Finally, if M = 1/2 this implies that S = 2/5 = 40 cents, W = 3/10 = 30 cents, which works. It follows that I am Molly. ================================ Brain Teaser No : 00210 There are 10 cups placed on a table such that 3 are face up and 7 are bottom up. A move is defined as inverting a pair (compulsorily) of cups. What is the minimum number of moves required to make all the cups face the same way? Answer It is not at all possible. There is no way that one can do that with given Move. A move is defined as inverting a pair of cups, compulsorily. Also, there are odd number of cups face up (3)and odd number of cups bottom up (7). Now whenever you make a move you have to invert 2 cups compulsorily. Hence, always odd number of cups will be face up and bottom up, whatever move you make. ==================================== Brain Teaser No : 00366 There are three bulbs on 19th floor and there are three switches X, Y and Z on the ground floor. Each switch belongs to one bulbs, not necessarily in order. You can switch on or off as many times you want but you can go on 19th floor only once. How will you find out which swich belongs to which bulb? Note that you are the only person over there. You can't go outside and can't use any tools. Answer The bulb becomes hot, if you keep it on for some time. This is the characteristics of the bulb and we can use it to solve the teaser. Switch on the switch X and wait for some time say 10 minutes, then switch it off. Now switch on the switch Y and go up to the 19th floor and check the bulbs. The bulb which is hot but not lit corresponds to switch X. The bulb which is lit corresponds to switch Y. The third bulb which is not lit and is cold corresponds to switch Z. ===================================== Brain Teaser No : 00650 Laloo, Baloo and Naloo were at the Health Club on the same day this month. 1. They all joined the health club last month. 2. One of them goes every 2 days, another one goes every 3 days and the third one goes every 7 days. 3. Laloo went to the health club for the first time this month on a Monday, Baloo went to the health club for the first time this month on a Wednesday and Naloo went to the health club for the first time this month on a Friday. 4. Exactly one of them was at the health club on the first day of this month. On which day of this month did Laloo, Baloo and Naloo meet? Note that here "this month" means any one month in general. Answer They met on the 27th of this month. Find out individual dates on which they were at the club. From (2), 2-day man went to the health club for the first time this month on the 1st or 2nd. Also, 3-day man went to the health club for the first time this month on the 1st, 2nd or 3rd. There are two possible cases: Case I : Laloo went on Monday, the 1st and every two days thereafter. Baloo went on Wednesday, the 3rd and every three days thereafter. Naloo went on Friday, the 5th and every seven days thereafter. Then, Laloo's dates - 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31 Baloo's dates - 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Naloo's dates - 5, 12, 19, 26 Case II : Baloo went on Wednesday, the 1st and every two days thereafter. Naloo went on Friday, the 3rd and every three days thereafter. Laloo went on Monday, the 6th and every seven days thereafter. Then, Baloo's dates - 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31 Naloo's dates - 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Laloo's dates - 6, 13, 20, 27 It is given that they were at the club on the same day this month, then Case I is not possible. And from Case II, they met on the 27th of this month. ======================================= Brain Teaser No : 00053 A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out $25,000. Since the man does not want to be seen carrying that much money, he places it in 15 evelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (i.e. no two tens in place of a twenty). At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes number(s) 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount. How many ones did the auctioneer find in the envelopes? Answer Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of bills possible of any available US currency" is only to misguide you. This is always possible for any amount !!! One more thing to notice here is that the man must have placed money in envelopes in such a way that if he bids for any amount less than $25000, he should be able to pick them in terms of envelopes. First envelope contains, 20 = $1 Second envelope contains, 21 = $2 Third envelope contains, 22 = $4 Fourth envelope contains, 23 = $8 and so on... Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617 Last envelope (No. 15) contains only $8617 as total amount is only $25000. Now as he bids for $8322 and gives envelope number 2, 8 and 14 which contains $2, $128 and $8192 respectively. Envelope No 2 conrains one $2 bill Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and one $1 bill Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one $2 bill Hence the auctioneer will find one $1 bill in the envelopes. ==================================== Brain Teaser No : 00140 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? Answer 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 can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further. =================================== Brain Teaser No : 00566 Here is the family tree of Mr. RAHUL RAHUL | --------------------------------------------- | | | | RATISH YASH OM TRILOK | | ? -------- ------------------ | | | | | AMAR AMIT RAM HARSH ASHOK | | ----- ------- | | | | | How many children does Mr. TRILOK have? Answer TRILOK have 5 children. Name of the person and number of his children are related by some pattern. Assign each vowel following values. A=0 E=1 I=2 O=3 U=4 The number of children to any person is the sum of the values represented by vowels in his name. RATISH = 0(A) + 2(I) = 2 OM = 3(O) = 3 AMIT = 0(A) + 2(I) = 2 ASHOK = 0(A) + 3(O) = 3 TRILOK = 2(I) + 3(O) = 5 Hence, TRILOK have 5 children. ====================================== Brain Teaser No : 00856 Find the least number which when divided by 35, leaves remainder 25; when divided by 45, leaves remainder 35 and when divided by 55, leaves remainder 45. Answer 3455 The answer is LCM of (35, 45, 55) minus 10. LCM of (35, 45, 55) is 3465. Hence, the answer is 3455. ============================= Brain Teaser No : 00094 There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner. What is the probability that they don't collide? Answer Let's mark the corners of the triangle as A,B,C. There are total 8 ways in which ants can move. 1. A->B, B->C, C->A
2. A->B, B->C, C->B
3. A->B, B->A, C->A
4. A->B, B->A, C->B
5. A->C, C->B, B->A
6. A->C, C->B, B->C
7. A->C, C->A, B->A
8. A->C, C->A, B->C
Out of which, there are only two cases under which the ants won't collide :
* A->B, B->C, C->A
* A->C, C->B, B->A
Therefore, probability of ants not colliding : 2/8 = 1/4
================================
Brain Teaser No : 00116
Complete the series :
5, 20, 24, 6, 2, 8, ?
Answer
12
Note the sequence of operations:
5 * 4 = 20
20 + 4 = 24
24 / 4 = 6
6 - 4 = 2
2 * 4 = 8
There is arithmatic operation on the current number to get next number in the series. The sequence of operations is *, +, / and - by 4.
So after multiplying by 4, next operation is addition of 4. So the next number is 8 + 4 = 12
=====================================
Brain Teaser No : 00219
The floor area of Milan's house is 1007 square feet. The space diagonal of the house s 59.75 feet.
What will be the height of the house? Note that floor is rectangle in shape. Also, length and width are integer.
Answer
20 feet
The floor area is 1007 square feet. The only possible floor dimension is 19 by 53. Thus the floor diagonal is 56.3 feet.
The space diagonal is 59.75 feet which makes right-angled triangle with floor diagonal and height. Hence, the height of the house is 20 feet.
===========================================
Brain Teaser No : 00456
B, J and P are related to each other.
1. Among the three are B's legal spouse, J's sibling and P's sister-in-law.
2. B's legal spouse and J's sibling are of the same sex.
Who is the married man?
Answer
J is the married man.
Note that a person's sister-in-law may be the wife of that person's brother or the sister of that person's spouse.
There are 2 cases:
1. If B's legal spouse is J, then J's sibling must be P and P's sister-in-law must be B.
2. If B's legal spouse is P, then P's sister-in-law must be J and J's sibling must be B.
It is given that B's legal spouse and J's sibling are of the same sex. Also, it is obvious
that P's sister-in-law is female. Then, B's legal spouse and J's sibling both must be
males.
B's spouse J's sibling P's sister-in-law
(male) (male) (female)
------------------------------------------------------
Case I J P B
Case II P B J
Case II is not possible as B & P are married to each other and both are male. Hence, J is
the married man.
====================================
Brain Teaser No : 00506
What is the last digit of 746? In other words, what will the remainder be, if 746 is
divided by 10?
Don't try to solve this on calculator, you may get the wrong answer. Also, do explain your
answer.
Answer
The last digit of 746 is 9.
The powers of any number have a repeating pattern for the last digit. It can be found
easily without performing the entire multiplication of each power.
Let's consider the powers of 7.
7N Value Last Digit
70 1 1
71 7 7
72 49 9
73 343 3
74 2401 1
75 16807 1
76 117649 1
Note that there is a repeating pattern of four numbers (1, 7, 9, 3) for the powers of 7.
Hence, the last digit of 744 will be 1, of 745 will be 7 and of 746 will be 9.
Also, there is no need to actually perform the entire multiplication. Start with 1,
multiply it by 7, discard all digits except units and multiply again by 7 and so on.
==============================
Brain Teaser No : 00259
What is the smallest number which when divided by 10 leaves a remainder of 9, when divided
by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7, when divided by 7
leaves a remainder of 6 and so on until when divided by 2 leaves a remainder of 1?
Answer
The smallest such number is 2519.
The easiest way is to find the Least Common Multiple (LCM) of 2, 3, 4, 5, 6, 7, 8 and 9.
And subtract 1 from it.
The LCM of 2, 3, 4, 5, 6, 7, 8 and 9 is given by 2520. Hence, the required number is 2519.
=================================
Brain Teaser No : 00262
What TWO number will come next in the series :
6, -1, 7, -2, 8, -3, 9, -4, 10, -5, _, _
Submitted by : David
Answer
11 and -6
The pattern is : odd numbers are counting up from 6 and even numbers are counting down from
-1.
The other way is : 6-7+8-9+10-11+12-13+14-15+16-17 and so on.
=================================
Brain Teaser No : 00392
The cricket match between India and Pakistan was over.
* Harbhajan scored more runs than Ganguly.
* Sachin scored more runs than Laxman but less than Dravid
* Badani scored as much runs as Agarkar but less than Dravid and more than Sachin.
* Ganguly scored more runs than either Agarkar or Dravid.
Each batsman scored 10 runs more than his immediate batsman. The lowest score was 10 runs.
How much did each one of them score?
Answer
A simple one. Use the given facts and put down all the players in order. The order is as
follow with Harbhajan, the highest scorer and Laxman, the lowest scorer.
1. Harbhajan
2. Ganguly
3. Dravid
4. Badani, Agarkar
5. Sachin
6. Laxman
Also, as the lowest score was 10 runs. Laxman must have scored 10, Sachin 20, Badani &
Agarkar 30 and so on.
1. Harbhajan - 60 runs
2. Ganguly - 50 runs
3. Dravid - 40 runs
4. Badani, Agarkar - 30 runs each
5. Sachin - 20 runs
6. Laxman - 10 runs
=================================
Brain Teaser No : 00025
What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?
Think carefully !!!
Answer
A tricky one.
7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder
left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0.
The only part to be consider is
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873
The remainder left after dividing 873 by 7 is 5
Hence, the remainder is 5.
====================================
Brain Teaser No : 00182
Subhash is 14 inches taller than Jatin. The difference between Subhash and Sanjeev is two
inches less than between Sanjeev and Jatin. Subhash at 6'6" is the tallest.
How tall are Sanjeev and Jatin?
Answer
Sanjeev is 6' and Jatin is 5'4"
It is given that Subhash at 6'6" is the tallest and also he is 14 inches taller than Jatin.
It means that Jatin is 5'4".
Now as the difference between Subhash and Sanjeev is two inches less than between Sanjeev
and Jatin, Sanjeev's height is more than Jatin's. And that is 6'.
Thus, Sanjeev is 6' and Jatin is 5'4".
==============================
Brain Teaser No : 00229
Difference between Bholu's and Molu's age is 2 years and the difference between Molu's and
Kolu's age is 5 years.
What is the maximum possible value of the sum of the difference in their ages, taken two at
a time?
Submitted by : Cake
Answer
The maximum possible value of the sum of the difference in their ages - taken two at a time
- is 14 years.
It is given that -
"Difference between Bholu's and Molu's age is 2 years"
"Difference between Molu's and Kolu's age is 5 years"
Now, to get the maximum possible value, the difference between Bholu's and Kolu's age
should be maximum i.e. Molu's age should be in between Bholu's and Kolu's age. Then, the
difference between Bholu's and Kolu's age is 7 years.
Hence, the maximum possible value of the sum of the difference in their ages - taken two at
a time - is (2 + 5 + 7) 14 years.
==================================
Brain Teaser No : 00574
Major Jasbir is forming five-person Special Task Group. The group must contain one leader,
two bomb-experts and two soldiers.
P, Q and R are possible bomb-experts. R, S and T are possible leaders. U, V and W are
possible soldiers. Also, P and R prefers to work with each other in the same team. T
prefers to work only if V works.
How many different possible Groups, Major Jasbir can make?
Answer
Major Jasbir can make 8 different possible groups.
As 2 bomb-experts to be selected from the given 3 and also P & R prefers to work together,
PR must be there in all the possible Groups. Also, T prefers to work only if V works. It
doesn't mean that V won't work without T.
Hence, possible groups are:
PR - S - UV
PR - S - VW
PR - S - WU
PR - T - UV
PR - T - VW
PQ - R - UV
PQ - R - VW
PQ - R - WU
Hence, there 8 different groups are possible.
=============================
Brain Teaser No : 00630
In a school, main-computer password changes after every hour based on set of words chosen
for each day. The following is the set of passwords for a particular day.
9AM-1st Password: is not ready cloth brain bath simple
10AM-2nd Password: ready not is cloth simple bath brain
11AM-3rd Password: cloth is not ready brain bath simple
12PM-4th Password: not is cloth ready simple bath brain
1PM-5th Password: ready cloth is not brain bath simple
Following the same rule, what will the 6th Password, if the 1st passoward is "Shy Shelly
Says She Shall Sew Sheets"?
Answer
The 6th Password will be "Shy She Says Shelly Sheets Sew Shall".
The pattern is - In the first step, the first three and the last three words are written in
a reverse order. Then, the first four and the last three words are written in a reverse
order. The process is repeated to obtain successive passwords.
9AM - 1st Password: Shy Shelly Says She Shall Sew Sheets
10AM - 2nd Password: Says Shelly Shy She Sheets Sew Shall
11AM - 3rd Password: She Shy Shelly Says Shall Sew Sheets
12PM - 4th Password: Shelly Shy She Says Sheets Sew Shall
1PM - 5th Password: Says She Shy Shelly Shall Sew Sheets
2PM - 6th Password: Shy She Says Shelly Sheets Sew Shall
==========================================
Brain Teaser No : 00655
Substitute digits for the letters to make the following addition problem true.
T H A T S
T H E
+ T H E O R Y
---------------------
A N Y W A Y
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one
mapping between digits and letters. e.g. if you substitute 3 for the letter Y, no other
letter can be 3 and all other Y in the puzzle must be 3.
Answer
T=8, H=6, A=9, S=7, E=3, O=2, R=4, Y=1, N=5, W=0
Its tough one and requires lots of trial-n-error. But it can be reduce from the following
deductions:
A = T + 1, T <> 1
S + E = 10
H + R = 10
one of (O, Y, N, W) is 0
H > 1
There are 8 possible values for (A,T) - (2,3), (3,4), (4,5), (5,6), (6,7), (7,8), (8,9)
Also, there are 4 possible value pairs for (S,E) and (H,R) - (1,9), (2,8), (3,7), (4,6),
but not necessarily in the same order.
Now, taking possible values of (A,T) one at a time and evaluating corresponding (S,E) and
(H,R) values, we can find the answer.
T H A T S 8 6 9 8 7
T H E 8 6 3
+ T H E O R Y + 8 6 3 2 4 1
--------------------- ---------------------
A N Y W A Y 9 5 1 0 9 1
===============================
Brain Teaser No : 00242
There is a shortage of tubelights, bulbs and fans in a village - Kharghar. It is found that
* All houses do not have either tubelight or bulb or fan.
* exactly 19% of houses do not have just one of these.
* atleast 67% of houses do not have tubelights.
* atleast 83% of houses do not have bulbs.
* atleast 73% of houses do not have fans.
What percentage of houses do not have tubelight, bulb and fan?
Answer
42% houses do not have tubelight, bulb and fan.
Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100
tubelights, 100 bulbs and 100 fans.
From the given data, we know that there is shortage of atleast (67+83+73) 223 items in
every 100 houses.
Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should
account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do
not have 2 items each, there would be a shortage of 162 items. But total of 204 items are
short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and
fan.
Thus, 42% houses do not have tubelight, bulb and fan.
=====================================
Brain Teaser No : 00248
A series comprising of alphabets contains only seven letters. The first five letters in the
given series are A, B, D, O, P
Can you find out the remaining two letters?
Answer
The remaining two letters are Q and R.
In the English Alphabet, there are only 7 uppercase letters with enclosed spaces or holes
in them. They are A, B, D, O, P, Q and R.
====================================
Brain Teaser No : 00255
Find next two numbers in the series:
60, 30, 20, 15, ?, ?
Answer
12 and 10
The pattern is 60 divide by the position of the number.
The first number is 60/1 = 60
The second number is 60/2 = 30
The third number is 60/3 = 20
The fourth number is 60/4 = 15
The fifth number is 60/5 = 12
The sixth number is 60/6 = 10
Hence the next two numbers in the series are 12 and 10.
======================================
Brain Teaser No : 00324
You have just built a house. However, you may only take certain items into the house. You
can take doors, but you aren't allowed any windows. You can't have, but you are allowed
coffee. A book is okay, but no paper. Finally, you can take a puppy or kitten, but you
can't take a dog or cat.
Give one more thing that you can/can't take, and explain why.
Submitted by : J W
Answer
Anything with double letters in spelling is allowed and without double letters is not
allowed.
If you you look at the spelling of the names of the items that are allowed in the house,
you will notice that they all have double letters i.e. DOOR, BOOK, PUPPY, KITTEN
So you can take items like Coffee, Balloon, Pillow.
================================
Brain Teaser No : 00620
Nicole and Sharon, together have more than 10 but fewer than 30 bodyguards. One day, one of
the bodyguards, Cruz, decided to leave Nicole and join Sharon. Now both the females had the
same number of bodyguards.
Eventually, Cruz rejoined Nicole. Also, Jim decided to leave Sharon and join Nicole. Now,
both the females had a prime number of bodyguards.
How many bodyguards did each have now?
Answer
Nicole have 11 bodyguards and Sharon have 7 bodyguards.
If one bodyguard is moved from one group to other, it decreases one group by one and
increases other by one.
Before Cruz joined Sharon, Nicole had two more bodyguards than Sharon. (Since Cruz's switch
made the number of bodyguards of both the females equal)
When Cruz switched back to Nicole, once again Nicole had two more bodyguards than Sharon.
Then, Jim moved from Sharon to Nicole decreasing Sharon's bodyguards by one and increasing
Nicole's bodyguards by one. Hence, Nicole had four more bodyguards than Sharon.
It is given that together had more than 10 but fewer than 30 bodyguards. Also, at the end
they both had prime number of bodyguards. The only numbers fit the given conditions are 7
and 11.
Hence, now Nicole have 11 bodyguards and Sharon have 7 bodyguards.
====================================
Brain Teaser No : 00784
The population of an island consists of two and only two types of people : the knights, who
invariably tell the truth and the knaves who always lie.
One day a stanger in the island, met four inhabitants. He asked the first one whether the
second was a Knight or a Knave. The reply was "Knave". Similarly, the second inhabitant
said that the third was a "Knave", and the third said the same about the fourth.
Now he asked the fourth inhabitant that what would the first have said about the third. The
reply once again was "Knave".
Who is the fourth inhabitant, Knight or Knave?
Answer
The fourth inhabitant is a Knave.
Take two cases.
Assume that the first inhabitant was a Knight. It means that second inhabitant was a Knave
as the first was telling the truth. Similarly, the third inhabitant was a Knight as the
second was lying. Inturn, the forth inhabitant was a Knave as the third was telling the
truth.
Now the first and the third both are Knight. So the first would have said "Knight" for the
third person. But the forth would have said "Knave" and he did. Hence, the fourth person
was Knave.
Tryout second case yourself i.e. assume that the first inhabitant was a Knave.
==============================
Brain Teaser No : 00085
You have 3 baskets, & each one contains exactly 4 balls, each of which is of the same size.
Each ball is either red, black, white, or purple, & there is one of each color in each
basket.
If you were blindfolded, & lightly shook each basket so that the balls would be randomly
distributed, & then took 1 ball from each basket, what chance is there that you would have
exactly 2 red balls?
Answer
There are 64 different possible outcomes, & in 9 of these, exactly 2 of the balls will be
red. There is thus a slightly better than 14% chance [(9/64)*100] that exactly 2 balls will
be red.
A much faster way to solve the problem is to look at it this way. There are 3 scenarios
where exactly 3 balls are red:
1 2 3
-----------
R R X
R X R
X R R
X is any ball that is not red.
There is a 4.6875% chance that each of these situations will occur.
Take the first one, for example: 25% chance the first ball is red, multiplied by a 25%
chance the second ball is red, multiplied by a 75% chance the third ball is not red.
Because there are 3 scenarios where this outcome occurs, you multiply the 4.6875% chance of
any one occurring by 3, & you get 14.0625%
==================================
Brain Teaser No : 00197
There are 70 employees working with BrainVista of which 30 are females. Also,
* 30 employees are married
* 24 employees are above 25 years of age
* 19 married employees are above 25 years, of which 7 are males
* 12 males are above 25 years of age
* 15 males are married.
How many unmarried females are there and how many of them are above 25?
Answer
15 unmarried females & none are above 25 years of age.
Simply put all given information into the table structure and you will get the answer.
Married Unmarried
Below 25 Above 25 Below 25 Above 25
Female 3 12 15 0
Male 8 7 20 5
=======================================
Brain Teaser No : 00370
You have 14 apples. Your Friend Marge takes away 3 and gives you 2. You drop 7 but pick up
4. Bret takes 4 and gives 5. You take one from Marge and give it to Bret in exchange for 3
more. You give those 3 to Marge and she gives you an apple and an orange. Frank comes and
takes the apple Marge gave you and gives you a pear. You give the pear to Bret in exchange
for an apple. Frank then takes an apple from Marge, gives it to Bret for an orange, gives
you the orange for an apple.
How many pears do you have?
Submitted by : Big Mike
Answer
None
Frank gave you a pear in exchange of the apple which Marge gave you. And you gave that pear
to Bret in exchange for an apple. All the others exchanges involved apples and/or organges.
================================
Brain Teaser No : 00403
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?
Answer
(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 reqiured for remaining (N - 4) secret agents.
Total telephone calls require are
= (N - 2) + 2 + (N - 4)
= 2N - 4
Let\'s take an example. Say there are 4 secret agents W, X, Y & Z. Divide them into two
groups of 2 each i.e. (W, X) and (Y, Z). Here, 4 telephone calls are required.
1. W will call up X.
2. Y will call up Z.
3. W, who knows WX will call up Y, who knows YZ.
4. X, who knows WX will call up Z, who knows YZ.
Take an another example. Say there are 5 secret agents J, K, L, M & N. Divide them into two
groups i.e. (J, K) and (L, M, N). Here, 6 telephone calls are required.
1. J will call up K.
2. L will call up M.
3. M will call up N. Now M and N know LMN.
4. J, who knows JK will call up M, who knows LMN.
5. K, who knows JK will call up N, who knows LMN.
6. L will call up to anyone of four.
=====================================
Brain Teaser No : 00450
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.
Answer
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.
==================================
Brain Teaser No : 00559
If you were to dial any 7 digits on a telephone in random order, what is the probability
that you will dial your own phone number?
Assume that your telephone number is 7-digits.
Submitted by : Alessandro Tabora
Answer
1 in 10,000,000
There are 10 digits i.e. 0-9. First digit can be dialed in 10 ways. Second digit can be
dialed in 10 ways. Third digit can be dialed in 10 ways. And so on.....
Thus, 7-digit can be dialed in 10*10*10*10*10*10*10 (=10,000,000) ways. And, you have just
one telephone number. Hence, the possibility that you will dial your own number is 1 in
10,000,000.
Note that 0123456 may not be a valid 7-digit telephone number. But while dialing in random
order, that is one of the possible 7-digit number which you may dial.
===================================
Brain Teaser No : 00200
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.
Answer
1, 2, 4, 8, 16, 32, 64, 128, 256, 489
Let's start from scratch.
* The apple vandor 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.
====================================
Brain Teaser No : 00170
A series comprising of alphabets contains 13 letters. The first seven letters in the given
series are A, E, F, H, I, L, M
Can you find the next two letters?
Answer
The next letters in the series are N, O, R, S, U, X.
The pattern is - letters whose English names (Phonetic Pronunciations) start with vowels.
=================================
Brain Teaser No : 00271
Substitute digits for the letters to make the following Division true
O U T
-------------
S T E M | D E M I S E
| D M O C
-------------
T U I S
S T E M
----------
Z Z Z E
Z U M M
--------
I S T
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one
mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other
letter can be 3 and all other M in the puzzle must be 3.
Submitted by : Calon
Answer
C=0, U=1, S=2, T=3, O=4, M=5, I=6, Z=7, E=8, D=9
It is obvious that U=1 (as U*STEM=STEM) and C=0 (as I-C=I).
S*O is a single digit and also S*T is a single digit. Hence, their values (O, S, T) must be
2, 3 or 4 (as they can not be 0 or 1 or greater than 4).
Consider, STEM*O=DMOC, where C=0. It means that M must be 5. Now, its simple. O=4, S=2,
T=3, E=8, Z=7, I=6 and D=9.
O U T 4 1 3
------------- -------------
S T E M | D E M I S E 2 3 8 5 | 9 8 5 6 2 8
| D M O C | 9 5 4 0
------------- -------------
T U I S 3 1 6 2
S T E M 2 3 8 5
---------- ----------
Z Z Z E 7 7 7 8
Z U M M 7 1 5 5
-------- --------
I S T 6 2 3
Also, when arranged from 0 to 9, it spells CUSTOMIZED.
=========================================
Brain Teaser No : 00180
Which number in the series does not fit in the given series:
1 4 3 16 6 36 7 64 9 100
Answer
This is a series with odd positions containing position number whereas even positions
containing square of the position.i.e. even position numbers are 4 16 36 64 100 and odd
position numbers are 1 3 5 7 9
Hence, 6 does not fit in the series. It should be 5.
=========================================
Brain Teaser No : 00194
Sita says to Gita, "I have 3 sons. They are all less than 10 years in age and all are of
different age. The product of the ages of the two younger brothers is equal to the age of
the eldest brother. Also, the sum of their ages is a Prime number."
How old is the eldest son?
Answer
The eldest son is 6 years old.
First list out all combinations of three numbers such that all of them are less than 10 and
product of first two is third number. There are only 3 combination.
2 2 4
2 3 6
2 4 8
3 3 9
Also, note that age of the younger brother can not be 1 year. As in that case other two
will be twins and it is mentioned in teaser that "the eldest brother" i.e. there is only
one brother with the eldest age.
Now check fo the second condition i.e. the sum of their ages is a Prime number.
2 + 2 + 4 = 8
2 + 3 + 6 = 11
2 + 4 + 8 = 14
3 + 3 + 9 = 15
11 is the Prime number. Hence, the required combination is 2, 3 and 6. The eldest son is 6
years old.
==================================
Brain Teaser No : 00754
Mark ate half of a pizza on Monday. He ate half of what was left on Tuesday and so on. He
followed this pattern for one week.
How much of the pizza would he have eaten during the week?
Submitted by : Ashley Krapf
Answer
Mark would have ate 127/128 (99.22%) of the pizza during the week.
Mark ate half the pizza on Monday. On Tuesday, he would have ate half of the remaining
pizza i.e. 1/4 of the original pizza. Similarly, he would have ate 1/8 of the original
pizza on Wednesday and so on for the seven days.
Total pizza Mark ate during the week is
= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
= 127/128
= 99.22% of the original pizza
================================
Brain Teaser No : 00087
Find the next number in the series 11 12 20 23 33 46 .....
Answer
All the numbers given are the fibonacci numbers in increasing bases starting 2 !!!
Therefore,
11 = 3 base 2
12 = 5 base 3
20 = 8 base 4
23 = 13 base5 etc ...
Answer = 67 which is 55 base 8
(Fibonacci series : 2, 3, 5, 8, 13, 21, 34, 55, 89..... where next number is sum of the
previous to numbers.)
===================================
Brain Teaser No : 00108
Substitute digits for the letters to make the following relation true.
S O M E
+ G O O D
----------
I D E A S
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one
mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other
letter can be 3 and all other S in the puzzle must be 3.
Answer
By just looking at the teaser, one can guess that I must be 1.
Now, take the right side of the teaser i.e. E + D = S and take the left side of the teaser
i.e. S + G = D. Here note that S + G should be greater than 11 as E or D can not be 0 (E +
D = S) and I is 1.
With trial and error, and also by applying such constraints, we get
8 7 9 5
+ 4 7 7 3
----------
1 3 5 6 8
==============================
Brain Teaser No : 00117
I bought a car with a peculiar 5 digit numbered licence plate which on reversing could
still be read. On reversing value is increased by 78633.
Whats the original number if all digits are different?
Answer
Only 0 1 6 8 and 9 can be read upside down. So on rearranging these digits we get the
answer as 10968.
================================
Brain Teaser No : 00120
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
Answer
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
===================================
Brain Teaser No : 00471
SlowRun Express runs between Bangalore and Mumbai, For the up as well as the down journey,
the train leaves the starting station at 10:00 PM everyday and reaches the destination at
11:30 PM after three days.
Mr. Haani once travelled by SlowRun Express from Mumbai to Bangalore. How many SlowRun
Express did he cross during his journey?
Answer
Mr. Haani crossed 7 SlowRun Expresses during his journey.
Let's say that Mr. Haani travelled by SlowRun Express on Wednesday 10:00PM from Mumbai. The
first train he would have crossed is the one scheduled to arrive at Mumbai at 11:30 PM the
same day i.e. the one that left Bangalore at 10:00 PM on last Sunday.
Also, he would have crossed the last train just before reaching Bangalore on Saturday.
Thus, Mr. Haani must have crossed 7 SlowRun Expresses during his journey.
===================================
Brain Teaser No : 00039
Find next number in the series :
3, 7, 31, 211, ?
Answer
1831
All the numbers in the series are Prime Numbers. So the next number will also be a prime
number.
Two consecutive numbers in the series
(A and B) Total Prime numbers bewtween A and B
(C) Prime number just before B
(D) E = (C + D)
3 and 7 1 5 6
7 and 31 6 29 35
31 and 211 35 199 234
* Number after 7 is the prime number on skipping 6 prime numbers after 7 i.e. 31
* Number after 31 is the prime number on skipping 35 prime numbers after 31 i.e. 211
* Hence, number after 211 is the prime number on skipping 234 prime numbers after 211
i.e. 1831
The other possible answer is 1891.
Subtract 1 from each number in the given series: 2, 6, 30, 210
First number = 2*1 = 2
Second number = 2*3 = 6
Third number = 6*5 = 30
Fourth number = 30*7 = 210
Fifth number = 210*9 = 1890
Sixth number = 1890*11 = 20790
Thus, the pattern is : multiply previous number by next odd number and add one to the
multiplication. Thus, the series is 3, 7, 31, 211, 1891, 20791, ...
Thanks to N. Anand for this much more simpler answer.
====================================
Brain Teaser No : 00047
The ratio of Boys to Girls is 6:4. 60% of the boys and 40% of the girls take lunch in the
canteen. What % of class takes lunch in canteen?
Answer
Assume there are 6X boys and 4X Girls
Total Students taking lunch in canteen
= (6X)(60/100) + (4X)(40/100)
= 36(X/10) + 16(X/10)
= 52(X/10)
Total students are = 6X + 4X = 10X
% of class taking lunch in canteen
= ((52X/10) * 100 ) / 10X
= 52 %
===================================
Brain Teaser No : 00699
Here in England McDonald's has just launched a new advertising campaign. The poster shows 8
McDonald's products and underneath claims there are 40312 combinations of the above items.
Given that the maximum number of items allowed is 8, and you are allowed to have less than
8 items, and that the order of purchase does not matter (i.e. buying a burger and fries is
the same as buying fries and a burger)
How many possible combinations are there? Are McDonald's correct in claiming there are
40312 combinations?
Submitted by : Alex Crosse
Answer
Total possible combinations are 12869.
It is given that you can order maximum of 8 items and you are allowed to have less than 8
items. Also, the order of purchase does not matter. Let's create a table for ordering total
N items using X products.
Items
Ordered
(N) Products Used (X)
1 2 3 4 5 6 7 8
1 1 - - - - - - -
2 1 1 - - - - - -
3 1 2 1 - - - - -
4 1 3 3 1 - - - -
5 1 4 6 4 1 - - -
6 1 5 10 10 5 1 - -
7 1 6 15 20 15 6 1 -
8 1 7 21 35 35 21 7 1
Total (T) 8 28 56 70 56 28 8 1
Ways to choose
X products from
8 products (W) 8C1 8C2 8C3 8C4 8C5 8C6 8C7 8C8
Total combinations
(T*W) 64 784 3136 4900 3136 784 64 1
Thus, total possible combinations are
= 64 + 784 + 3136 + 4900 + 3136 + 784 + 64 + 1
= 12869
=================================
Brain Teaser No : 00003
In a hotel, rooms are numbered from 101 to 550. A room is chosen at random. What is the
probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?
Answer
There are total 450 rooms.
Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not
there) Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6
So the probability is 90/450 i.e. 1/5 or 0.20
=================================
Brain Teaser No : 00015
In the town called Alibaug, the following facts are true:
* No two inhabitants have exactly the same number of hairs.
* No inhabitants has exactly 2025 hairs.
* There are more inhabitants than there are hairs on the head of any one inhabitants.
What is the largest possible number of the inhabitants of Alibaug?
Answer
2025
It is given that no inhabitants have exactly 2025 hairs. Hence there are 2025 inhabitants
with 0 to 2024 hairs in the head.
Suppose there are more than 2025 inhabitants. But these will violate the condition that
"There are more inhabitants than there are hairs on the head of any one inhabitants." As
for any number more than 2025, there will be same number of inhabitants as the maximum
number of hairs on the head of any inhabitant.
=======================================
Brain Teaser No : 00035
In a sports contest there were m medals awarded on n successive days (n > 1).
1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.
2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so
on.
3. On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?
Answer
Total 36 medals were awarded and the contest was for 6 days.
On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6
I got this answer by writing small program. If anyone know any other simpler method, do
submit it.
==================================
Brain Teaser No : 00330
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?
Answer
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.
==================================
Brain Teaser No : 00520
Somebody marked the six faces of a die with the numbers 1, 2 and 3 - each number twice. The
die was put on a table. Four people - Abu, Babu, Calu and Dabu - sat around the table so
that each one was able to see only three sides of the die at a glance.
* Abu sees the number 1 and two even numbers.
* Babu and Calu can see three different numbers each.
* Dabu sees number 2 twice and he can't remember the third number.
What number is face down on the table?
Answer
Number 3 is face down on the table.
If Abu can see two even numbers i.e. number 2 twice, and if Dabu can see number 2 twice,
then number 2 must be facing up.
Now everything else is simple. (see the following diagram)
Dabu Abu
1
3 2 2
1
Calu Babu
Thus, the number hidden from the view is number 3 and hence the answer.
========================================
Brain Teaser No : 00291
A drinks machine offers three selections - Tea, Coffee or Random (Either tea or Coffee) but
the machine has been wired up wrongly so that each button does not give what it claims.
If each drink costs 50p, how much minimum money do you have to put into the machine to work
out which button gives which selection?
Submitted by : Alex Crosse
Answer
You have to put just 50p.
Put 50p and push the button for Random. There are only 2 possibilities. It will give either
Tea or Coffee.
* If it gives Tea, then the button named Random is for Tea. The button named Coffee is
for Random selection. And the button named Tea is for Coffee.
* If it gives Coffee, then the button named Random is for Coffee. The button named Tea
is for Random selection. And the button named Coffee is for Tea.
Thus, you can make out which button is for what by putting just 50p and pressing Random
selection first.
===============================
Brain Teaser No : 00298
Which of the following numbers is the odd one out, and why?
1, 2, 3, 5, 9, 13, 21
Note that 2 is not the odd one.
Submitted by : Brett Hurrell
Answer
The odd number is 9.
It is a Fibonacci Series - a series in which next number is summation of previous 2
numbers.
The first two numbers are 1 and 2.
Third number is = 1 + 2 = 3
Fourth number is 3 + 5 = 8
Fifth number is 5 + 8 = 13
Sixth number is 8 + 13 = 21
Hence 9 is the odd number It should 8.
===================================
Brain Teaser No : 00315
A farmer built a fence around his 25 cows in a square region. He built it in such a way
that one can see 5 poles from either of the four sides.
What are the minimum number of poles the farmer must have used?
Answer
16 poles
X X X X X
X X
X X
X X
X X X X X
One pole at each corner and three poles along each side so that one can always see 5 poles
from either of the four sides. The corner pole is shared by two sides and hence reducing
the number of poles to 16.
==============================
Brain Teaser No : 00579
Mr. Wagle goes to work by a bus. One day he falls asleep when the bus still has twice as
far to go as it has already gone.
Halfway through the trip he wakes up as the bus bounces over some bad potholes. When he
finally falls asleep again, the bus still has half the distance to go that it has already
travelled. Fortunately, Mr. Wagle wakes up at the end of his trip.
What portion of the total trip did Mr. Wagle sleep?
Answer
Mr. wagle slept through half his trip.
Let's draw a timeline. Picture the bus route on a line showen below:
---------------- ________ -------- ________________
Start 1/3 1/2 2/3 End
----- shows time for which Mr. Wagle was not sleeping
_____ shows time for which Mr. Wagle was sleeping
When Mr. Wagle fell asleep the first time, the bus sill had twice as far to go as it had
already gone, that marks the first third of his trip.
He wake up halfway through the trip i.e slept from 1/3 mark to the 1/2 mark. He fell sleep
again when the bus still had half the distance to go that it had already traveled i.e 2/3
mark.
Adding up, all sleeping times,
= (1/2 - 1/3) + (1 - 2/3)
= 1/6 + 1/3
= 1/2
Hence, Mr. wagle slept through half his trip.
============================
Brain Teaser No : 00153
In a Road Race, one of the three bikers was doing 15km less than the first and 3km more
than the third. He also finished the race 12 minutes after the first and 3 minutes before
the third.
Can you find out the speed of each biker, the time taken by each biker to finish the race
and the length of the course?
Assume that there were no stops in the race and also they were driving with constant speeds
through out the race.
Answer
Let us assume that
Speed of First biker = V1 km/min
Speed of Second biker = V2 km/min
Speed of Third biker = V3 km/min
Total time take by first biker = T1 min
Total distance = S km
Now as per the data given in the teaser, at a time T min
X1 = V1 * T ----> 1
X1 - 15 = V2 * T ----> 2
X1 - 18 = V3 * T ----> 3
At a Distance S Km.
S = V1 * T1 ----> 4
S = V2 * (T1 + 12) ----> 5
S = V3 * (T1 + 15) ----> 6
Thus there are 6 equations and 7 unknown data that means it has infinite number of
solutions.
By solving above 6 equations we get,
Time taken by first biker, T1 = 60 Min.
Time taken by Second biker, T2 = 72 Min.
Time taken by first biker, T3 = 75 Min.
Also, we get
Speed of first biker, V1 = 90/T km/min
Speed of second biker, V2 = (5/6)V1 = 75/T km/min
Speed of third biker, V3 = (4/5)V1 = 72/T km/min
Also, the length of the course, S = 5400/T km
Thus, for the data given, only the time taken by each biker can be found i.e. 60, 72 and 75
minutes. For other quantities, one more independent datum is required i.e. either T or V1
or V2 or V3
Thanks to Theertham Srinivas for the answer !!!
==========================================
Brain Teaser No : 00313
A group of friends went on a holiday to a hill station. It rained for 13 days. But when it
rained in the morning, the afternoon was lovely. And when it rained in the afternoon, the
day was preceded by clear morning.
Altogether there were 11 very nice mornings and 12 very nice afternoons. How many days did
their holiday last?
Answer
The holiday last for 18 days.
Let's assume the number of days as follows:
Rain in the morning and lovely afternoon = X days
Clear morning and rain in the afternoon = Y days
No rain in the morning and in the afternoon = Z days
Number of days with rain = X + Y = 13 days
Number of days with clear mornings = Y + Z = 11 days
Number of days with clear afternoons = X + Z = 12 days
Solving above 3 equations, we get X = 7, Y = 6 and Z = 5
Hence, total number of days on holiday = 18 days