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Logical puzzles & answers – For Interviews, Placement, Competitive and Entrance Examinations
In More InfoQuestion: 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
Question: Find sum of digits of D.
Let
A= 1999^{1999}
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 of D is 1.
Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
Consider,
A = 1999^{1999}
< 2000^{2000}
= 2^{2000} * 1000^{2000}
= 1024^{200} * 10^{6000}
< 10^{800} * 10^{6000}
= 10^{6800}
i.e. A < 10^{6800}
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E <= 9
i.e. E is a single digit number.
Also, 1999 = 1 (mod 9) so 1999^{1999} = 1 (mod 9)
Therefore we conclude that E=1.
Question: 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?
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(50X)] X meters whereas the platoon moved (50X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50X)
(50+X)*(50X) = 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.
Question: If you take a marker & start from a corner on a cube, what is the maximum number of edges you can trace across if you never trace across the same edge twice, never remove the marker from the cube, & never trace anywhere on the cube, except for the corners & edges?
Answer: 9.
To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube. There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but is useful for solving many types of spatial puzzles.
Question: 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.
Question: Three men – Sam, Cam and Laurie – are married to Carrie, Billy and Tina, but not necessarily in the same order. Sam’s wife and Billy’s Husband play Carrie and Tina’s husband at bridge. No wife partners her husband and Cam does not play bridge. Who is married to Cam?
Answer: Carrie is married to Cam.
“Sam’s wife and Billy’s Husband play Carrie and Tina’s husband at bridge.” It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina. As Cam does not play bridge, Billy’s husband must be Laurie. Hence, Carrie is married to Cam.
Question: There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24. Find the tractors each originally had?
Answer: X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing. It’s given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48). Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24).
Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12).
Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
Question: A certain street has 1000 buildings. A signmaker is contracted to number the houses from 1 to 1000. How many zeroes will he need?
Answer: The signmaker will need 192 zeroes.
Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ……. (901..1000)
For the first group, signmaker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192
Question: There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?
Answer
It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter. Take 8 coins and weigh 4 against 4. If both are not equal, goto step 2. If both are equal, goto step 3. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H’s is heavier or one of L’s is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing. If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3. If both are equal, L4 is the odd coin and is lighter. If L2 is light, L2 is the odd coin and is lighter. If L3 is light, L3 is the odd coin and is lighter. If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2. If both are equal, there is some error. If H1 is heavy, H1 is the odd coin and is heavier. If H2 is heavy, H2 is the odd coin and is heavier. If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4. If both are equal, L1 is the odd coin and is lighter. If H3 is heavy, H3 is the odd coin and is heavier. If H4 is heavy, H4 is the odd coin and is heavier. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing. If both are equal, there is some error.If X is heavy, X is the odd coin and is heavier.If X is light, X is the odd coin and is lighter.
Question: In a sports contest there were m medals awarded on n successive days (n > 1).
 On the first day 1 medal and 1/7 of the remaining m – 1 medals were awarded.
 On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
 On the n^{th} 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
Question: 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 19 in some order.
Find the number?
Answer: 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.
Question: 1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.What part of the contents of the container is left at the end of the second day?
Answer
Assume that contents of the container is X
On the first day 1/3rd is evaporated.
(1 – 1/3) of X is remaining i.e. (2/3)X
On the Second day 3/4th is evaporated.
Hence,
(1 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X
Hence 1/6th of the contents of the container is remaining.
Question:Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one. For how long did Vipul study in candle light?
Answer
Vipul studied for 3 hours in candle light.
Assume that the initial lenght of both the candle was L and Vipul studied for X hours.
In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4
After X hours, total thick candle remaining = L – XL/6
After X hours, total thin candle remaining = L – XL/4
Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L – XL/6) = 2(L – XL/4)
(6 – X)/6 = (4 – X)/2
(6 – X) = 3*(4 – X)
6 – X = 12 – 3X
2X = 6
X = 3
Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.
Question: If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on. How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?
Answer: Rs.333,062,500
To begin with, you want to know the total number of days: 365 x 50 = 18250.
By experimentation, the following formula can be discovered, & used to determine the amount earned for any particular day: 1 + 2(x1), with x being the number of the day. Take half of the 18250 days, & pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, & you will see that if you add these pairs together, they always equal Rs.36500.
Multiply this number by the total number of pairs (9125), & you have the amount you would have earned in 50 years.
Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11…..upto 18250 terms)
Question: A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, & now his salary is Rs. 22702.68. What was his salary to begin with?
Answer: Rs.22176.
Assume his salary was Rs. X
He earns 5% raise. So his salary is (105*X)/100
A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68
Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68
X = 22176
Question: At 6’o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12’o.
Answer: 66 seconds.
It is given that the time between first and last ticks at 6’o is 30 seconds.
Total time gaps between first and last ticks at 6’o = 5
(i.e. between 1 & 2, 2 & 3, 3 & 4, 4 & 5 and 5 & 6)
So time gap between two ticks = 30/5 = 6 seconds.
Now, total time gaps between first and last ticks at 12’o = 11
Therefore time taken for 12 ticks = 11 * 6 = 66 seconds (and not 60 seconds)
Question: 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.
Question: In Mr. Mehta’s family, there are one grandfather, one grandmother, two fathers, two mothers, one fatherinlaw, one motherinlaw, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughterinlaw. How many members are there in Mr. Mehta’s family? Give minimal possible answer.
Answer
There are 7 members in Mr. Mehta’s family. Mother & Father of Mr. Mehta, Mr. & Mrs. Mehta, his son and two daughters.
Mother & Father of Mr. Mehta


Mr. & Mrs. Mehta


One Son & Two Daughters
Question: When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, “If you tell a lie, you will put to the sword, and if you tell the truth you will be hanged.” The soldier could make only one statement. He made that statement and went free. What did he say?
Answer
The soldier said, “You will put me to the sword.”
The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!
Question: 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
Question: The game of TicTacToe is being played between two players. Only the last mark to be placed in the game as shown. Who will win the game, O or X? Can you tell which was the sixth mark and at which position? Do explain your answer.
Answer: O will win the game. The sixth mark was X in square 9.
The 7th mark must be placed in square 5 which is the win situation for both X and O. Hence, the 6th mark must be placed in a line already containing two of the opponents marks. There are two such possibilities – the 6th mark would have been either O in square 7 or X in square 9.
As we know both the players are intelligent enough, the 6th mark could not be O in square 7. Instead, he would have placed O in square 5 and would have won.
Hence, the sixth mark must be X placed in square 9. And the seventh mark will be O. Thus O will win the game.
Question: At the Party:
 There were 9 men and children.
 There were 2 more women than children.
 The number of different manwoman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 manwoman couples possible.Also, of the three groups – men, women and children – at the party:
 There were 4 of one group.
 There were 6 of one group.
 There were 8 of one group.
Exactly one of the above 6 statements is false. Can you tell which one is false? Also, how many men, women and children are there at the party? Assume that both the players are intelligent enough.
Answer: Statement (4) is false. There are 3 men, 8 women and 6 children.
Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false.
So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true.
From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases – either there are 8 men and 3 women or there are 3 men and 8 women.
If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible.
Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.
Question: 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 (22319) 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 (204162) 42 houses do not have all 3 items – tubelight, bulb and fan.
Thus, 42% houses do not have tubelight, bulb and fan.
Question: Mr. Subramaniam rents a private car for AndheriColabaAndheri trip. It costs him Rs. 300 everyday. One day the car driver informed Mr. Subramaniam that there were two students from Bandra who wished to go from Bandra to Colaba and back to Bandra. Bandra is halfway between Andheri and Colaba. Mr. Subramaniam asked the driver to let the students travel with him. On the first day when they came, Mr. Subramaniam said, “If you tell me the mathematically correct price you should pay individually for your portion of the trip, I will let you travel for free.” How much should the individual student pay for their journey?
Answer
The individual student should pay Rs. 50 for their journey.
Note that 3 persons are travelling between Bandra and Colaba.
The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs Rs. 150.
For AndheriBandraAndheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.
For BandraColabaBandra, three persons i.e Mr. Subramaniam and two students, are travelling. Hence, each student would pay Rs. 50.
Question: 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 onetoone 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 beAnswer
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 IC=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).
Question: 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. At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?Answer: 4:21:49.5
Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.
For every minute, minute hand travels 6 degrees.
Hence, for X minutes it will travel 6 * X degrees.
For every minute, hour hand travels 1/2 degrees.
Hence, for X minutes it will travel X/2 degrees.
At 4.00 PM, the angle between minute hand and hour hand is 120 degrees. Also, after X minutes, minute hand and hour hand are exactly aligned. So the angle with respect to 12 i.e. Vertical Plane will be same. Therefore,
6 * X = 120 + X/2
12 * X = 240 + X
11 * X = 240
X = 21.8182
X = 21 minutes 49.5 seconds
Hence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.
Question: A soldier looses his way in a thick jungle. At random he walks from his camp but mathematically in an interesting fashion. First he walks one mile East then half mile to North. Then 1/4 mile to West, then 1/8 mile to South and so on making a loop. Finally how far he is from his camp and in which direction?
Answer
The soldier is 0.8944 miles away from his camp towards EastNorth.
It is obvious that he is in EastNorth direction.
Distance travelled in North and South directions
= 1/2 – 1/8 + 1/32 – 1/128 + 1/512 – 1/2048 + and so on… (a geometric series with r = (1/4) )
(1/2) * ( 1 – (1/4)^{n} )
= —————————
( 1 – (1/4) )
= 1 / ( 2 * ( 1 – (1/4) ) )
= 2/5
Similarly in East and West directions
= 1 – 1/4 + 1/16 – 1/64 + 1/256 – and so on… (a geometric series with r = (1/4) )
(1) * ( 1 – (1/4)^{n} )
= —————————
( 1 – (1/4) )
= 1 / ( ( 1 (1/4) )
= 4/5
So the soldier is 4/5 miles away towards East and 2/5 miles away towards North. So using right angled triangle, soldier is 0.8944 miles away from his camp.
Question: Raj has a jewel chest containing Rings, Pins and Earrings. The chest contains 26 pieces. Raj has 2 1/2 times as many rings as pins, and the number of pairs of earrings is 4 less than the number of rings. How many earrings does Raj have?
Answer: 12 earrings
Assume that there are R rings, P pins and E pair of earrings.
It is given that, he has 2 1/2 times as many rings as pins.
R = (5/2) * P or P = (2*R)/5
And, the number of pairs of earrings is 4 less than the number of rings.
E = R – 4 or R = E + 4
Also, there are total 26 pieces.
R + P + 2*E = 26
R + (2*R)/5 + 2*E = 26
5*R + 2*R + 10*E = 130
7*R + 10*E = 130
7*(E + 4) + 10*E = 130
7*E + 28 + 10*E = 130
17*E = 102
E = 6
Hence, there are 6 pairs of Earrings i.e. total 12 Earrings
Question: How many ways are there of arranging the sixteen black or white pieces of a standard international chess set on the first two rows of the board? Given that each pawn is identical and each rook, knight and bishop is identical to its pair.
Answer: 6,48,64,800 ways
There are total 16 pieces which can be arranged on 16 places in ^{16}P_{16} = 16! ways.
(16! = 16 * 15 * 14 * 13 * 12 * ….. * 3 * 2 * 1)
But, there are some duplicate combinations because of identical pieces.
 There are 8 identical pawn, which can be arranged in ^{8}P_{8} = 8! ways.
 Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in ^{2}P_{2} = 2! ways.
Hence, the require answer is
= (16!) / (8! * 2! * 2! * 2!)
= 6,48,64,800
Question: A person with some money spends 1/3 for cloths, 1/5 of the remaining for food and 1/4 of the remaining for travel. He is left with Rs 100/. How much did he have with him in the begining?
Answer: Rs. 250/
Assume that initially he had Rs. X
He spent 1/3 for cloths =. (1/3) * X
Remaining money = (2/3) * X
He spent 1/5 of remaining money for food = (1/5) * (2/3) * X = (2/15) * X
Remaining money = (2/3) * X – (2/15) * X = (8/15) * X
Again, he spent 1/4 of remaining maoney for travel = (1/4) * (8/15) * X = (2/15) * X
Remaining money = (8/15) * X – (2/15) * X = (6/15) * X
But after spending for travel he is left with Rs. 100/ So
(6/15) * X = 100
X = 250
Question: 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 (BA),
(60 * r) – (24 * r) = (30 * 60 * y) – (70 * 24 * y)
36 * r = 120 * y ——— D
Solving for (CB),
(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.
Question: 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.
Question: 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 onethird 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.
Question: Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5. How many such 3digit numbers are there?
Answer: There are 45 different 3digit 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 3digit 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 3digit numbers, where last digit is the sum of first two digits.
Question: Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.
Answer: 285714.
If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.
The simplest way is to write a small program. And the other way is trial and error !!!
Question: Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up. If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?
Answer: 52 / 2703
There are two cases to be considered.
CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B
Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53
So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)
CASE 2 : King of Hearts is not drawn from Pack A
Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53
So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)
Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
Question: 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.
 A>B, B>C, C>A
 A>B, B>C, C>B
 A>B, B>A, C>A
 A>B, B>A, C>B
 A>C, C>B, B>A
 A>C, C>B, B>C
 A>C, C>A, B>A
 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
Question: Find all sets of consecutive integers that add up to 1000.
Answer
There are total 8 such series:
 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  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  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  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  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  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  Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202.
198 + 199 + 200 +201 + 202 = 1000  Sum of 1 number starting from 1000.
1000 = 1000
Question: There is a 4character 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 casesensitive.
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 ^{4}C_{2} 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.
Question: 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?
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.
Question: 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 onetoone 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 trialnerror 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 1Question: 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.
 If the singer and the dancer are the same sex, then the dancer is older than the singer.
 If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer.
 If the singer is a man, then the singer and the dancer are the same age.
 If the singer and the dancer are of opposite sex then the man is older than the woman.
 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.
Question: 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 ;))
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%
Question: 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
Question: There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24. Find the tractors each originally had?
Answer
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing. It’s given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48). Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24). Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12). Hence, initially
X had 39 tractors, Y had 21 tractors and
Z had 12 tractors.
Question: 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.
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(50X)] X meters whereas the platoon moved (50X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50X)
(50+X)*(50X) = 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.
Question:Assume that you have enough coins of 1, 5, 10, 25 and 50 cents. How many ways are there to make change for a dollar? Do explain your answer.
Answer: There are 292 ways to make change for a dollar using coins of 1, 5, 10, 25 and 50 cents.
Let’s generalised the teaser and make a table as shown above.
If you wish to make change for 75 cents using only 1, 5, 10 and 25 cent coins, go to the .25 row and the 75 column to obtain 121 ways to do this.
The table can be created from lefttoright and toptobottom. Start with the top left i.e. 1 cent row. There is exactly one way to make change for every amount. Then calculate the 5 cents row by adding the number of ways to make change for the amount using 1 cent coins plus the number of ways to make change for 5 cents less using 1 and 5 cent coins.
Let’s take an example:
To get change for 50 cents using 1, 5 and 10 cent coins.
* 50 cents change using 1 and 5 cent coins = 11 ways
* (5010) 40 cents change using 1, 5 and 10 cent coins = 25 ways
* 50 cents change using 1, 5 and 10 cent coins = 11+25 = 36 ways
Let’s take another example:
To get change for 75 cents using all coins up to 50 cent i.e. 1, 5, 10, 25 and 50 cents coins.
* 75 cents change using coins upto 25 cent = 121 ways
* (7550) 25 cents change using coins upto 50 cent = 13 ways
* 75 cents change using coins upto 50 cent = 121+13 = 134 ways
For people who don’t want to tease their brain and love to do computer programming, there is a simple way. Write a small multiloop program to solve the equation: A + 5B + 10C + 25D + 50E = 100
where,
A = 0 to 100
B = 0 to 20
C = 0 to 10
D = 0 to 4
E = 0 to 2
The program should output all the possible values of A, B, C, D and E for which the equation is satisfied.
Question: 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
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 !!!
Question: What is the fourdigit number in which the first digit is 1/3 of the second, the third is the sum of the first and second, and the last is three times the second?
Answer: The 4 digit number is 1349.
It is given that the first digit is 1/3 of the second. There are 3 such possibilities.
 1 and 3
 2 and 6
 3 and 9
Now, the third digit is the sum of the first and second digits.
 1 + 3 = 4
 2 + 6 = 8
 3 + 9 = 12
It is clear that option 3 is not possible. So we are left with only two options. Also, the last digit is three times the second, which rules out the second option. Hence, the answer is 1349.
Question: 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?
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.
If it is given that:
25 – 2 = 3
100 x 2 = 20
36 / 3 = 2
Question: What is 144 – 3 = ?
Answer
There are 3 possible answers to it.
Answer 1 : 9
Simply replace the first number by its square root.
(25) 5 – 2 = 3
(100) 10 x 2 = 20
(36) 6 / 3 = 2
(144) 12 – 3 = 9
Answer 2 : 11
Drop the digit in the tens position from the first number.
(2) 5 – 2 = 3
1 (0) 0 x 2 = 20
(3) 6 / 3 = 2
1 (4) 4 – 3 = 11
You will get the same answer on removing left and right digit alternatively from the first number i.e remove left digit from first (2), right digit from second (0), left digit from third (3) and right digit from forth (4).
(2) 5 – 2 = 3
10 (0) x 2 = 20
(3) 6 / 3 = 2
14 (4) – 3 = 11
Answer 3 : 14
Drop left and right digit alternatively from the actual answer.
25 – 2 = (2) 3 (drop left digit i.e. 2)
100 * 2 = 20 (0) (drop right digit i.e. 0)
36 / 3 = (1) 2 (drop left digit i.e. 1)
144 – 3 = 14 (1) (drop right digit i.e. 1)
A 3 digit number is such that it’s unit digit is equal to the product of the other two digits which are prime. Also, the difference between it’s reverse and itself is 396.
Question:What is the sum of the three digits?
Answer: The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers – 224, 236, 326 and 339. Now, it is also given that – the difference between it’s reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.
Question: 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.
Question: At University of Probability, there are 375 freshmen, 293 sophomores, 187 juniors, & 126 seniors. One student will randomly be chosen to receive an award. What percent chance is there that it will be a junior? Round to the nearest whole percent
Answer: 19%
This puzzle is easy. Divide the number of juniors (187) by the total number of students (981), & then multiply the number by 100 to convert to a percentage.
Hence the answer is (187/981)*100 = 19%
Question: 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 7digits.
Answer: 1 in 10,000,000
There are 10 digits i.e. 09. 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, 7digit 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 7digit telephone number. But while dialing in random order, that is one of the possible 7digit number which you may dial.
Question: An anthropologist discovers an isolated tribe whose written alphabet contains only six letters (call the letters A, B, C, D, E and F). The tribe has a taboo against using the same letter twice in the same word. It’s never done. If each different sequence of letters constitues a different word in the language, what is the maximum number of sixletter words that the language can employ?
Answer
The language can employ maximum of 720 sixletter words.
It is a simple permutation problem of arranging 6 letters to get different sixletter words. And it can be done in in 6! ways i.e. 720 ways.
In otherwords, the first letter can be any of the given 6 letters (A through F). Then, whatever the first letter is, the second letter will always be from the remaining 5 letters (as same letter can not be used twice), and the third letter always be from the remaining 4 letters, and so on. Thus, the different possible sixletter words are 6*5*4*3*2*1 = 720
Question: Kate, Demi, Madona, Sharon, Britney and Nicole decided to lunch together in a restaurant. The waiter led them to a round table with six chairs. How many different ways can they seat?
Answer: There are 120 different possible seating arrangments.
Note that on a round table ABCDEF and BCDEFA is the same.
The first person can sit on any one of the seats. Now, for the second person there are 5 options, for the third person there are 4 options, for the forth person there are 3 options, for the fifth person there are 2 options and for the last person there is just one option.
Thus, total different possible seating arrangements are
= 5 * 4 * 3 * 2 * 1
= 120
Question: 3 blocks are chosen randomly on a chessboard. What is the probability that they are in the same diagonal?
AnswerThere are total of 64 blocks on a chessboard. So 3 blocks can be chosen out of 64 in ^{64}C_{3} ways.
So the sample space is = 41664There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.
3 blocks out of 8 blocks in diagonal can be chosen in ^{8}C_{3} ways.
But there are 2 such diagonals, hence favourables = 2 * ^{8}C_{3} = 2 * 56 = 112The require probability is
= 112 / 41664
= 1 / 372
= 0.002688
Question: What is the area of the triangle ABC with A(e,p) B(2e,3p) and C(3e,5p)?where p = PI (3.141592654)
AnswerA tricky ONE.Given 3 points are colinear. Hence, it is a straight line.
Hence area of triangle is 0.
Question: 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
Question: 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.
Question: A man has Ten Horses and nine stables as shown here.
[] [] [] [] [] [] [] [] [] The man wants to fit Ten Horses into nine stables. How can he fit Ten horses into nine stables?
Answer
The answer is simple. It says the man wants to fit “Ten Horses” into nine stables. There are nine letters in the phrase “Ten Horses”. So you can put one letter each in all nine stables.
Question: A man is at a river with a 9 gallon bucket and a 4 gallon bucket. He needs exactly 6 gallons of water. How can he use both buckets to get exactly 6 gallons of water? Note that he cannot estimate by dumping some of the water out of the 9 gallon bucket or the 4 gallon bucket
Answer
For the sack of explanation, let’s identify 4 gallon bucket as Bucket P and 9 gallon bucket as Bucket Q.
Operation  4 gallon bucket (Bucket P)  9 gallon bucket (Bucket Q) 
Initially  0  0 
Fill the bucket Q with 9 gallon water  0  9 
Pour 4 gallon water from bucket Q to bucket P  4  5 
Empty bucket P  0  5 
Pour 4 gallon water from bucket Q to bucket P  4  1 
Empty bucket P  0  1 
Pour 1 gallon water from bucket Q to bucket P  1  0 
Fill the bucket Q with 9 gallon water  1  9 
Pour 3 gallon water from bucket Q to bucket P  4  6 
9 gallon bucket contains 6 gallon of water, as required.
Question: Each of the five characters in the word BRAIN has a different value between 0 and 9. Using the given grid, can you find out the value of each character? B R A I N 31 B B R B A 31 N I A B B 32 N I B A I 30 I R A A A 23 37 29 25 27 29 The numbers on the extreme right represent the sum of the values represented by the characters in that row. Also, the numbers on the last raw represent the sum of the values represented by the characters in that column. e.g. B + R + A + I + N = 31 (from first row) Answer: B=7, R=6, A=4, I=5 and N=9 Make total 10 equations  5 for rows and 5 for columns  and sovle them. From Row3 and Row4, N + I + A + B + B = N + I + B + A + I + 2 B = I + 2 From Row1 and Row3, B + R + A + I + N = N + I + A + B + B  1 R = B  1 From Column2, R + B + I + I + R = 29 B + 2R + 2I = 29 B + 2(B  1) + 2I = 29 3B + 2I = 31 3(I + 2) + 2I = 31 5I = 25 I = 5 Hence, B=7 and R=6 From Row2, B + B + R + B + A = 31 3B + R + A = 31 3(7) + 6 + A = 31 A = 4 From Row1, B + R + A + I + N = 31 7 + 6 + 4 + 5 + N = 31 N = 9 Thus, B=7, R=6, A=4, I=5 and N=9Question: There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?
Answer
It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
 Take 8 coins and weigh 4 against 4.
 If both are not equal, goto step 2
 If both are equal, goto step 3
 One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H’s is heavier or one of L’s is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
 If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
 If both are equal, L4 is the odd coin and is lighter.
 If L2 is light, L2 is the odd coin and is lighter.
 If L3 is light, L3 is the odd coin and is lighter.
 If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
 If both are equal, there is some error.
 If H1 is heavy, H1 is the odd coin and is heavier.
 If H2 is heavy, H2 is the odd coin and is heavier.
 If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
 If both are equal, L1 is the odd coin and is lighter.
 If H3 is heavy, H3 is the odd coin and is heavier.
 If H4 is heavy, H4 is the odd coin and is heavier.
 The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
 If both are equal, there is some error.
 If X is heavy, X is the odd coin and is heavier.
 If X is light, X is the odd coin and is lighter.
In a sports contest there were m medals awarded on n successive days (n > 1).
 On the first day 1 medal and 1/7 of the remaining m – 1 medals were awarded.
 On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
 On the n^{th} and last day, the remaining n medals were awarded.
Question: 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.
Question: 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 19 in some order.
Find the number.
Answer: 381654729
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What’s New in the Chess NX v4.0 serial key or number?
Screen Shot
System Requirements for Chess NX v4.0 serial key or number
 First, download the Chess NX v4.0 serial key or number

You can download its setup from given links: