Chess NX v4.0 serial key or number

Chess NX v4.0 serial key or number

Chess NX v4.0 serial key or number

Chess NX v4.0 serial key or number

  • Agricultural machinery company to use software solutions from Siemens including product lifecycle management (PLM) to expand technological leadership
Siemens Digital Industries Software has announced that Maschinenfabrik Bernard Krone GmbH & Co. KG, a leading agricultural machinery company based in Germany, has selected the Siemens Digital Innovation Platform to help establish its digitalization strategy. Using solutions including the Teamcenter ® portfolio and NX™ software, Krone will launch a new digitalization program that will support processes over the entire product lifecycle as well as collaborative work between departments by leveraging the broad, integrated solutions that are part of the Siemens Digital Innovation Platform.
Augmented Reality (AR) has arrived in the workplace: For the last three years, a consortium of six companies and institutions chaired by Siemens has been researching the use of augmented reality (AR) in industry. The aim of the project, known as Glass@Service, was to be able to use intelligent data glasses as personalized information systems by combining them with new types of interactions, such as eye and gesture control, and innovative IT services. The first practical tests in actual production and logistics processes have now been successfully completed.
The family-run business Movanis develops and sells Automated Guided Vehicles, or more precisely, driverless forklift trucks. For optimum capture of their surroundings, these vehicles are fitted with cameras and scanners. In order to process the data and signals extremely efficiently and quickly, the vehicles are equipped with Siemens Industrial PC (IPC) control and communication technology. These vehicles, which represent state-of-the-art technology, allow Movanis’ customers to make optimum use of the benefits of an automated production and logistics environment. In addition to this, the vehicles compensate for staff shortages if, for example, there are not enough forklift truck drivers available.
Flottweg SE is a leading manufacturer of decanters, separators, and belt presses. The company took an unusual approach for a plant manufacturer to ensure operation of its plants and machines is extremely user-friendly: In collaboration with renowned design experts and based on user feedback, Flottweg developed a new interface, including visualization, which works on any plant. The new solution, which is easy to configure and maintain, was implemented using the Siemens Simatic WinCC Professional visualization software. Amongst other things, the focus was on using standard components which reduced the engineering workload by up to 20 percent during configuration. Flottweg’s customers benefit from an interface which has a clear and open structure, allowing easy interaction with the machine or plant. This saves time and reduces the probability of error during operation.
Founded in 1993, Jinyu Bio-technology Co. Ltd. is a maker of animal vaccines. To enjoy the benefits of the increased reliability, traceability, and safety that digitalized production offers, Jinyu Bio has embarked on an Industry 4.0 journey with Siemens. The experienced Siemens team has drawn up a ten-year digitalization solution covering the entire life cycle. The plan calls for integration at three levels – that is, end-to-end, vertical, and horizontal integration.
OCP Group is a world leader in the phosphate industry and one of the world’s largest fertilizer producers. Present throughout the value chain, OCP extracts, values and commercializes phosphate and phosphate products. The Group operates several mining sites and processing plants in Morocco. Each of these sites has a large stockyard to ensure even material flow. OCP has decided to equip some of these stockyards with a material tracking and management system (MAQ) from Siemens. As part of the industrial digitalization program, all reclaimers will be operated autonomously by MAQ. This new solution will ensure the autonomous driving of the machines without local operator, leading to increased efficiency of the reclaiming process. In addition, the Group will have access to operating data, such as the status of machinery and overall stock levels.
Having produced over six million radiators, the Stelrad Radiator Group (Stelrad) is one of the largest manufacturers in the heating, ventilation and air conditioning industry in Europe. The company focuses on providing high quality products, innovative and individual product development and a wide product range. To maintain its competitive edge, Stelrad has developed a carefully designed, long-term investment plan, which includes modernization of the automation system. The Dutch company is relying on control and drive technology from Siemens to achieve this. With the new solution, which was implemented in just 14 days, the company was able to reduce radiator lead times by 50 percent, shorten delivery times and increase the energy efficiency of the production line by optimizing the heating of the paint ovens through process automation.  
For over 300 years, Martell has been producing and exporting high quality cognac matured in oak casks. In order to improve coordination of the production teams and optimize material flow, Martell looked for a new time management and planning software. The company found what it was looking for with Siemens PLM Software. The new, user-friendly Simatic IT Preactor APS production and planning software allowed Martell to increase production by twelve percent and reduce delayed deliveries of specific ingredients to almost zero. At the same time, the company was able to significantly reduce the planning effort by 20 percent, the time required for tank filling by 30 percent, and even the team’s travel time between the individual production and warehousing sites.
Norway’s largest ferry connection, Moss-Horten, will run on electric power starting in 2021, powered by new battery technologies from Siemens. The ferry will be equipped with the largest battery pack that has ever been installed on a ferry of this size. Siemens’ water-cooled batteries allow record-high charging power and will reduce battery cost and C02 emissions. 
With the Sirius Asset Monitor MindSphere app, Siemens offers an app that can be used for the transparent mapping of low-voltage equipment in plants. The Simocode motor management system is also included. Simocode pro provides detailed information that contributes to optimal plant operation, efficient maintenance planning and successful energy monitoring. The app enables access to the motor management system anytime and anywhere. It provides the user with detailed information on the equipment status as well as error messages and alerts.
Источник: [https://torrent-igruha.org/3551-portal.html]
, Chess NX v4.0 serial key or number

Blog

Logical puzzles & answers – For Interviews, Placement, Competitive and Entrance Examinations

In More Info

Question: 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= 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 of 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
< 10800 * 106000
= 106800
i.e. A < 106800
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 19991999 = 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-(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.

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 sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?

Answer: The sign-maker 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, sign-maker 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).

  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

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 1-9 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(x-1), 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 father-in-law, one mother-in-law, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughter-in-law. 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 Tic-Tac-Toe 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:

  1. There were 9 men and children.
  2. There were 2 more women than children.
  3. The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.Also, of the three groups – men, women and children – at the party:
  4. There were 4 of one group.
  5. There were 6 of one group.
  6. 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 (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.

Question: Mr. Subramaniam rents a private car for Andheri-Colaba-Andheri 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 Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.

For Bandra-Colaba-Bandra, 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 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

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).

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 East-North.

It is obvious that he is in East-North 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 Ear-rings. 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 ear-rings.

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 Ear-rings i.e. total 12 Ear-rings

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 16P16 = 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 8P8 = 8! ways.
  • Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in 2P2 = 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 (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.

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 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.

Question: 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.

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.

  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

Question: Find all sets of consecutive integers that add up to 1000.

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

Question: 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.

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.

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 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

Question: 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.

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 G1People from G2No of ways to hire G1 without a single pair of indentical twinsNo of ways to hire G2Total ways
0510C010C5252
1410C110C42100
2310C2 * 8/910C34800
3210C3 * 8/9 * 6/810C23600
4110C4 * 8/9 * 6/8 * 4/710C1800
5010C5 * 8/9 * 6/8 * 4/7 * 2/610C032
Total11584

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-(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.

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 left-to-right and top-to-bottom. 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
* (50-10) 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
* (75-50) 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 multi-loop 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 four-digit 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. 1 and 3
  2. 2 and 6
  3. 3 and 9

Now, the third digit is the sum of the first and second digits.

  1. 1 + 3 = 4
  2. 2 + 6 = 8
  3. 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 7-digits.

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.

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 six-letter words that the language can employ?

Answer
The language can employ maximum of 720 six-letter words.
It is a simple permutation problem of arranging 6 letters to get different six-letter 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 six-letter 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 64C3 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 8C3 ways.
But there are 2 such diagonals, hence favourables = 2 * 8C3 = 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.

[T] [E] 1589 [H] [O] [R] [S] [E] [S]

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.

Operation4 gallon bucket
(Bucket P)
9 gallon bucket
(Bucket Q)
Initially00
Fill the bucket Q with 9 gallon water09
Pour 4 gallon water from bucket Q to bucket P45
Empty bucket P05
Pour 4 gallon water from bucket Q to bucket P41
Empty bucket P01
Pour 1 gallon water from bucket Q to bucket P10
Fill the bucket Q with 9 gallon water19
Pour 3 gallon water from bucket Q to bucket P46

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=9

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.

  1. Take 8 coins and weigh 4 against 4.
    • If both are not equal, goto step 2
    • If both are equal, goto step 3
  2. 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.
  3. If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
  4. If both are equal, L4 is the odd coin and is lighter.
  5. If L2 is light, L2 is the odd coin and is lighter.
  6. 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.

 

  1. 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).

  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.

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 1-9 in some order.

Find the number.

Answer: 381654729

Источник: [https://torrent-igruha.org/3551-portal.html]
Chess NX v4.0 serial key or number
prolific usb to serial xp

ps2 to 9pin serial adapter

lookup of car wheel serial number

nissan serial numbers

bs1 pro 2 58 serial crack

serial number igre

serial numbers altiris inventory wyse

windows vista ultimate serial crack

the sims serial killer

3dmark 06 serial

dvd next copy serial

sony vegas 5.0 serial key

goodsync pro v 6 serial

crown forklift serial numbers

kansas state serial rapist

eia-485 serial cable to pc

redhat enterprise linux 5 serial

cpu serial identifier

serial killers kill because

simple serial number tracking

office xp pro 2001 serial

techumseh engine serial numbers

serial cracks and registration numbers

serial codes dvdfab

serial number on a car

office 2007 serial

j 4 0a serial crack

wall street guru serial crack keygen

zonealarm security suite serial keygen

serial port email

rapid resizer 2.5.2 serial

product serial codes hacked spyware doctor

fruity loops 8 xxl serial

6 pin round serial port

autoshutdown 4 10 serial crack

magic iso v5.4 build 256 serial

serial to usbcable

serial number 1click dvdcopy5

komatsu models and serial numbers

serial search engines

serial port to usb adaptor driver

sams v4.2.2 serial

netcom 8 port serial

birth of america serial

silicon craft serial display modules

trane byc serial numbers

hp serial number from network

location free player serial

annuity serial contract

black widow thriller video tv serial

dvd x player serial number

quick heal antivirus 10 serial

xdata 7.3 serial ssg

winchester 94 ae serial numbers

elite keylogger 2.5 serial

serial killer soundtrack

mulit-port serial port dell

route66 navigate7 serial

serial attacks network

multimedia fusion 2 serial

how long serial cable

free serial key numbers

alcohol 120 1.9.2 serial

serial printer cable pinouts

get serial number in solaris

chevy motor serial number identication

astro 2.0 serial key

belkin usb to serial cable

usb serial 422

myof 1 0 20 serial crack

serial number roller coaster

serial killes want media coverage

boat hull serial number seach

final cut express 1 serial number

serial look up for software programs

vintage blaupunkt serial numbers

corel x4 crack serial keygen

color of 9 pin serial cable

homeworld 2 serial number

dvd next copy serial product key

find disk serial number xp

pro tools serial

sims serial number

extensis suitcase serial number

pci express serial cards

amethyst cadwizz serial

5.1 blind serial window

9 pins serial cable straight

serial killer movies tv

serial killer photos and biographies

newname pro 1 52 serial crack

adobe flash cs4 serial code

3rd planit serial

physic serial killers

serial killer england

gun's serial numbers

serial number bills

serial number font folio 9

malayalam serial actress sangeetha mohan

remington serial number year

equipment manager 1 0 serial crack

disc 2428.3 serial

galactic trader serial keygen

internet download accelerator 401 815 serial

newname pro 1 51 serial crack

bobcat serial number location

free serial numbers generator

serial cable dell axim

find xp serial

serial number for adobe reader 5

top serial site

pci x serial port

drums set serial number

serial number guides

serial keygens ultimate mobsters bot 2.0.6e

serial null connector

serial number for photo shop

sketchup 6 serial html

serial thief of baghdad

mitsumi serial mouse drivers

palm pilot software serial 20

ford 8n serial number location

nero 8 trial serial

photoshop cs3 extended serial number

spyware remover 11 serial crack

cute site builder serial

tribes 2 serial generator

peacemaker v1.05 serial number

winchester serial number identification

grandfather clock serial 2777

colt 1873 serial numbers

diskeeper 10 serial keygen patch

pcpitstop 2.0 serial warez

multisim 10 serial

proactive system password recovery serial

actual contacts for outlook serial

ray and faye copeland serial killers

the price is right serial

serial google earth

tally 9 serial

office xp 2002 genuine serial

play buddy serial numbers

littlesnitch 2.0.2 serial crack

serial eudora 7.1

fender squire serial number

doom 3 serial number

obtain hp serial number windows

adobe flash cs4 professional serial key

adobe creative suite free serial number

freeware pc serial connectio

no adware serial code

proscan 4 serial

r-wipe clean 7.7 serial

1975 ford f250 serial number codes

caterpillar ten 10 serial number list

mavis beacon deluxe 17 serial

daio station explorer serial crack keygen

acoustic guitar serial

yamaha cornet serial number

clinical research serial subject systematic sampling

hp-28c serial no firmware version

command and conquer generals serial code

intel universal serial bus driver

tunebite free serial

scanner serial port

warez serial photomatix pro

vee serial port

dvd cloner 5 crack serial

sims 2 seasons serial code

remind me 1 5 serial crack

barclays etf serial

henry portrait of serial killer movie

comm pci serial 8 port

ocd personality disorders in serial killers

audio notes v1.32 serial

magna excitor serial number

serial ata drive connector

registry cleaner v6.0 serial key

mws 0.94f key serial

serial numbers sibelius 4

upgrade serial number nero 7

xcopy serial dvd

identify hardware by serial number nvidia

nikon capture nx v1.2 serial

xp pro serial finder

network interface card parallel serial speeds

washburn guitar serial number code

crafted in japan fender serial number

once upon a knight serial

vmware workstation 6 serial key

henry lee allen serial killer

thunder popper and serial

barefoot ipmonitor 4 0 serial crack

serial number imprinters

serial number product key cracks

serial ata drive enclosure

serial killers box 3 disc

serial mouse instructions

windows washer 5 serial

monster truck stunt rally serial

free ares ultra serial code

antidote rx v2 serial number

serial numbers with green star

mozilla password recovery serial

prolific usb to serial cable

tv serial mahabharta download

uranium backup serial

getflv 5.8 serial crack torrents

nero 6 ultra serial number

jpeg recovery serial

b2k serial killer

lefthandedness and serial murders

ion multiport serial

microsoft windows xp serial number

usb to serial vista driver

imahustlababay serial killer

win97 reg code serial

steganos safe 5 serial

peachtree 2007 serial number

mystery case files prime suspect serial

serial vs parallel wiring diagram

trojan remover 7 serial

ide eide scsi serial ata

true stories of serial killers

radio serial numbering system motorola

cables to go usb to serial

ross bicycle serial no

emulator speculator 6.30 serial

solaris serial ports

quicktime 7.1.6 serial crack

serial mgt default password v890

cables to go usb to serial

serial w9w84 crack

open adder serial

reimage serial number

electronics serial port manager rs422

m1911 45 serial numbers

nursing central serial number

pinnacle studio 5 serial number

vegas movie studio 8 serial number

ultraedit 14 serial number

serial crack asus dvd

guc232awm serial to usb

find jeep serial number

boat serial number locater

edelbrock carburetor serial number

serial ms office 2000

apature serial number

backup exec 9.0 serial numbers

cuteftp pro 7 serial

photo brush 4 serial

tracing gibson guitae serial number

atapi ata serial adapter dangle

usb to serial adapter configuration

serial output programs

serial killer fred west

admanager plus 4.4 serial

labels serial number

vextractor 1 2 serial crack

tivo directv r10 serial without phone

zero popup 3 5 serial crack

serial microsoft visio 2003

4030 john deere tractors serial numbers

serial connector pinout

configuring port pci serial card

rm to mp3 wav convertor serial

ulead photoimpact x3 serial

quartus 8.1 serial

capture one 4 serial

morphvox classic serial key please

ocx virtual serial port

adobe acrobat 5 serial number

business plan pro serial activation code

spyware doctor 5.0.0.179 serial

garfiels typing pal serial number

ebp devils et factorization serial

avs dvd copy serial key crack

what is a serial dilution test

garand by serial number

crack serial corel draw 12 suite

chevy serial numbers

taylormade golf club serial number verification

ar-15 pre ban serial numbers

driver prolific usb serial

serial pocket tunes

stock neuromaster 1.31 serial

ultra mp3 mobile serial

solaris 10 serial console screen

ez architect serial key

snappy fax 4.11 serial

serial numbers for springfield 1903 rifles

excel serial number

mongoose 1989 serial numbers

Источник: [https://torrent-igruha.org/3551-portal.html]
.

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

Add a Comment

Your email address will not be published. Required fields are marked *