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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
You and your friend save as much money as possible by participating in the contest and fly to Athens. Exploring Athens you and all other teams come to Parthenon, the temple of Athena, who is the goddess of wisdom and intelligence. You also see many images of an owl. You learn from the guide that it is the “Athena Noctua”, also called the Minerva owl or the little owl. The guide also mentions some interesting stories of Athena’s wisdom and cleverness. The tourist guide at Parthenon takes you to an abandoned tunnel. And here you find an inscribed stone and a bowl of colored flowers. Suddenly you hear a deep and ancient gong ringing as you start reading the Stone Of Instructions. The instructions read as follows: Every person entering the Temple of Athena will be given a bowl with 15 red flowers and 12 yellow flowers. Each time the Gong of Time rings, he/she must do one of two things: 1. Exchange: If he has at least 3 red flowers in his bowl, then he may exchange 3 red flowers for 2 yellow flowers. 2. Swap: He may replace each yellow flower in his bowl with a red flower and replace each red flower in his bowl with a yellow flower. That is, if he starts with i red flowers and j yellow flowers, then after he performs this operation, he will have j red flowers and i yellow flowers. [You can leave the tunnel only when you make all the combinations (no. of red, no. of yellow) of flowers that are possible]. What is the number of combinations you need to make to leave the tunnel?A. 27B. 52C. 54D. None of these |
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Answer» B Exchange : `R_(n) -Y_(n)=(R_(n-1)-3)-(Y_(n-1)+2)=(R_(n-1)-Y_(n-1)-5) Swap : `R_(n)-Y_(n) =(Y_(n-1)-R_(n-1))` From the above two statements it can be declared that (R, Y) can never become (5,5) because the initial value of R-Y is 3. Hence the value of ((R-Y) mod 5) remains either 3 or 2 but never goes to 0. |
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| 2. |
You and your friend save as much money as possible by participating in the contest and fly to Athens. Exploring Athens you and all other teams come to Parthenon, the temple of Athena, who is the goddess of wisdom and intelligence. You also see many images of an owl. You learn from the guide that it is the “Athena Noctua”, also called the Minerva owl or the little owl. The guide also mentions some interesting stories of Athena’s wisdom and cleverness. The tourist guide at Parthenon takes you to an abandoned tunnel. And here you find an inscribed stone and a bowl of colored flowers. Suddenly you hear a deep and ancient gong ringing as you start reading the Stone Of Instructions. The instructions read as follows: Every person entering the Temple of Athena will be given a bowl with 15 red flowers and 12 yellow flowers. Each time the Gong of Time rings, he/she must do one of two things: 1. Exchange: If he has at least 3 red flowers in his bowl, then he may exchange 3 red flowers for 2 yellow flowers. 2. Swap: He may replace each yellow flower in his bowl with a red flower and replace each red flower in his bowl with a yellow flower. That is, if he starts with i red flowers and j yellow flowers, then after he performs this operation, he will have j red flowers and i yellow flowers. [You can leave the tunnel only when you make all the combinations (no. of red, no. of yellow) of flowers that are possible]. After how many operations will you have 5 each of red and yellow flowers?A. 5B. 10C. 15D. None of these |
| Answer» Correct Answer - D | |
| 3. |
After winning the challenge fair and square, and winning a Snorlax, Max finally reaches Viridian City. As he enters viridian city, he notices that there is a Pokémon trainer in front of the gym. I will only let you pass if you beat me in my favourite game, he says. This is a game played with a sequence of tiles, each labelled with two numbers. You start at the first the in the sequence and choose one number from each tile that you stop at, according to the following rules: • At tile i, if you pick up the smaller number, you move on to the next tile, i+1, in the sequence. • At tile i, if you pick up the larger number, you skip the next tile and move to tile i+2 in the sequence. The game ends when your next move takes you beyond the end of the sequence. Your score is the sum of all the numbers you have picked up. For example, suppose you have a sequence of four tiles as follows: Then the maximum score you can achieve is 3, by choosing the numbers that are circled. Now, you are given the following tiles. What is the maximum score you can make to ensure you beat the security guard? |
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Answer» 1,4,1,3,3,4,-1 Max score = 15 |
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| 4. |
After a day filled with adventures in Tehran, you now reach the bank of Hari river. There you find a box that contains the key to the boat which can help you cross the river quickly. The box can be opened only if the correct password is entered. On the top part of the box there is a number carved. “101111830140311” The box you have got can be opened by deciphering the code number given on the top of the box which would provide you the password. The password consists only English alphabets (capital). Each alphabet is coded by a three digit number in which the ones-place is given by the number of curved lines in that alphabet and the tens- place is given by the number of straight lines in the alphabet. To find the hun- dredths-place digit first place all the alphabets with the same last two digits together in their alphabetical order and interchange the first half with the second half. Now number them 1 onwards and thus the corresponding number will take the hundredths-place. Example: Let these letters (of some unknown alphabet) be the ones having one curved line and one straight line in ascending order then they will be coded as- (Consider “B” to have 2 curves and 1 straight line whereas “U” to have only a curved line. Also consider these letter as shown here: GIJQRY) What is the correct password ?A. SQYWDB. SPIWDC. SQIMJD. UPYMJ |
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Answer» B The procedure has been clearly described in the question. So, find out the 3 digit alphabet corresponding to all the English alphabets. Therefore, the final answer should be SPIWD. |
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| 5. |
You are running out of time but have no idea about the route from Nepal to IIT Guwahati. The various cut-out parts of the map that can help you reach IIT Guwahati are hidden in the houses shown in the figure below. You will get the parts of the map only on delivering the right article(one per house) to all the houses. (i) An ‘X’ coloured ARTICLE should be delivered to an ‘X’ coloured HOUSE only by using an ‘X’ coloured TRUCK. (ii) A truck can also carry other coloured articles so that it can place them at any of the CHECKPOINTS from where other truck can later carry it to the destination. (iii) Trucks can carry any number of articles at a time. (iv) The main objective is to start all the trucks at the same time, with same speed and deliver the articles without collision of trucks. Also, a truck picks up every article that comes on its way. (v) Last and the most importantly, the PATHS of the trucks should not overlap at any point other than junctions. At junctions, the paths can though cross each other (but both trucks should not reach that junction at same time which leads to collision). The trucks can move only forward. Draw the final path of all the trucks and give the number of turns taken by the blue truck. A. 8B. 7C. 6D. None of the above |
| Answer» Correct Answer - C | |
| 6. |
You say the code word, and the door opens. You walk in, and look around the empty room. It is a dead end, but you see a poem on the wall. It reads: As you dance each _______around, through once, once only, Me, when you shall have found, do not hide your glee ! One of you notices that the code word fits in the blank of the poem. As you think about the poem, you see an arrow which directs your eyes downward. To your amazement, you find the whole floor filled with letters and symbols! You realize that you need to use the poem to find a solution to the puzzle under them. The floor looked like this: Rearrange the clues from the puzzle above, using the poem, to find a question and then answer it to get out of the cave. What is the answer which can take you out of the cave?A. 7B. 70C. 18D. 81 |
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Answer» D After decoding the poem, you realize that you must go around ‘H’ in the matrix in the direction given by arrow and note down the word formed. On finding out all such words and combining the you will get the following question squre/ of/larg est/one /digit/ number/ So, the answer is `=9^(2) =81 ` |
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| 7. |
Now suppose, the staff toggles 15 consecutive doors then with how many doors can you acheive his aim of toggling a state of a single door ?A. 53B. 54C. 55D. None of these |
| Answer» Number of doors and 15 should not have any common factor. So, 53 is the answer | |
| 8. |
Mario moves on to find a way to get through the main gates of Bowser’s chamber. He suddenly finds falling through a trapdoor and lands on something hard. The Devil Ghost Banshee , comes out and says, “This room is a standard 8×8 chessboard. Each of its 64 square is assigned a weight. These weights are assigned in such a manner that weight of a square is an average of the weight of the square that it is surrounded by.” Mario now has to determine the weights of all 64 squares and for that Banshee tells him the weights of X squares. What could be the minimum value of X with which Mario can deduce the weights of every other square?A. 8B. 32C. 56D. None of these |
| Answer» All squares have to be of equal weight (Think !). Suppose they have unequal weights, then there has to be a maximum weight present in those 64 squares.Then that square must be surrounded by smaller weights, but the smaller weights cannot average to a higher number (since average of any numbers is less than their maximum). So only possiblity is for the weights to be equal. | |
| 9. |
Mario’s aim is to apply this flickable staff several times to toggle the state of a single door. What are the possible number of times Mario would have to use this staff to achieve his aim?A. 25B. 32C. 40D. 49 |
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Answer» Toggle all the doors except 5th, 10th, 15th, 20th, 25th, 30th, 35th, 40th, 45th, 49th This way, Only Door 1 will be opened and all others closed. That means minimum steps required are 39 ….. So any odd number equal to or greater than 39 is the possible steps So 49 steps is the answer |
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| 10. |
Tintin unlocks the chest, and plugs the key into the main door. A dial emerges on the door. Beginning with the dial set at zero, the dial must be turned counter-clockwise to the first combination number, (then clockwise back to zero), and clockwise to the second combina- tion number, (then counter-clockwise back to zero), and counter-clockwise again to the third and final number, where upon the door shall immediately spring open. There are 40 numbers on the dial, including the zero. Without knowing the combination numbers, what is the maximum number of trials required to open the safe (one trial equals one attempt to dial a full three-number combination)?A. 64000B. 1600C. 63999D. 1599 |
| Answer» We get `40*40*1 = 1600` (Since the last round, it opens immediately) 1 (Case of 000) = 1599 | |
| 11. |
Now suppose that the tool can toggle 15 consecutive lamps then with how many lamps can Tintin achieve the same?A. 53B. 54C. 55D. none of these |
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Answer» Number of doors and 15 should not have any common factor. So, 53 is the answer |
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| 12. |
The spy sent Tintin four messages which consists of only the letters from A to D, that are encoded in the following way: A, B, C, D are equivalent to 00, 01, 10, 11 respectively. An operation ‡ is defined over the numbers 0 and 1 as All messages that Tintin receives are encoded as (Message)`‡`(Key) where Key is a fixed word that is used for encoding all the Messages. But, sadly, Tintin does not know the key. BDBDBD, AAAA, BDBBBB, DAACCC . You forgot the messages corresponding to these encodings but you do remember that all of these four messages had ADC as its substring. Few seconds later, Tintin receives another message that was encoded as BBDBD. Can you tell what the actual message could possibly be? Not clear? Want an example – let us have two messages AB and CD, and let the key be BC, then `(({:(" "AB),(+BC):})/(BD))/` `(({:(" "CD),(+BC):})/(DB))/` are the corresponding cypher texts.[Hint: if `("word"1)‡("word"2)=("word"3) " then " ("word"1)=("word"2)‡("word"3)]`A. ABACDB. ABADCC. ABABCD. ABBDC |
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Answer» From the hint, we realize that the 3 words are cyclic. Therefore, `{:(M,,,K,,E),(,ul(K),ul(E),,ul(M),),(,E,,M,,K):}` Where, M: Message, K: Key, E: Encoding. Observe that, this function is unary in nature. Since, we only had the last 3 alphabets of the key, let us only focus on the last 3 alphabets of the messages and the encodings. Every message ended with ADC. Therefore, using the 3rd formula, DBD ADC DCB = the key. Now, using the key, the encoding’s BBDBD last 3 words DBD and the 2nd formula, DCB DBD ADC Hence, the message ends with ADC, which is Option B. |
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| 13. |
After successfully infiltrating into the castle, Tintin has to send a word back to his island for his army to prepare, so he uses his secret cube technique. He made an hollow glassy rubik’s cube that can be unfolded in any fashion like any normal cube. He inscribed letters on the centre square of each face indicating the orientation of cube which she always keep the same( F-front , B-back, L-left, R-right, U-up , D-down) and also inscribed digits from 1-8 on rest of the squares.The unfolded image of the current configuration is shown in fig_1. He wrote one word on each face in the gaps and rotated left and right face clockwise and also makes an additional rotation. Unfolding the cube in some another way, which is shown in fig_2 , he sends it with his pigeon. What was the additional rotation that Tintin made? A. Down face Clockwise RotationB. Up face counter-clockwise rotationC. Down Face counter clockwise RotationD. Up face clockwise rotation |
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Answer» To check which face is rotated check the orientation of center letter of each face which has been changed after the two rotation have been made. Observe along the line UBDF . We see that the orientation of the U face is been changed clockwise. The moves that are made in order are : 1. LEFT face clockwise 2. RIGHT face clockwise 3. UP face clockwise We need to trace back the letters to get them in their original position. For this we have to see the effect of following moves on the current configuration. 1. UP face anticlockwise. 2. RIGHT face anticlockwise. 3. LEFT face anticlokwise. Trace the changes in the letters on each face. i.e. When Up face is rotated clockwise it will change letters in right , left , front , back face will change. Postions of the letters in the Up face will change and no effect will be observe on the Down face. Similarly Left and Right face rotation wont effect letters of the other one. Letters can be guessed even when they are rotated. But some letters can have different possiblities like W and M which can be found by observing the orientation of a nearby lettter. Hence after all the moves are made letters on the face are:: Front face : F D R E A M S Back face : B W O R K Hence the common letter is only R. |
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| 14. |
While in Rastapopolous’s Island, Tintin notices many corrupt practices being held there, one of them being gambling. He decides to bring an end to it and walks up to The Trick- ler, the self-proclaimed biggest gambler of the island. After an angry conversation with him, they decide if Tintin beats him, he would stop gambling forever. They decide to play a game. The Trickler controls three ‘rat’ pieces, while Tintin controls a single ‘snake’ piece. Initially, all four pieces are placed somewhere on a two-dimen- sional plane. They take turns making moves, with The Trickler going first. Each move, a player is allowed to move one of her pieces a distance of at most one unit along the straight line. Tintin wins if his ‘snake’ piece can catch one of the rabbit pieces.A. YesB. NoC. Depends on PositionD. None of these |
| Answer» Bonus for All …..Want to know the solution ?? Check Juniors paper for correct Question and Solution ! | |
| 15. |
Moving further ahead, they reach the castle walls where Tintin sees this mysterious poster (shown below) pasted on the wall. He realised that an entrance through a secret tunnel lay behind it which can be opened by climbing the “spiraling green leaves”. He realises that the poster is hinting towards a word. So help Tintin crack the logic of the poster, which demands the 4th letter in the word. A. HB. BC. ED. D |
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Answer» The hint was the term, ‘Spiralling’ which indicated that, it was a square inside square kind of spiral. Start from the outermost “SQUARE”, To get the alphabet corresponding to that particular square, Connect each white square to the nearest x white squares in that chosen SQUARE, where x corresponds to the number in that square For the 6th square (the central square of the whole box) use the directions given to figure the path. Hence the answer is option A)H. |
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| 16. |
After successfully infiltrating into the castle, Tintin has to send a word back to his island for his army to prepare, so he uses his secret cube technique. He made an hollow glassy rubik’s cube that can be unfolded in any fashion like any normal cube. He inscribed letters on the centre square of each face indicating the orientation of cube which she always keep the same( F-front , B-back, L-left, R-right, U-up , D-down) and also inscribed digits from 1-8 on rest of the squares.The unfolded image of the current configuration is shown in fig_1. He wrote one word on each face in the gaps and rotated left and right face clockwise and also makes an additional rotation. Unfolding the cube in some another way, which is shown in fig_2 , he sends it with his pigeon. How many letters are common on Front and Back face before making any rotations ? A. 2B. 3C. 4D. None of These |
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Answer» To check which face is rotated check the orientation of center letter of each face which has been changed after the two rotation have been made. Observe along the line UBDF . We see that the orientation of the U face is been changed clockwise. The moves that are made in order are : 1. LEFT face clockwise 2. RIGHT face clockwise 3. UP face clockwise We need to trace back the letters to get them in their original position. For this we have to see the effect of following moves on the current configuration. 1. UP face anticlockwise. 2. RIGHT face anticlockwise. 3. LEFT face anticlokwise. Trace the changes in the letters on each face. i.e. When Up face is rotated clockwise it will change letters in right , left , front , back face will change. Postions of the letters in the Up face will change and no effect will be observe on the Down face. Similarly Left and Right face rotation wont effect letters of the other one. Letters can be guessed even when they are rotated. But some letters can have different possiblities like W and M which can be found by observing the orientation of a nearby lettter. Hence after all the moves are made letters on the face are:: Front face : F D R E A M S Back face : B W O R K Hence the common letter is only R. |
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| 17. |
Tintin reaches Rastapopoulos’s island, but they are informed he has moved out, and the situation is pretty explosive there. Alonso, Bianca, and Coco are running for the post of the island head. On the ballot, each voter lists the three candidates in order of their preference. Counting only the first preferences results, dramatically, in a three-way tie. To break the deadlock, the second preferences are counted, but again there is a three-way tie. Alonso notes that, since the number of voters is odd, they can make two-way decisions without ties. He proposes that the voters first choose between Bianca and Coco, then the winner faces Alonso for the position. Bianca thinks it’s a good resolution, since they only want to identify the winner, not the runner-up. Coco disagrees and complains that this is giving Alonso an advantage. Who is right? Assuming the voters never change their preferences, what is Alonso’s chance of winning under his proposed voting system?A. AlonsoB. CocoC. Both are wrongD. None of these |
| Answer» From the conditions of the problem, the number of voters is 3n for some odd positive integer n. Each of the candidates A, B and C received n first preferences and n second preferences. Let the number of voters who voted A gt B gt C be x and the number of voters who voted `A gt C gt B` be y. Then x + y = n and the size of the remaining voting groups can be easily computed: `A gt B gt C : x, A gt C gt B : y, B gt C gt A: x, B gt A gt C : y, C gt A gt B : x, C gt B gt A: y`. When only comparing B against C, we have `B gt C : x + x + y = n + x, C gt B : y + x + y = n + y`. Similarly (noting the cyclic symmetry), we have `A gt B : n + x, B gt A: n + y, A gt C : n + y, C gt A: n + x`. Since n is odd, `x 6= y. If x gt y`, then B wins against C but goes on to lose against A. If `y gt x`, then C beats B then also goes on to lose against A. Therefore Christie is right, Adrian will always win under the proposed system. | |
| 18. |
Realising he needs money, Tintin reaches inside the bank situated on the island, and sees a row of lockers which start with 0 and go on till infinity. A locker can contain any number of coins. A coin is placed in the locker with index 7 (i.e. the 8th square). The aim is to move this coin to the locker with index 1. (i.e. the 2nd square) and all other squares empty. There are two rules to this: 1) Fission Rule: A coin may be replaced by a pair of coins by placing one in each of the immediately adjacent squares. Eg. Coin #3 can be replaced by Coin`#2` and Coin`#4` where Coin#(number) rep- resents coin in position with index number. 2) Fusion Rule: A pair of coins separated by exactly one intervening square can be replaced by a single coin in that middle square. Eg. Coin`#2` and Coin`#4` can be replaced by Coin#3. Minimum number of moves required to achieve our aim ?A. 18B. 19C. 17D. 16 |
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Answer» This puzzle seems rather difficult to solve by trying random moves that appear to get one closer to the solution. However, consider the board a sequence of integers, each being the number of pennies at that location, and the moves as adding or subtracting (+1, 1, +1) to three consecutive values. Then by inspection we can come up with a solution, not worrying about whether the number of pennies is always positive, by adding the columns: 0 0 0 0 0 0 +1 (start) 0 0 0 0 1 +1 1 0 0 0 1 +1 1 0 0 +1 1 +1 0 0 0 +1 1 +1 0 0 0 0 +1 0 0 0 0 0 0 (end) This gives you the four critical moves that must be in any solution. However, applying them directly is illegal. [But you can] just [keep] splitting the left penny (7 times), then [do the] “critical 4 moves” and you come up with the mirror image position, so you now combine to the final position. Ergo, 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 2 1 1 1 1 1 lt +1 1 +1 1 1 1 1 2 1 1 1 1 lt +1 1 +1 1 1 1 1 1 2 0 1 1 1 1 1 1 1 1 1 0 1 lt 1 +1 1 1 1 1 1 1 1 0 1 0 lt 1 +1 1 1 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 |
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| 19. |
Number of days: 20 ltbr. Daily Expenses: 1 Azkaban Unit per day Maximum cash he can have at any time: 4 Azkaban Units Daily Exhange rate: Igor hopes that Viktor will solve the puzzles before he returns from Azkaban and he prom- ises to train Viktor with more challenging puzzles ahead.You have to report the sum of the product of digits in the OMR sheet i.e if minimum expenditure is 26, report your answer as `3 ( 2 xx 6 = 12 rArr 1+2 = 3)` |
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Answer» `{:(56,rArr,3),(39,rArr, 9):}` Best possible way is hit and trial while using logic. |
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| 20. |
Number of days: 15 Daily Expenses: 2 Azkaban Units per day Maximum cash he can have at any time: 5 Azkaban Units Daily Exchange rate: |
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Answer» `{:(56,rArr,3),(39,rArr, 9):}` Best possible way is hit and trial while using logic. |
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| 21. |
Igor returns from Azkaban 2 days later and finds out that Viktor had solved the puzzles quite easily. So he decides to raise the stakes. He takes Viktor to the room with the life sized chessboard. Viktor was a rather famous chess player, so he didn’t mind. Igor was aware of this, so he brings in a new piece, a SuperQueen.It can perform all moves of a queen as well as a knight. Igor gives Viktor 10 SuperQueens and increases the chess board’s dimensions to `10 xx 10`. He then asks for the number of possible ways that all 10 pieces can be placed on the board without coming in each other’s path. What is the answer? |
| Answer» This question is a modification of the N-queens problem. For more information regarding N-queens problem visit | |
| 22. |
Another band of loyalist join them in the morning, taking the total to 60 followers. For breakfast, they decide to have bamboo sticks. Unfortunately, he has only 47 bamboo sticks, all of them identical cylinders. Anxious to evenly distribute the 47 sticks amongst them- selves, Tintin decides to cut the sticks with his sword. What is the minimum number of times that Tintin has to use his sword to share the 47 bamboo sticks between all 60 of them, know- ing that each of them must receive identical parts. Note : A sword can be used to break mutliple sticks at a time. |
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Answer» Since we have to divide 47 bamboo sticks to 60 people in equal quantities, every person must end up with `47//60` of the initial piece. Let us look at, `47//60 = 1//3 + 1//4 + 1//5`. For the first cut, Tintin puts the 47 sticks side by side in a way that he will cut 20 sticks at `1//3` of the length, 15 sticks at `1//4` of the length and 12 sticks at `1//5` of the length. He obtaines 20 parts of length `1//3` (that he keeps for the moment) 20 parts of length `2//3` (to be cut later) 15 parts of length `1//4` (that he keeps for the moment) 15 parts of length `3//4` (to be cut later) For the 2nd cut, Tintin puts side by side 20 parts of length 2/3 in the middle, 15 parts of length 3/4 at 1/3rd it’s length(i.e., 1/4th of total length) and 12 parts of length 4/5 that he will cut in the middle. He obtaines 40 additional parts of length `1//3`, that he adds to the 20 parts obtained previously. He can now distribute these 60 pieces of length `1//3`. 15 additional fragments of length `1//4` (which he adds to the existing ones) and 15 fragments of `2//4` length (to be cut later) 24 fragments of length 2/5 (to be cut later) On the same principle, Tintin needs to cut the rest of the sticks once more to divide the 15 fragments of length `2//4` in 30 fragments of length `1//4` and 24 fragments of length `2//5` into 48 fragments of length `1//5`. He can now distribute all the 60 pieces of `¼` and 60 pieces of `1//5`. Hence the Answer is 3. |
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| 23. |
He then enters a mushroom market. In the market, whenever two people greet each other, they have to swap their mushrooms. There are 64 people present in the Market.In order to save time, each pair of people is only allowed to greet each other atmost once. After a lot of greetings, Mario notices that it is no longer possible to return all mushrooms to their respective owners through more greetings. To sensibly resolve this maddening confusion, he decides to bring in even more people (with more mushrooms), to allow for even more greetings and mushroom swappings. How many extra people are needed to return all mushrooms (including the extra ones) to their rightful owners? Assume a person has exactly one mushroom. |
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Answer» Two extra people are needed to return correct mushrooms using the following method Let the extra people have names E1 and E2. Step 1: Arrange the 64 people into lines such that In each line, every person is having the mushroom of the person directly behind. Except the last person whose mushroom is with the first person in that line. Note : If two people have each others’ mushrooms, they will make a separate line of only two people. Step 2:: For a single line, first E1 swaps mushrooms with the last person. Then E2 swaps mushrooms with each person in middle one at a time from front to back. Finally E1 will swap mushrooms with the first person.Then E1 & E2 swap each others’ mushrooms. This way everyone in this line along with E1 & E2 will have their own mushrooms. Step 3: Repeat Step 2 for each line. |
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| 24. |
As they were moving through the jungle, they met an old drunkard who insisted everyone should to swap their hats. The troop contains 64 people. In order to save time, each pair within the troop swaps hat with each other only once (i.e. no swapping is repeated). After a round of madness, Tintin notices that it is no longer possible to return all hats to their respective owners through more swappings. To sensibly resolve this maddening confusion, he decides to bring in even more hat-wearing guests in his troop, to allow for even more hat swappings. How many extra guests are needed to return all hats (including the extra ones) to their rightful owners? Note : Assume a person has exactly one mushroom. |
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Answer» Two extra people are needed to return correct hats using the following method Let the extra people have names E1 and E2. Step 1: Arrange the 64 people into lines such that In each line, every person is having the hat of the person directly behind. Except the last person whose hat is with the first person in that line. Note : If two people have each others’ hats, they will make a separate line of only two people. Step 2:: For a single line, first E1 swaps hats with the last person. Then E2 swaps hats with each person in middle one at a time from front to back. Finally E1 will swap hats with the first person.Then E1 & E2 swap each others’ hats. This way everyone in this line along with E1 & E2 will have their own hats. Step 3: Repeat Step 2 for each line. |
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| 25. |
Al Capone and his (n-1) friends are sitting in a circular fashion when Tintin meets him(To- tal of n people). Each of them owns a coin. First person passes 1 coin to the person sitting on his left side. The second person in turn passes 2 coins to the person sitting on the left. Third person passes 1 coin to the left, 4th passes 2 coin to his left and so on. So each person receiving 1 coin from the right has to pass 2 coins to the left. Similarly, a person receiving 2 coins from the right has to pass 1 coin to the left. If at any point of time, a person runs out of coin he is thrown out of the game. The game will terminate if at the end only 1 person is left with all the coins in his posses- sion. There might be values of n where the game may not terminate e.g. 3 persons left with 4 4 4 coins respectively. For how many different values of n from 4 to 100 does the game terminate?A. 16B. 7C. 10D. 46 |
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Answer» If you start observing behaviour of games starting from n=1, you can observe patterns which you can generalize. It is easy to see that in the first round each person sitting at an even numbered position will be thrown out of the game as he/she has 1 coin initially, receives 1 coin from right but has to now pass both the coins to the left. Also observe that the 1st person will be always thrown out of the game. If n = 2k, after the first round k1 will remain. If n = 2k+1, then k will remain. Case 1 : n is even for n=2k , after the first round configuration would look like 3(4) 5(2) 7(2) … 2k1(2) where a(b) denotes position a with b coins. at this point of time. Number of remaining people are k, if k is odd the game will not terminate because a person who looses one coin in a round will gain one coin in the next round. Case 2 : n is odd for n=2k+1, after the first round configuration would look like 3(3) 5(2) … `2k+1(2)` If k is odd, the game will not terminate. If both the cases if k is even, at the end of two rounds exactly half will remain. Because a person who looses a coin will keep loosing in subsequent round. After this elimination you again have to deal with cases where remaining people are odd or even. Thus you can observe that if after the first round people remain then the game will always terminate. This can happen if initially there are or person are there. Hence, the Answer is option: C)10 |
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It’s time for one of the most awaited events of the Tri-Wizard Tournament, The Yule Ball. Viktor Krum decides to ask out Hermione as his date for the ball. Knowing how smart she is, he decides to impress Hermione by scoring the highest ever score in one the most played games at Hogwarts The game is described as follows: This is a game played with a sequence of tiles, each labelled with two numbers. You start at the first tile in the sequence and choose one number from each tile that you stop at, according to the following rules: • At tile i, if you pick up the smaller number, you move on to the next tile, i+1, in the sequence. • At tile i, if you pick up the larger number, you skip the next tile and move to tile i+2 in the sequence. The game ends when your next move takes you beyond the end of the sequence. Your score is the sum of all the numbers you have picked up. The goal is to maximize the final score. For example, suppose you have a sequence of four tiles as follows: Then, the maximum score you can achieve is 3, by choosing the numbers that are circled. In each of the cases (a) and (b) below, compute the maximum store that Viktor can achieve by picking up numbers according to the rules given above. A. 14,14B. 13,14C. 14,15D. 14,16 |
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Answer» C) 14, 15 (a) 2, -1, 1 (b) 1,4,1,3,3,4, -1 |
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