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(a)
Solutions to Questions 1 to 5: After reading the first paragraph you would get the following
structure for the grid:
Your first focus in the second paragraph should be to first fill in the direct clues into the grid. You get
the following grid after using the clues:
Qamar represents Italy in Hockey. Sachin represents India. Virbahdra competes in Judo. Ojasvi
represents Sri Lanka in Cycling. Zavi competes in Weight Lifting.
At this point the following clues are left to be used:
Sachin represents India (used) but not in Judo and neither in Weight Lifting. Someone from amongst
the seven athletes represents Egypt and he/she competes in Swimming. Virbahdra competes in Judo
(used) but not for Malaysia. The athlete from Pakistan competes in Wrestling. Xenon does not
represent Malaysia or Egypt.
Also, the countries/sports and country/sport combinations still to be filled in are: Malysia; Pakistanwrestling;
Egypt- Swimming; Nepal; Basketball.
Also, from the grid we have reached at this point of our solving, we know that the two combinations
of country sport listed above have to be allocated to Pandey and Xenon. Based on this, we can make
the following deductions which would help us complete the solution grid:
Deduction i): Sachin cannot be matched with Swimming or Wrestling. This leaves only Basketball for
Sachin.
Deduction ii): Zavi and Virbhadra cannot be from either Pakistan or Egypt. Hence, they must be from
either Malaysia or Nepal. Further, it is given to us that Virbahdra is not from Malaysia. Hence, we
can pair Virbhadra for Nepal and Zavi for Malaysia.
Deduction iii): At this point, we are only left with deciding about the combinations of Egypt—
Swimming AND Pakistan—Wrestling and allocating these between Pandey and Xenon.
Since, we know that Xenon does not represent Egypt, we can easily decide the following
combinations:
Pandey–Egypt–Swimming & Xenon–Pakistan- Wrestling.
The answers for questions 1 to 5 can be read off the table:
1. Zavi who competes in weightlifting represents Nepal.
2. Option (d) can be seen to be correct.
3. It can be seen that it is Sachin who represents India. Option (d) is correct.
4. Sachin competes in Basketball. Option (b) is correct.
5. Virbhadra represents Nepal. Option (d) is correct.
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(d)
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(d)
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(b)
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(4)
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(1)
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(1)
Solutions to Questions 6 to 10: Based on the information in the question we can start off with a circular figure with 7 blank seats as shown below. Based on the clue “Anil is third to the left of Yukta and to the immediate right of Rashmi” we get: From this point we can use the clue: “Pritam is second to the left of Gitali who is not an immediate neighbour of Mitali.” This gives us only one way to complete the seating arrangement as shown below: The answers can then be read off the final seating grid. 6. Mitali is to the immediate left of Pritam. Option (1) is correct. 7. Mitali is second to the right of Rashmi. Option (2) is correct. 8. None of the given pairs satisfies the condition. Option (4) is correct. 9. Sudesh is sitting between Rashmi and Gitali. Option (2) is correct. 10. Anil is second to the left and fifth to the right of Pritam. Option (4) is correct.
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(2)
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(4)
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(2)
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(4)
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(1)
Solutions to Questions 11 to 15: The starting grid for this question would look as follows: After using the direct clues we get the following position in the solution grid: At this point in our solution, we have to still fit in the following sport/musical instrument or Sportmusical instrument combination. Swimming; Chess ; golf; Sarod and Sarangi. From this point the following deductions help us get to the final answers: Deduction 1: Since Esha does not play Sarod, she must be playing Sarangi. Consequently, Dev plays Sarod. Deduction 2: Since Freni does not swim or play golf, her sport must be Chess. The grid would become as follows: The answers can now be read off the above table: 11. Dev plays Sarod. Option (1) is correct.
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(4)
Esha plays Kabaddi. Option (d) is correct
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(4)
Option (d) is correct
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(4)
Freni plays Chess. Option (d) is correct
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(3)
Dev is the swimmer. Option (c) is correct.
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(4)
Solutions to Questions 16 to 20: Since there are 8 people sitting around the circle, we can look at creating the following picture: From this point, if we use the clue: Danish sits third to the right of Charlie & Hina is not the neighbour of Danish and Charlie we get only the following possible placement for the eight people. The answers can be read off the above figure. 16. Hina sits to the right of Furqan. Option (d) is correct.
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(1)
Apart from the pair of Charlie- Hina, in each of the other pairs the second person is sitting 3 places to the left of the first person. Option (a) is the one that does not belong to the group and hence option (a) is correct.
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(4)
Option (d) is the correct answer.
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(1)
Option (a) is correct.
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(4)
Option (d) is correct
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(4)
Solutions to Questions 21 to 26: From the clue: Anu sits fourth to the right of Hitesh while second to the left of Farman we get: From this point if we further use the clues that: Chintu is not the neighbour of Farman and Bhavesh & Durgesh sits third to the right of Chintu we would have only one position to place Chintu and Durgesh in. Subsequently, the other people would each have only a single position to get placed in. The solution would be as shown by the figure given below: The answers to Questions 21 to 26 can then be read off the final solution figure: 21. Anu sits between Ganesh and Durgesh. Option (d) is correct.
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(1)
Hitesh and Bhavesh sit between Esha and Farman. Option (a) is correct.
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(1)
Apart from the Durgesh Chintu pair, all other pairs are separated only by one person in between them. Hence, option (a) is correct.
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(3)
Ganesh is to the immediate left of Anu. Option (c) is correct.
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(3)
Bhavesh sits second to the right of Durgesh. Option (c) is correct.
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(4)
Farman is third to the left of Esha and is also fifth to the right of Esha. Option (d) is correct.
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(4)
Solutions to Questions 27 to 31: The starting grid in this question (after using the direct information in the question) would look as follows: At this point we have to place the following: Kolkata, Lucknow, Agra, Bhopal, Auditor, Painter and Teacher. Amongst these, we also know the following pairs are mandatory: Teacher- Lucknow; Painter-Bhopal (& these have to be placed under Javed and Nitin). The following deductions help us finalise the grid: Deduction 1: Javed does not work in Lucknow – hence the Teacher – Lucknow pair has to be filled under Nitin and the Painter- Bhopal pair has to be filled under Javed. Deduction 2: Kandarp does not work in Kolkata- hence he can only work in Agra while Gimi must be working in Kolkata. The finalized grid would look as below: The answers are: 27. Moni is the auditor. Option (d) is correct.
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(4)
None of the given combinations is correct. Option (d) is correct.
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(4)
Nitin is working in Lucknow. Option (d) is correct.
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(2)
Option (b) is correct
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(4)
. Luv is the builder. Option (d) is correct.
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(4)
Solutions to Questions 32 to 35: After putting the given data in, the appropriate grid for this question would look as follows: At this point, the missing pieces pertain to the following cities/ Airlines: Cities left: Agra, Indore, Gandhinagar Airlines left: Indigo Airlines, MDLR Airlines Express, Jetlite. Since we know that Agra cannot be Om and neither Sanjay, it leaves us only Usman who can go to Agra. Also, the one traveling to Gandhinagar does not travel by Air Deccan. This can only mean that Air Deccan (and Sanjay) must be going to Indore and hence, Om would be going to Gandhinagar. Since, Om is going to Gandhinagar, he cannot be going by MDLR Airlines Express. The table converts to the following (Note: We are not able to resolve the exact airline used by Om, Qadir and Usman.) The answers can be read off this table: 32. It can be either Qadir or Usman. Hence, option (d) is correct.
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(1)
Om goes to Gandhinagar. Option (a) is correct.
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(2)
Sanjay who goes by Air Deccan visits Indore. Option (b) is correct.
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(3)
Option (c) is correct.
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(4)
Solutions to Questions 36 to 40: The ideal starting grid structure for this question would be: We have direct information about the shifts of Mini, Pia, Rita and Sumant. We also have direct information about the weekly offs of Sumant, Rita and Pia. Placing this information into the above table we would get: At this point we should be thinking about the following: (i) The shifts and weekly offs for Lina, Nita and Queen & (ii) the weekly off for Mini (iii) We also realize that of the remaining 3 people who have to be allocated to shifts, each of them has to be allocated only to either shift I or shift III (as it is given that Shift II has only Mini and Rita.) (iv) Of the remaining 3 people whose shifts have to be decided, two would need to go into any one of the two remaining shifts while the third would go to the other shift. Thus, between shifts I and III, one of them would have to have 2 people working in it while the other shift would have 3 people working in it. (v) The weekly offs for 4 people – hence for four days is still not known. These are Tuesday, Wednesday, Thursday and Saturday. In order to move ahead in completing the table we have: Lina and Pia do not work in the same shift – meaning that Lina should be working in Shift I and her weekly off is Tuesday. Also, Queen’s weekly off being on the next day to Lina’s weekly off, her weekly off must be on Wednesday. Further Queen should be working in a shift different from Lina— so since Lina is in shift I, Queen would be in Shift III. (Remember that shift II has been completely filled up). With this information placed the solution grid would become as below: We still need to place Nita and her weekly off, as well as decide on Mini’s weekly off. We still need to decide whose weekly off between Nita and Mini is on Thursday and whose is on Sunday. We also need to decide the one shift that has 3 people working. (Note: Since there are 7 officers and 3 shifts with a minimum of 2 officers per shift- it means that one of the shifts would have 3 officers and the other two would have 2 officers each.) The clue that helps us decide on these unknowns in the table is: The one who has a weekly off on Sunday works in the first shift. This means that Mini cannot have her weekly off on Sunday—hence her weekly off is on Thursday. Also, Nita must be a first shift worker and her weekly off is on Sunday. The solution grid gets completed as below: The answers can now be read off the above table: 36. It is Nita whose weekly off is on Sunday. Option (d) is correct
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(4)
Option (d) is correct.
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(2)
Mini’s weekly off is on Thursday. Option (b) is correct.
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(1)
Option (a) is correct.
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(1)
Option (a) is correct.
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(3)
Solutions for 1 to 5: Solve using constraints and checking the options for each constraint: Constraint 1: The constraints are U and X not together Constraint 2: A and C not together. Constraint 3: C not with V 1. ABUXY contradicts constraint 1. Hence, this team is not possible. Thus Option c is correct. ABC UVWXY
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(2)
2. C AND W are compulsory. Hence, of the senior faculties we need to take B. The teams possible then are: BC-WUY and BC-WXY. Hence, only 2 teams are possible. Option (b) is correct.
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(3)
3. Constraint 3 clearly defines that V cannot go with C. Hence, Option (c) is correct.
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(1)
B must compulsorily go to the new office as we can select only one out of A and C. Hence, Option (a) is correct.
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(3)
Both i and iii are true. Hence, Option (c) is correct.
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(1)
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(1)
cubes). It can be visualized that the top edges of the cube would have 7 + 7 + 3 + 3 = 20 cubes and the bottom edges would also have the same number (i.e. 20 cubes). The vertical edges would have a further 7 + 7 + 7 + 7 = 28 cubes. Out of the total of 68 cubes which are on the edges, 8 cubes would be corner cubes. Hence, the number of cubes painted on two sides would be 68 – 8 = 60. Option (a) is correct.
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(1)
The number of one face colored cubes would be: On the front and back surfaces: 7 ¥ 5 + 7 ¥ 5 = 70 cubes. (you can remember this as (m-2) ¥ (n-2) where m and n are the number of rows and columns on the surface in question) On the top and bottom surfaces: 5 ¥ 3 + 5 ¥ 3 = 30 cubes. On the lateral surfaces: 7 ¥ 3 + 7 ¥ 3 = 42 cubes. Thus a total of 142 cubes would have one side painted. Option (a) is correct.
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(3)
again be remembered as (m-2)(n-2)(p-2) where m, n and p are the number of parts in which the cube surfaces have been cut – in this case m=9, n=7 and p=5]. Alternately, you could also have solved this using: All cubes – cubes with 3 sides colored – cubes with 2 sides colored – cubes with 1 side colored = 315 – 8 – 60 – 142 = 105. Option (c) is correct.
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(2)
The cubes along the vertical edges would be painted both red and green. There are 7 such cubes (with 2 faces colored) on each of the 4 edges of the cuboid. Thus, the total number of cubes which would be painted on two sides (in red and green) would be 7 ¥ 4 = 28. Option (b) is correct.
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(2)
The statements can be mapped as follows: Number 1: Black Hair Long tail Number 2: Short tail wore a collar Number 3: White Hair No Collar Scenario 1: If statement 2 of Shopkeeper 1 is correct & his statement 1 is false: Statement 1 of shopkeeper would be false. Then his (shopkeeper 2’s) statement 2 would be correct and shopkeeper 3’s statement 2 would be false.This would mean shopkeeper 3’s first statement would be true. It can be concluded that: The dog has white hair, he wears a collar and has a long tail. Scenario 2: If statement 1 of Shopkeeper 1 is correct & his statement 2 is false: Statement 1 of shopkeeper would be true. Then his (shopkeeper 2’s) statement 2 would be false while his statement 1 would be true. Shopkeeper 3’s statement 2 would be true.This would mean shopkeeper 3’s first statement would be false. It can be concluded that: The dog has black hair, wears no collar and has a short tail. Option (b) is a possible description of the dog
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(2)
Looking at the statements after the initial condition, it can be easily seen that Dipesh is lying when he says that he is a fishmonger- as nobody can say that he/she is a fishmonger truthfully. Also, Dipesh would not be a fishmonger—and he cannot be a Greenie. Hence, Dipesh would be a Wishy Washy, thus making his second statement true. This means that Tarun is a greenie. This in turn means that Harish’s second statement about Tarun is false. This also means that Harish cannot be a greenie which renders his first statement also false. Thus, Harish must be a Fish monger. Option (b) is correct.
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(3)
In this question, the following would be the starting grid: Profession (Barber, Carpenter, Stockbroker, Tanner) Shirt colour (Black, Crimson, Salmon, Topaz) Mr. Barber Mr. Carpenter Mr. Stockbroker Mr. Tanner After using the clues, we get: The clues are used in the following manner to reveal the stockbroker’s profession: Since Mr. Carpenter is the barber and the stockbroker is friends with the tanner – the stockbroker must be neither the baker nor the tanner. Also, since we know that no person has a profession which equals his name. Hence, Mr. Stockbroker must be the Carpenter:
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(1)
The pairs who were successful getting pictures were: P and Q photographed the anteater. P and T photographed the bear. R and S photographed the crocodile. S and T photographed the deer. Q and R photographed the emu. There are 10 possible pairs. Each person went out four times, each time with a different partner. Each person got a photograph twice and got nothing twice. First, to determine which pairs photographed an animal and which didn’t, make a 5x5 matrix with row and column headings of P, Q, R, S and T. Since we don’t always know what animal was photographed, assign the animals a number, 1 through 5. Everybody took photographs of two animals. So you can assign animal 2 to P [(P, RST) = 2] and you can assign animal 3 to Q [(Q, RST) = 3] in the matrix. Then you are left with assigning animal 4 to S and animal 5 to T [(R, S) = 4 and (S, T) = 5.] Since S has both animals 4 and 5, S cannot have animals 2 and (3) That means that we can assign 2 to R and 3 to T. The pairs who were successful were—(P, Q), (P, T), (Q, R), (R, S) and (S, T). Next, we need to determine which pair photographed which animal. We know that • The Emu is animal 1 or (3) • The Anteater is not 2 and 5. • The Bear is not 4 and 5. Therefore, the bear is 2 or (3). Consider these cases separately: (Emu, Bear) = (1, 2), (1, 3) and (3, 2). The first two cases lead to contradictions. Therefore, (Emu, Bear) = (3, 2) which leads to Crocodile = (4) Then, since T did not photograph the Anteater, the Anteater cannot be 2 or 5. So the Anteater must be (1) Then the Deer is 5.
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(2)
The garden path is 15 metre long. In the second part of the journey, once they passed the azalea, A and C beat B by 1 metre. They are a half hour behind D, but remember that C rested for an hour. C would have beaten D by half an hour if she had not rested. Therefore, the slower speed is 1 metre in .5 hour, or 2 m/hr. That makes the first part of the journey 8 + 2 = 10 metre. The second part must be half as long, or 5 metre.
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(3)
The best strategy is to not pick the first boy, then choose the next boy who is taller than the first. The probability is then 11/24 that he will pick the tallest boy. Let’s call the boys: A, B, C and D, tallest to shortest. If any of them are the first to leave the house, then the probability of picking the tallest boy is 0, 1, 1/2 or 1/3, respectively. That results in a combined probability of P = (0 + 1 + 1/2 + 1/3) / 4 = 11/24.
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(3)
Fanny and Gopal played in the 13th set. The total number of sets played = (15 + 14 + 9) / 2 = 19. Harish played 9 sets and must have sat out 10 sets. Since no one sat out two sets in a row, he must have played and lost in all the even numbered sets and sat out all the odd numbered sets.
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(4)
Jimmy is married to Tender. B is married to Rashmi. First, assume “Haan” means “Yes” and “Nahin” means “No.” Then Chander said that Bander is married to Tender. Pammi said that she is married to Chander. And Jimmy said that he is not married to Rashmi. Pammi, then, can’t be single and must be married to Jimmy or Bander. But if Pammi is married to Jimmy, Jimmy couldn’t have said what he did. And if Pammi is married to Bander, then Chander lied and he is married and Jimmy is single. But you still wouldn’t know if Tender or Rashmi is single. So assume that “Haan” means “No” and “Nahin” means “Yes.” Then Chander said that Bander is not married to Tender. Pammi said that she is not married to Chander. And Jimmy said that he is married to Rashmi. Jimmy must be single. Pammi is either married to Chander or she is single. Chander is either single and Bander is not married to Tender OR Chander is married and Bander is married to Tender. It isn’t possible for Chander to be married. If Chander is single, then Pammi is single. Jimmy is married to Tender and Bander is married to Rashmi.
Ans .
(2)
All of B’s statements are true. All of A’s and C’s statements are false. If all of A’s statements were true, they would lead to contradictions. His second statement would conflict with his third and with C’s first statement. If all of C’s statements were true, they would also lead to contradictions. If they were all true, B’s second, third and fourth statements would be false, which contradicts C’s fourth statement. If you assume all of B’s statements are true, there are no contradictions. If they are all true, you can determine that all of A’s and C’s statements are false.
Ans .
(2)
My neighbour tells the truth on Tuesdays. Take them one at a time and assume he is telling the truth on each of the three days that he makes a statement. Then assume he is not telling the truth on any of the days that he makes a statement. The only case that works is the one where he is telling the truth on the third day. On that third day, from the statements he made, he lies on Wednesdays and Fridays. He tells the truth on Mondays and Tuesdays. Yesterday was not Thursday, Saturday or Sunday. That means that he couldn’t be telling the truth Monday because yesterday can’t be Sunday and therefore, he must tell the truth on Tuesdays.
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(4)
The maximum number of knights is 32
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(2)
Armani is 21 He is the tallest and squares numbers.Birianni is 8. He is the thinnest and reverses numbers. Chutanmi is 16. He is the shortest and divides numbers in half.
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(4)
Deepika’s car entry code is 39876. Concentrate on the digits which have multiple potential matches per position because there are 10 numbers listed, but the access code is only 5 digits long. That means 5 of the random numbers in the list must be duplicates or triplicates so they can be ignored. The first digit in random number list does not have any duplicates... It must be left to the end to solve. The second digit has 4’s and 9’s duplicated. The third digit has 1’s triplicated and 8’s duplicated. The forth digit has 3’s duplicated and 7’s triplicated. The last digit has 1’s and 6’s duplicated. Because the first digit has no duplicates, one or more triplicates must be used. Because the 9th random number, 98174 uses both triplicates digits, we know that the answer cannot contain xx17x, but must be either xx13x or xx87x because we know that the other digit positions must all be duplicates. In other words, the solution will contain 1 unduplicated row (the first row), 3 duplicated rows, and 1 triplicated row. The 5th random row, 63136 contains both the 1 and 3, leaving us with xx87x as the only possible solution. By keeping track of which rows have been used, and which columns contain duplicates that must be used, the two duplicates are quickly found: x9876. This leaves the 7th row which has not been used, which results in the final answer: 39876.
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(3)
A 16 B 64 C and E 59 D 49
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(4)
Your poker hand consists of • 2 of spades • 5 and 9 of hearts • 4 of diamonds • 8 of clubs Since all suits are present, there must be at least two red cards and two black cards. And there is only one suit with two cards. Since the hearts total 14 and there are no aces or face cards and all cards have a different value, the heart combination can only be (4, 10), (5, 9) or (6, 8). The sum of the odd cards equals the sum of the even cards, therefore, there must be two odd and three even cards. The maximum of the odd cards is 7 + 9 = 16. The minimum of the even cards is 2 + 4 + 6 = 1(2). Therefore, the sums must be 12, 14, or 16. The only combination of cards that meets all these criteria is the one shown above.
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(2)
During the first reading of the questions one needs to react to the chain of events taking place in the questions: Divayabh submitted the application form on Monday. Tuesday was a holiday. Clearance from clerk on Wednesday and consecutive clearance from senior clerk on Wednesday itself. Finally the application reached to the head clerk on Thursday. Hence option (b) is the correct answer.
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(2)
The questions are approached by tabulating all the indirect and direct clues in the following table: Order Manager Duration of Meeting Gap taken by Mark 1 C 9:00 AM TO 9:30 AM 25 2 G 9:55 AM TO 10:25 AM 25 3 F 10:50 AM TO 11:20 AM 25 4 A 11:45 AM TO 12:15 PM 25 5 E 12:40 PM TO 1:10 PM 50 6 D/B 2:00 PM TO 2:30 PM 25 7 B/D 2:55 PM TO 3:25 PM - Now we can answer the questions: Hence option (b) is the correct answer.
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(3)
The time will be 3:50 pm. Hence option (d) is the correct answer.
Ans .
(1)
30. In these questions there are two types of clues given, since there are multiple objects in the questions which are to be arranged in a particular fashion. First are the primary clues and then are the secondary clues. To solve the questions first use all the primary clues to create a clear structure of the situation in the following manner: Let heights be H1 to H6. Where H1 being the tallest and H6 being the shortest. Since Suresh is wearing orange coloured jacket, he is standing between Sachin and Yuvraj, positions of all the three can be [Sachin, Suresh, Yuvraj] or [Yuvraj, Suresh, Sachin]. Now since Rahul is wearing yellow-coloured jacket and is exactly opposite Sachin and Saurav, the person wearing the green coloured jacket is exactly opposite Suresh, the positions of the persons may be given as: Possibility 1 Possibility 2 Rahul Saurav Mahendra Mahendra Saurav Rahul (Yellow) (Green) (Blue) (Blue) (Green) (Yellow) ROAD ROAD Sachin Suresh Yuvraj Yuvraj Suresh Sachin (White) (Orange) (Red) (Red) (Orange) (White) It is evident that Mahendra is opposite to Yuvraj. Now since Mahendra is opposite to the person wearing red-coloured jacket, it is given that Sachin is wearing white coloured jacket hence, by elimination Mahendra is the person wearing blue-coloured jacket. Now delving into the heights of the six enemies, the height of Mahendra is H(1). The shortest person is Suresh because he is opposite to the person wearing green jacket. Hence Suresh is the shortest i.e. H6. Since Sachin is taller than Rahul but shorter than Yuvraj and Saurav this means that Rahul and Sachin are H5 and H(4). But heights of Saurav and Yuvraj are still unknown. Now we can tabulate all the information into a table: Person Colour Height Sachin White H4 Saurav Green H2/H3 Rahul Yellow H5 Yuvraj Red H3/H2 Mahendra Blue H1 Suresh Orange H6 29. Hence option (a) is the correct answer.
Ans .
(3)
Hence option (c) is the correct answer.
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(2)
31–32. The clues in the questions that can be decoded directly for tabulation will lead to the following table: Table 1 for Questions 31-32 Station Get In Get Down Base Station ???? XXX Kalyan ???? XXX Thane XXX ???? Kurla Only Gavli Biswas and Devendra Dadar Amar Only Ekta Mumbai Central XXX Amar, Gavli an Calvin Now from second clue Firoz gets down at Thane and he got in either at the base station or at Kalyan. Now since Firoz got down at Thane and he had got in with Calvin then it means that both Calvin and Firoz got in either at base station or at Kalyan. Since Biswas and Devendra got down at station Kurla then they too have got in either at the Base station or at Kalyan. It is given that Ekta got in with two other persons i.e. in a group of three persons. So, Ekta must have got in at the Base station. Now we can tabulate the conclusions as: Table 2 for Questions 31 and 32 Station Get In Get Down Base Station Ekta and (Calvin, Firoz) or (Biswas , Devendra) XXX Kalyan (Calvin, Firoz) or (Biswas , Devendra) XXX Thane XXX Only Firoz Kurla Only Gavli Biswas and Devendra Dadar Amar Only Ekta Mumbai Central XXX Amar, Gavli and Calvin Now we can answer the questions. 31. Hence option (c) is the correct answer.
Ans .
(3)
Hence option (d) is the correct answer.
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(4)
32. Hence option (d) is the correct answer. 33–35. From the sequence of statements, we can decode the clues to reach the following table: Table for Questions 33 and 34 House Truth Lie Dharma Yog Karma Devotion Salvation School Now we answer the Questionss. 33. Hence option (c) is the correct answer.
Ans .
(3)
Hence option (c) is the correct answer.
Ans .
(3)
The factors of 18 being 36, 18, 12, 9, 6, 4, 3, 2 and 1 we can clearly visualize that the possible products for the three numbers to get a multiple of 36 are: 18, 2, 1(sum of ages = 21); 12, 3, 1(sum of ages = 16); 6, 3, 2(sum of ages = 11); 4, 3, 3(sum of ages = 10); 9, 4, 1(sum of ages = 14 ); 9, 2, 2(sum of ages = 13); 36, 1, 1(sum of ages = 38); 6, 6, 1(sum of ages = 13). We further know that the sum of the ages is not enough information to identify the ages. This can only mean that the sum of the ages must be 13 – since that is the only sum of ages that still leaves confusion, once the sum of the ages is known (the confusion only gets resolved when it is known that there is an oldest child.
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(2)
Solutions 1 to 5 In this question there are essentially three clues. When we use the first clue (The shorter of Sachin and Laxman is the older of the two century hitters) we get two main possibilities depending on who is shorter between Sachin and Laxman. We get the following options: (In the above table Y denotes the younger century hitter) At this stage and the constraint of clue 2 (viz: The younger of Laxman and Rahul is the shorter of the two century hitters.) Looking at individual possibilities of following deductions would emerge: Possibility 1: Ruled out – Sachin is the older as well as the younger century hitter – not possible Possibility 2 and 3: Rule out – The shorter century hitter has to come out of Laxman or Rahul, but in this case the shorter century hitter is Sachin. Possibility 6: Ruled out: The younger of Rahul and Laxman has to be the shorter century hitter. But in this case Rahul is the younger century hitter while Laxman is the older century hitter (thus Rahul is younger than Laxman) yet is the taller one of the 2 century hitters also. Possibilities 4 and 5 cannot be eliminated as they obey all 3 clues. Thus Sachin is the younger century hitter while Laxman is the older century hitter (a conclusion which is consistent in both possibility 4 and 5). The answers are: (1) (b) Laxman
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(1)
Sachin
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(1)
Sachin
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(2)
Laxman
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(3)
Rahul
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(4)
Solutions 6 to 8 If we see clue 1 we get that the brother and daughter of Steffi are directly across the net from each other. Once we place these two we have two ways of placing Steffi and her son. The answers are: 6. (d) Since we do not know about the worst player (as it changes from possibility 1 to possibility 2).
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(1)
The brother emerges as the best player in both possibilities
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(4)
Data inadequate again as the answer would change based on which possibility we consider. Since both possibilities are possible, hence (4).
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(1)
Solutions 9 to 11 There are originally (by default) eight possibilities to distribute the three dishes amongst A, B and C. (PT = Paneer Tikka, RC = Roasted Chicken) These are: Hence the answers are: The three constraints are: 1. If A = PT, B= C 2. If B = PT, Aπ C 3. If C = RC, A= B 9. (a) Abdul always orders RC
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(1)
If C orders RC (Final possibility 3), B orders RC. Hence option (a) is correct.
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(1)
If B = PT (Final possibility 1) then Charu would also eat PT. Hence (a)
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(2)
Solutions 12 and 13 For convenience the squares can be numbered as: 1 2 3 4 5 6 7 8 9 It is obvious that the person who plays next will win by putting his mark in square 5. We need to hence find who will make the next move. Obviously the 7th move (winning one in this case) would be placed by the person who has not played the previous move. Hence, we need to see as to which of the 6 moves in the grid could possibly be the sixth and last one. A little bit of introspection will give you that the cross in the ninth square is the only possible 6th move. Hence, the next move (7th) will be that of a face. Thus the answers are: 12. (2)
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(2)
(2)
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(1)
Solutions 14 to 19 We start off by making a grid as follows (Note: in this grid we have used clues numbered 3 and 7) 416 Madhuri Kareena Aishwarya Shilpa Juhi Amit Shahrukh Anil Sunil 417 Akshay ¥ From clue 4 we get that Anil and Amit are in different films than Aishwarya. We also know from clue 2 that Shahrukh and Aishwarya are not together and from clue 1 that Aishwarya is not with Akshay too. Also since Aishwarya is between Akshay and Amit, Aishwarya’s number could only by 418 (As 416 is Madhuri) and Amit must be 419). Also, clue 6 gives us that neither Shilpa nor Juhi have acted with Amit. Based on these realisations, the grid above transforms into: 416 419 418 Madhuri Kareena Aishwarya Shilpa Juhi 419 Amit ¥ ✓ ¥ ¥ ¥ Shahrukh ¥ ¥ Anil ¥ ¥ 418 Sunil ¥ ¥ ✓ ¥ ¥ 417 Akshay ¥ ¥ ¥ At this point we need to concentrate on the time of film completion to move further in the question. From clue 2 Shahrukh’s film takes 3/4th the time of Sunil’s film. (But we know that Sunil’s film has Aishwarya). Hence, we conclude (from clue 2 and clue 4) If we add the information from clue 6 to this, we realise that Shilpa and Juhi must be after Amit. Thus, only Madhuri is left for Shahrukh. Also, since Anil’s film took maximum number of days Anil must be with Shilpa. We then get the following chronological figure: Also from clue 8, we get that x must be 32 days (Sunil’s film) and Shahrukh’s film must be 24 days and Anil’s film 48 days. The answers are: 14. Amit (a)
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(4)
Sunil and Aishwarya
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(2)
416
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(3)
Juhi
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(3)
32 days
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(2)
Shilpa
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(4)
Solutions 20 to 25 For this question if you try to refer to the information with respect to the country of the player you will not get any headway, but if you try to reference the information with respect to the room number you will find it easier. An obvious question that might crop up in your mind will obviously be – “But how do I know that?” and indeed “How do i decide what variable to reference for solving a complex question like this? The answer is simple really while reading the question for the first time, try to focus on spotting the variable with respect to which the maximum number of direct clues are associated. In this question it is clear that the maximum number of direct clues here are associated with the room number and hence we try to associate things with the room number. From clue number 5: Room No. 143 Æ 12 m and 8 Sixes (meal = m and sixes = S) From clue no. 7: Room No. 145 Æ 8 m and (P–2S) (Where P is the number of sixes hit by the Pakistan player). From clue 8: Pakistan, England and Australia can either be 141, 143 and 145 or 142, 144 and 146. But from question 20 (and it’s options) it is clear that Australia cannot be in Room 146. Thus, we get the following situation at this stage. 141 Pakistan 142 143 England (12 m, 8 s) 144 145 Australia (8 m, P–2s) 146 We also have the following information: From clue 1: Room No. 142 Æ 24 m (twice the player who hits 8 sixes). From clue 3 we have that the Indian is 8 less than the English (Meals) and 10 more than the English (Sixes). This means that the Indian is 4 m and 18 S. From clue 2: NZ + Room No. 146 = 40 (Meals). From clue 6: someone has (16 m and 24 S). Now since we have to adjust New Zealand, India and Sri Lanka (which we discover from options to question 24) and all the above facts the only way to do so is as shown below. 141 Pakistan 142 New Zealand (24m,) 143 England (12m, 8S) 144 India (4m, 18S) 145 Australia (8m, P – 2S) 146 Sri Lanka (16m, 24S) This arrangement also satisfies clue 4. The answer: 20. 145
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(3)
18
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(2)
12
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(3)
Sri Lanka
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(4)
)Sri Lanka
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(2)
New Zealand
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(1)
Solutions 26 to 31 We can make the following visual representations in order to represent the information: Clue 1: Coke – Pepsi – Sprite OR Sprite – Pepsi – Coke Clue 2:Coke ≠ Sprite ≠ Pepsi Clue 3: Limca Æ ¥ R Clue 4: Limca Æ ¥ 2.2 Clue 5: R > P Clue 6: (and clue 2) Sprite and Limca will share 1 litre and 1.6 litre amongst themselves (we get the values of the 4 quantities from clue 7 and from the options to question 26). Clue 7: 0.6 litre – Coke – 1 litre Clue 8: S Æ ¥ Coke and S Æ 0.6 litre At this stage we can make the first deduction: Since Coke is neither 0.6 nor 1 litre nor 1.6 litre, Coke must be 2.2 and since 1 litre and 1.6 litre are shared between Limca and Sprite, Pepsi must be 0.6 litre. This gives us 2 possibilities for distributing the 4 quantities: Possibility 1: Pepsi (0.6), Limca (1.6 litre), Sprite (1 litre) and Coke (2.2) OR Possibility 2: Pepsi (0.6), Limca (1 litre), Sprite (1.6 litre) and Coke (2.2). We now try to place both arrangements in order in terms of the bottles P, Q, R, S by obeying the constraints of the question. Considering possibility 1 above and clue 7, we get: Pepsi Coke Sprite or Sprite Coke Pepsi as the two possible arrangements such that Coke comes between 0.6 litres and 1 litre. However, according to clue 1, Pepsi is between Sprite and Coke and possibility 1 clearly violates this condition. Hence, we reject possibility 1. Thus, we have possibility 2 as the only possibility left i.e.: Pepsi = 0.6 Limca = 1 Sprite = 1.6 Coke = 2.2 We just need to arrange the placement of these 4 cold drinks in the right order. We need to realise at this point that both Coke and Pepsi cannot be extremes. Hence, the extremes have to be made up of Limca and Sprite. This can be done in two ways: (A) P Q R S Limca Sprite (1) 1.6 In this case we need to place Coke between 0.6 and 1. Thus we get: P Q R S Limca Coke Pepsi Sprite (1) 2.2 0.6 1.6 But we can see that clue 5 is violated. (R > P) Hence, we reject this arrangement and go to the next possibility: (B) P Q R S Sprite Pepsi Coke Limca 1.6 0.6 2.2 1 This arrangement meets all constraints and gives us the answers as: 26. 0.6 litre
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(1)
Bottle P
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(2)
Coke
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(3)
1 ltr
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(3)
R
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(3)
1.6 ltr
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(4)
40 can be scored by: (1) 50 – 30 + 20 or (2) 10 + 10 + 20. Hence option (d) is correct.
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(2)
There are 3 ¥ 3 ¥ 3 = 27 possible scores. Out of these three there are only 2 ways of scoring 100. Hence 2/27. option (2)
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(4)
There are only two ways of scoring 20 from the table viz : 50 + 10 – 40 or 10+50 – 40. In both cases there are three right answers. Hence, the required probability is zero. option (d) is correct.
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(2)
The maximum Penalty for 2 incorrect answers would occur when they happen in the last round. Hence, the score would be 50 +50 – 40 = 60 Hence option (b) is correct.
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(4)
None of these
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(2)
Chopra and Daya
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(2)
Mr. Bhatia
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(1)
Ram Shyam
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(3)
Mr. Ahluwalia and Mr. Eeshwar are both named Ram Shyam. Hence (3)
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(3)
There will be 8C2 + 8C2 + 4 quarter finals + 2 semifinals + 1 final = 63 matches.
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(2)
A team can be eliminated even if it wins 5 of its 7 matches (as 5 teams can possibly win 5 matches each) The logic for this is that in each group there are 8C2 = 28 matches, hence 28 winners. Thus, it is not possible for 5 teams to win 6 matches each. Hence if a team wins six matches it is safe. However, if a team wins 5 matches, 4 more teams could also have won 5 matches each and in that case the team would get eliminated. Hence, advancement is not guaranteed with 5 wins.
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(1)
In this question we need to guarantee elimination. With 2 wins a team could still qualify (And hence elimination is not guaranteed) since there could be a scenario of (7 + 6 + 5 + 2 + 2 + 2 + 2 +2) wins giving a situation wherein at least one of the teams with 2 wins needs to qualify. Hence, only if a team has 1 win, it’s elimination is guaranteed. Option (1) is correct.
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(3)
Quarter finals, semifinals and finals. Hence 3, option (1) is correct.
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(3)
Option (c) is possible. It is not necessary for the winner to have more wins than any other team. Since it is possible that a team with 2 wins in the first phase and 3 more wins (Quarterfinal, semifinal and final) in the 2nd stage wins the tournament and a team with 7 wins in the first phase loses the Quarterfinal. Hence (a) is not true. Option (b) is also eliminated as it could happen that in one group a team with 5 wins is eliminated while in the other group a team with 2 wins advances. Option (d) is also incorrect, since in the second stage of the tournament 4 teams who lose the quarterfinal will have no wins, 2 teams which lose the semifinals will have 1 win, 1 team which loses the final will have 2 wins and the team which wins the tournament will have 3 wins. Option (c) could happen if a team is eliminated with 5 wins in the first stage and another team from the other group goes through to the second stage with 2 wins and then wins the 3 rounds in the second stage to emerge the winner with a total of 5 wins.