CAT 2019 DI and Reasoning - slot 01
Directions for Question 01 to 4:
A supermarket has to place 12 items (coded A to L) in shelves numbered 1 to 16. Five of these items are types of biscuits, three are types of candies and the rest are types of savouries. Only one item can be kept in a shelf. Items are to be placed such that all items of same type are clustered together with no empty shelf between items of the same type and at least one empty shelf between two different types of items. At most two empty shelves can have consecutive numbers. The following additional facts are known.
1. A and B are to be placed in consecutively numbered shelves in increasing order.
2. I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
3. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
4. K is to be placed in shelf number 16.
5. L and J are items of the same type, while H is an item of a different type.
6. C is a candy and is to be placed in a shelf preceded by two empty shelves.
7. L is to be placed in a shelf preceded by exactly one empty shelf.
In how many different ways can the items be arranged on the shelves?
a. 1
b. 2
c. 4
d. 8
The twelve items (A to L) are 5 biscuits, 3 candies and 4 savouries. Each one type of things will be placed in consecutive shelves and there will be a gap of at least one shelf between two different types of things. Total 16 shelves are there that means 4 empty shelves will be there.
Item A and B will be placed together and in same order. Item I and J will be placed after A and B but they will be placed together in any order. D, E and F are savouries and they will be placed in same order after biscuits and candies, means they will be placed at last place. As K is placed in Shelf number 16 that means it is a savoury and the shelves number of D, E and F are 13, 14 and 15 respectively. So the 4 savouries are D, E, F and K.
L, J and I are same type of item but H is of different type. Item C is a candy.
Now either A and B are of same type as L, I and J or A and B are of different type from L, I and J.
If A and B are of different type from L, I and J then H will be of same type as A and B. Now item C, which is a candy, cannot be with any group as there are only 3 candies. So this possibility is ruled out.
Hence A, B, L, I and J are Biscuits and C, G and H are candies.
These 12 items can be arranged in 16 shelves in the following manners:
Place of H can be interchanged with G and place of I can be interchanged with J. So each case has four ways. Total 8 ways are there to place these items.
The correct answer is: 8
Question 02: Which of the following items is not a type of biscuit?
a. G
b. B
c. A
d. L
The twelve items (A to L) are 5 biscuits, 3 candies and 4 savouries. Each one type of things will be placed in consecutive shelves and there will be a gap of at least one shelf between two different types of things. Total 16 shelves are there that means 4 empty shelves will be there.
Item A and B will be placed together and in same order. Item I and J will be placed after A and B but they will be placed together in any order. D, E and F are savouries and they will be placed in same order after biscuits and candies, means they will be placed at last place. As K is placed in Shelf number 16 that means it is a savoury and the shelves number of D, E and F are 13, 14 and 15 respectively. So the 4 savouries are D, E, F and K.
L, J and I are same type of item but H is of different type. Item C is a candy.
Now either A and B are of same type as L, I and J or A and B are of different type from L, I and J.
If A and B are of different type from L, I and J then H will be of same type as A and B. Now item C, which is a candy, cannot be with any group as there are only 3 candies. So this possibility is ruled out.
Hence A, B, L, I and J are Biscuits and C, G and H are candies.
These 12 items can be arranged in 16 shelves in the following manners:
Items A, B, L, I and J are Biscuits. So G is not Biscuit.
The correct answer is: G
b. 1, 2, 8, 12
c. 1, 7, 11, 12
d. 1, 5, 6, 12
The twelve items (A to L) are 5 biscuits, 3 candies and 4 savouries. Each one type of things will be placed in consecutive shelves and there will be a gap of at least one shelf between two different types of things. Total 16 shelves are there that means 4 empty shelves will be there.
Item A and B will be placed together and in same order. Item I and J will be placed after A and B but they will be placed together in any order. D, E and F are savouries and they will be placed in same order after biscuits and candies, means they will be placed at last place. As K is placed in Shelf number 16 that means it is a savoury and the shelves number of D, E and F are 13, 14 and 15 respectively. So the 4 savouries are D, E, F and K.
L, J and I are same type of item but H is of different type. Item C is a candy.
Now either A and B are of same type as L, I and J or A and B are of different type from L, I and J.
If A and B are of different type from L, I and J then H will be of same type as A and B. Now item C, which is a candy, cannot be with any group as there are only 3 candies. So this possibility is ruled out.
Hence A, B, L, I and J are Biscuits and C, G and H are candies.
These 12 items can be arranged in 16 shelves in the following manners:
In case-1 shelves 1, 2, 6, 12 are empty, while in case-2 shelves 1, 7, 8, 12 are empty.
The correct answer is: 1, 2, 6, 12
Question 04: Which of the following statements is necessarily true?
a. There are two empty shelves between the biscuits and the candies.
b. All biscuits are kept before candies.
c. All candies are kept before biscuits.
d. There are at least four shelves between items B and C
The twelve items (A to L) are 5 biscuits, 3 candies and 4 savouries. Each one type of things will be placed in consecutive shelves and there will be a gap of at least one shelf between two different types of things. Total 16 shelves are there that means 4 empty shelves will be there.
Item A and B will be placed together and in same order. Item I and J will be placed after A and B but they will be placed together in any order. D, E and F are savouries and they will be placed in same order after biscuits and candies, means they will be placed at last place. As K is placed in Shelf number 16 that means it is a savoury and the shelves number of D, E and F are 13, 14 and 15 respectively. So the 4 savouries are D, E, F and K.
L, J and I are same type of item but H is of different type. Item C is a candy.
Now either A and B are of same type as L, I and J or A and B are of different type from L, I and J.
If A and B are of different type from L, I and J then H will be of same type as A and B. Now item C, which is a candy, cannot be with any group as there are only 3 candies. So this possibility is ruled out.
Hence A, B, L, I and J are Biscuits and C, G and H are candies.
These 12 items can be arranged in 16 shelves in the following manners:
Biscuits may be placed before candies and may be placed after candies but there are at least four shelves between items B and C in both the cases.
The correct answer is: There are at least four shelves between items B and C
Question 05: Directions for questions 5 to 8:
A new game show on TV has 100 boxes numbered 1, 2, . . . , 100 in a row, each containing a mystery prize. The prizes are items of different types a, b, c, . . . , in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes
What is the minimum possible number of different types of prizes?
There are total 100 items of different type and there is only one item of type ‘a’ that is diamond ring. As it is said that the number of items at least doubles as we move to the next type. So the number of items of type ‘b’ can be minimum 2 and maximum 99. Number of items of type ‘c’ can be minimum 4 and maximum 97. Number of items of type ‘d’ can be minimum 8 and maximum 93.
Hence the number of items of each type can be
In item ‘g’ minimum is more than maximum hence it is not possible. So maximum six items and minimum two items can be there.
Minimum two items can be there.
Question 06: What is the maximum possible number of different types of prizes?
There are total 100 items of different type and there is only one item of type ‘a’ that is diamond ring. As it is said that the number of items at least doubles as we move to the next type. So the number of items of type ‘b’ can be minimum 2 and maximum 99. Number of items of type ‘c’ can be minimum 4 and maximum 97. Number of items of type ‘d’ can be minimum 8 and maximum 93.
Hence the number of items of each type can be
In item ‘g’ minimum is more than maximum hence it is not possible. So maximum six items and minimum two items can be there.
Maximum six items can be there.
Question 07: Which of the following is not possible?
a. There are exactly 75 items of type e.
b. There are exactly 45 items of type c.
c. There are exactly 30 items of type b.
d. There are exactly 60 items of type d.
There are total 100 items of different type and there is only one item of type ‘a’ that is diamond ring. As it is said that the number of items at least doubles as we move to the next type. So the number of items of type ‘b’ can be minimum 2 and maximum 99. Number of items of type ‘c’ can be minimum 4 and maximum 97. Number of items of type ‘d’ can be minimum 8 and maximum 93.
Hence the number of items of each type can be
In item ‘g’ minimum is more than maximum hence it is not possible. So maximum six items and minimum two items can be there.
If there are exactly 45 items of type ‘c’ then minimum 90 items of ‘d’ must be there and that will make total more than 100. If there are only three items a, b and c then the total number of items will be less than 100 as in this case item ‘b’ cannot be more than 22. Remaining all other options are possible.
The correct answer is: There are exactly 45 items of type c.
Question 08: What is the maximum possible number of different types of items?
a. 3
b. 4
c. 6
d. 5
There are total 100 items of different type and there is only one item of type ‘a’ that is diamond ring. As it is said that the number of items at least doubles as we move to the next type. So the number of items of type ‘b’ can be minimum 2 and maximum 99. Number of items of type ‘c’ can be minimum 4 and maximum 97. Number of items of type ‘d’ can be minimum 8 and maximum 93.
Hence the number of items of each type can be
In item ‘g’ minimum is more than maximum hence it is not possible. So maximum six items and minimum two items can be there.
There are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100. It means total 75 ( 31 + 43 + 1) items are there of this type. Hence it is the last type of item because next type must have 150 items which is not possible. It also means that the total number of item before this must be 25. Now to maximize the different type of items we must take minimum number of items at initial stage. There are many ways to make the sum 25.
1, 2, 4, 18, 75 or 1, 3, 6, 15, 75 or 1, 2, 5, 17, 75 or 1, 2, 6, 16, 75 or 1, 2, 7, 15, 75 and many more.
Hence in this case maximum 5 items can be there.
The correct answer is: 5
Question 09: Directions for questions 9 to 12:
Five vendors are being considered for a service. The evaluation committee evaluated each vendor on six aspects – Cost, Customer Service, Features, Quality, Reach, and Reliability. Each of these evaluations are on a scale of 0 (worst) to 100 (perfect). The evaluation scores on these aspects are shown in the radar chart. For example, Vendor 1 obtains a score of 52 on Reliability, Vendor 2 obtains a score of 45 on Features and Vendor 3 obtains a score of 90 on Cost.
On which aspect is the median score of the five vendors the least?
a. Quality
b. Customer Service
c. Cost
d. Reliability
From the given radar chart we can formulate a table of vendors and their score in each aspect.
Median score of the five vendors Cost, Customer Service, Features, Quality, Reach, and Reliability are 78, 50, 55, 62, 62 and 52 respectively. It is minimum 50 in Customer Service.
The correct answer is: Customer Service
Question 10: A vendor's final score is the average of their scores on all six aspects. Which vendor has the highest final score?
a. Vendor 2
b. Vendor 3
c. Vendor 4
d. Vendor 1
From the given radar chart we can formulate a table of vendors and their score in each aspect.
Final score of these five vendors is 62.83, 55.66, 65.66, 57 and 56.83 respectively. So the highest final score is of vendor-3.
The correct answer is: Vendor 3
Question 11: List of all the vendors who are among the top two scorers on the maximum number of aspects is:
a. Vendor 2, Vendor 3 and Vendor 4
b. Vendor 1 and Vendor 5
c. Vendor 1 and Vendor 2
d. Vendor 2 and Vendor 5
From the given radar chart we can formulate a table of vendors and their score in each aspect.
Vendor-1 and vendor-5 are three times (maximum) among the top two scorers.
The correct answer is: Vendor 1 and Vendor 5
Question 12: List of all the vendors who are among the top three vendors on all six aspects is:
a. Vendor 3
b. None of the Vendors
c. Vendor 1
d. Vendor 1 and Vendor 3
From the given radar chart we can formulate a table of vendors and their score in each aspect.
Vendor-3 is six times among the top three scorers.
Question 13: Directions for questions 13 to 16:
The figure below shows the street map for a certain region with the street intersections marked from a through l. A person standing at an intersection can see along straight lines to other intersections that are in her line of sight and all other people standing at these intersections. For example, a person standing at intersection g can see all people standing at intersections b, c, e, f, h, and k. In particular, the person standing at intersection g can see the person standing at intersection e irrespective of whether there is a person standing at intersection f.
Six people U, V, W, X, Y, and Z, are standing at different intersections. No two people are standing at the same intersection.
The following additional facts are known.
1. X, U, and Z are standing at the three corners of a triangle formed by three street segments.
2. X can see only U and Z.
3. Y can see only U and W.
4. U sees V standing in the next intersection behind Z.
5. W cannot see V or Z.
6. No one among the six is standing at intersection d.
Who is standing at intersection a?
a. V
b. Y
c. W
d. No one
No one
X, U and Z can be either at b, f and g or at b, c and g in any order as there are only these two triangles. U and Z cannot be at b and g because U sees V standing in the next intersection behind Z. It means X is either at b or at g. If X is at g then all the conditions are not satisfying. Now by trial and error we can say that X is at b, Z is at f, U is at g, V is at e, Y is at K and W is at l.
No one is standing at a.
The correct answer is: No one
Question 14: Who can V see?
a. U, W and Z only
b. U and Z only
c. Z only
d. U only
X, U and Z can be either at b, f and g or at b, c and g in any order as there are only these two triangles. U and Z cannot be at b and g because U sees V standing in the next intersection behind Z. It means X is either at b or at g. If X is at g then all the conditions are not satisfying. Now by trial and error we can say that X is at b, Z is at f, U is at g, V is at e, Y is at K and W is at l.
V can see only U and Z.
The correct answer is: U and Z only
Question 15: What is the minimum number of street segments that X must cross to reach Y?
a. 2
b. 3
c. 4
d. 1
X, U and Z can be either at b, f and g or at b, c and g in any order as there are only these two triangles. U and Z cannot be at b and g because U sees V standing in the next intersection behind Z. It means X is either at b or at g. If X is at g then all the conditions are not satisfying. Now by trial and error we can say that X is at b, Z is at f, U is at g, V is at e, Y is at K and W is at l.
X will go b to g and then g to k to reach Y. Hence he has to cross minimum 2 street segments.
The correct answer is: 2
Question 16: Should a new person stand at intersection d, who among the six would she see?
a. U and Z only
b. W and X only
c. U and W only
d. V and X only
X, U and Z can be either at b, f and g or at b, c and g in any order as there are only these two triangles. U and Z cannot be at b and g because U sees V standing in the next intersection behind Z. It means X is either at b or at g. If X is at g then all the conditions are not satisfying. Now by trial and error we can say that X is at b, Z is at f, U is at g, V is at e, Y is at K and W is at l.
If a new person stand at intersection d then she can see W and X only.
The correct answer is: W and X only
Question 17: Directions for questions 17 to 20:
Six players – Tanzi, Umeza, Wangdu, Xyla, Yonita and Zeneca competed in an archery tournament. The tournament had three compulsory rounds, Rounds 1 to 3. In each round every player shot an arrow at a target. Hitting the centre of the target (called bull’s eye) fetched the highest score of 5. The only other possible scores that a player could achieve were 4, 3, 2 and 1. Every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6. The possible scores in Rounds 4 to 6 were identical to the first three. A player’s total score in the tournament was the sum of his/her scores in all rounds played by him/her. The table below presents partial information on points scored by the players after completion of the tournament. In the table, NP means that the player did not participate in that round, while a hyphen means that the player participated in that round and the score information is missing.
The following facts are also known.
1. Tanzi, Umeza and Yonita had the same total score.
2. Total scores for all players, except one, were in multiples of three.
3. The highest total score was one more than double of the lowest total score.
4. The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
5. Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.
What was the highest total score?
a. 25
b. 21
c. 24
d. 23
Rounds 1 to 3 are compulsory rounds so every player has to play in these rounds. If a player hits a bull’s eye (score 5 points) in first three rounds then only he will get a chance to shoot in rounds 4 to 6 which are bonus rounds.
Tanzi is playing only once in Bonus rounds that means he has scored one bull’s eye in first three rounds. Similarly Umeza has scored 2 bull’s eyes, Wangdu has scored no bull’s eye, Xyla has scored 3 bull’s eyes, Yonita has scored 1 bull’s eye and Zeneca has scored 2 bull’s eyes in first three rounds. Total 9 bull’s eyes are scored in first three rounds.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
If there is 1 bull’s eye in round-3, then there will be 2 bull’s eyes in round-2 and 6 bull’s eyes in round-1, which is not possible because Wangdu has not scored any bull’s eye.
Now if there are 2 bull’s eyes in round-3, then there will be 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1.
If there are 3 bull’s eyes in round-3, then there will be 6 bull’s eyes in round-2 and 0 bull’s eye in round-1, which is again not possible as Xyala has scored bull’s eyes in all the first three rounds.
So there are 2 bull’s eyes in round-3, 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1. As Tanzi and Wangdu have scored 4 points in round-2 that means remaining all have scored bull’s eye in this round.
The score of Tanzi and Zeneca is same in round-1 either both will hit bull’s eye or both will not hit bull’s eye. If both do not hit bull’s eye then there will not be 3 bull’s eyes in round-1. hence both Tanzi and Zeneca hit the bull’s eye in round-1.
Total score of Wangdu can be 12 or less, which is the least possible total score among six players. It is given that the highest total score was one more than double of the lowest total score. So any one of these two is not multiple of 3. Hence total score of Wangdu has to be 12 and remaining all total score will be multiple of 3 except the highest score (25 points). Also the total score of Tanzi, Umeza and Yonita is same and that has to be 15.
Xyla scored the highest total points 25 because if Zeneca scores 25 then she has to score 5 points in round-3, which is not possible.
Zeneca total score should also be multiple of 3 but it cannot be 21 because Tanzi and Zeneca scored different scores in round-3. So the total score of Zeneca has to be 24.
The highest total score is 25.
The correct answer is: 25
Question 18: What was Zeneca's total score?
a. 22
b. 24
c. 23
d. 21
Rounds 1 to 3 are compulsory rounds so every player has to play in these rounds. If a player hits a bull’s eye (score 5 points) in first three rounds then only he will get a chance to shoot in rounds 4 to 6 which are bonus rounds.
Tanzi is playing only once in Bonus rounds that means he has scored one bull’s eye in first three rounds. Similarly Umeza has scored 2 bull’s eyes, Wangdu has scored no bull’s eye, Xyla has scored 3 bull’s eyes, Yonita has scored 1 bull’s eye and Zeneca has scored 2 bull’s eyes in first three rounds. Total 9 bull’s eyes are scored in first three rounds.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
If there is 1 bull’s eye in round-3, then there will be 2 bull’s eyes in round-2 and 6 bull’s eyes in round-1, which is not possible because Wangdu has not scored any bull’s eye.
Now if there are 2 bull’s eyes in round-3, then there will be 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1.
If there are 3 bull’s eyes in round-3, then there will be 6 bull’s eyes in round-2 and 0 bull’s eye in round-1, which is again not possible as Xyala has scored bull’s eyes in all the first three rounds.
So there are 2 bull’s eyes in round-3, 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1. As Tanzi and Wangdu have scored 4 points in round-2 that means remaining all have scored bull’s eye in this round.
The score of Tanzi and Zeneca is same in round-1 either both will hit bull’s eye or both will not hit bull’s eye. If both do not hit bull’s eye then there will not be 3 bull’s eyes in round-1. hence both Tanzi and Zeneca hit the bull’s eye in round-1.
Total score of Wangdu can be 12 or less, which is the least possible total score among six players. It is given that the highest total score was one more than double of the lowest total score. So any one of these two is not multiple of 3. Hence total score of Wangdu has to be 12 and remaining all total score will be multiple of 3 except the highest score (25 points). Also the total score of Tanzi, Umeza and Yonita is same and that has to be 15.
Xyla scored the highest total points 25 because if Zeneca scores 25 then she has to score 5 points in round-3, which is not possible.
Zeneca total score should also be multiple of 3 but it cannot be 21 because Tanzi and Zeneca scored different scores in round-3. So the total score of Zeneca has to be 24.
Zeneca’s total score is 24.
The correct answer is: 24
Question 19: Which of the following statements is true?
a. Xyla was the highest scorer.
b. Zeneca was the highest scorer.
c. Zeneca’s score was 23.
d. Xyla’s score was 23.
Rounds 1 to 3 are compulsory rounds so every player has to play in these rounds. If a player hits a bull’s eye (score 5 points) in first three rounds then only he will get a chance to shoot in rounds 4 to 6 which are bonus rounds.
Tanzi is playing only once in Bonus rounds that means he has scored one bull’s eye in first three rounds. Similarly Umeza has scored 2 bull’s eyes, Wangdu has scored no bull’s eye, Xyla has scored 3 bull’s eyes, Yonita has scored 1 bull’s eye and Zeneca has scored 2 bull’s eyes in first three rounds. Total 9 bull’s eyes are scored in first three rounds.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
If there is 1 bull’s eye in round-3, then there will be 2 bull’s eyes in round-2 and 6 bull’s eyes in round-1, which is not possible because Wangdu has not scored any bull’s eye.
Now if there are 2 bull’s eyes in round-3, then there will be 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1.
If there are 3 bull’s eyes in round-3, then there will be 6 bull’s eyes in round-2 and 0 bull’s eye in round-1, which is again not possible as Xyala has scored bull’s eyes in all the first three rounds.
So there are 2 bull’s eyes in round-3, 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1. As Tanzi and Wangdu have scored 4 points in round-2 that means remaining all have scored bull’s eye in this round.
The score of Tanzi and Zeneca is same in round-1 either both will hit bull’s eye or both will not hit bull’s eye. If both do not hit bull’s eye then there will not be 3 bull’s eyes in round-1. hence both Tanzi and Zeneca hit the bull’s eye in round-1.
Total score of Wangdu can be 12 or less, which is the least possible total score among six players. It is given that the highest total score was one more than double of the lowest total score. So any one of these two is not multiple of 3. Hence total score of Wangdu has to be 12 and remaining all total score will be multiple of 3 except the highest score (25 points). Also the total score of Tanzi, Umeza and Yonita is same and that has to be 15.
Xyla scored the highest total points 25 because if Zeneca scores 25 then she has to score 5 points in round-3, which is not possible.
Zeneca total score should also be multiple of 3 but it cannot be 21 because Tanzi and Zeneca scored different scores in round-3. So the total score of Zeneca has to be 24.
Xyla is the highest scorer.
The correct answer is: Xyla was the highest scorer.
Question 20: What was Tanzi's score in Round 3?
a. 5
b. 4
c. 3
d. 1
Rounds 1 to 3 are compulsory rounds so every player has to play in these rounds. If a player hits a bull’s eye (score 5 points) in first three rounds then only he will get a chance to shoot in rounds 4 to 6 which are bonus rounds.
Tanzi is playing only once in Bonus rounds that means he has scored one bull’s eye in first three rounds. Similarly Umeza has scored 2 bull’s eyes, Wangdu has scored no bull’s eye, Xyla has scored 3 bull’s eyes, Yonita has scored 1 bull’s eye and Zeneca has scored 2 bull’s eyes in first three rounds. Total 9 bull’s eyes are scored in first three rounds.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
If there is 1 bull’s eye in round-3, then there will be 2 bull’s eyes in round-2 and 6 bull’s eyes in round-1, which is not possible because Wangdu has not scored any bull’s eye.
Now if there are 2 bull’s eyes in round-3, then there will be 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1.
If there are 3 bull’s eyes in round-3, then there will be 6 bull’s eyes in round-2 and 0 bull’s eye in round-1, which is again not possible as Xyala has scored bull’s eyes in all the first three rounds.
So there are 2 bull’s eyes in round-3, 4 bull’s eyes in round-2 and 3 bull’s eyes in round-1. As Tanzi and Wangdu have scored 4 points in round-2 that means remaining all have scored bull’s eye in this round.
The score of Tanzi and Zeneca is same in round-1 either both will hit bull’s eye or both will not hit bull’s eye. If both do not hit bull’s eye then there will not be 3 bull’s eyes in round-1. hence both Tanzi and Zeneca hit the bull’s eye in round-1.
Total score of Wangdu can be 12 or less, which is the least possible total score among six players. It is given that the highest total score was one more than double of the lowest total score. So any one of these two is not multiple of 3. Hence total score of Wangdu has to be 12 and remaining all total score will be multiple of 3 except the highest score (25 points). Also the total score of Tanzi, Umeza and Yonita is same and that has to be 15.
Xyla scored the highest total points 25 because if Zeneca scores 25 then she has to score 5 points in round-3, which is not possible.
Zeneca total score should also be multiple of 3 but it cannot be 21 because Tanzi and Zeneca scored different scores in round-3. So the total score of Zeneca has to be 24.
Tanzi’s score in round-3 was 1 point.
The correct answer is: 1
Question 21: Directions for questions 21 to 24:
Princess, Queen, Rani and Samragni were the four finalists in a dance competition. Ashman, Badal, Gagan and Dyu were the four music composers who individually assigned items to the dancers. Each dancer had to individually perform in two dance items assigned by the different composers. The first items performed by the four dancers were all assigned by different music composers. No dancer performed her second item before the performance of the first item by any other dancers. The dancers performed their second items in the same sequence of their performance of their first items.
The following additional facts are known.
(i) No composer who assigned item to Princess, assigned any item to Queen.
(ii) No composer who assigned item to Rani, assigned any item to Samragni.
(iii) The first performance was by Princess; this item was assigned by Badal.
(iv) The last performance was by Rani; this item was assigned by Gagan.
(v) The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.
Which of the following is true?
a. The third performance was composed by Ashman.
b. The third performance was composed by Dyu.
c. The second performance was composed by Dyu.
d. The second performance was composed by Gagan.
The first four performances are performed by four different dancers and assigned by four different composers. Similarly next four performances are performed by four different dancers and assigned by four different composers. The items assigned by Ashman were performed consecutively. Hence items assigned by Ashman were performed at 4th and 5th number. The number of performances between items assigned by each of the Badal, Gagan and Dyu were the same. The first performance is assigned by Badal and the last performance is assigned by Gagan. The next performance assigned by Badal can be at 6th place or 7th place. Similarly, the 1st performance assigned by Gagan can be at 2nd or 3rd place. As we are to make equal difference between their performances so it can be only if Badal assigned 1st and 6th performances, Gagan assigned 3st and 8th performances and Dyu assigned 2st and 7th performances.
The second performance was composed by Dyu.
The correct answer is: The second performance was composed by Dyu.
Question 22: Which of the following is FALSE?
a. Queen did not perform in any item composed by Gagan.
b. Princess did not perform in any item composed by Dyu.
c. Rani did not perform in any item composed by Badal.
d. Samragni did not perform in any item composed by Ashman.
The first four performances are performed by four different dancers and assigned by four different composers. Similarly next four performances are performed by four different dancers and assigned by four different composers. The items assigned by Ashman were performed consecutively. Hence items assigned by Ashman were performed at 4th and 5th number. The number of performances between items assigned by each of the Badal, Gagan and Dyu were the same. The first performance is assigned by Badal and the last performance is assigned by Gagan. The next performance assigned by Badal can be at 6th place or 7th place. Similarly, the 1st performance assigned by Gagan can be at 2nd or 3rd place. As we are to make equal difference between their performances so it can be only if Badal assigned 1st and 6th performances, Gagan assigned 3st and 8th performances and Dyu assigned 2st and 7th performances.
Queen did not perform in any item composed by Gagan is false.
The correct answer is: Queen did not perform in any item composed by Gagan.
Question 23: The sixth performance was composed by:
a. Ashman
b. Dyu
c. Gagan
d. Badal
The first four performances are performed by four different dancers and assigned by four different composers. Similarly next four performances are performed by four different dancers and assigned by four different composers. The items assigned by Ashman were performed consecutively. Hence items assigned by Ashman were performed at 4th and 5th number. The number of performances between items assigned by each of the Badal, Gagan and Dyu were the same. The first performance is assigned by Badal and the last performance is assigned by Gagan. The next performance assigned by Badal can be at 6th place or 7th place. Similarly, the 1st performance assigned by Gagan can be at 2nd or 3rd place. As we are to make equal difference between their performances so it can be only if Badal assigned 1st and 6th performances, Gagan assigned 3st and 8th performances and Dyu assigned 2st and 7th performances.
Badal composed the 6th performance.
The correct answer is: Badal
Question 24: Which pair of performances were composed by the same composer?
a. The first and the seventh
b. The third and the seventh
c. The second and the sixth
d. The first and the sixth
The first four performances are performed by four different dancers and assigned by four different composers. Similarly next four performances are performed by four different dancers and assigned by four different composers. The items assigned by Ashman were performed consecutively. Hence items assigned by Ashman were performed at 4th and 5th number. The number of performances between items assigned by each of the Badal, Gagan and Dyu were the same. The first performance is assigned by Badal and the last performance is assigned by Gagan. The next performance assigned by Badal can be at 6th place or 7th place. Similarly, the 1st performance assigned by Gagan can be at 2nd or 3rd place. As we are to make equal difference between their performances so it can be only if Badal assigned 1st and 6th performances, Gagan assigned 3st and 8th performances and Dyu assigned 2st and 7th performances.
The first and the sixth performances were composed by same composer.
The correct answer is: The first and the sixth
Question 25: Directions for questions 25 to 28:
The Ministry of Home Affairs is analysing crimes committed by foreigners in different states and union territories (UT) of India. All cases refer to the ones registered against foreigners in 2016.
The number of cases – classified into three categories: IPC crimes, SLL crimes and other crimes – for nine states/UTs are shown in the figure below. These nine belong to the top ten states/UTs in terms of the total number of cases registered. The remaining states (among top ten) is West Bengal, where all the 520 cases registered were SLL crimes.
The table below shows the ranks of the ten states/UTs mentioned above among ALL states/UTs of India in terms of the number of cases registered in each of the three category of crimes. A state/UT is given rank r for a category of crimes if there are (r ‐ 1) states/UTs having a larger number of cases registered in that category of crimes. For example, if two states have the same number of cases in a category, and exactly three other states/UTs have larger numbers of cases registered in the same category, then both the states are given rank 4 in that category. Missing ranks in the table are denoted by *.
What is the rank of Kerala in the ‘IPC crimes’ categories?
West Bengal is ranked first in the total number of crime registered as there 520 cases were registered, all in SLL crimes. Telangana is ranked 10th in total number of cases registered. Total 25 cases are registered in Telangana including all the three categories. So no other state/UT can have more than 25 cases registered in one or other categories. Hence Delhi is ranked 1st and Goa is ranked 2nd in IPC crime. It is given that Karnataka and Maharashtra have rank 3 in IPC crime so Kerala will be ranked 5th in IPC crime.
In SLL crime West Bengal is ranked 1st and Delhi will be ranked 3rd. In other crime categories also we can rank only those states which have more than 25 (total number of cases registered in Telangana) number of cases registered. So Delhi is ranked 1st, Tamil Nadu is ranked 2nd , Puducherry is ranked 3rd and Karnataka is ranked 4th in Other crime category.
Kerala is ranked 5th in IPC crime.
Question 26: In the two states where the highest total number of cases are registered, the ratio of the total number of cases in IPC crimes to the total number in SLL crimes is closest to
a. 11 : 10
b. 19 : 20
c. 1 : 9
d. 3 : 2
West Bengal is ranked first in the total number of crime registered as there 520 cases were registered, all in SLL crimes. Telangana is ranked 10th in total number of cases registered. Total 25 cases are registered in Telangana including all the three categories. So no other state/UT can have more than 25 cases registered in one or other categories. Hence Delhi is ranked 1st and Goa is ranked 2nd in IPC crime. It is given that Karnataka and Maharashtra have rank 3 in IPC crime so Kerala will be ranked 5th in IPC crime.
In SLL crime West Bengal is ranked 1st and Delhi will be ranked 3rd. In other crime categories also we can rank only those states which have more than 25 (total number of cases registered in Telangana) number of cases registered. So Delhi is ranked 1st, Tamil Nadu is ranked 2nd , Puducherry is ranked 3rd and Karnataka is ranked 4th in Other crime category.
Highest total number of cases are registered in West Bengal and Delhi. So the ratio of IPC crime these two states together is to total number of cases SLL crimes in these two states is
64 : (520 + 26) or 1 : 9.
The correct answer is: 1 : 9
Question 27: Which of the following is DEFINITELY true about the ranks of states/UT in the ‘other crimes’ category?
(i) Tamil Nadu : 2
(ii) Puducherry : 3
a. both (i) and (ii)
b. only (i)
c. only (ii)
d. neither (i) , nor (ii)
West Bengal is ranked first in the total number of crime registered as there 520 cases were registered, all in SLL crimes. Telangana is ranked 10th in total number of cases registered. Total 25 cases are registered in Telangana including all the three categories. So no other state/UT can have more than 25 cases registered in one or other categories. Hence Delhi is ranked 1st and Goa is ranked 2nd in IPC crime. It is given that Karnataka and Maharashtra have rank 3 in IPC crime so Kerala will be ranked 5th in IPC crime.
In SLL crime West Bengal is ranked 1st and Delhi will be ranked 3rd. In other crime categories also we can rank only those states which have more than 25 (total number of cases registered in Telangana) number of cases registered. So Delhi is ranked 1st, Tamil Nadu is ranked 2nd , Puducherry is ranked 3rd and Karnataka is ranked 4th in Other crime category.
Both the statements are true.
The correct answer is: both (i) and (ii)
Question 28: What is the sum of the ranks of Delhi in the three categories of crimes?
West Bengal is ranked first in the total number of crime registered as there 520 cases were registered, all in SLL crimes. Telangana is ranked 10th in total number of cases registered. Total 25 cases are registered in Telangana including all the three categories. So no other state/UT can have more than 25 cases registered in one or other categories. Hence Delhi is ranked 1st and Goa is ranked 2nd in IPC crime. It is given that Karnataka and Maharashtra have rank 3 in IPC crime so Kerala will be ranked 5th in IPC crime.
In SLL crime West Bengal is ranked 1st and Delhi will be ranked 3rd. In other crime categories also we can rank only those states which have more than 25 (total number of cases registered in Telangana) number of cases registered. So Delhi is ranked 1st, Tamil Nadu is ranked 2nd , Puducherry is ranked 3rd and Karnataka is ranked 4th in Other crime category.
Sum of ranks of Delhi is = 1 + 3 + 1 = 5.
Directions for Question 29 to 32:
The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.
Question 29: Which digit does the letter A represent?
Here A, B, C, D, E, F, G, H, J, K are coded 0 to 9 in any order with distinct letters representing distinct digits.
A has to be 1 because in sum of two digits carry over if any can be 1 only. Now F + F = F, hence F has to be 0. A + F = C means 1 + 0 = C but C cannot be 1 because A is 1, hence C has to be 2 and there is 1 carry over from previous addition that is G + K.
Now H + H = F means H + H = 0, So H has to be 5 and again there is no carry over from previous addition that is A + J = G. There is 1 carry over from H + H = 10, that means B has to be 9.
Now we can say J + 1 = G and G + K = 11. From these two equations we can say J + K = 10. We can tabulate the multiple cases formed by these equations.
Question 30: Which digit does the letter B represent?
Question 31: Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?
Question 32: Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?