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# CAT 2019 DILR Question Paper with Solutions [Slot 1]

#### [SLOT 1]

INSTRUCTION

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 1:
Which digit does the letter A represent?

Question 2:
Which digit does the letter B represent?

Question 3:
Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?

Question 4:
Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?

F + F = F.

This implies that F has to be 0. Since all the digits are less than 10, the maximum sum of any two letters will be 17 (i.e. 9 + 8).

Therefore, a maximum of 1 can be carried over to the digits place on the left. In the ten thousands place, either the units digit of H + H = F or the units digit of H + H + 1 = F.

It cannot be H + H + 1 because 2H + 1 is an odd number and the units digit of that cannot be 0. From this, we can infer that the units digit of 2H must be 0.

This implies that H = 5. Sum of both the 6-digit numbers is a 7-digit number. This implies that the leftmost digit in the 7-digit number has to be 1. Therefore. A = 1.

Since H = 5, H + H = 10. This means that B + A + 1 = AA = 10A + A. A = 1. Therefore. B = 9.

In the hundreds place the units place of the sum A + F is C. Now A = 1 and F = 0.

But C cannot be 1. Hence, C has to be A + 1, i.e. C = 2. Therefore, G + K should be greater than 10.

The units digit of the sum of G + K is 1. This implies that G and K are either 3 and 8 or 4 and 7. not necessarily in that order.

We now have the following:

 9 5 1 1 G 0 1 5 J 0 K 0 1 1 0 G 2 1 0

Since G is a single digit, J + 1 is less than 9.

If J = 3. G = 4 and K = 7. In that case. D and E will be 6 and 8. not necessarily in that order.

If j = 4. G = 5. which is not possible because H = 5 and each letter has a distinct number.

If J = 6. G = 7 and K = 4. In that case. D and E will be

3 and 8. not necessarily in that order.

If j = 7. G = 8 and K = 3. In that case, D and E will be

4 and 6 not necessarily in that order.

If J = 8, G = 9, which is not possible because B = 9 and each letter has a distinct number.

INSTRUCTION

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.

1. No composer who assigned item to Princess, assigned any item to Queen.
2. No composer who assigned item to Rani, assigned any item to Samragni.
3. The first performance was by Princess; this item was assigned by Badal.
4. The last performance was by Rani; this item was assigned by Gagan.
5. The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Question 5:
Which of the following is true?
1. The second performance was composed by Dyu.
2. The third performance was composed by Ashman.
3. The second performance was composed by Gagan.
4. The third performance was composed by Dyu.

Question 6:
Which of the following is FALSE?
1. Queen did not perform in any item composed by Gagan.
2. Rani did not perform in any item composed by Badal.
3. Samragni did not perform in any item composed by Ashman.
4. Princess did not perform in any item composed by Dyu.

Question 7:
The sixth performance was composed by:
1. Gagan
2. Ashman
4. Dyu

Question 8:
Which pair of performances were composed by the same composer?
1. The first and the sixth
2. The second and the sixth
3. The first and the seventh
4. The third and the seventh

From the information given, there are a total of 8 dance items.

It is given that the first performance was by Princess and it was composed by Badal and the last performance was by Rani and composed by Gagan. No dancer performed her second item before the performance of the first item by all the other dancers and the dancers performed their second item in the same sequence as their first.

Therefore, the fourth and the fifth performances are by Rani and Princess respectively.

Items assigned by Ashman were consecutively numbered. Because the first four performances were by four different composers, the only possibility is that Ashman composed the fourth and the fifth performances.

If the sixth performance was composed by Badal, then the seventh would have been composed by Dyu. It is given that the number of performances between items assigned by remaining composers is the same. In that case, performances two. three and four would have been composed by Badal. Dyu and Gagan, in that order.

If the sixth performance was composed by Dyu, then the seventh would have been composed by Badal. In that case, the number of performances between items assigned by Badal is 5. Therefore, the number of items assigned by Dyu should also be 5. But that is not possible.

Therefore, the second, third, sixth and seventh performances were composed by Dyu, Gagan. Badal and Dyu respectively.

Since no composer who assigned music to Princess is assigned music to Queen, the sixth performance cannot be by Queen. Therefore, the sixth and seventh performances will be by Samragni and Queen respectively. Since the sequence is the same for their first items as well, the second and third performances will be by Samragni and Queen respectively.

Therefore, the overall sequence will be as follows:

INSTRUCTION

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.

Question 9:
On which aspect is the median score of the five vendors the least?
1. Customer Service
2. Reliability
3. Cost
4. Quality

Question 10:
A vendor's final score is the average of their scores on all six aspects. Which vendor has the highest final score?
1. Vendor 1
2. Vendor 2
3. Vendor 4
4. Vendor 3

Question 11:
List of all the vendors who are among the top two scorers on the maximum number of aspects is:
1. Vendor 1 and Vendor 5
2. Vendor 2, Vendor 3 and Vendor 4
3. Vendor 2 and Vendor 5
4. Vendor 1 and Vendor 2

Question 12:
List of all the vendors who are among the top three vendors on all six aspects is:
1. Vendor 1
2. Vendor 1 and Vendor 3
3. Vendor 3
4. None of the Vendors

From the graph, we get the following values:

 Cost Reliability Reach Quality Features Customer Service Vendor 1 75 55 80 75 40 55 Vendor 2 80 40 55 70 45 40 Vendor 3 90 75 65 65 55 50 Vendor 4 70 25 45 40 90 70 Vendor 5 60 60 70 50 75 30

INSTRUCTION

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.

Question 13:
What was the highest total score?
1. 24
2. 21
3. 25
4. 23

Question 14:
What was Zeneca's total score?
1. 22
2. 23
3. 21
4. 24

Question 15:
Which of the following statements is true?
1. Zeneca’s score was 23.
2. Xyla was the highest scorer.
3. Zeneca was the highest scorer.
4. Xyla’s score was 23.

Question 16:
What was Tanzi's score in Round 3?
1. 4
2. 3
3. 1
4. 5

Since Tanzi played another round, he/she must have scored 5 in either Round 1 or Round 3. In the other round, let us say that Tanzi scored x. So Tanzi’s total score would be 14 + x.

Umeza played Round 4 and Round 5. This means Umeza scored 5 in two of the first three rounds. In the remaining round, let Umezas score be y. Umeza's total score would be 13 + y.

Since the total score was not a multiple of 3 for only one person and Tanzi, Umeza and Yonita had the same total score, both 14 + x and 13 + y should be multiples of 3.

14 + x will be a multiple of 3 if x = 1 or 4. In that case the total score will be 15 or 18.

13 + y will be a multiple of 3 if y = 2 or 5. But if y = 5. then Umeza would have played Round 6 but that did not

happen. Therefore, y = 2 and x = 1. The total score of Umeza. Tanzi and Yonita is 15.

Since Wangdu did not play any round after Round 3, the maximum score that Wangdu can get is 12 when he

scores 4 in both Round 1 and Round 3.

Since Xyla played all the rounds, Xyla must have scored 5 in each of the first three rounds. So Xyla's minimum total score is 22, if Xyla scored 1 in Round 6.

Zeneca played Round 4 and Round 5. So Zeneca must have scored 5 in two of the first three rounds. So Zeneca's minimum and maximum total scores are 21 and 24 respectively. Therefore, Wangdu had the lowest score.

If Wangdu scored 12. then the highest score would be 25. Only Xyla can score 25 (5 in the first three rounds and 4 in Round 6).

If Wangdu scored 11. then the highest score would be 23. This is not possible because there will be two total scores that are not multiples of 3.

If Wangdu scored 10. then the highest score would be 21. But we know that Xyla's minimum score is 22. Therefore, this is not possible.

Any score of Wangdu less than 10 would mean the highest score is less than 20 but we know that Xyla's minimum score is 22. Therefore, Wangdu scored 12 and Xyla scored 25. This implies Wangdu scored 4 in each of Round 1 and Round 3 and Xyla scored 4 in Round 6.

Xyla's total score is not a multiple of 3. Hence. Zeneca's total score must be a multiple of 3. Zeneca would have scored 21 or 24.

Tanzi and Zeneca scored the same in Round 1. Tanzi's score in Round 1 is either 1 or 5. If Tanzi scored 1 in Round 1, then Zeneca would also have scored 1 in Round 1. But in this case, both Zeneca and Tanzi would have scored 5 in Round 3. But it is given that their scores in Round 3 are different. Therefore. Tanzi scored 5 in Round 1 and 1 in Round 3.

The number of players hitting bull s eye in Round 2 is either 2 or 4. If it is 2, then the total number of 5s in Round 2 and Round 3 combined should be 3. Two of those 5s were scored by Xyla. Umeza and Zeneca would each have scored at least one 5 in Rounds 2 and 3 combined but in this case, the number of 5s in Round 2 and 3 combined would be at least 4. which is not possible. Therefore, the number of players hitting bull's eye in Round 2 are 4. Since Tanzi and Wangdu scored 4 in Round 2, all the other players would have hit bull's eye in Round 2. This means that the number of players hitting bull’s eye in Round 3 are 2. Xyla is one of them and the other one has to be either Umeza or Zeneca. But if Zeneca had scored 5 in Round 3. then Zeneca would have played Round 6, which Zeneca didn't. Therefore, Umeza is the other person who scored 5 in Round 3.

Since Umeza s total score is 15, Umeza scored 2 in Round 1.

Since Yonita’s total score is 15, Yonita scored 2 in Round 1.

Zeneca s total score cannot be 21 because in that case, both Zeneca and Tanzi would have scored the same in Round 3, but they had different scores.

Therefore. Zeneca scored 4 in Round 3.

 Round 1 Round 2 Round 3 Round 4 Round 5 Round 6 Total Tanzi 5 4 1 5 NP NP 15 Umeza 2 5 5 1 2 NP 15 Wangdu 4 4 4 NP NP NP 12 Xyla 5 5 5 1 5 4 25 Yonita 2 5 3 5 NP NP 15 Zeneca 5 5 4 5 5 NP 24

INSTRUCTION

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.

Question 17:
Who is standing at intersection a?
1. No one
2. V
3. W
4. Y

Question 18:
Who can V see?
1. U only
2. U, W and Z only
3. U and Z only
4. Z only

Question 19:
What is the minimum number of street segments that X must cross to reach Y?
1. 2
2. 3
3. 1
4. 4

Question 20:
Should a new person stand at intersection d, who among the six would she see?
1. U and Z only
2. V and X only
3. W and X only
4. U and W only

It is given that X, U and Z are standing at the three corners of the triangle formed by three street segments. This means X, U and Z are standing at b, c. g or b, f, g. not necessarily in that order.

It is given that U sees V standing in the next intersection behind Z. This means U, V and Z are in a straight line with Z being in between U and V. The possibilities for this are:

(i) U, Z and V standing at b. f and j respectively. In this case, X is at g. Since X can only see U and Z, no one is standing at e. h and k. For Y to be able to see U and W, Y has to be at a and W has to be at i. But in this case, W will be able to see V, which is not valid. Hence, this case is not possible.

(ii) U, Z and V standing at b, c and d respectively. This is not possible because nobody is standing at d.

(iii) U, Z and V standing at c. b and a respectively. In this case. X is at g. Since X can only see U and Z. no one is standing at e, f. h and k. In this case. Y will not be able to see U. Hence, this is not possible.

(iv) U, Z and V standing at c. g and k respectively. In this case, X will be at b. Since X can only see U and Z, no one is standing at a. f and j. In this case. Y will not be able to see U. Hence, this is not possible.

(v) U, Z and V standing at g. f and e respectively. In this case, X will be at b. Since X can only see U and Z. no one is standing at a, c and j. W cannot see V or Z. Therefore. W cannot be at h or i. So W will be at k or I.

If W is at k, Y will be at h, i or I. In none of these three places can Y see both U and W.

Therefore. W is at I. For Y to be able to see only U and W, Y has to be at k.

(vi) U, Z and V standing at f, g and h respectively. In this case, X will be at b. Since X can only see U and Z, no one is standing at a, c and j. For Y to be able to see U. Y has to be at e. But in this case, Y will be able to see Z and V as well. But Y can only see U and W. Hence, this is not possible.

People are standing in the following way:

INSTRUCTION

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 state (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 *.

Question 21:
What is the rank of Kerala in the ‘IPC crimes’ category?

Question 22:
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
1. 1:9
2. 11:10
3. 3:2
4. 19:20

Question 23:
Which of the following is DEFINITELY true about the ranks of states/UT in the ‘other crimes’ category?

ii) Puducherry: 3

1. both i) and ii)
2. neither i) , nor ii)
3. only ii)
4. only i)

Question 24:
What is the sum of the ranks of Delhi in the three categories of crimes?

First, let us rank the states for which data has been given in each of the three categories.

IPC:

From the graph, the states/UTs with the highest number of cases in descending order are:

1. Delhi
2. Goa
3. Karnataka and Maharashtra
4. Kerala
5. Telangana
6. Haryana
8. Puducherry
9. West Bengal

SLL:

It is mentioned that in West Bengal, there are 520 cases registered as SLL crimes. No other state or union territory in the graph has more than West Bengal. The states/UTs with the highest number of cases in descending order are:

1. West Bengal
2. Karnataka
3. Delhi
4. Maharashtra and Goa
5. Haryana
7. Telangana and Kerala
8. Puducherry

Other Crimes:

From the graph, the states/UTs with the highest number of cases in descending order are:

1. Delhi
3. Puducherry
4. Karnataka
5. Goa
6. Kerala
7. Haryana
8. Telangana and Maharashtra
9. West Bengal

INSTRUCTION

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.

Question 25:
What is the minimum possible number of different types of prizes?

Question 26:
What is the maximum possible number of different types of prizes?

Question 27:
Which of the following is not possible?
1. There are exactly 30 items of type b.
2. There are exactly 45 items of type c.
3. There are exactly 75 items of type e.
4. There are exactly 60 items of type d.

Question 28:
You ask for the type of item in box 45. Instead of being given a direct answer, you are told that 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.

What is the maximum possible number of different types of items?

1. 6
2. 3
3. 5
4. 4

INSTRUCTION

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.

Question 29:
In how many different ways can the items be arranged on the shelves?
1. 2
2. 8
3. 4
4. 1

Question 30:
Which of the following items is not a type of biscuit?
1. G
2. B
3. L
4. A

Question 31:
Which of the following can represent the numbers of the empty shelves in a possible arrangement?
1. 1,7,11,12
2. 1,5,6,12
3. 1,2,6,12
4. 1,2,8,12

Question 32:
Which of the following statements is necessarily true?
1. There are two empty shelves between the biscuits and the candies.
2. All biscuits are kept before candies.
3. There are at least four shelves between items
4. All candies are kept before biscuits.

There are 12 items such that, 5 of them are biscuits, 3 of them are candies and 4 of them are savories.

It is given that K is in the shelf numbered 16.

D, E and F are placed in consecutively numbered shelves in the ascending order. They are placed after biscuits and candies. From this, it can be inferred that D, E and F are savories and they are placed at the end. Since K is in the last shelf, K is also a savory and all items of the same type are placed in consecutively numbered shelves, D, E, F and K will be in shelves numbered 13, 14, 15 and 16 respectively.

Now L and J are same type of items and since I and J are in consecutively numbered shelves, L, I and J are either biscuits or candies. It is given that C is a candy. If L, I and J are candies, then there would be a total of four candies. But it is not possible because there are only three candies. Therefore, L, I and J are biscuits. H is different from L. Therefore, H has to be a candy. Both A and B are in consecutively numbered shelves. This implies that they are of the same type. If A and B are candies, then the total number of candies will be 4. But this is not possible because there are only three candies. Therefore, A and B are biscuits.

We have two cases here.

Case 1: Candies are placed after biscuits

It is given that L is placed after exactly one empty shelf. Since biscuits are placed before candies, L is to be placed in shelf 2. Hence, the shelves numbered 2, 3, 4, 5 and 6 will have biscuits in them. I and J are in shelves numbered higher than those of A and B. Therefore. A and B will be in shelves 3 and 4 respectively. I and J will be in shelves 5 and 6. not necessarily in that order.

C is placed in a shelf preceded by two empty shelves. Therefore. C will be in shelf 9. Both H and G will be in shelves 10 and 11 in no particular order. Shelf numbered 12 will be empty.

 Biscuits 2. L3.A4. B5.1/J6. J/l Candies 9. C10.H/G11.G/H Savories 13.D14.E15.F16.K

The shelves numbered 1, 7,8 and 12 are empty.

Case 2: Biscuits are placed after candies

C is placed in a shelf preceded by two empty shelves. Since candies are placed before biscuits, C will be in shelf 3. Both H and G will be in shelves 4 and 5, not necessarily in that order. Shelf 6 will be empty.

It is given that L is placed after exactly one empty shelf. Therefore. L will be in shelf 7. Hence, the shelves numbered 7, 8, 9,10 and 11 will have biscuits in them.

I and J are in shelves numbered higher than those of A and B. Therefore. A and B will be in shelves 8 and 9 respectively. I and J will be in shelves 10 and 11. not necessarily in that order. Shelf numbered 12 will be empty.

The arrangement will be as follows:

 Candies 3. C4. H/G5. G/H Biscuits 7. L8. A9. B10. I/J11. J/l Savories 13. D14. E15. F16. K

The shelves numbered 1. 2, 6 and 12 are empty.

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