CAT 2019 Question Paper

Directions for questions 63 to 66:

Answer the following question based on the information given below.

Questions 63 to 66 carry 3 marks each.

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.

X, U, and Z are standing at the three corners of a triangle formed by three street segments.
X can see only U and Z.
Y can see only U and W.
U sees V standing in the next intersection behind Z.
W cannot see V or Z. 6. No one among the six is standing at intersection d.

Q. 1.

Should a new person stand at intersection d, who among the six would she see?

A).
V and X only
B).
W and X only
C).
U and Z only
D).
U and W only

Explanation :

Refer to the explanation in the solution of the first question.

A new person standing at d would see W and X only.

Hence option 2.

Directions for questions 63 to 66:

Answer the following question based on the information given below.

Questions 63 to 66 carry 3 marks each.

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.

X, U, and Z are standing at the three corners of a triangle formed by three street segments.
X can see only U and Z.
Y can see only U and W.
U sees V standing in the next intersection behind Z.
W cannot see V or Z. 6. No one among the six is standing at intersection d.

Q. 2.

What is the minimum number of street segments that X must cross to reach Y?

A).
2
B).
3
C).
1
D).
4

Explanation :

Refer to the explanation in the solution of the first question.

X must reach Y via g so that he would cross minimum street segments. i.e., he would cross 2 street segments.

Hence option 1.

Directions for questions 63 to 66:

Answer the following question based on the information given below.

Questions 63 to 66 carry 3 marks each.

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.

X, U, and Z are standing at the three corners of a triangle formed by three street segments.
X can see only U and Z.
Y can see only U and W.
U sees V standing in the next intersection behind Z.
W cannot see V or Z. 6. No one among the six is standing at intersection d.

Q. 3.

Who can V see?

A).
Z only
B).
U only
C).
U and Z only
D).
U, W and Z only

Explanation :

Refer to the explanation in the solution of the first question.

V can see only U and Z.

Hence option 3

Directions for questions 63 to 66:

Answer the following question based on the information given below.

Questions 63 to 66 carry 3 marks each.

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.

X, U, and Z are standing at the three corners of a triangle formed by three street segments.
X can see only U and Z.
Y can see only U and W.
U sees V standing in the next intersection behind Z.
W cannot see V or Z. 6. No one among the six is standing at intersection d.

Q. 4.

Who is standing at intersection a?

A).
No one
B).
W
C).
Y
D).
V

Explanation :

From (1), X, U, Z are at b,c, g or at b, f, g in some order. Thus, X, U or Z is definitely at g.

Let X is at g.

Case (i): U and Z at b and f.

From (4), U has to be at b, Z at f and V at j. From (2), no one is at c, e, k and h. As Y sees both U and W, Y must be at a and W at i. But then W sees V, which contradicts (5).
Thus, this case is not valid.

Case (ii): U and Z at b and c.

From (4), U has to be at c, Z at b and V at a. From (2), no one is at c, e,f, k and h. But then Y must be at I, j or l. But in that case Y cannot see U. Thus, this case is not valid.

Therefore, X cannot be at g.

Let Z is at g.

From (4), U is at c or f.

Case (i) U is at c and hence V is at k and X is at a. Again there is no place for V. This case is invalid.

Case (ii) U is at f, X is at b and V at h. V will be at e as he sees U. But then he will be able to see Z also. So this case is also invalid.

Thus, U is at g. Therefore, Z is at f, V at e and X at b. So, from (2) no one will be at a, c and j. From (3) and (5), it can be concluded that Y is at k and W at l. Thus, we have

No one is standing at intersection a.

Hence option 1.

Directions for questions 59 to 62:

Answer the following question based on the information given below.

Questions 59 to 62 carry 3 marks each.

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.

Q. 5.

List of all the vendors who are among the top three vendors on all six aspects is:

A).
Vendor 1 and Vendor 3
B).
None of the Vendors
C).
Vendor 1
D).
Vendor 3

Explanation :

Following table lists down the top three vendors (highlighted in blue) in each aspect

As observed, only Vendor 3 is among the top three in all the six aspects.

Hence option 4.

Directions for questions 59 to 62:

Answer the following question based on the information given below.

Questions 59 to 62 carry 3 marks each.

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.

Q. 6.

List of all the vendors who are among the top two scorers on the maximum number of aspects is:

A).
Vendor 1 and Vendor 2
B).
Vendor 1 and Vendor 5
C).
Vendor 2, Vendor 3 and Vendor 4
D).
Vendor 2 and Vendor 5

Explanation :

Following table lists down the top two vendors (highlighted in blue) in each aspect.

As observed, Vendor 1 is among the top two in 3 aspects (Reach, Quality and Customer Service). Vendor 5 is also among top two in 3 aspects (Reliability, Reach and Features).

All other vendors are among top two in 2 aspects only.

Hence option 2.

Directions for questions 59 to 62:

Answer the following question based on the information given below.

Questions 59 to 62 carry 3 marks each.

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.

Q. 7.

A vendor's final score is the average of their scores on all six aspects. Which vendor has the highest final score?

A).
Vendor 1
B).
Vendor 4
C).
Vendor 2
D).
Vendor 3

Explanation :

The average score of that vendor is highest who has the highest total score across all the six aspects. The totals of all vendors are shown in the table below.

So,Vendor 3 (396) has the highest final score.

Hence option 4.

Directions for questions 59 to 62:

Answer the following question based on the information given below.

Questions 59 to 62 carry 3 marks each.

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.

Q. 8.

On which aspect is the median score of the five vendors the least?

A).
Cost
B).
Customer Service
C).
Reliability
D).
Quality

Explanation :

The approximate data is tabulated as follows.

For each aspect, the five values are arranged in ascending order (from left to right) as shown below. Also shown is the median value as shaded below.

The least median score is 50 and that corresponds to 'Customer Service'

Hence option 2.

Directions for questions 55 to 58:

Answer the following question based on the information given below.

Questions 55 to 58 carry 3 marks each.

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.

Q. 9.

Which pair of performances were composed by the same composer?

A).
The second and the sixth
B).
The third and the seventh
C).
The first and the seventh
D).
The first and the sixth

Explanation :

Refer to the explanation in the solution of the first question.

The first and the sixth performances were composed by the same composer.

Hence option 4.

Directions for questions 55 to 58:

Answer the following question based on the information given below.

Questions 55 to 58 carry 3 marks each.

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.

Q. 10.

The sixth performance was composed by:

A).
Dyu
B).
Gagan
C).
Ashman
D).
Badal

Explanation :

Refer to the explanation in the solution of the first question.

The sixth performance was composed by Badal.

Hence, option 4.

Directions for questions 55 to 58:

Answer the following question based on the information given below.

Questions 55 to 58 carry 3 marks each.

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.

Q. 11.

Which of the following is FALSE?

A).
Rani did not perform in any item composed by Badal
B).
Princess did not perform in any item composed by Dyu
C).
Samragni did not perform in any item composed by Ashman
D).
Queen did not perform in any item composed by Gagan

Explanation :

Refer to the explanation in the solution of the first question.

Queen performed third item composed by Gagan. So statement given in option 4 is FALSE.

Hence, option 4.

Directions for questions 55 to 58:

Answer the following question based on the information given below.

Questions 55 to 58 carry 3 marks each.

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.

Q. 12.

Which of the following is true?

A).
The second performance was composed by Gagan
B).
The third performance was composed by Ashman
C).
The second performance was composed by Dyu
D).
The third performance was composed by Dyu

Explanation :

Four music composers assigned one item; among items 1 to 4; to each of the four finalist. These four composers assigned one item; among items 5 to 8; to different finalists.
From (v), as items assigned by Ashman were performed consecutively, these must be the fourth and the fifth performances.

From (iii), (iv) and (v), the first, the fourth, the fifth and the eighth performances were composed by Badal, Ashman, Ashman, Gagan respectively. Also as the number of performances between items assigned by each of the remaining composers was the same, Badal and Gagan must have composed 6^th and 3^rd items. Therefore, the second and the seventh items were composed by Dyu. As Ashman and Gagan composed items for Rani, using (ii), we can concluded that Gagan’s other composition was performed by Queen. Also, using (i), Badal’s other composition was performed by Samragni.

Thus, we have

It can be seen that the second performance was composed by Dyu.

Hence, option 3.

Directions for questions 51 to 54:

Answer the following question based on the information given below.

Questions 51 to 54 carry 3 marks each.

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.

Q. 13.

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

Explanation :

Refer to the explanation in the solution of the first question.

The digit 6 cannot be represented by the letter G.

Answer: 6.

Directions for questions 51 to 54:

Answer the following question based on the information given below.

Questions 51 to 54 carry 3 marks each.

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.

Q. 14.

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

Explanation :

Refer to the explanation in the solution of the first question.

The digit 6 cannot be represented by the letter G.

Answer: 6.

Directions for questions 51 to 54:

Answer the following question based on the information given below.

Questions 51 to 54 carry 3 marks each.

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.

Q. 15.

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

Explanation :

Refer to the explanation in the solution of the first question.

The digit 7 cannot be represented by the letter D.

Answer: 7.

Directions for questions 51 to 54:

Answer the following question based on the information given below.

Questions 51 to 54 carry 3 marks each.

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.

Q. 16.

Which digit does the letter B represent?

Explanation :

Refer to the explanation in the solution of the first question.

B represents the digit 9.

Answer: 9.

Directions for questions 51 to 54:

Answer the following question based on the information given below.

Questions 51 to 54 carry 3 marks each.

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.

Q. 17.

Which digit does the letter A represent?

Explanation :

Units place of (F + F) is F; F = 0 or 5 Units place of (H + H) or (H + H + 1) is also F. This is possible only when H is 5 and there is no carry over.

So, F = 0 and H = 5

∴ (1 + B + A) = (10A + A) ⇒ A = 1 and B = 9

As units place of (G + K) = 1, actual value of G + K = 11

So, 1 + F + A = 1 + 1 + 0 = 2 = C

Now, 1 + J = G and G + K = 11

Also, G, J and K ∈ {3, 4, 6, 7, 8}

Therefore, (G, J, K) = (8, 3, 7) or (7, 4, 6) or (4, 7, 3), accordingly D and E can be represented by (4,6) or (3, 8) or (6, 8) in some order.

The letter A represents digit 1.

Answer: 1.

Directions for questions 47 to 50:

Answer the following question based on the information given below.

Questions 47 to 50 carry 3 marks each.

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.

Q. 18.

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?

A).
5
B).
3
C).
6
D).
4

Explanation :

Considering the given options, the maximum number of different types can be 6.

Assume that there are 6 items.

Now number of items of same type as the one in box 45 = 1 + 31 + 43 = 75

So number of remaining items = 25

1 + 2 + 4 + 8 + 16 = 31. If there are 5 types of items, the minimum number of items of 5 types = 31.

31 + 75 >100

So, there cannot be 6 types of items.

Now consider that there are 5 types of items.

Now number of items of same type as the one in box 45 = 1 + 31 + 43 = 75

So number of remaining items = 25

Now, 25 = (1 + 2 + 4 + 17) or (1 + 3 + 6 + 16) ,… etc

So there can be 5 types of different items.

Hence, option 1.

Directions for questions 47 to 50:

Answer the following question based on the information given below.

Questions 47 to 50 carry 3 marks each.

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.

Q. 19.

Which of the following is not possible?

A).
There are exactly 30 items of type b
B).
There are exactly 75 items of type e
C).
There are exactly 60 items of type d
D).
There are exactly 45 items of type c

Explanation :

There is exactly one prize of type a.

[1]. If there are 30 items of type b, then items of type c = 100 – 30 −1 = 69. So this case is possible.

[2]. There are 75 items of type e, then items of type b, c and d = 100 – 75 −1 = 24.

Some of the values of (b, c, d) are (2, 4, 18) or (2, 5, 17) or (3, 6, 15). So this case is possible.

[3]. If there are 60 items of type d, then items of type b and c = 100 – 60 −1 = 39. So this case is possible as we can find many combinations for (b, c).

[4]. If there are 45 items of type c, then items of type a, b and c in all cannot be more than 1 + 22 + 45 = 68. Now items of type d has to be more than 90. But then total number of items exceed 100. So this case is not possible.

Hence, option 4.

Directions for questions 47 to 50:

Answer the following question based on the information given below.

Questions 47 to 50 carry 3 marks each.

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.

Q. 20.

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

Explanation :

There is exactly one prize of type a.

As we need to find maximum possible different types of prizes, number of prizes of type b has to be minimum possible and hence must be 2, number of items of type c = 4 …and so on.

1(type a) + 2(type b) + 4(type c) + 8(type d) + 16(type e) + 32(type e) = 63

Suppose there is prize of type f then number of items has to be at least 64. But then there are more than 100 items, which is not true. So there cannot be prize of type f.

Answer: 6.

Directions for questions 47 to 50:

Answer the following question based on the information given below.

Questions 47 to 50 carry 3 marks each.

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.

Q. 21.

What is the minimum possible number of different types of prizes?

Explanation :

There is exactly one prize of type a.

There can be 99 items of type b. Thus, there can be only two types of items.

Answer: 2.

Directions for questions 43 to 46:

Answer the following question based on the information given below.

Questions 43 to 46 carry 3 marks each.

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.

Tanzi, Umeza and Yonita had the same total score.
Total scores for all players, except one, were in multiples of three.
The highest total score was one more than double of the lowest total score.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Q. 22.

What was Tanzi's score in Round 3?

A).
1
B).
5
C).
3
D).
4

Explanation :

Refer to the explanation in the solution of the first question.

Tanzi’s score in Round 3 was 1.

Hence, option 1.

Directions for questions 43 to 46:

Answer the following question based on the information given below.

Questions 43 to 46 carry 3 marks each.

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.

Tanzi, Umeza and Yonita had the same total score.
Total scores for all players, except one, were in multiples of three.
The highest total score was one more than double of the lowest total score.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Q. 23.

Which of the following statements is true?

A).
Xyla was the highest scorer
B).
Zeneca’s score was 23
C).
Zeneca was the highest scorer
D).
Xyla’s score was 23

Explanation :

Refer to the explanation in the solution of the first question.

Xyla was the highest scorer.

Hence, option 1.

Directions for questions 43 to 46:

Answer the following question based on the information given below.

Questions 43 to 46 carry 3 marks each.

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.

Tanzi, Umeza and Yonita had the same total score.
Total scores for all players, except one, were in multiples of three.
The highest total score was one more than double of the lowest total score.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Q. 24.

What was Zeneca's total score?

Explanation :

Refer to the explanation in the solution of the first question.

Zeneca’s total score was 24.

Hence, option 4.

Directions for questions 43 to 46:

Answer the following question based on the information given below.

Questions 43 to 46 carry 3 marks each.

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.

Tanzi, Umeza and Yonita had the same total score.
Total scores for all players, except one, were in multiples of three.
The highest total score was one more than double of the lowest total score.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Q. 25.

What was the highest total score?

A).
23
B).
21
C).
24
D).
25

Explanation :

Tanzi, Umeza, Xyla, Yonita and Zeneca got 1,2,3,1,2 chances to shoot in the bonus rounds respectively. Therefore, in the compulsory round, 9 bull’s eye were hit. From (4), it can be concluded that number of bull’s eye hit in Rounds 1, 2 and 3 were 3, 4 and 2 respectively.

Note that Xyla got three chances in the bonus round. So she must have hit three bull’s eye in the compulsory rounds. Therefore, Xyla’s minimum score is 5 × 4 + 1 + 1 = 22.

Zeneca’s maximum score could be 5 × 4 + 4 = 24. Her minimum score could be 5 × 4 + 1 = 21. If her score was the maximum score, it must be an odd number. So, 23 is the maximum score. Now using (3), the lowest score = 11. But both 11 and 23 are not divisible by 3. Using (2), we can conclude that 23 is not the maximum score. So, Xyla’s must have scored maximum. And Zeneca’s score was either 21(5 × 4 + 1) or 24(5 × 4 + 4).

Thus, Xyla scored 5 in each of the compulsory rounds and 4 in round 6.

Tanzi hit one bull’s eye either in round 1 or in round 3. So her minimum score = 5 + 4 + 1 + 5 = 15 and her maximum score = 5 + 4 + 4 + 5 = 18.

Yonita’s maximum score = 5 + 4 + 3 + 5 = 17.

So from (1) and (2), Tanzi, Umeza and Yonita each had total score 15. And hence, Wangdu scored least points i.e., 12 points. The only possible combination is 4 points in each of the compulsory rounds.

So, Tanzi scored 1 and 5 in rounds 1 and 3 in some order. Assume that she scored 1 in round 1 and 5 in round 3. From (5), Zeneca scored 1 in round 1. But then she must also have scored 5 in round 3 as she hit bull’s eye twice in the compulsory rounds. But this is contradiction to (5). So, Tanzi and Zeneca scored 5 in round 1. Tanzi scored 1 in round 3.

Thus, Tanzi, Zeneca and Xyla hit bull’s eye in round 1. Therefore Yonita (total score =15) must have hit bull’s eye in the second round and scored 2 points in the first round.

Umeza must have hit two bull’s eye in rounds 2 and 3. Also, she must have scored 2 points in the first round.

Thus, Umeza and Xyla hit the bull’s eye in the third round. Therefore, Zeneca hit bull’s eye in the second round and scored 4 points in the third round.

Thus, we have

So, the highest total score was 25.

Hence, option 4.

Directions for questions 39 to 42:

Answer the following question based on the information given below.

Questions 39 to 42 carry 3 marks each.

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 *.

Q. 26.

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

Explanation :

The approximate values in the bar chart can be tablated as follows.

Following table is sorted list for all three crime categories where in each category, states are arrranged in descending order (from top to bottom) according to the number of cases. The 'Rank' column determines the rank of each state among all states/UTs of India.

Observe that in IPC crimes category, Maharashtra and Karnataka are ranked 3. So Delhi and Goa would be ranked 1 and 2 respectively.

In Other crimes category, Delhi is clearly ranked 1 as it has the highest number of cases.

In SLL crimes category, West Bengal and Karnataka are ranked 1 and 2 respectively. Since Goa and Maharashtra have been ranked 4. Also Delhi's number of cases (36) is lesser than that of Karnataka (49), so Delhi cannot be ranked 2. So it has to be ranked 3. Please note that Delhi having 36 cases was an approximate value.

Required sum = 1 + 3 + 1 = 5.

Answer: 5.

Directions for questions 39 to 42:

Answer the following question based on the information given below.

Questions 39 to 42 carry 3 marks each.

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 *.

Q. 27.

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

Tamil Nadu: 2
Puducherry: 3

A).
neither i , nor ii
B).
only ii
C).
only i
D).
both i and ii

Explanation :

The approximate values in the bar chart can be tablated as follows.

All the above states/UTs are arranged in descending order according to the number of cases. The 'Rank' column in the table lists the rank given to each state/UT among all states/UTs of India.

Since Telangana and Maharashtra are both ranked 8, so all the states having more than 6 cases (Other crimes) are ranked 1 to 7.

So the final ranking is as follows.

So, Tamil Nadu is ranked 2 and Puducherry is ranked 3. Hence both statements (i) and (ii) are true.

Hence option 4.

Directions for questions 39 to 42:

Answer the following question based on the information given below.

Questions 39 to 42 carry 3 marks each.

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 *.

Q. 28.

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).
19 : 20
B).
11 : 10
C).
1 : 9
D).
3 : 2

Explanation :

The two states where the highest total number of cases are registered are West Bengal and Delhi.

Total number of cases across both these cities in the category;

IPC = 0 + 62 = 62.

SLL = 520 + 35 = 555.

Required ratio = 62/555 = 0.1117 ≈ 1/9.

Hence option 3.

Directions for questions 39 to 42:

Answer the following question based on the information given below.

Questions 39 to 42 carry 3 marks each.

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 *.

Q. 29.

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

Explanation :

The approximate values in the bar chart can be tablated as follows.

All the above states/UTs are arranged in descending order according to the number of cases. The 'Rank' column in the table lists the rank given to each state/UT among all states/UTs of India.

Note that Maharashtra and Karnataka are ranked 3. This means the two states with higher number of cases than Maharashtra and Karnataka have to be ranked as 1 and 2. So Delhi and Goa are ranked 1 and 2 respectively.

Since two states have been given the same rank (3), so the next state will be ranked 5.

So Kerela is ranked 5.

Answer: 5.

Directions for question 38:

Five sentences related to a topic are given below. Four of them can be put together to form a meaningful and coherent short paragraph. Identify the odd one out. Choose its number as your answer and key it in.

Q. 30.

Which of the following statements is necessarily true?

A).
There are at least four shelves between items B and C
B).
There are two empty shelves between the biscuits and the candies
C).
All biscuits are kept before candies
D).
All candies are kept before biscuits

Explanation :

Refer to the table and explanation in the solution of the first question.

Considering case (ii), 1, 2, 6, 12 are empty shelves.

Hence, option 4.

Directions for question 37:

The four sentences (labelled 1, 2, 3, and 4) given in this question, when properly sequenced, form a coherent paragraph. Decide on the proper order for the sentences and key in this sequence of four numbers as your answer.

Q. 31.

Which of the following can represent the numbers of the empty shelves in a possible arrangement

A).
1,2,8,12
B).
1,7,11,12
C).
1,5,6,12
D).
1,2,6,12

Explanation :

Refer to the table and explanation in the solution of the first question.

G is not a type of biscuit. It is a candy.

Hence, option 4.

Directions for questions 35 and 36:

Answer the following question based on the information given below.

Questions 35 to 38 carry 3 marks each.

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.

A and B are to be placed in consecutively numbered shelves in increasing order.
I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
K is to be placed in shelf number 16.
L and J are items of the same type, while H is an item of a different type.
C is a candy and is to be placed in a shelf preceded by two empty shelves.
L is to be placed in a shelf preceded by exactly one empty shelf

Q. 32.

Which of the following items is not a type of biscuit?

A).
A
B).
B
C).
L
D).
G

Explanation :

Refer to the table and explanation in the solution of the first question.

G is not a type of biscuit. It is a candy.

Hence, option 4.

Directions for questions 35 and 36:

Answer the following question based on the information given below.

Questions 35 to 38 carry 3 marks each.

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.

A and B are to be placed in consecutively numbered shelves in increasing order.
I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
K is to be placed in shelf number 16.
L and J are items of the same type, while H is an item of a different type.
C is a candy and is to be placed in a shelf preceded by two empty shelves.
L is to be placed in a shelf preceded by exactly one empty shelf

Q. 33.

In how many different ways can the items be arranged on the shelves?

A).
1
B).
8
C).
2
D).
4

Explanation :

There are 5 types of biscuits, 3 types of candies and 4 types of savouries. Among 16 shelves, there are 4 empty shelves.

It is given that all items of same type are clustered together with no empty shelf between items of the same type.

From (3) and (4), it can be concluded that D, E, F and K are savouries.

From (2) and (5), L, I and J are of one type and H is the other type. Therefore from (6), as C is a candy, L, I J must be types of biscuits and H is a type of candy. Now using (1), we can conclude that A and B are of one type but not candies as there are only 3 types of candies.

Therefore,

Biscuits: A, B, I, J, L Candies: C, H, G Savouries: D, E, F, K

From (3), (4), (6) and (7), there shelf number 12 must be an empty shelf. Also, D, E, F and K are placed in shelves numbered 13, 14, 15 and 16 respectively.

Now from (1), (2) and (7), the sequence (from left to right) in which biscuits are kept is:

(Empty shelf), L, A, B, (I/J), (J/I).

From (6), the candies must be in the following order: (Empty shelf), (Empty shelf), C, (H/G), (G/H)

Thus, we have

In each case, J and I can be arranged in 2 ways and G and H can be arranged among them in 2 ways. Thus, 2 × 2 = 4 ways.

Total number of ways the items can be arranged on the shelves = 4 + 4 = 8

Hence, option 2.

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