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# Test: Analytical Reasoning

## 50 Questions MCQ Test Logical Reasoning (LR) and Data Interpretation (DI) | Test: Analytical Reasoning

Description
This mock test of Test: Analytical Reasoning for CAT helps you for every CAT entrance exam. This contains 50 Multiple Choice Questions for CAT Test: Analytical Reasoning (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Analytical Reasoning quiz give you a good mix of easy questions and tough questions. CAT students definitely take this Test: Analytical Reasoning exercise for a better result in the exam. You can find other Test: Analytical Reasoning extra questions, long questions & short questions for CAT on EduRev as well by searching above.
QUESTION: 1

### Directions for Questions: 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. (2019) Q. Who is standing at intersection a?

Solution:

From the given data, X, U and Z are standing at the three corners of the triangle formed by three street so, they can be at any of b, c, f, g intersection.
Now, X cannot be at position g, since X can see only two people.
U, Z and V are standing in a line and Z and V are standing next to each other. Also Z and X are not at b and g as then V won’t be standing in the same row. Hence X is on b.
Since X ca n 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 l.
If W is at k, Y will be at h, i or l. In none of these three places Y can see both U and W.
Therefore, W is at l. For Y to be able to see only U and W, Y has to be at k.
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.
Hence, People are standing in the following way:

No one is standing at intersection a.

QUESTION: 2

### Directions for Questions: 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. (2019) Q. Who can V see?

Solution:

From the given data, X, U and Z are standing at the three corners of the triangle formed by three street so, they can be at any of b, c, f, g intersection.
Now, X cannot be at position g, since X can see only two people.
U, Z and V are standing in a line and Z and V are standing next to each other. Also Z and X are not at b and g as then V won’t be standing in the same row. Hence X is on b.
Since X ca n 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 l.
If W is at k, Y will be at h, i or l. In none of these three places Y can see both U and W.
Therefore, W is at l. For Y to be able to see only U and W, Y has to be at k.
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.
Hence, People are standing in the following way:

V can see U and Z only.

QUESTION: 3

### Directions for Questions: 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. (2019) Q. What is the minimum number of street segments that X must cross to reach Y?

Solution:

From the given data, X, U and Z are standing at the three corners of the triangle formed by three street so, they can be at any of b, c, f, g intersection.
Now, X cannot be at position g, since X can see only two people.
U, Z and V are standing in a line and Z and V are standing next to each other. Also Z and X are not at b and g as then V won’t be standing in the same row. Hence X is on 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 l.
If W is at k, Y will be at h, i or l. In none of these three places Y can see both U and W.
Therefore, W is at l. For Y to be able to see only U and W, Y has to be at k.
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.
Hence, People are standing in the following way:

X must cross two segments (b-g-k) to reach Y.

QUESTION: 4

Directions for Questions: 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.

(2019)

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

Solution:

From the given data, X, U and Z are standing at the three corners of the triangle formed by three street so, they can be at any of b, c, f, g intersection.
Now, X cannot be at position g, since X can see only two people.
U, Z and V are standing in a line and Z and V are standing next to each other. Also Z and X are not at b and g as then V won’t be standing in the same row. Hence X is on b.
Since X ca n 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 l.
If W is at k, Y will be at h, i or l. In none of these three places Y can see both U and W.
Therefore, W is at l. For Y to be able to see only U and W, Y has to be at k.
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.
Hence, People are standing in the following way:

When a new person is standing at d, that person can see only W and X.

QUESTION: 5

Directions for Questions: 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.

(2019)

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

Solution:

From condition (4): K is placed in shelve no 16 From condition (3): D, E and F are placed after all Biscuits and Candies. So, D, E, F and K are savouries and placed at position 13, 14, 15 and 16.
From condition (1, 2 and 5) : A and B are placed in consecutive number, So are of same type, like that I and J of same type. Again L and J are of same type, different from type of H.
From condition (6) : C is a candy. Number of candies are three and that are C, G and H Thus, A, B, I, J and L are biscuits.
Combining all conditions, we get
Biscuits : A, B, I, J and L.
Candy : C, G and H.
Savouries : D, E, F and K.
As L is preceded by one empty shelve and C is preceded by two empty shelves. So, possible arrangement of biscuits are L A B I J or LABJI. and possible arrangement of candies are C G H or C H G.
Let the income of Bimala be ₹100 Then Amla’s income = 1.2 × 100 = ₹120 and arrangement of savouries are D, E, F, K at position 13, 14, 15, 16 Now, one of the possible arrangement of 12 items in 16 shelves position are like this :

Total number of different ways in which biscuits can arrange = 2 {By changing I and J position} Total number of different ways in which candy can arrange = 2 {by changing G and H position}
total number of different ways in which empty at 1 and (7 and 8 together) can arrange = 2.
Hence, total number of arrangement = 2 × 2 × 2 = 8

QUESTION: 6

Directions for Questions: 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.

(2019)

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

Solution:

From condition (4): K is placed in shelve no 16 From condition (3): D, E and F are placed after all Biscuits and Candies. So, D, E, F and K are savouries and placed at position 13, 14, 15 and 16.
From condition (1, 2 and 5) : A and B are placed in consecutive number, So are of same type, like that I and J of same type. Again L and J are of same type, different from type of H.
From condition (6) : C is a candy. Number of candies are three and that are C, G and H Thus, A, B, I, J and L are biscuits.
Combining all conditions, we get
Biscuits : A, B, I, J and L.
Candy : C, G and H.
Savouries : D, E, F and K.
As L is preceded by one empty shelve and C is preceded by two empty shelves. So, possible arrangement of biscuits are L A B I J or LABJI. and possible arrangement of candies are C G H or C H G.
Let the income of Bimala be ₹100 Then Amla’s income = 1.2 × 100 = ₹120 and arrangement of savouries are D, E, F, K at position 13, 14, 15, 16 Now, one of the possible arrangement of 12 items in 16 shelves position are like this :

G is a candy not a biscuit.

QUESTION: 7

Directions for Questions: 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.

(2019)

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

Solution:

From condition (4): K is placed in shelve no 16 From condition (3): D, E and F are placed after all Biscuits and Candies. So, D, E, F and K are savouries and placed at position 13, 14, 15 and 16.
From condition (1, 2 and 5) : A and B are placed in consecutive number, So are of same type, like that I and J of same type. Again L and J are of same type, different from type of H.
From condition (6) : C is a candy. Number of candies are three and that are C, G and H Thus, A, B, I, J and L are biscuits.
Combining all conditions, we get
Biscuits : A, B, I, J and L.
Candy : C, G and H.
Savouries : D, E, F and K.
As L is preceded by one empty shelve and C is preceded by two empty shelves. So, possible arrangement of biscuits are L A B I J or LABJI. and possible arrangement of candies are C G H or C H G.
Let the income of Bimala be ₹100 Then Amla’s income = 1.2 × 100 = ₹120 and arrangement of savouries are D, E, F, K at position 13, 14, 15, 16 Now, one of the possible arrangement of 12 items in 16 shelves position are like this :

There are two possible ways for empty shelves 1, 7, 8, 12 or 1, 2, 6, 12

QUESTION: 8

Directions for Questions: 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.

(2019)

Q. Which of the following statements is necessarily true?

Solution:

From condition (4): K is placed in shelve no 16 From condition (3): D, E and F are placed after all Biscuits and Candies. So, D, E, F and K are savouries and placed at position 13, 14, 15 and 16.
From condition (1, 2 and 5) : A and B are placed in consecutive number, So are of same type, like that I and J of same type. Again L and J are of same type, different from type of H.
From condition (6) : C is a candy. Number of candies are three and that are C, G and H Thus, A, B, I, J and L are biscuits.
Combining all conditions, we get
Biscuits : A, B, I, J and L.
Candy : C, G and H.
Savouries : D, E, F and K.
As L is preceded by one empty shelve and C is preceded by two empty shelves. So, possible arrangement of biscuits are L A B I J or LABJI. and possible arrangement of candies are C G H or C H G.
Let the income of Bimala be ₹100 Then Amla’s income = 1.2 × 100 = ₹120 and arrangement of savouries are D, E, F, K at position 13, 14, 15, 16 Now, one of the possible arrangement of 12 items in 16 shelves position are like this :

In the given arrangement number of shelve between B and C = 4
There is other way of arrangement in which ‘C’ is at position ‘3’ and ‘B’ is at position ‘9‘
Then number of shelves between B and C = 5.

QUESTION: 9

Directions for Questions: 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.

(2019)

Q. Which of the following is not possible?

Solution:

There can be 30 items of type b, when there are 69 items of type c.
There can be 75 items of type e, when the number of items of type b, c and d are 3, 7 and 14 respectively. There can be 60 items of type d, when the number of items of type b and c are 11 and 28 respectively.
But there cannot be 45 items of type c. Because in that case, the maximum number of items of type b can be 22. Since the total number of items is not 100, we have to go to the next type. i.e. d, The number of items of d would have to be at least 90. This is not possible because the total number of items will be at least 138, but we only have 100. Hence cannot be 45 items of type c.

QUESTION: 10

Directions for Questions: 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.

(2019)

Q. 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?

Solution:

The maximum possible number of prizes can be 5. In that case, the maximum possible number of items should belong to type e. There can 75 items of type e, when the number of items of type b, c and d are 2, 4 and 18 respectively.
Therefore, the maximum possible number of different type of items is 5.

QUESTION: 11

Directions for Questions: 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.

(2019)

Q. What was the highest total score?

Solution:

From given data, we get that total score of Tanzi = Umeza = Yonita Since Tanzi played another round, he/she must have hit bulls’eye 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 have hit bull’s eye in two of the first three rounds. In the remaining round, let Umeza’s, score be y. Umeza’s total score would be 13 + y.
Yonita played round 4, he/she must have hit bull’s eye in round 1 or round 2. Let Yonita scored z in either of round 1 or 2, then, his/her total score = 5 + Z + 3 + 5 = 13 + Z.
Since the total score was not a multiple of 3 for only one person and Tanzi, Umeza and Yonita had the same total score 14 + x, 13 + y and 13 + z should be multiples of 3. 14 + x will be a multiple of 3 if x = 1 or 4. In that case the total score wil be 15 or 18. 13 + y will be multiple of 3 if y = 2 or 5. But if y = 5, then Umeza would had been played Round 6. Therefore, y = z = 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 and maximum total score is 26, based on Xyla scored (1 to 5) in round 6.
Zeneca played round 4 and round 5. So Zeneca must have hit bull’s eye and 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 socred 12 and Xyla scored 25 is only possible value.
Xyla’s total score is 25, which 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 socred 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. Therfore, Tanze 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 toal number of 5 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 roudns 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.
After adding all, we get that table :

The highest total score was 25.

QUESTION: 12

Directions for Questions: 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.

(2019)

Q. What was Zeneca’s total score?

Solution:

From given data, we get that total score of Tanzi = Umeza = Yonita Since Tanzi played another round, he/she must have hit bulls’eye 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 have hit bull’s eye in two of the first three rounds. In the remaining round, let Umeza’s, score be y. Umeza’s total score would be 13 + y.
Yonita played round 4, he/she must have hit bull’s eye in round 1 or round 2. Let Yonita scored z in either of round 1 or 2, then, his/her total score = 5 + Z + 3 + 5 = 13 + Z.
Since the total score was not a multiple of 3 for only one person and Tanzi, Umeza and Yonita had the same total score 14 + x, 13 + y and 13 + z should be multiples of 3. 14 + x will be a multiple of 3 if x = 1 or 4. In that case the total score wil be 15 or 18. 13 + y will be multiple of 3 if y = 2 or 5. But if y = 5, then Umeza would had been played Round 6. Therefore, y = z = 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 and maximum total score is 26, based on Xyla scored (1 to 5) in round 6.
Zeneca played round 4 and round 5. So Zeneca must have hit bull’s eye and 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 socred 12 and Xyla scored 25 is only possible value.
Xyla’s total score is 25, which 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 socred 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. Therfore, Tanze 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 toal number of 5 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 roudns 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.
After adding all, we get that table :

Zeneca’s total score was 24.

QUESTION: 13

Directions for Questions: 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.

(2019)

Q. Which of the following statements is true?

Solution:

From given data, we get that total score of Tanzi = Umeza = Yonita Since Tanzi played another round, he/she must have hit bulls’eye 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 have hit bull’s eye in two of the first three rounds. In the remaining round, let Umeza’s, score be y. Umeza’s total score would be 13 + y.
Yonita played round 4, he/she must have hit bull’s eye in round 1 or round 2. Let Yonita scored z in either of round 1 or 2, then, his/her total score = 5 + Z + 3 + 5 = 13 + Z.
Since the total score was not a multiple of 3 for only one person and Tanzi, Umeza and Yonita had the same total score 14 + x, 13 + y and 13 + z should be multiples of 3. 14 + x will be a multiple of 3 if x = 1 or 4. In that case the total score wil be 15 or 18. 13 + y will be multiple of 3 if y = 2 or 5. But if y = 5, then Umeza would had been played Round 6. Therefore, y = z = 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 and maximum total score is 26, based on Xyla scored (1 to 5) in round 6.
Zeneca played round 4 and round 5. So Zeneca must have hit bull’s eye and 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 socred 12 and Xyla scored 25 is only possible value.
Xyla’s total score is 25, which 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 socred 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. Therfore, Tanze 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 toal number of 5 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 roudns 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.
After adding all, we get that table :

The statement, “Xyla was the highest scorer”, true.

QUESTION: 14

Directions for Questions: 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.

(2019)

Q. What was Tanzi’s score in Round 3?

Solution:

From given data, we get that total score of Tanzi = Umeza = Yonita Since Tanzi played another round, he/she must have hit bulls’eye 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 have hit bull’s eye in two of the first three rounds. In the remaining round, let Umeza’s, score be y. Umeza’s total score would be 13 + y.
Yonita played round 4, he/she must have hit bull’s eye in round 1 or round 2. Let Yonita scored z in either of round 1 or 2, then, his/her total score = 5 + Z + 3 + 5 = 13 + Z.
Since the total score was not a multiple of 3 for only one person and Tanzi, Umeza and Yonita had the same total score 14 + x, 13 + y and 13 + z should be multiples of 3. 14 + x will be a multiple of 3 if x = 1 or 4. In that case the total score wil be 15 or 18. 13 + y will be multiple of 3 if y = 2 or 5. But if y = 5, then Umeza would had been played Round 6. Therefore, y = z = 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 and maximum total score is 26, based on Xyla scored (1 to 5) in round 6.
Zeneca played round 4 and round 5. So Zeneca must have hit bull’s eye and 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 socred 12 and Xyla scored 25 is only possible value.
Xyla’s total score is 25, which 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 socred 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. Therfore, Tanze 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 toal number of 5 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 roudns 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.
After adding all, we get that table :

Tanzi’s score in Round 3 was 1.

QUESTION: 15

Directions for Questions: 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.

(2019)

Q. Which of the following is true?

Solution:

From the information given, total number of performance 4 × 2 = 8
From fact (i) to (iv), Queen got its dance items assigned by either, Dyu, Gagan or Ashman and Samragni get its dance items assigned by either Badal, Dyu or Gagan.
First item performed by Princess and assigned by badal Last item performed by Rani and assigned by Ashiman As sequence of performance was same. So, Princess performed at number 1 and 5, while Rani performed at number 4 and 8 Now from (v) item assigned by Ashman were performed continuously. So, Ashman assigned item Rani at 4th performance and Princess at 5th performance.
Again, number of performance between item assigned by other three are same, so Badal assigned performance at 6th number. And Samragni was performed at 6th place followed by Queen performance assigned by Dyu.
Gagan assigned item to Rani at 8th performance.

Only the statement, The second performance was composed by Dyu”, is true.

QUESTION: 16

Directions for Questions: 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.

(2019)

Q. Which of the following is FALSE?

Solution:

From the information given, total number of performance 4 × 2 = 8
From fact (i) to (iv), Queen got its dance items assigned by either, Dyu, Gagan or Ashman and Samragni get its dance items assigned by either Badal, Dyu or Gagan.
First item performed by Princess and assigned by badal Last item performed by Rani and assigned by Ashiman As sequence of performance was same. So, Princess performed at number 1 and 5, while Rani performed at number 4 and 8 Now from (v) item assigned by Ashman were performed continuously. So, Ashman assigned item Rani at 4th performance and Princess at 5th performance.
Again, number of performance between item assigned by other three are same, so Badal assigned performance at 6th number. And Samragni was performed at 6th place followed by Queen performance assigned by Dyu.
Gagan assigned item to Rani at 8th performance.

Queen did not perform in any item composed by Gagan is false.

QUESTION: 17

Directions for Questions: 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.

(2019)

Q. The sixth performance was composed by:

Solution:

From the information given, total number of performance 4 × 2 = 8
From fact (i) to (iv), Queen got its dance items assigned by either, Dyu, Gagan or Ashman and Samragni get its dance items assigned by either Badal, Dyu or Gagan.
First item performed by Princess and assigned by badal Last item performed by Rani and assigned by Ashiman As sequence of performance was same. So, Princess performed at number 1 and 5, while Rani performed at number 4 and 8 Now from (v) item assigned by Ashman were performed continuously. So, Ashman assigned item Rani at 4th performance and Princess at 5th performance.
Again, number of performance between item assigned by other three are same, so Badal assigned performance at 6th number. And Samragni was performed at 6th place followed by Queen performance assigned by Dyu.
Gagan assigned item to Rani at 8th performance.

Sixty performance was composed by Badal.

QUESTION: 18

Directions for Questions: 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.

(2019)

Q. Which pair of performances were composed by the same composer?

Solution:

The 1st and 6th performances are composed by the same composer.

QUESTION: 19

Directions for Questions: The first-year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions.
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15- mark questions followed the 10-mark questions.
The following additional facts are known.
i. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
ii. Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT.
iii. All questions prepared by a faculty member appeared consecutively in MT as well as ET.
iv. Chelan prepared the third question in both MT and ET: and Esha prepared the eighth question in both.
v. Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.

(2019)

Q. The second question in ET was prepared by:

Solution:

From given data,
Let number of questions worth 5 marks, 10 marks and 15 marks was x, y and z respectively, then, 4 ≤ x, 3 ≤ y and 2 ≤ z
Case-1: Total possible combination that gives 100  marks Minimum total marks that are from 10 marks question = 3 × 10 = 30 marks
Minimum total marks that are from 15 marks questions = 2 × 15 = 30 marks
Maximum total marks that are from 5 marks questions = 5 × 8 = 40 marks
∴ 4 ≤ x ≤ 8
Case-II : When x and z are minimum
Minimum number of 5 marks questions = 4 Total marks = 5 × 4 = 20 marks
Minimum number of 15 marks questions = 2
total marks = 2 × 15 = 30
Maximum number of 10 marks questions

So, 3 ≤ y ≤ 5.
Case-III : When x and y are minimum
Minimum number of 5 marks questions = 4
total marks = 5 × 4 = 20 marks
Minimum number of 10 marks questions = 3
total marks = 3 × 10 = 30 marks.
Maximum number of 15 marks questions

∴ 2 ≤ z ≤ 3
Now, tabulating the values, we have

Again from question, number of questions in ET is more than MT. So, ET has maximum 5 marks questions than MT
For ET : x = 8, y = 3 and z = 2
⇒ Total marks = 8 × 5 + 3 × 10 + 2 × 15 = 100
Total number of questions = 8 + 3 + 2 = 13
For MT (i) x = 5, y = 3 and z = 3
⇒ Total marks = 5 × 5 + 3 × 10 + 3 × 15 = 100
Total number of questions = 5 + 3 + 3 = 11
(ii) x = 4, y = 5 and z = 2
Total marks = 4 × 5 + 5 × 10 + 2 × 15 = 100
total number of questions = 4 + 5 + 2 = 11
Now, considering MT and ET together, each faculty member prepared the same number of questions total number of questions = 13 + 11 = 24
So, number of questions prepared by each faculty = 24/6 = 4 questions
Here, Annie prepared only one question for MT, So, rest 10 questions of MT is prepared by other 5 faculty such that 2 questions are prepared by each.
Again, each faculty, prepare 4 questions.
So, each faculty except Annie prepared 2 questions for ET and Annie prepared three questions for ET.
Here each members prepared, questions continuously. Based on that, we get the following table:
Order of questions prepared by the faculty

Second question in ET was prepared by Dave.

QUESTION: 20

Directions for Questions: The first-year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions.
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15- mark questions followed the 10-mark questions.
The following additional facts are known.
i. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
ii. Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT.
iii. All questions prepared by a faculty member appeared consecutively in MT as well as ET.
iv. Chelan prepared the third question in both MT and ET: and Esha prepared the eighth question in both.
v. Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.

(2019)

Q. How many 5-mark questions were there in MT and ET combined?

Solution:

From given data,
Let number of questions worth 5 marks, 10 marks and 15 marks was x, y and z respectively, then, 4 ≤ x, 3 ≤ y and 2 ≤ z
Case-1: Total possible combination that gives 100  marks Minimum total marks that are from 10 marks question = 3 × 10 = 30 marks
Minimum total marks that are from 15 marks questions = 2 × 15 = 30 marks
Maximum total marks that are from 5 marks questions = 5 × 8 = 40 marks
∴ 4 ≤ x ≤ 8
Case-II : When x and z are minimum
Minimum number of 5 marks questions = 4 Total marks = 5 × 4 = 20 marks
Minimum number of 15 marks questions = 2
total marks = 2 × 15 = 30
Maximum number of 10 marks questions

So, 3 ≤ y ≤ 5.
Case-III : When x and y are minimum
Minimum number of 5 marks questions = 4
total marks = 5 × 4 = 20 marks
Minimum number of 10 marks questions = 3
total marks = 3 × 10 = 30 marks.
Maximum number of 15 marks questions

∴ 2 ≤ z ≤ 3
Now, tabulating the values, we have

Again from question, number of questions in ET is more than MT. So, ET has maximum 5 marks questions than MT
For ET : x = 8, y = 3 and z = 2
⇒ Total marks = 8 × 5 + 3 × 10 + 2 × 15 = 100
Total number of questions = 8 + 3 + 2 = 13
For MT (i) x = 5, y = 3 and z = 3
⇒ Total marks = 5 × 5 + 3 × 10 + 3 × 15 = 100
Total number of questions = 5 + 3 + 3 = 11
(ii) x = 4, y = 5 and z = 2
Total marks = 4 × 5 + 5 × 10 + 2 × 15 = 100
total number of questions = 4 + 5 + 2 = 11
Now, considering MT and ET together, each faculty member prepared the same number of questions total number of questions = 13 + 11 = 24
So, number of questions prepared by each faculty = 24/6 = 4 questions
Here, Annie prepared only one question for MT, So, rest 10 questions of MT is prepared by other 5 faculty such that 2 questions are prepared by each.
Again, each faculty, prepare 4 questions.
So, each faculty except Annie prepared 2 questions for ET and Annie prepared three questions for ET.
Here each members prepared, questions continuously. Based on that, we get the following table:
Order of questions prepared by the faculty

As Annie prepared 5th question of 5 marks for MT then, number of 5 marks questions for MT = 5
and, number of 5 marks question for ET = 8
Total number of 5 marks questions = 5 + 8 = 13

QUESTION: 21

Directions for Questions: The first-year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions.
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15- mark questions followed the 10-mark questions.
The following additional facts are known.
i. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
ii. Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT.
iii. All questions prepared by a faculty member appeared consecutively in MT as well as ET.
iv. Chelan prepared the third question in both MT and ET: and Esha prepared the eighth question in both.
v. Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.

(2019)

Q. Who prepared 15-mark questions for MT and ET?

Solution:

From given data,
Let number of questions worth 5 marks, 10 marks and 15 marks was x, y and z respectively, then, 4 ≤ x, 3 ≤ y and 2 ≤ z
Case-1: Total possible combination that gives 100  marks Minimum total marks that are from 10 marks question = 3 × 10 = 30 marks
Minimum total marks that are from 15 marks questions = 2 × 15 = 30 marks
Maximum total marks that are from 5 marks questions = 5 × 8 = 40 marks
∴ 4 ≤ x ≤ 8
Case-II : When x and z are minimum
Minimum number of 5 marks questions = 4 Total marks = 5 × 4 = 20 marks
Minimum number of 15 marks questions = 2
total marks = 2 × 15 = 30
Maximum number of 10 marks questions

So, 3 ≤ y ≤ 5.
Case-III : When x and y are minimum
Minimum number of 5 marks questions = 4
total marks = 5 × 4 = 20 marks
Minimum number of 10 marks questions = 3
total marks = 3 × 10 = 30 marks.
Maximum number of 15 marks questions

∴ 2 ≤ z ≤ 3
Now, tabulating the values, we have

Again from question, number of questions in ET is more than MT. So, ET has maximum 5 marks questions than MT
For ET : x = 8, y = 3 and z = 2
⇒ Total marks = 8 × 5 + 3 × 10 + 2 × 15 = 100
Total number of questions = 8 + 3 + 2 = 13
For MT (i) x = 5, y = 3 and z = 3
⇒ Total marks = 5 × 5 + 3 × 10 + 3 × 15 = 100
Total number of questions = 5 + 3 + 3 = 11
(ii) x = 4, y = 5 and z = 2
Total marks = 4 × 5 + 5 × 10 + 2 × 15 = 100
total number of questions = 4 + 5 + 2 = 11
Now, considering MT and ET together, each faculty member prepared the same number of questions total number of questions = 13 + 11 = 24
So, number of questions prepared by each faculty = 24/6 = 4 questions
Here, Annie prepared only one question for MT, So, rest 10 questions of MT is prepared by other 5 faculty such that 2 questions are prepared by each.
Again, each faculty, prepare 4 questions.
So, each faculty except Annie prepared 2 questions for ET and Annie prepared three questions for ET.
Here each members prepared, questions continuously. Based on that, we get the following table:
Order of questions prepared by the faculty

15 marks questions for both ET and MT prepared by only Dave, Esha and Fakir

QUESTION: 22

Directions for Questions: The first-year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions.
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15- mark questions followed the 10-mark questions.
The following additional facts are known.
i. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
ii. Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT.
iii. All questions prepared by a faculty member appeared consecutively in MT as well as ET.
iv. Chelan prepared the third question in both MT and ET: and Esha prepared the eighth question in both.
v. Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.

(2019)

Q. Which of the following questions did Beti prepare in ET?

Solution:

From given data,
Let number of questions worth 5 marks, 10 marks and 15 marks was x, y and z respectively, then, 4 ≤ x, 3 ≤ y and 2 ≤ z
Case-1: Total possible combination that gives 100  marks Minimum total marks that are from 10 marks question = 3 × 10 = 30 marks
Minimum total marks that are from 15 marks questions = 2 × 15 = 30 marks
Maximum total marks that are from 5 marks questions = 5 × 8 = 40 marks
∴ 4 ≤ x ≤ 8
Case-II : When x and z are minimum
Minimum number of 5 marks questions = 4 Total marks = 5 × 4 = 20 marks
Minimum number of 15 marks questions = 2
total marks = 2 × 15 = 30
Maximum number of 10 marks questions

So, 3 ≤ y ≤ 5.
Case-III : When x and y are minimum
Minimum number of 5 marks questions = 4
total marks = 5 × 4 = 20 marks
Minimum number of 10 marks questions = 3
total marks = 3 × 10 = 30 marks.
Maximum number of 15 marks questions

∴ 2 ≤ z ≤ 3
Now, tabulating the values, we have

Again from question, number of questions in ET is more than MT. So, ET has maximum 5 marks questions than MT
For ET : x = 8, y = 3 and z = 2
⇒ Total marks = 8 × 5 + 3 × 10 + 2 × 15 = 100
Total number of questions = 8 + 3 + 2 = 13
For MT (i) x = 5, y = 3 and z = 3
⇒ Total marks = 5 × 5 + 3 × 10 + 3 × 15 = 100
Total number of questions = 5 + 3 + 3 = 11
(ii) x = 4, y = 5 and z = 2
Total marks = 4 × 5 + 5 × 10 + 2 × 15 = 100
total number of questions = 4 + 5 + 2 = 11
Now, considering MT and ET together, each faculty member prepared the same number of questions total number of questions = 13 + 11 = 24
So, number of questions prepared by each faculty = 24/6 = 4 questions
Here, Annie prepared only one question for MT, So, rest 10 questions of MT is prepared by other 5 faculty such that 2 questions are prepared by each.
Again, each faculty, prepare 4 questions.
So, each faculty except Annie prepared 2 questions for ET and Annie prepared three questions for ET.
Here each members prepared, questions continuously. Based on that, we get the following table:
Order of questions prepared by the faculty

10th question of 10 marks prepared by Beti for ET

QUESTION: 23

Directions for Questions: Students in a college are discussing two proposals -
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.
A student does not necessarily support either of the two proposals.
In an upcoming election for student union president, there are two candidates in fray: Sunita and Ragini. Every student prefers one of the two candidates.
A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.
1. 250 students supported proposal A and 250 students supported proposal B.
2. Among the 200 students who preferred Sunita as student union president. 80% supported proposal A.
3. Among those who preferred Ragini, 30% supported proposal A.
4. 20% of those who supported proposal B preferred Sunita.
5. 40% of those who did not support proposal B preferred Ragini.
6. Every student who preferred Sunita and supported proposal B also supported proposal A.
7. Among those who preferred Ragini, 20% did not support any of the proposals.

(2019)

Q. What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?

Solution:

As per question it is not clear that a student can not prefer both Ragini and Sunita simultaneously
From point 1, 250 students supported proposal A and 250 students supported proposal B
From point 2, it is clear that 200 students preferred Sunita so number of students who preferred Ragini = 500 – 200 = 300
Number of students who preferred Sunita and supported proposal A = 80% of 200 = 160
From point 3, Number of students who preferred Ragini and supported proposal A = 30% of 300 = 90
From point 4, Number of students who preferred Sunita and supported proposal B = 20% of 250 = 50
From point 6, Number of students who preferred Sunita and supported proposal A and B both = 50
From point 7, Number of students who preferred Ragini and didn’t supported any proposal = 20% of 300 = 60
Now the complete table can be formed as given below :

Number of students who supported both proposals A and B = 100
Number of the students surveyed who supported both proposals A and B and preferred Sunita as student union president = 50
∴ Required % =

QUESTION: 24

Directions for Questions: Students in a college are discussing two proposals -
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.
A student does not necessarily support either of the two proposals.
In an upcoming election for student union president, there are two candidates in fray: Sunita and Ragini. Every student prefers one of the two candidates.
A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.
1. 250 students supported proposal A and 250 students supported proposal B.
2. Among the 200 students who preferred Sunita as student union president. 80% supported proposal A.
3. Among those who preferred Ragini, 30% supported proposal A.
4. 20% of those who supported proposal B preferred Sunita.
5. 40% of those who did not support proposal B preferred Ragini.
6. Every student who preferred Sunita and supported proposal B also supported proposal A.
7. Among those who preferred Ragini, 20% did not support any of the proposals.

(2019)

Q. How many of the students surveyed supported proposal B, did not support proposal A and preferred Ragini as student union president?

Solution:

As per question it is not clear that a student can not prefer both Ragini and Sunita simultaneously
From point 1, 250 students supported proposal A and 250 students supported proposal B
From point 2, it is clear that 200 students preferred Sunita so number of students who preferred Ragini = 500 – 200 = 300
Number of students who preferred Sunita and supported proposal A = 80% of 200 = 160
From point 3, Number of students who preferred Ragini and supported proposal A = 30% of 300 = 90
From point 4, Number of students who preferred Sunita and supported proposal B = 20% of 250 = 50
From point 6, Number of students who preferred Sunita and supported proposal A and B both = 50
From point 7, Number of students who preferred Ragini and didn’t supported any proposal = 20% of 300 = 60
Now the complete table can be formed as given below :

Number of students surveyed supported proposal B, did not support proposal A and preferred Ragini as student union president.
= Number of students who supported proposal B and preferred Ragini
– Number of students who supported proposal A and B and preferred Ragini = 200 – 50 = 150

QUESTION: 25

Directions for Questions: Three doctor, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges ₹ 100/-, Dr. Kane sees each patient for 15 minutes and charges ₹ 200/-, while Dr. Wayne sees each patient for 25 minutes and charges ₹ 300/-.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.

The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.

(2019)

Q. What is the maximum number of patients that the clinic can cater to on any single day?

Solution:

Since clinic is open for from 9 a.m. to 11.30 a.m. i.e. for 150 mins, hence maximum number of patients for each doctor will be as follows
Ben = 150/10 = 15
Kane = 150/15 = 10
Wayne = 150/25 = 6
Total = 15 + 10 + 6 = 31

QUESTION: 26

Directions for Questions: Three doctor, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges ₹ 100/-, Dr. Kane sees each patient for 15 minutes and charges ₹ 200/-, while Dr. Wayne sees each patient for 25 minutes and charges ₹ 300/-.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.

The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.

(2019)

Q. The queue is never empty on one particular Saturday. Which of the three doctors would earn the maximum amount in consultation charges on that day?

Solution:

When quene is never empty on one particular Saturday, then
Ben would earned = 15 × 100 = 1500
Kane would earned = 10 × 200 = 2000
Wayne would earned = 6 × 300 = 1800
Hence, maximum amount would be earned by Dr. Kane

QUESTION: 27

Directions for Questions: Three doctor, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges ₹ 100/-, Dr. Kane sees each patient for 15 minutes and charges ₹ 200/-, while Dr. Wayne sees each patient for 25 minutes and charges ₹ 300/-.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.

The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.

(2019)

Q. Mr. Singh visited the clinic on Monday, Wednesday, and Friday of a particular week, arriving at 8:50 a.m. on each of the three days. His token number was 13 on all three days. On which day was he at the clinic for the maximum duration?

Solution:

To stay for the maximum time in the clinic the 13th numbered patient must go in the room of Dr. Wayne.
As 12th and 13th number of patient will visited by one of the doctors Dr. Ben or Dr. Wayne. So for 13th no of patient to visit Dr. Wayne, Dr. Wayne must be in the room which is numbered higher than Dr. Ben’s room number. It is happing only on Monday when Dr. Ben is in room 1 and Dr. Wayne is in room 3.
So Monday is correct answer.

QUESTION: 28

Directions for Questions: Three doctor, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges ₹ 100/-, Dr. Kane sees each patient for 15 minutes and charges ₹ 200/-, while Dr. Wayne sees each patient for 25 minutes and charges ₹ 300/-.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.

The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.

(2019)

Q. On a slow Thursday, only two patients are waiting at 9 a.m. After that two patients keep arriving at exact 15 minute intervals starting at 9:15 a.m. – i.e. at 9:15 a.m., 9:30 a.m., 9:45 a.m. etc. Then the total duration in minutes when all three doctors are simultaneously free is

Solution:

If  you continue filling the  above table till    11.30  a.m. as filled  in the table, you can easily see that all 3 doctors will never be free at same time.
Hence 0 is the correct option.

QUESTION: 29

Directions for Questions: Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2, . . . ., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10. 1 and 2 in Round 7 and so on.
The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.

The following information is known about Rounds 1 through 6:
1. Gordon did not score consecutively in any two rounds.
2. Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
1. Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
2. Joshin scored in Round 7, while Amita scored in Round 10.
3. No player scored in all the four rounds.

(2019)

Q. What were the scores of Chen, David, and Eric respectively after Round 3?

Solution:

Presenting the point situation as per the conditions given in the Problem:

Chen = 3 (scored either in Round 1/ 2);
David = 3 (scored either in Round 2/ 1);
E = 3 (Round 3)

QUESTION: 30

Directions for Questions: Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2, . . . ., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10. 1 and 2 in Round 7 and so on.
The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.

The following information is known about Rounds 1 through 6:
1. Gordon did not score consecutively in any two rounds.
2. Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
1. Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
2. Joshin scored in Round 7, while Amita scored in Round 10.
3. No player scored in all the four rounds.

(2019)

Q. Which three players were in the last three positions after Round 4?

Solution:

Presenting the point situation as per the conditions given in the Problem:

Amita = 7, Bala = 2, Chen = 3; David = 6; Eric = 3, Fatima = 7, Gordon = 14, Hansa = 1, Ikea = 1; Joshin does not play any of the 4 rounds.

QUESTION: 31

Directions for Questions: Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2, . . . ., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10. 1 and 2 in Round 7 and so on.
The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.

The following information is known about Rounds 1 through 6:
1. Gordon did not score consecutively in any two rounds.
2. Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
1. Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
2. Joshin scored in Round 7, while Amita scored in Round 10.
3. No player scored in all the four rounds.

(2019)

Q. Which player scored points in maximum number of rounds?

Solution:

Presenting the point situation as per the conditions given in the Problem:

From above table Ikea scored 5 times (maximum)

QUESTION: 32

Directions for Questions: Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2, . . . ., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10. 1 and 2 in Round 7 and so on.
The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.

The following information is known about Rounds 1 through 6:
1. Gordon did not score consecutively in any two rounds.
2. Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
1. Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
2. Joshin scored in Round 7, while Amita scored in Round 10.
3. No player scored in all the four rounds.

(2019)

Q. Which players scored points in the last round?

Solution:

Presenting the point situation as per the conditions given in the Problem:

Eric, Amita & Chen scored in Round 10.

QUESTION: 33

Directions for Questions: In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.

These five people form three teams. Team 1. Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
(1) Each team speaks exactly four languages and has the same number of members.
(2) English and Chinese are spoken by all three teams. Basque and French by exactly two teams and the other languages by exactly one team.
(3) None of the teams include both Quentin and Robert.
(4) Paula and Sally are together in exactly two teams.
(5) Robert is in Team 1 and Quentin is in Team 3.

(2019)

Q. Who among the following four is not a member of Team 2?

Solution:

From points (1) and (5), the persons in Team 1 speak English, Chinese, Arabic and French. (Robert speaks both Arabic and French).
Again from points (1) and (5), the persons in Team 3 speak, English, Chinese  and Dutch. (Quentin speaks Dutch and English). Since each person speaks two Languages and each team speaks exactly four languages, we need to find one person for Team 3, who speaks one language among English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from point (5) and the above, Paula, Quentin and Sally, (Basque, Chinese, Dutch and English) are in Team 3. Since there are three persons in Team 3, Teams 1 and 2 should also have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic and French) is one of the team members. Now, two more persons, who speak languages among the above four are to be selected. It is possible only with Paula and Terence.
From point (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two languages Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to find one more person, who speaks one of these three languages and French. It is possible with only Terence.

Quentin is not a member of Team 2.

QUESTION: 34

Directions for Questions: In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.

These five people form three teams. Team 1. Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
(1) Each team speaks exactly four languages and has the same number of members.
(2) English and Chinese are spoken by all three teams. Basque and French by exactly two teams and the other languages by exactly one team.
(3) None of the teams include both Quentin and Robert.
(4) Paula and Sally are together in exactly two teams.
(5) Robert is in Team 1 and Quentin is in Team 3.

(2019)

Q. Who among the following four people is a part of exactly two teams?

Solution:

From points (1) and (5), the persons in Team 1 speak English, Chinese, Arabic and French. (Robert speaks both Arabic and French).
Again from points (1) and (5), the persons in Team 3 speak, English, Chinese  and Dutch. (Quentin speaks Dutch and English). Since each person speaks two Languages and each team speaks exactly four languages, we need to find one person for Team 3, who speaks one language among English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from point (5) and the above, Paula, Quentin and Sally, (Basque, Chinese, Dutch and English) are in Team 3. Since there are three persons in Team 3, Teams 1 and 2 should also have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic and French) is one of the team members. Now, two more persons, who speak languages among the above four are to be selected. It is possible only with Paula and Terence.
From point (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two languages Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to find one more person, who speaks one of these three languages and French. It is possible with only Terence.

Sally is part of exactly two teams 2 and 3

QUESTION: 35

Directions for Questions: In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.

These five people form three teams. Team 1. Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
(1) Each team speaks exactly four languages and has the same number of members.
(2) English and Chinese are spoken by all three teams. Basque and French by exactly two teams and the other languages by exactly one team.
(3) None of the teams include both Quentin and Robert.
(4) Paula and Sally are together in exactly two teams.
(5) Robert is in Team 1 and Quentin is in Team 3.

(2019)

Q. Who among the five people is a member of all teams?

Solution:

From points (1) and (5), the persons in Team 1 speak English, Chinese, Arabic and French. (Robert speaks both Arabic and French).
Again from points (1) and (5), the persons in Team 3 speak, English, Chinese  and Dutch. (Quentin speaks Dutch and English). Since each person speaks two Languages and each team speaks exactly four languages, we need to find one person for Team 3, who speaks one language among English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from point (5) and the above, Paula, Quentin and Sally, (Basque, Chinese, Dutch and English) are in Team 3. Since there are three persons in Team 3, Teams 1 and 2 should also have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic and French) is one of the team members. Now, two more persons, who speak languages among the above four are to be selected. It is possible only with Paula and Terence.
From point (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two languages Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to find one more person, who speaks one of these three languages and French. It is possible with only Terence.

Paula is a member of all the teams.

QUESTION: 36

Directions for Questions: In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.

These five people form three teams. Team 1. Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
(1) Each team speaks exactly four languages and has the same number of members.
(2) English and Chinese are spoken by all three teams. Basque and French by exactly two teams and the other languages by exactly one team.
(3) None of the teams include both Quentin and Robert.
(4) Paula and Sally are together in exactly two teams.
(5) Robert is in Team 1 and Quentin is in Team 3.

(2019)

Q. Apart from Chinese and English, which languages are spoken by Team 1?

Solution:

From points (1) and (5), the persons in Team 1 speak English, Chinese, Arabic and French. (Robert speaks both Arabic and French).
Again from points (1) and (5), the persons in Team 3 speak, English, Chinese  and Dutch. (Quentin speaks Dutch and English). Since each person speaks two Languages and each team speaks exactly four languages, we need to find one person for Team 3, who speaks one language among English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from point (5) and the above, Paula, Quentin and Sally, (Basque, Chinese, Dutch and English) are in Team 3. Since there are three persons in Team 3, Teams 1 and 2 should also have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic and French) is one of the team members. Now, two more persons, who speak languages among the above four are to be selected. It is possible only with Paula and Terence.
From point (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two languages Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to find one more person, who speaks one of these three languages and French. It is possible with only Terence.

Apart from Chinese and English, Team 1 speaks Arabic and French

QUESTION: 37

Directions for Questions: You are given an n×n square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.

(2018)

Q. Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5×5 matrix?

Solution:

Even if one mistake is allowed, then also we required 4 elements.

QUESTION: 38

Directions for Questions: You are given an n×n square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.

(2018)

Q. Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5×5 square matrix?

Solution:

Given that all the cells adjacent to any particular cell must have different numerals, which is satisfied only when there are at least 9 numerals.

QUESTION: 39

Directions for Questions: Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.
1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

(2018)

Q. Which of the following MUST be true?

Solution:

According to 1, 2 and 3, we get one case for P1 to P6 and 2 cases for P7 and P8.

Also, from 4, we get 2 cases:

From (5), we get
If total number of low (L) pipes = 3, then
number of high (H) pipes = 6
and number of medium (M) pipes = 11
Also if number of low (L) pipes = 4, then
number of high (H) pipes = 8
and number of medium (M) pipes = 8
∴ Two cases arise for P1 to P10

On combining the above all results getting from (1), (2), (3), (4) and (5) we get the following possible cases for P1 to P 20
Case 1: H M H M H L H H M H M H M H M L M L M L
No. of L = 4
No. of H = 8
No. of M = 8
Case 2: H M H M H L H H M H L M H M H M L M L M
No. of L = 4,
No. of H = 8,
and No. of  M = 8
Case 3: H M H M H L H H M H M L H M H M L M L M
No. of L = 4
No. of H = 8
No. of M = 8
The contamination level at P10 was recorded as high.

QUESTION: 40

Directions for Questions: Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.
1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

(2018)

Q. What best can be said about the number of pumps at which the contamination levels were recorded as medium?

Solution:

According to 1, 2 and 3, we get one case for P1 to P6 and 2 cases for P7 and P8.

Also, from 4, we get 2 cases:

From (5), we get
If total number of low (L) pipes = 3, then
number of high (H) pipes = 6
and number of medium (M) pipes = 11
Also if number of low (L) pipes = 4, then
number of high (H) pipes = 8
and number of medium (M) pipes = 8
∴ Two cases arise for P1 to P10

On combining the above all results getting from (1), (2), (3), (4) and (5) we get the following possible cases for P1 to P 20
Case 1: H M H M H L H H M H M H M H M L M L M L
No. of L = 4
No. of H = 8
No. of M = 8
Case 2: H M H M H L H H M H L M H M H M L M L M
No. of L = 4,
No. of H = 8,
and No. of  M = 8
Case 3: H M H M H L H H M H M L H M H M L M L M
No. of L = 4
No. of H = 8
No. of M = 8
At exactly 8 pumps contamination levels recorded as medium.

QUESTION: 41

Directions for Questions: Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.
1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

(2018)

Q. If the contamination level at P11 was recorded as low, then which of the following MUST be true?

Solution:

According to 1, 2 and 3, we get one case for P1 to P6 and 2 cases for P7 and P8.

Also, from 4, we get 2 cases:

From (5), we get
If total number of low (L) pipes = 3, then
number of high (H) pipes = 6
and number of medium (M) pipes = 11
Also if number of low (L) pipes = 4, then
number of high (H) pipes = 8
and number of medium (M) pipes = 8
∴ Two cases arise for P1 to P10

On combining the above all results getting from (1), (2), (3), (4) and (5) we get the following possible cases for P1 to P 20
Case 1: H M H M H L H H M H M H M H M L M L M L
No. of L = 4
No. of H = 8
No. of M = 8
Case 2: H M H M H L H H M H L M H M H M L M L M
No. of L = 4,
No. of H = 8,
and No. of  M = 8
Case 3: H M H M H L H H M H M L H M H M L M L M
No. of L = 4
No. of H = 8
No. of M = 8
If the contaminated level at P11 was recorded as low, then the contamination level at P14 was recorded as medium.

QUESTION: 42

Directions for Questions: Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.
1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

(2018)

Q. If contamination level at P15 was recorded as medium, then which of the following MUST be FALSE?

Solution:

According to 1, 2 and 3, we get one case for P1 to P6 and 2 cases for P7 and P8.

Also, from 4, we get 2 cases:

From (5), we get
If total number of low (L) pipes = 3, then
number of high (H) pipes = 6
and number of medium (M) pipes = 11
Also if number of low (L) pipes = 4, then
number of high (H) pipes = 8
and number of medium (M) pipes = 8
∴ Two cases arise for P1 to P10

On combining the above all results getting from (1), (2), (3), (4) and (5) we get the following possible cases for P1 to P 20
Case 1: H M H M H L H H M H M H M H M L M L M L
No. of L = 4
No. of H = 8
No. of M = 8
Case 2: H M H M H L H H M H L M H M H M L M L M
No. of L = 4,
No. of H = 8,
and No. of  M = 8
Case 3: H M H M H L H H M H M L H M H M L M L M
No. of L = 4
No. of H = 8
No. of M = 8
If contamination level at P15 was recorded as medium, then contamination levels at P11 and P16 are recorded as medium and low respectively.

QUESTION: 43

Directions for Questions: 1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:
1 . The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

(2018)

Q. What best can be said about the number of satellites serving C?

Solution:

It is given that the satellites serving either B, C or S do not serve O.
From (1), let the number of satellites serving B, C and S be 2K, K, K respectively.
Let the number of satellites exclusively serving B be x.
From (3), the number of satellites exclusively serving C and exclusively serving S will each be 0.3x
From (4), the number of satellites serving O is same as the number of satellites serving C and S but not B. Let the number be y.
Since the number of satellites serving C is same as the number of satellites serving S, we can say that (number of satellites serving only B and C) + 0.3x + 100 + y = (number of satellites serving only B and S) + 0.3x + 100 + y
∴ The number of satellites serving only B and C = the number of satellites serving only B and S = Z (let)
Therefore, the venn diagram will be as follows

Given that there are a total of 1600 satellites
∴ x + z + 0.3x + z + 100 + y + 0.3x + y = 1600
⇒  1.6x + 2y + 2z = 1500   …(i)
Also k = 0.3x + z + y +100
and 2k = x + 2z + 100
⇒ 2(0.3x + z + y + 100) = x + 2z + 100
⇒  0.4x = 2y + 100
⇒ x = 5y + 250 …(ii)
From equation (i) and (ii) we will get
1.6 (5y + 250) + 2y + 2z = 1500
Z = 550 – 5y … (iii)
The number of satellites serving C
= z + 0.3x + 100 + y
= (550 – 5y) + 0.3(5y + 250) + 100 + y
= 725 – 2.5y
This number will be maximum when y is minimum.
Minimum value of y is 0.
∴ The maximum number of satellites serving C will be 725.
From (iii), z = 550 – 5y
Since the number of satellites cannot be negative,
z ≥ 0, ⇒ 550 – 5y ≥ 0
y ≤ 110.
∴ Maximum value of y is 110.
When y = 110, the number of satellites serving C will be 725 – 2.5 × 110 = 450. This will be the minimum number of satellites serving C.
The number of satellites serving C must be between 450 and 725.

QUESTION: 44

Directions for Questions: 1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:
1 . The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as th e number of satellites serving both C and S but not B.

(2018)

Q. What is the minimum possible number of satellites serving B exclusively?

Solution:

It is given that the satellites serving either B, C or S do not serve O.
From (1), let the number of satellites serving B, C and S be 2K, K, K respectively.
Let the number of satellites exclusively serving B be x.
From (3), the number of satellites exclusively serving C and exclusively serving S will each be 0.3x
From (4), the number of satellites serving O is same as the number of satellites serving C and S but not B. Let the number be y.
Since the number of satellites serving C is same as the number of satellites serving S, we can say that (number of satellites serving only B and C) + 0.3x + 100 + y = (number of satellites serving only B and S) + 0.3x + 100 + y
∴ The number of satellites serving only B and C = the number of satellites serving only B and S = Z (let)
Therefore, the venn diagram will be as follows

Given that there are a total of 1600 satellites
∴ x + z + 0.3x + z + 100 + y + 0.3x + y = 1600
⇒  1.6x + 2y + 2z = 1500   …(i)
Also k = 0.3x + z + y +100
and 2k = x + 2z + 100
⇒ 2(0.3x + z + y + 100) = x + 2z + 100
⇒  0.4x = 2y + 100
⇒ x = 5y + 250 …(ii)
From equation (i) and (ii) we will get
1.6 (5y + 250) + 2y + 2z = 1500
Z = 550 – 5y … (iii)
From (ii), the number of satellites serving B exclusively = x = 5y + 250
This is minimum when y is minimum.
Minimum value of y = 0.
∴ The minimum number of satellites serving B exclusively
= 5 × 0 + 250 = 250.

QUESTION: 45

Directions for Questions: 1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:
1 . The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as th e number of satellites serving both C and S but not B.

(2018)

Q. If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S?

Solution:

It is given that the satellites serving either B, C or S do not serve O.
From (1), let the number of satellites serving B, C and S be 2K, K, K respectively.
Let the number of satellites exclusively serving B be x.
From (3), the number of satellites exclusively serving C and exclusively serving S will each be 0.3x
From (4), the number of satellites serving O is same as the number of satellites serving C and S but not B. Let the number be y.
Since the number of satellites serving C is same as the number of satellites serving S, we can say that (number of satellites serving only B and C) + 0.3x + 100 + y = (number of satellites serving only B and S) + 0.3x + 100 + y
∴ The number of satellites serving only B and C = the number of satellites serving only B and S = Z (let)
Therefore, the venn diagram will be as follows

Given that there are a total of 1600 satellites
∴ x + z + 0.3x + z + 100 + y + 0.3x + y = 1600
⇒  1.6x + 2y + 2z = 1500   …(i)
Also k = 0.3x + z + y +100
and 2k = x + 2z + 100
⇒ 2(0.3x + z + y + 100) = x + 2z + 100
⇒  0.4x = 2y + 100
⇒ x = 5y + 250 …(ii)
From equation (i) and (ii) we will get
1.6 (5y + 250) + 2y + 2z = 1500
Z = 550 – 5y … (iii)
Given that at least 100 satellites serve 0; we can say in this case that y ≥ 100.
Number of satellites serving S
= Number of satellites serving C = 725 – 2.5y
Number of satellites serving S is minimum when y is maximum, i.e. 110
∴ Minimum number of satellites serving
S = 725 – 2.5 × 110 = 450.
Number of satellites serving S is maximum when y is minimum, i.e., 100.
∴ Maximum number of satellites serving
S = 725 – 2.5 × 100 = 475
Therefore, the number of satellites serving S is at most 475

QUESTION: 46

Directions for Questions: 1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:
1 . The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as th e number of satellites serving both C and S but not B.

(2018)

Q. If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE?

Solution:

It is given that the satellites serving either B, C or S do not serve O.
From (1), let the number of satellites serving B, C and S be 2K, K, K respectively.
Let the number of satellites exclusively serving B be x.
From (3), the number of satellites exclusively serving C and exclusively serving S will each be 0.3x
From (4), the number of satellites serving O is same as the number of satellites serving C and S but not B. Let the number be y.
Since the number of satellites serving C is same as the number of satellites serving S, we can say that (number of satellites serving only B and C) + 0.3x + 100 + y = (number of satellites serving only B and S) + 0.3x + 100 + y
∴ The number of satellites serving only B and C = the number of satellites serving only B and S = Z (let)
Therefore, the venn diagram will be as follows

Given that there are a total of 1600 satellites
∴ x + z + 0.3x + z + 100 + y + 0.3x + y = 1600
⇒  1.6x + 2y + 2z = 1500   …(i)
Also k = 0.3x + z + y +100
and 2k = x + 2z + 100
⇒ 2(0.3x + z + y + 100) = x + 2z + 100
⇒  0.4x = 2y + 100
⇒ x = 5y + 250 …(ii)
From equation (i) and (ii) we will get
1.6 (5y + 250) + 2y + 2z = 1500
Z = 550 – 5y … (iii)
The number of satellites serving at least two of B, C or S = number of satellites serving exactly two of B, C or S + Number of satellites serving all the three
= z + z + y + 100
= 2(550 – 5y) + y + 100
= 1200 – 9y.
Given that this is equal to 1200
∴1200 – 9y = 1200   ⇒ y = 0
If y = 0, then x = 5y + 250 = 250
and z = 550 – 5y = 550
No. of satellites serving C k = z + 0.3x + 100 + y
= 550 + 0.3 × 250 + 100 + y = 725
No. of satellites serving B = 2k = 2 × 725 = 1450.
From the given options, we can say that the option “the number of satellites serving C cannot be uniquely determined” must be False.

QUESTION: 47

Directions for Questions: An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

(2018)

Q. What is the maximum number of customers that the ATM can serve with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?

Solution:

Since there are a limited number of 500 rupee notes, we can minimize the number of 500 rupee notes dispensed to each customer, while ensuring that each customer is served at most 20 notes.
If no 500 rupee notes is dispensed, the minimum number of notes that must be dispensed is 25 (all 200 rupee notes). This is not possible.
If one 500 rupee note is dispensed, the minimum number of notes is 24 (one 500 rupee note, twenty two 200 rupee notes and one 100 rupee note). This is also not possible.
If two 500 rupee notes are dispensed, the minimum number of notes is 22 (two 500 rupee notes and twenty 200 rupee notes).
If three 500 rupee notes are dispensed, the minimum number of notes is 21 (three 500 rupee notes, seventeen 200 rupee notes and one 100 rupee note).
If four 500 rupee notes are dispensed, the minimum number of notes is 19 (four 500 rupee notes and fifteen 200 rupee notes).
Hence, the minimum number of 500 rupee notes that can be dispensed to any person is 4.
With fifty 500 rupee notes, a maximum of 12 persons can be served.

QUESTION: 48

Directions for Questions: An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

(2018)

Q. What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?

Solution:

To dispense the smallest possible number of notes to a person with 500 rupee notes as his/her preference, the ATM should dispense all 500 rupee notes.
Hence, minimum number of notes required to serve any person with 500 rupee notes as his/her preference = 10 (all of 500 rupees).
Total number of 500 rupee notes required to serve 50 customers with 500 rupee notes as his/her preference = 10 × 50 = 500
To minimize the number of notes to be served to a person with 100 rupee notes as his/her preference, we can maximize the number of 500 rupee notes served to him, keeping the 100 rupee notes more than the sum of the other two denominations.
This is possible if the machine serves eight 500 rupee notes and ten 100 rupee notes.
Hence, the total number of 500 rupee notes required to serve 50 customers with 100 rupee notes as his/her preference = 8 × 50 = 400
Total number of 500 rupee notes required in the given scenario = 500 + 400 = 900

QUESTION: 49

Directions for Questions: Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:
1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politician s are in the administration committee, and 20% are in the teaching committee.

(2018)

Q. Based on the given information, which of the following statements MUST be FALSE?

Solution:

Total = 24
Bureaucrats are in the ratio 3 : 3 : 4 only value will be 3, 3, 4.
So x = 1
Politicians are in ratio 1 : 1 : 3, only value will be 1, 1, 3.
p + q + r = 9

q < p < r
Possible value of p, q, r are 3, 2, 4 and 3, 1, 5.
Hence, educationalist

The size of the research committee is less than teaching committee is false.

QUESTION: 50

Directions for Questions: Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:
1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the numbe r of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politician s are in the administration committee, and 20% are in the teaching committee.

(2018)

Q. Which of the following CANNOT be determined uniquely based on the given information?

Solution:

Total = 24
Bureaucrats are in the ratio 3 : 3 : 4 only value will be 3, 3, 4.
So x = 1
Politicians are in ratio 1 : 1 : 3, only value will be 1, 1, 3.
p + q + r = 9

q < p < r
Possible value of p, q, r are 3, 2, 4 and 3, 1, 5.
Hence, educationalist

Size of the teaching committee cannot be determined uniquely.