#### Team Selection

Importance of Team Selection in various exams

Team selection or logical grouping forms a part of Data Interpretation and Logical Reasoning part of the CAT question paper. Apart from CAT, in the previous years, this topic is also covered in the form of questions in other MBA/PGDM entrance exams like XAT, SNAP, IIFT etc. The CAT questions on this concept are asked repeatedly, and from the trends of previous years CAT papers, it is evident that one or two questions are asked every year.

Below are the details about the concept for different competitive MBA exam:

#### CAT :

 Team Selection (Logical grouping) Year No of Questions Good attempt Difficulty 2019 Slot 1 1 0 to 1 Moderate to difficult Slot 2 1 0 to 1 Moderate to difficult 2018 Slot 1 1 1 Moderate Slot 2 1 1 Easy 2017 Slot 1 1 1 Moderate Slot 2 1 1 Moderate
XAT :

 Team Selection (Logical grouping) Year Number of Questions Good attempt Difficulty 2019 1 1 Moderate 2018 1 1 Moderate 2017 2 1 to 2 Difficult
SNAP :

 Team Selection (Logical grouping) Year Number of Questions Good attempt Difficulty 2019 2 1 to 2 Moderate to difficult 2018 1 1 Easy 2017 1 1 Easy to Moderate

### List of Concepts in this Chapter

The questions on Team selection or logical grouping usually ask for a selection of a team of say ‘p’ members from ‘q’ (q>p) available for selection or it can be the selection of a committee of a certain number of members.

Certain given constraints or restrictions drive the grouping of the team.  To solve these questions, the following approach can be kept in mind:

• There will always be data hidden in the statements given that should be deciphered properly
• Constraints should be kept in mind when segregating the data
• Moving step by step with the help of flowcharts will make the problem easier to solve
• After a result is reached, the back-calculation should be done to check if the solution is correct
• There is a unique solution to these problems and if the solution obtained is wrong, one or two constrained would be disobeyed which will help the candidate know that this is not the solution he is looking for

Let us understand this concept with the help of an example:

Among 5 students of group 1– A, B, C, D, E and 6 students of group 2 – U, V, W, X, Y, Z, a team of 5 students is selected such that it consists exactly 3 students from group 2. It is known that:

• C and V should not be selected together.
• If B is selected, U and V cannot be selected.
• Amongst A, D, E and Y exactly two persons should be selected
• If E is in the team, either U or W can be in the team and not both
• If A is selected, X should be selected.
• Z will be in the team only if C is selected.

How many such teams are possible?

Solution:

Upon reading the question, it is evident that the team will always have 2 students from group 1 since 3 students have to come from group 2.

Group 1 has 5 members and we have to select 2 members. Hence, the possible ways to do so are 5C2 = 10. But we should note that all of the 10 cases cannot result in a team and we have to solve individually for all 10 cases. For example, we can choose B and C as the required members from group 1 but from 3rd constraint, we know that at least one among A, D or E should be in the team thus making the above selection useless.

Selection of 3 from a group of 6 can be done in 6C3 = 20 ways but since we know, many of those selections will not be of use (keeping in mind above constraints), solving for them in a limited time exam is not the best approach.

We will try to solve the question basis constraint 3 since there are 6 ways to do so and that is less time consuming as well.

Teams:

A and D are selected in the team (from condition 3), which implies that 2 members of group 1 are already selected and we need 3 members from group 2. From condition 5, we can say that X will be in the team because if A is selected then X has to be selected. Y and Z cannot be selected because Y belongs to group 1 and C is not selected which is a mandatory condition to select Z. Left members are U, V and W. We can select 2 among U, V and W which can be done in 3 ways.

2. Suppose, selection of A and E is done from group 1. Time to choose 3 from group 2. X will be one of them (condition 5). Because E is in the team either one of U or W can be in the team and not both (condition 4). Therefore, only two options are possible: U and V or W and V.

3. Now, on the selection of A from 1 and Y from 2. We need 1 more member from group 1 and 2 from group 2. From group 1, we can select B or C (D or E cannot be selected because of condition 3).

With B, we cannot select both U and V in the team (condition 2) and since A is in the team, X has to be there. Without C, Z cannot be in the team. The only option left is W. Hence, only 1 possible case.

4. Next, if we select D and E. Both are from group 1, leaving space for 3 people from group 2. Condition 4 allows only one among U or W because E is in the team. Thus, the only option left available to us is V and X.

Note that X can be in the team even if A is not in the team because it is vice versa in this case.

5. Next, we select D from 1 and Y from 2. One more from group 1 can be B or C. Similar to the 3rd case where we selected A and Y but here, we are not restricted to the selection of X alone as there is no A. X will be in some of the teams but not in all teams. With C in the team, Z will also be in the team but V cannot be in the team (condition 1). With B in the team, neither U nor V can be in the team so we should select X and W.

6. The Last case is with E and Y selection. With E in the team, only one of U or W can be in the team. From group 1 we can select either B or C. Rest will be the same as 5th case.

### Some Questions from Previous Papers

Some important questions on team selection and logical grouping from previous year papers are given below which can help in the preparation of CAT and similar exams.

Q1. 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 the 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. (CAT 2019-Slot 1)

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 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.

Answer the questions that follow:

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

1. 4

2. 2

3. 8

4. 1

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

1. A

2. L

3. B

4. G

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

1. 1,7,11,12

2. 1,5,6,12

3. 1,2,6,12

4. 1,2,8,12

Which of the following statements is necessarily true?

1. There are two empty shelves between the biscuits and the candies

2. All candies are kept before biscuits

3. All biscuits are kept before candies.

4. There are at least four shelves between items B and C.

Total number of Biscuits = 5

Total number of Candies = 3

So Total number of Savouries = 12 – 5 – 3 = 4

From point iii) and iv) it is clear that D, E, F and K are 4 savouries and are kept in shelves numbered 13,14, 15 and 16 as there is no empty shelf between items of the same type. From point II), V and VII) I, J and L are of the same type L being in the least numbered shelf among 3. As from point VI,) C is candy so I, J and L must be Biscuits as there are only 3 candies.

From point V) H is not Biscuits so it must be a candy thus A and B must be Biscuits.

Now Item can be placed as given below.

Case 1) when all biscuits are placed after candies. Case 2) When all candies are placed after biscuits.

Total 8 cases are possible. All questions can be answered now.
1. There are two empty shelves between the biscuits and the candies.
2. All candies are kept before biscuits.
3. All biscuits are kept before candies.
4. There are at least four shelves between items B and C

Q2. In the table below the checkmarks indicate all languages are 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. (CAT 2019-Slot 2)

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.

Answer the questions that follow:

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

1. Paula

2. Sally

3. Quentin

4. Terence

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

1. Sally

2. Paula

3. Robert

4. Quentin

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

1. Paula

2. Terence

3. Sally

4. No one

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

1. Arabic and Basque

2. Basque and French

3. Arabic and French

4. Basque and Dutch

First , let represent all the five people and language spoken by then by first letter of alphabet of their name. For example English will be represented as E.

From the point (1) and (2) it is clear that there will be three members in each team and each team will speak exactly 4 languages.

As E and C are spoken by all the teams so either P or T (only these two can speak C) must be part of each team.

From point 5) R is part of T1 and Q is part of T3. P and S are together in exactly two teams. If P and S are there in Team 1, then the three members of T1 will be P, S and R and together they can speak 5 languages (all except D) so they can’t form a team .

Same is the case of R, S and T. Thus P and S are together in Team 2 and Team 3. And R, P and T are part of Team 1.

Q3. 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: (CAT 2018-Slot 1)

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 politicians are in the administration committee, and 20% are in the teaching committee.

Answer the questions that follow:

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

1. The size of the research committee is less than the size of the administration committee

2. In the teaching committee, the number of educationalists is equal to the number of politicians

3. In the administration committee, the number of bureaucrats is equal to the number of educationalists

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

What is the number of bureaucrats in the administration committee?

What is the number of educationalists in the research committee?

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

1. The total number of educationalists in the three committees

2. The total number of bureaucrats in the three committees

3. The size of the research committee

4. The size of the teaching committee

Total = 24

Bureaucrats are in the ratio 3 : 3 : 4 only value will be 3, 3, 4. So x = 1

Educationalist n <m < o and m=(o+n)/2

Politicians are in ratio 1 : 1 : 3 only value will be 1, 1, 3.

Possible value of m, n, o are 3, 2, 4 and 3, 1, 5.

All questions can be answered now.
A similar pattern can be expected for CAT 2021

### How to deal with that topic preparation

Level 1:

For level 1 preparation candidates need to gain a basic understanding of the logical grouping topic. Data interpretation and logical reasoning topics are all about practice because no formulas need to be remembered. Therefore, to approach the topic, take a question and try to extract all the information given. Pay heed to the constraints and try different permutations and combinations to see what works. Take the help of the basic framework explained above.

Level 2:

When the candidate reaches this level, the expectation is that he has solved the basic questions and is clear with the concept and approach. Now, the focus should be on the shortcuts and structured thinking which helps the student reduce the number of cases possible through the process of elimination.

Level 3:

This level focusses on time management and solving the previous year CAT questions by timing oneself. Here, the candidate should be ready for approaching questions that have a high level of difficulty. Optimization and strengthening of the ability to visualize the question is advised.