Wednesday, July 22nd, 2020
In this post, we will learn about logical reasoning concepts on selection and group formation that is frequently asked in CAT exam. This topic generally deals with the selection of a team of say ‘r’ members from ‘n’ (n>r) available for selection or it can be the selection of committee of certain number of members. Certain number of constraints drives this selection. In order to understand these constrains and the implicit details related to them, let us start the discussion with an example.
Question: Among five students of group I – A, B, C, D, E and six students of group II – U, V, W, X, Y, Z, a team of five students is selected such that it consists exactly three students from group II. It is also known that:
Find the total number of such teams possible.
Let us take each condition one by one and try to get the obvious and hidden details about each information available.
‘If and Only If’ is the most important sentence out of as it implies that either select both or none, thus making them a pair.
Checking all the possibilities let us try to make all the possible teams.
There can be a number of ways to start the puzzle. Three of the easiest approaches are:
Group I has 5 members and we need to select 2 members from group I so total possible cases can be 5C2 = 10. There is a possibility that not all of the 10 cases will result in a team and we have to solve individually for all 10 cases. For instance, we select B and C as two members from group I but from condition 3 we know that at least one among A, D or E (all three belongs to group 1) would be in the team thus making this case useless.
Selecting three from a group of 6 is 6C3 = 20. Solving 20 cases would be even more time consuming.
We know that among A, D, E and Y, exactly two will be in the team. Number of possible cases are six as already discussed before. Solving 6 cases would definitely be less time consuming than other approaches. In addition, since we already abiding to one of the conditions so now we need to check for only five remaining conditions.
Let us start making teams by selecting two members as per condition 3.
2. Next we select A and E from A. D. E and Y. A and E belong to group I and so we need three more from group II. X will be one of them (condition 5). Now, since E is in the team either one of U or W can be in the team (condition 4). Y and Z cannot be in the team because of the reason explained in previous point. Therefore, we have two options. U and V or W and V.
3. Now, we select A and Y. A belongs to group I and Y belongs to group II. We need one more member from group I and two more from group II. From group I, we can select B or C (D or E cannot be selected because of condition 3).
With B, we cannot have either U or V in the team (condition 2) and since A is in the team, X has to be in the team. Without C, Z cannot be in the team. The only option left is W.
With C, Z will also be in the team (condition 6). Thus, all 5 are selected and no more case possible.
4. Next we select D and E. Both are from group I, leaving space for 3 group II members. Condition 4 allows only one among U or W because E is in the team. Again, Y and Z cannot be in the team (explained earlier). Thus, the only option left is with V and X.
Note that X can be in the team even if A is not in the team.
5. Next, we select D and Y. D is from group I and Y from group II. One more from group I can be B or C. Similar to the third case when 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. Last case is with E and Y selection. With E in the team, either U or W can be in the team. From group I we can select either B or C. Rest of the case is similar to the last case.
Did we miss any case? No. How can we be so sure? Because we discussed all the 6 cases coming out of condition number 3. Any team would include exactly two among A, D, E, Y and we made a case for all possible combinations. This is a much better practice, as it would avoid any confusion related to number of cases possible. We should always check subcases for further combinations with each member one by one. Following a flow diagram surely helps in determining the combinations with given restrictions. When there is no such condition which gives us any concrete information related to number of cases, we should go with the approach involving least number of cases.
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Logical Reasoning – Set 1: A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands.
Logical Reasoning – Set 2: Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata by a flight operated by Cheap Air.
Logical Reasoning – Set 3: A new airlines company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with.
Logical Reasoning – Set 4: In a square layout of size 5m × 5m, 25 equal sized square platforms of different heights are built.
Logical Reasoning – Set 5: Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N).
Data Interpretation – Set 1: Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET).
Data Interpretation – Set 2: A study to look at the early teaming of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South (S).
Data Interpretation – Set 3: There were seven elective courses – El to E7 – running in a specific term in a college.
Data Interpretation – Set 4: Funky Pizzaria was required to supply pizzas to three different parties.
Data Interpretation – Set 5: At a management school, the oldest 10 dorms, numbered 1 to 10, need to be repaired urgently, The following diagram represents the estimated repair costs (in Rs. Crores) for the 10 dorms.
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