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CAT: Overcoming barriers in Logical Reasoning

October 08, 2008 10:24 IST

Logical Reasoning (LR) probably is the closest to the real life problems students would face as managers. In CAT, the emphasis on logical reasoning has increased over the past few years, which makes it an important topic from a test preparation point of view.

Here are some guidelines which you would find useful in the countdown to CAT.

~ Interpreting given information correctly, and precisely

A common mistake students make is interpreting the given information incorrectly. The precise meaning of a statement should be understood for it to be used in conjunction with other statements correctly. For example, when referring to averages, "the average of the group reduces to" is not the same as "the average of the group reduces by".

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Consider the following question from CAT 07 where quite a few students interpreted the question incorrectly.

This was the first question in a caselet on 'Low cost airline company connecting ten Indian cities A to J'. The question read as follows:

Q. What is the lowest price, in rupees, a passenger has to pay for travelling by the shortest route from A to J?
(1) 2275
(2) 2850
(3) 2890
(4) 2930
(5) 3340

Note that the question requires you to find the shortest route first and then the lowest cost route. It is not the same as the lowest cost route between A to J.

The correct answer for this question is option (4) 2930. Whereas the answer by the lowest cost route approach is option (1) 2275.

Those who are interested to know the complete solution can download the complete solution key for CAT 2007 here

~ Develop your own ways of representing the data by practice

Often the data given needs to be written in a summarised form so that you can refer to it or remember it while solving questions. Develop your own methods of summarising the data. For example, let us try to represent the condition 'if P and Q are in the team then R cannot be there in the team'.

You may represent this as 'P and Q => x R'. This is a one way condition. There can be a two way condition such as 'if M is in the team then Q has to be there in the team and vice versa'. This may be represented as 'M<=>Q'.

~ Practice to intuitively solve problems

LR questions need to be solved with a particular line of thinking. You might know all the conditions necessary to solve the problems, but it is important to know which condition to use when and arrive at the solution in the minimum time possible. This is what we mean by intuitive thinking or the sixth sense that you need to develop to solve the questions. This comes only by practice. You can still develop it with fewer days in hand.

~ Tackle LR differently

LR is quite different in nature compared to other sections. Hence one needs to highlight some general strategies specifically applicable to LR.

Accuracy is far more important in LR. There is no bigger mistake than getting the complete set wrong. Avoid making guesses or reading in a hurry as it would lead to decreased accuracy.

As far as possible do not attempt LR in the beginning. If you get the set wrong or if it is tough and takes more time, that would affect your performance subsequently.

Use the option substitution method judiciously. LR questions are typically not meant to be solved using substitution of options. The most effective way is to understand the logic behind the question and solve it. There are exceptions to this rule though and you should use them wisely.

Read and understand the common data and questions thoroughly:

LR questions appear lengthy as compared to DI sets as they typically offer more information in text. While reading them, one might feel a lot of time is wasted. Yet, it helps to understand every condition thoroughly before attempting the questions. You may need to revisit the conditions while solving questions. Consider it time utilised not wasted. To illustrate these points, let's look at an LR caselet from CAT 2006, which is one of the more difficult caselets to have appeared in CAT. You could try the caselet on your own, before proceeding to the suggested solution method:

Erdos Number
Mathematicians are assigned a number called Erdos number (named after the famous mathematician, Paul Erdos). Only Paul Erdos himself has an Erdos number of zero. Any mathematician who has written a research paper with Erdos has an Erdos number of 1. For other mathematicians, the calculation of his/her Erdos number is illustrated below:

Suppose a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdos number. Let the Erdos number of Y be y. Then X has an Erdos number of y+1. Hence any mathematician with no co-authorship chain connected to Erdos has an Erdos number of infinity.

In a seven day long mini conference organized in memory of Paul Erdos, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdos number. Nobody had an Erdos number less than that of F.

On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdos number of the group of eight mathematicians to 3. The Erdos numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdos number of the group of eight to as low as 3.

At the end of the third day, five members of this group had identical Erdos numbers while the other three had Erdos numbers distinct from each other.

On the fifth day, E co-authored a paper with F which reduced the group's average Erdos number by 0.5. The Erdos numbers of the remaining six were unchanged with the writing of this paper.

No other paper was written during the conference.

Q1. The person having the largest Erdos number at the end of the conference must have had Erdos number (at that time):

(1) 5
(2) 7
(3) 9
(4) 14
(5) 15

Q2. How many participants in the conference did not change their Erdos number during the conference?

(1) 2
(2) 3
(3) 4
(4) 5
(5) cannot be determined

Q3. The Erdos number of C at the end of the conference was:

(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

Q4. The Erdos number of E at the beginning of the conference was:

(1) 2
(2) 5
(3) 6
(4) 7
(5) 8

Q5. How many participants had the same Erdos number at the beginning of the conference?

(1) 2
(2) 3
(3) 4
(4) 5
(5) cannot be determined

Methodology

This LR caselet is full of data and information. One must take care to go through the questions and understand them thoroughly. Let us summarize the important points of this caselet. For the sake of simplicity let's use the notation EN for Erdos Number.

General Information

Only Paul Erdos has an EN of zero....1

Any mathematician who has written a paper with Erdos has an EN of 1....2

When a group of mathematicians co-author a paper, every mathematician except the one with the lowest EN gets an EN equal to the lowest EN in the group + 1....3

Specific Information

Initially A had infinite EN and F has the lowest EN....4

F co-authored a paper with A and C which reduced the average EN of the group to 3. No other combination of three members would have brought down the average to 3....5

After this five members had the same EN and the remaining three ENs were distinct....6

E co-authored a paper with F which reduced the group's average by 0.5....7

No other paper was authored....8

We need to solve the caselet completely in order to answer the questions.

For the sake of simplicity let us refer to the ENs of the members by their names in small letters. For example, let us denote the EN of A as a.

a = Infinity

Since F has the lowest EN, after coauthoring the paper with F, the ENs of A and C would become f + 1 each.

Since this brings the average of the group to 3 it means that the total of all the ENs now is 24.

=> a + b + c + d + e + f + g + h = 24 ....9

From condition 6 we know that five of the ENs are same and the remaining ENs are distinct. Since A and C have the same EN, they cannot be the ones with distinct ENs and they are two of the five members with the same EN which is f + 1.

Therefore 5 members have the same EN as (f + 1) ....10

From condition 7, we know that E has co-authored a paper with F and that would make the EN of E as f + 1. Since it reduces the average EN of the group by 0.5 we can conclude

e - (f + 1) = 8 x 0.5 = 4

=> e = f + 5....11

So now we have five members with EN (f + 1) and three other members with distinct ENs. One of them is E with EN as (f + 5) and the other being F. Let the remaining distinct EN be x.

Therefore from condition 9 we can conclude,

5(f + 1) + f + 5 + f + x = 24

7f + 10 + x = 24

7f + x = 14 ....12

Since f cannot be zero, the only possible value for f is 1. From this we can see that x = 7.

Since we know the value of f, we can go on to solve the complete caselet. Keeping the length of the article in mind, it is unfeasible to give the complete solution here. Users interested to know the same can download the complete solution key for CAT 2006 here.

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