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The simplicity of the CAT is essentially reflected in the fact that all the questions asked in the test have simple straight-line solutions.

Here we take a look at how the questions in the Data Interpretation (DI) section of the CAT can be easily solved through simplistic solutions. To illustrate this, let us look at the following question asked in the CAT 2006 paper. There are 5 parts of this question with 4 marks each - a total of 20 marks can be gained from this question.

Before referring to the clues below, try to solve the question on your own. The ideal time for each question is approximately 8-10 minutes (a goal which you can achieve if you can identify the straight-line method to solve this question).

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 that 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 organised 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’. The following additional points are also known:

Clue a)On the third day of the mini 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.

Clue b)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.

Clue c)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.

Question& Answers:

1. The person having the largest Erdos number at the end of the conference must have had an Erdos number (at that time):
1) 5                 2) 7                            3) 9                 4) 14               5) 15

2. 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

3. The Erdos number of‘C’ at the end of the conference was:
1) 1                 2)2                             3)3                  4)4                  5)5

4. The Erdos number of‘E’ at the beginning of the conference was:
1) 2                             2) 5                            3) 6                 4) 7                 5) 8

5. 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


The first three paragraphs explain to you the mathematical situation and just give you the following deduction:

Deduction 1)‘A’ had an infinite Erdos number and‘F’ had the least Erdos number (could be 1, 2, 3 etc we do not know at this point of time).

Further deductions that come out naturally from the next few clues would be:
Deduction 2) From Clue a: When‘F’ co-authored with‘A’ and‘C’ on Day 3, the average reduced to 3. Hence, the total reduced to 24. Since, no other co-authorship would have reduced the average to as low as 3, we can deduce that A and C initially represent the maximum Erdos numbers amongst the given group. Since, A has the maximum Erdos number in the group, C must have had the second highest number.

Deduction 3) Also, after the third day the Erdos numbers would be: A and C- f+1, B-b, D- d, E- e, F- f , G- g and H- h. Also, since five people have the same Erdos numbers and the remaining three have distinct numbers at the end of day 3, there must be 5 people having their Erdos numbers as f+1 (otherwise this condition would not be possible).
From this point if you move into the logic of what values are possible for f and f+1 given that the total of all eight numbers is 24, the following analyses would emerge:

Possibility a)If f= 3, f+1= 4, then the total would come out to 3+4x5= 23 + two more values, which have to be distinct and greater than 4 but add up to 1.This would not be possible. Also, deduce that f cannot be greater than 3, since the situation would only get worse in such a case.

Possibility b) If f= 2, f+1= 3, then the total would come out to 2+3x5= 17 + two more values, which have to be distinct and greater than 3 but should add up to 7. It can be easily seen that this would not be possible.

Possibility c) If f = 1, f+1= 2, then the total would come out to 1+ 2x5= 11+ two more values, which are distinct and greater than 2 and add up to 13 (since the overall total is 24). This can be done in three ways as: 7+6, 8+5 or 9+4. Also, since this is the only possible value of f, we can realise that the eight numbers at the end of day 3 would have one 1 (for‘F’), five 2’s (for‘A’,’C’ and three more people. Who? We do not know yet!) and two more distinct values.

Deduction 4) Further it is said in Clue c that when‘E’ co-authored with‘F’, the group average reduced by 0.5. This means that the group total reduced by 4. Also‘E’’s co- authoring with‘F’ must have had the result of reducing the total of the group by 4 (8x0.5). Also,‘E’’s number, after co-authoring with‘F’ would have come down to 2. Hence,‘E’’s number initially would have been 6. Consequently, the other distinct number must have been 7. Hence, the answer to the questions would have been:

1. The highest number at the end of the conference would be 7.
2. Of‘A’,‘B’,‘C’,‘D’,‘E’,‘F’,‘G’,‘H’ it is known that‘A’,’C’ and‘E’ changed their Erdos numbers. The others did not. Hence, the answer is 5.
3. 2
4. 6
5. After the third day, when‘A’ and‘C’’s Erdos number came down to 2 each, there were 5 people on 2. The other three would have had different Erdos numbers. Hence, at the start of the conference, there would be three people with identical Erdos numbers.

In all, if we sum up there are only 3-4 critical reactions which determine the solution of the above question. If you take the straight-line path to solving this question, the thinking and solving time will not cross 5-7 minutes.

This question carried 20 marks out of 100 in the DI section of the CAT 2006 paper. If you analyse this from the perspective that the cut off for the DI section was approximately 25-30, and there was a total of 150 minutes for the three sections of the paper, one can see why CAT could be perceived as an easy paper Analysis such as the above can be shown for any of the five sets that appeared in the CAT 2006.

Before we end, a couple of points about DI that you should know:

1) Linear Equations: All DI questions are based on linear equations. While trying to solve questions on DI, always remember that all you need to do is to identify the linear relationship between the variables. Hence, through your preparation for DI, be sure to create a thought pattern which allows you to identify and crack the linear relationship between the variables in the question.

2) Move out of mathematical solving of equations: More often than not it would help you to attempt solving DI questions by trying to stick to the logic of the question rather than resorting to mathematical processes. Look closely at how the equations were solved in the above question - that will explain.