DI-LR Strat
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Are you prepared for Mathematical Reasoning in DI-LR?

In the previous post, we covered the Dos and Don’ts of representing or structuring data and how to prioritise conditions.

In this post, we will take a look at the type of reasoning sets that pose major challenges to the average test-taker.

Be ready for Numerical and Algebraic Reasoning

When test-takers say they are finding a DI or LR a set tough, what they mean is that they are not seeing the following in the sets they encounter:

  • closed DI sets around pie-charts, graphs, tables
  • LR sets around arrangements with simple plugin conditions

What stumps most people, is sets that are NOT direct calculation and NOT direct arrangement.

These sets are usually, Open Sets, which we defined in the previous post, that blur the line between DI and LR and require you to be able to venture into territory beyond what is typical DI and LR — numerical & algebraic reasoning.

Numerical Reasoning sets require you to test out the various numbers that a particular variable, say production in a particular month, can take given the conditions. The only way you can proceed by listing, testing & eliminating possibilities given the conditions, making the solving of these sets very similar to the solving of Sudoku.

Sets that involve Algebraic Reasoning, require you to take one unknown value to be X and then use the conditions to write the other values in terms of X and use the conditions to determine things about X:

  • The precise value of X or
  • The maximum and minimum values it can take

The key to cracking such sets is to be open to two things:

  • the use of algebra or Sudoku style reasoning
  • the possibility that the set will remain open even after solving and questions might not be direct but those involve that ranges or inequalities

Once you have changed your outlook and are willing to explore these non-standard lines of reasoning and explore the use of algebra you will take your DI-LR skills to the next level.

Let’s take a DI set from a SimCAT that we had classified as a must-solve.

SimCAT 2 - DI 2

The only thing you need to crack this is to use algebra and be comfortable with the set remaining an OPEN Set even after the solving.

Once you use the basic conditions and take one of the values as X, your representation should look like this.

photo 4.JPG

Once you calculate the value of X and fill in the remaining values the table should look like this.

photo 5.JPG

From here on you should be able to answer all the questions correctly by just ensuring that you read what is asked for, properly, without being in a hurry to rush to the next set.

Decoding my favourite CAT LR set of all time

As I mentioned in my previous post my favourite LR set is from CAT 2006 — The Erdös Number set. Before I wax eloquent about it, go ahead read the set and give it a solid try.

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

Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y.

Then X has an Erdös number of y + 1. Hence any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity.

In a seven day long mini-conference organized in memory of Paul Erdös, 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 Erdös number.
  • Nobody had an Erdös 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 Erdös number of the group of eight mathematicians to 3. The Erdös 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 Erdös 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 Erdös numbers while the other three had Erdös numbers distinct from each other.
  • On the fifth day, E co-authored a paper with F which reduced the group‘s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper.
  • No other paper was written during the conference.

What makes this set so unique?

  • The concept is absolutely novel in the context of CAT Logical Reasoning Sets — something that is not remotely related to anything that one has seen before.
  • There is no table and hence one has to really give some thought on how to represent the data.
  • Not a single condition is a plugin condition, every single condition is a deductive condition.
  • It is an open set and not a closed set.

After solving it in 2006, recently I tried to solve this again, albeit with a small challenge that I set myself — to solve this completely without putting pen on paper.

So here is how one can go about logically breaking open this set.

  • No one had an Erdos number lower than F so we can take his Erdos number as X and proceed since he is the only one who authored papers with the others during the conference
  • When F authors papers with A and C, on the 3rd day, their Erdos numbers become X+1 and X+1
  • When he does this the average of the group comes down to 3. So the total of the group becomes 24 since there are 8 people in the group.
  • When F authors a paper with E, on the 5th day, the average of the group comes down by 0.5, which means that the total decreases by 4 (average from 3 to 2.5 so total from 24 to 20, or directly by multiplying the decrease, 0.5, with 8)
  • This decrease is only due to the decrease in the Erdos number of E after writing a paper with F since no other papers were written.
  • After writing a paper with F his Erdos number of E would have become X + 1, since it decreased by 4, before writing it should have been X + 5.

So now we have some of the values for each of these days.

After Day 3 we know that

A = X + 1, C = X + 1, E = X + 5, F = X; TOTAL = 24

After Day 5

A = X + 1, C = X + 1, E = X + 1, F = X; TOTAL = 20

  • We also know that at the end of Day 3 five of people had the same Erdos number that means five of the values were the same.
  • We know 4 values and we do not know 4 values, B, D, G, H
  • What can be the equal value?
  • It has to be among the three values we know, X, X+1 or X+5 since there are only 4 values we don’t know and there are 5 equal values.
  • Also, we know that apart from the equal values all the remaining three values are different, so 5 equal values, 3 different values.
  • So the equal value has to be X+1, otherwise the apart from the five equal values, there will be two X + 1’s, the values of A and C.
  • So the values we now know at the end of Day 3 are X, X+1, X+1, X+1, X+1, X+1, X + 5 and one unknown value.
  • The total at the end of the Day 3 is 24. So, 7x + 10 + Unknown Value = 24, or 7x + Unknown Value = 14, hence X has to be 1 since unknown value cannot be 0 or a negative value
  • If X is 1 then the Unknown value is 7.
  • The values of A and C are 2 and E is 6 to begin with and changes to 2 on the 5th day.
  • The catch is that we do not know who has the Erdos number of 7

Now you can answer the set.

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

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

How many participants in the conference did not change their Erdös number during the conference?

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

The Erdös number of E at the beginning of the conference was:
(1) 2
(2) 5
(3) 6
(4) 7
(5) Cannot be determined

How many participants had the same Erdös number at the beginning of the conference?
(1) 5
(2) 8
(3) 2
(4) 3
(5) Cannot be determined

If you see it has all the atypical qualities of tough LRs — the need to use algebra and at some point, the need to test and eliminate numbers (7X+ Unknown = 14).

But you also realize that if you are open to viewing the set for what it is and do not expect it to yield to you automatically, you can solve the set.

The best part about this set is that it is based on a true story!

Paul Erdos was a famous, eccentric mathematician who believed that mathematics is a social activity and hence always co-authored, or rather solved, mathematical problems with this friends and the Erdos number was instituted by his friends as a homage to him. You should read up the Wiki Entry on him, here is an excerpt from the same.

Possessions meant little to Erdős; most of his belongings would fit in a suitcase, as dictated by his itinerant lifestyle. Awards and other earnings were generally donated to people in need and various worthy causes. He spent most of his life as a vagabond, traveling between scientific conferences, universities and the homes of colleagues all over the world. He earned enough in stipends from universities as a guest lecturer, and from various mathematical awards to fund his travels and basic needs; money left over he used to fund cash prizes for proofs of “Erdős problems”. He would typically show up at a colleague’s doorstep and announce “my brain is open”, staying long enough to collaborate on a few papers before moving on a few days later. In many cases, he would ask the current collaborator about whom to visit next.

Another roof, another proof – Paul Erdos

Let’s keep our brains open!


  1. RANJITH G says

    Sir this is Simcat 8 question. Can you tell how to solve this without pen and paper by using ratios method?

    Ajay initially starts working on an assignment. After he completes half the work, he quits and Vijay completes the remaining work. The work is completed in 18 days. If Ajay quits after completing 1/3rd work and Vijay completes the remaining work, the work is completed in 16 days. In how many days will the work be completed if both Ajay and Vijay work together right from the beginning and complete the work together?


    • The difference of 2 days stems from Ajay finishing 1/3 instead of 1/2 of the work, that means, for the difference 1/6 of the work, Ajay takes 2 days more than Vijay, so for half the work he will take 6 days more than Vijay.

      They do half the work each in 18 days, so Ajay does it in 12 days Vijay in 6 days (x + x – 6 = 18, solved mentally). The full job they will do in 24 and 12 days.

      Vijay does the full job in half the time, so he has twice the capacity, so when he joins Ajay, Ajay’s capacity becomes 3 times, and time 1/3, 8 days.


      • RANJITH G says

        thanks a lot for replying sir. I have a doubt. You have mentioned the extra 2 days came from ajay doing his 1 by 3rd instead of 1 by 2 and thats why the time reduced by 2 days. But it can also be vijay doing 2 by 3rd of the work in the latter case instead of 1 by 2 work in the former case. so why do you consider only ajay and ignored vijay?


      • Scenario 1: Ajay works for 1/2, Vijay, 1/2
        Scenario 2: Ajay works for 1/3, Vijay 2/3 (1/6 + 1/2)

        What is the difference between scenario one and 1 and 2? Vijay started working earlier.

        The work from 33.33% to 50% or 1/6 of the work is being done by Vijay instead of Ajay and it is this that is reducing the time by 2 days.

        The rest of the work, 50%, is anyway being done by Vijay in both the cases and so should take the same time in both cases.

        So for that 1/6 of the work which both are doing alone – Ajay in Scenario 1, and Vijay in Scenario 2, that Vijay takes 2 fewer days.


  2. tanya says

    This was an extremely helpful post. Mathematical reasoning has always troubled me but I couldnt put my finger on why. It would be very helpful if you could make more posts explaining how to solve similar sets.
    Thank you!



    sir please start your youtube chanel we will be happy to get knowledge from you.


  4. Pingback: How to improve your DI-LR Percentile – I | The CAT Writer

  5. Pingback: How to crack the DI-LR section of the CAT – I | The CAT Writer

  6. Manash Biswas says

    Sir i am facing problem to process a sets at first glance.I mean it really takes long for me to process the amount of information a typical DI-LR set provides.Could you please give me any advise on it.

    It feels really great to know such a story behind a set. I would have seen it as a normal set (Erdos Number) but thanks to you for bringing this story down.



    • Hi Manash,

      Good to know that people appreciate the backstory of the Erdos number set — the profs who set the paper bring a lot more to the making of a set that meets the eye.

      To reduce the processing time you need to first ensure that you are reading only once and not twice with the first reading just happening passively.

      Secondly, the only way I can think of is for you to be both the horse and the rider, whipping yourself to push yourself faster and think and concentrate harder. I cannot think of any other way.

      Hope this helps,

      All the best!


  7. Shiva Nirmal says

    Sir, in the Erdos set, how can you reach the conclusion whether or not C had a change in her Erdos number? From what I understood the answer to the 2nd question in that set, “How many participants in the conference did not change their Erdös number during the conference?” should be cannot be determined.
    Correct me if I’m wrong.


    • Hi Shiva,

      Your assumption is that C and F can have the same Erdos number.

      Read these two conditions clearly and you will see why that cannot be the case:

      Nobody had an Erdös 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 Erdös number of the group of eight mathematicians to 3. The Erdös 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 Erdös number of the group of eight to as low as 3.

      All the best!


      • Shiva Nirmal says

        But C could also have an Erdos number of X+1 before co-authoring the paper with F. Isn’t this a possible scenario?


      • That would mean that C could already have co-authored a paper with F before the conference.

        The sentence also says that F authoring of papers with A and C brought the greatest reduction in the average that was possible, no other co-authorship would have reduced the average by so much.

        Any reduction to have the possibility of being the greatest must-have F as a co-author.

        If, F with A and C produces the maximum reduction and C is X+1 before the co-authoring then it means that all the rest should also be not greater than X+1 since this is the greatest reduction (E was X + 6, so F, A, and E should produce the greatest reduction)

        So, all the remaining 5 would also need to be X+1, so including C there would be six (X+1)s at the end of day 3, which violates the next condition.


  8. Tushar says

    Sir I am having a tough time with solving reasoning based DI questions. It is mostly due to the fact that I lack basic concepts of at least and at most questions. I can solve the easy ones but it really take a toll to solve the others. Could please suggest a book that would help me in solving such questions as find the solutions written tough to comprehend.
    Thank you.


    • Hi Tushar,

      I am not sure what you mean by basic concepts.

      The best way to improve you DI is by looking at all the video solutions of the SimCATs again and going through all the posts and videos starting with the oldest one on this page — https://cat100percentile.com/tag/di/

      There are no books that I know that can be of help.

      All the best!


  9. Sir , In cat 2006 question
    Why Erdös number of F was considered constant ?
    Because before the conference Erdös number for A was infinite and for C it was finite number , So among A and C , Erdös number of C is smaller , Thus the Erdös number for F should have changed to C+1 ?


  10. Sir , In cat 2006 question
    Why Erdös number of F was considered constant ?
    Because before the conference Erdös number for A was infinite and for C it was finite number , So among A and C , Erdös number of C is smaller , Thus the Erdös number for F should have changed to C+1 ?


    • Hi Ayush,

      If you really want o get better at DI-LR you should really think hard about this. It is not tough at all to understand why and if you do not understand this then you will not understand the set at all.

      I can give you the answer but then that will make no difference to your ability.

      Read the set once more, twice more, thrice more, ten times more but figure it out on your own.

      All the best!


  11. Ayushi says

    The way you write your blogs really pushes me to love the process and enjoy solving a set rather than getting bogged down by it. I hope to cultivate this skill for other sections as well.
    It was a great read.
    Thankyou 🙂


  12. Bhargav Joshi says

    Sir can you please provide the answers to questions 2 and 4 of the books set i.e. the first one.


  13. sarthak says

    Hello Sir,

    In the Erdos number set, how do we rule out C having Erdos number f+1 at the start.



    • Because it is stated that no other combination of co-authoring would have brought down the average by 3.

      If C was F+1 to begin with then there are other combos that could have reduced it further.


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