CAT 2016: How to improve your DI-LR percentile – Part II

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 are that 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.

In Numerical Reasoning sets you have 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 is proceed by listing, testing & eliminating possibilities given the conditions, making the soving of these sets very similar to the solving of Sudoku.

In sets that involve Algebraic Reasoning sets, at some point you have 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 SimCAT 2 that we had classified as a must-solve.

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

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

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

From here on you should be able to answer all the questions correctly by just ensuring that you read what being 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 thought about 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 when he 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 he 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, other wise the apart from the five equal values, there will be two X + 1 values of A and C.
• So the values we now know at the end of Day 3 are 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 realise 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 is famous, eccentric mathematician who believed that mathematics is a social activity and hence always co-authored or rather saved 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, 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 generallydonated 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” (see below). 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. Pranav says

Hello sir,

Can you please put some light on how to improve quant.

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2. Rajesh Kanjan says

Hello Sir, Waiting for your fourth (and third in the series) post on QA.
Thank You.

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3. Faraz Ahmed says

Sir, i have a doubt. how can we be certain about C`s ergos number in the beginning being 2 or not. Say, it was 2 in the beginning, after collaborating with F, the ergos number will still remain the same i.e. 2. The average will be brought down to 3 after F`s interaction woth A &C, ergo all the henceforth conditions will hold true, then ergos number of 3 scientists is equal to 2 in the beginning other than C. How can we be sure about C`s ergo number in the beginning. It very well could be 2 and thus the answer to the last qn that how many had same ergos number in the beginning could be either 3 scientists or 4 scientists. So it should be Cannot be determined. Kindly point out the mistake i am making.

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• Hi,

If C’s Erdos number already 2 then the condition that no other co-authoring would have brought down the average Erdos number will get violated — F could have co-authored a paper with A and E (X+5) to reduce the Erdos number by a greater margin.

Good question though, hope this clarifies.

All the best!

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