Mock CAT 3 Analysis

Overview:


In this paper, Verbal Ability was on the easier side, whereas Quantitative Ability section was tough, and Logical Reasoning based Data Interpretation sections was moderate.
The information the cover page made available to you was:

• There were 75 questions in all, distributed over three sections.



• Each section had 25 questions and carried 4 marks for each question.



• Wrong answers carried negative marks equal to one-fourth of the marks allotted to the question.



• Total time available to answer all the sections was two and a half hours.

Executive Summary:
A synopsis on how this paper could have been attempted is:

*Note: The number of attempts and the score has been worked out on the basis of the experts’ insight on how the students would have taken the test and what score IIMs have been considering for dispatching the call letters. The correct cut-offs can be confirmed statistically only after seeing the actual performance of all of you.

SECTION I: VERBAL ABILITY

In this paper, Verbal Ability was of easy-moderate level of difficulty. This section must have been a morale booster for you after Mock –1 and Mock-2. Let us do a specific analysis of this section.

A glance at the section would have revealed to you the order of attempts.

You could have attempted the section in the following order
1. Fact Inference Judgement - 7-8 mins
2. Para-completion – 15 mins max
3. RC Passages- Rest of the time!

1. Fact Inference Judgement: Qs 16-20

A good starter. You could have attempted all 5 questions. Q 17 was probably a bit tricky. The others were easy. In 7-8 minutes a score of 12-16 was a sure possibility.

2. Paracompletion:Qs 1-5

These were a mix of levels of difficulty. You could have attempted all –probably getting 2-3qs correct and in the process notching up a net score of 8-12.

3. Reading Comprehension:

You would have found out that passage 3 was an easy passage. The other two were of moderate difficulty. Assuming that you attempted one passage reasonably accurately , a score of 8-12 was a possibility here.

Taking account of test conditions a score of 24-28 looks very much possible. If your English and reading are up to the mark then you could score even above 30 in this section.

An attempt of 14-15 could have yielded a score in the range of 24-28.

PS: The above attempts and possible score have been derived based on sample simulation of the paper. They are subject to variance under actual test conditions.

SECTION II: QUANTITATIVE ABILITY

The quantitative ability section of this Mock CAT was tough in its difficulty index. Few questions demanded clarity in concepts as maximum portion of the paper constituted number theory, Algebra and geometry and there were only few problems, which could be done by smart work.

In Q. 28 realize that straight lines at a distance of 1 unit from the origin with slopes 1 form nothing but a square of side length 2. Hence, the area is 4 sq. units.

Q. 29 was the one that could have been solved by smart work. Observe that T100 = 98/100 * 101/103. Since, this is the last term and contains 103 in the denominator, then so should S and hence the only feasible answer is option 1.

In Q.32 It effectively means the remainder when S is divided by 1000. Lets remove a factor of 50.So now we have 883*881*871*873*35*439 when divided by 20=>3*1*11*13*(-5)*(-1) => 5 when divided by 20.Since we had removed a factor of 50 so we multiply it back to get the final result i.e. 5*50 = 250

In Q. 38 was a conceptual problem, which required you to know total product of the factors of N and total product of the factors of N, which are not multiples of 5.So, total product of the factors of N, which are multiples of 5 could be calculated.

Q. 39 required you to arrange the given equation to yield (m-1)(n-1) = 37.Now it could be seen that for (m, n) = (38,2) and (2,38), option 1 & 3 satisfy. Also, we could have m = 0 or n = 0 so that we get –1 & 2 also as the values for the expression.

Q. 41 could have been solved by elimination of options. For x = -0.5, y = 1-3 = -2
Option 2, 3 & 4 are eliminated. For x = -1.5, y = 3-6 = -3, which also satisfies option 1 and hence the answer.

In Q. 42 one could have put the value of one variable from the linear equation into the quadratic and evaluated the discriminant of the resulting quadratic and made it greater than equal to 0 to obtain the desired range. A must attempt for sure.

Q. 45 also required you to plug in values to establish relations within functional values at different points. Putting x = 0 & x = 1, we get f(3) – f(1) = 2.Putting x = 2 & x = -1 and solving we get f(0) + f(1) – f(3) – f(4) = -6=> f(0) – f(4) = -4 => f(4) = 5.

In Q. 48 is another good question. In case the similarity condition did not strike you, sine rule could have made life easy. Let CM = b & CN = a, BM = 5x, DN = 5y, z = 3a = 2b

Q. 49 and 50 were not that tough a nut to crack if you have a concrete knowledge base. If N = abc, K = 20+81(a+c)/(a+b+c). To minimise K we minimise a+c & maximise b. Maximum value of b = 9 & minimum possible value of a+c = 3+0 = 3.Hence, minimum K = 20+81/4 = 40.25.
Also, a+b+c = 17 such that 101a+20b+101c = 96*17+4 = 1636.The only possible value in the given range is for N = 719.719 when divided by 17 gives a remainder of 5.
Hence, a considerable number of questions could have been attempted. With 8 attempts and a decent accuracy of 75%, a score of 22-24 was definitely attainable.

SECTION III: LOGICAL REASONING AND DATA INTERPRETATION

LR/DI section in this paper was of moderate difficulty level. Proper selection of easy sets would have really boosted your score. Lets see them one by one.

Set 1(Q.51 to 55)

Basically one needs to establish a tabular relation among the batches, days, subjects and the name of faculties on the basis of the conditions given and taking care that none of them gets violated. Once the table is ready, the questions can be tackled one by one on the basis of the table. But attempting this set could have proved wrong as it may have taken longer time to comprehend and formulate the right table.

Set 2(Q.56 to 60)

Relation could be established among the number of students who wrote exactly one, exactly two and exactly three names and then their possible values could be observed.56 and 57 could have been answered on this. For the rest of the problems the additional information was to be used which could have given you 2 cases in which the name of candidates along with the number of times their names were written in the first choice and also in the remaining two choices. It would be smart to attempt 2 sitters from this set and proceed to the next set.

Set 3(Q.61 to 65)

This set was a cakewalk. One could have formed a tabular relation of the possibilities. From statement 1 Shane received 0 cookies. From statement 3, Shane and Chris, and Matthew and Graham can have [(0,1)(2,3)] or [(0,2)(1,3)] or [(0,3)(1,4)] cookies respectively. Using statements 2 and 4 we can conclude that only possible case is [(0,3)(1,4)]. So Greame must have got 2 cookies. Using statement 1 and statement 7 we can say that Graham and Greame like wrestling shows, as Shane likes a sports show and there are only two sports shows. From statement 5, Matthew likes Smack down and Shane likes Sports Center. So Chris likes Hitz.From statement 6 Graham likes Krackjack. From statement 8 Greame likes Hide n Seek or Good Day. Shane likes Good Day or Bon-Bon. Matthew likes Hide n Seek or Bon-Bon or Good Day. It would have been criminal not to have attempted this set.

Set 4(Q.66 to 70)

Another easy set in the offering. From the table since A, C, E, F, G, H, L all have more than 1 entries as same which means they can never be the one who predicted all correct. Now, we needed to check only for the remaining five friends. I happens to be the one who counted all-correct as we could have all 12 combinations mentioned as in the question. All the questions could be attempted one by one after finding out the right combination of counts for all the baskets. This set was also one among those that must have been attempted.

Set 5(Q.71 to 75)

This is the easiest of all sets. It should not be a surprise if one is told that it was very much a 6-7 minute set. Approach could have been slightly different here. Assume A = 4 be the first mine. Then rest of the mines are given by: B = 14, G = 7, D = 12, I = 19, E = 20, J = 26, C = 29, F =31, H = 36. As per question 71 if A = 4 is not a mine then none of the above hold for a mine. Hence the right answer is 12.As per the question 72 block 14 & 16 are mines. It is obvious that a shift of 2 blocks will result in all the mines so that the first mine is A = 6. Since, 27 is not in the above list 29 cannot be the block which contains mine. For 73, 1-6-> 6-11-> 11-17-> 17-23-> 23-28-> 28-34-> 34-40-> 40-42.In 74,total number of mine counts if removed from total number of blocks would yield maximum number of throws i.e. 42-10=32.For,75 E can be maximum when A = 6 => E = 22.Also J can be minimum when A = 1 => J = 23. Hence, the difference can be 1.
This set was a must attempt.

This translates to an attempt of about 10-12 questions. With a realistic accuracy of 70%, a score of 26-28 could have been achieved.


All the best!!