Overview:
In this paper, Verbal Ability was on the moderate side, Quantitative Ability section was tough, whereas Logical Reasoning based Data Interpretation section was above average.
The information the cover page made available to you was:
| There were 120 questions in all, distributed over three sections. Each section had 40 questions and carried 4 marks for each question. Wrong answers carried negative marks equal to one-third 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:
| Section | Topic | Total Qs | Suggested Time (in min.) | Possible number of attempts | Possible Score* |
| I | Verbal Ability | 40 | 50 | 15-17 | 11-13 |
| II | Quantitative Ability | 40 | 50 | 10 | 12 |
| III | Logical Reasoning and Data Interpretation | 40 | 50 | 12 | 14 |
| | Total | 120 | 150 | 37 to 40 | 38+ |
*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
The Verbal Ability Section in this paper was of moderate level of difficulty.
Q1-30 carried 1 mark and Q 31-40 carried 2 marks.
| Q 1,2 Fill in the blanks | Two questions of moderate to higher level of difficulty. You could have attempted both questions getting atleast 1 correct |
| Q 3,4 Word Usage | These two questions were easy. You could have attempted both the questions and got both correct. |
| Q5-8 RC Passage | A moderate level passage. If attempted you could have got at least 3 Qs correct. |
| Q9-12 Fact, Inference, Judgement | These 4 questions were sitters and you could have got at least 3 correct from an attempt of 4. |
| Q13-17 Para-jumbles | All 5 questions were difficult and you could have probably attempted one to two questions selectively. You could have got 2 correct from an attempt of 3 Qs. |
| Q 18-21 | 4 questions in which you had to identify the Grammatically incorrect sentences. These were of moderate level of difficulty. You could have attempted 3 and got 2 correct. |
| Q 22,23 | These were 2 easy-medium level questions, which required you to read to some extent. You may have left this sensing that it was a new question type. |
| Q 24-26 | These were 3 questions on phrasal verbs. Q 24 was easy. But Q25 and Q26 were difficult. You may have attempted 2 getting 1 correct. |
| Q 27-30 RC Passage | This passage was of moderate to higher level of difficulty. The questions tested your in depth understanding of the passage. You could have left this passage. |
| Q31-35 RC Passage | This was a moderate passage to read but the questions were tricky. If attempted you could have got 2 Qs correct from an attempt of 4. |
| Q 36,37 Para-completion | Two Para-completions, which were tricky. You could have attempted both getting 1 correct. |
| Q 38-40 Odd word out | You could have left this chunk except q 40. You could have got 1 correct out of 1 attempt. |
Overall in Section 1-A you could have ended up attempting 12-15 Qs in 20-25 minutes getting a score of 9-10. In Section I-B, you could have attempted 3 Qs getting a score of 2-4. Overall a score of 11-13 was possible from an attempt of 15-17 in 50mins.
SECTION II: QUANTITATIVE ABILITY
The quantitative ability section of this Mock CAT was high on difficulty index. One could have picked up the easy questions and then attacked them rather than going randomly for the questions, as there were only a few sitters scattered across the section.
In Q. 41 one could have connected the center of H1 and the meeting point of H2 and H1 and then simply used Pythagoras theorem to calculate the radius of H3 in terms of radius of H1.Then, one could have calculated the cube of their ratio.
For Q. 42 one could have just ignored the greatest integer function in any case one needed to find integral values. The least value of x can be –7 and the greatest can be 1 and these are 9 in number.
In Q. 44 one could have started jotting down numbers 1 onwards taking care of the fact that no two numbers in the set have HCF more than 3, hence we neglect 8,10,12,14,15,16,18and 20.So, the max possible number of elements =20-8=12
Q. 50 was another sitter in the offering. The values were corresponding to
(|x|, |y|) = (7,1) or (1,7) or (5,5). Hence the total integral solutions are 3 x 4=12
For Q. 52, one could have followed the wordings and drawn the tree building scenario. Let the height of the tree from where it breaks be a, the height of building be a + b, and the separation between the foot of building and that of tree be z, then b + a=z and z =a x 31/2.So, tan of the required angle = b/z= (31/2 - 1)/ 31/2.
In Q. 56 one could have converted PQR and RQP in decimal and taken the difference. The resulting value will be independent of Q and the values of P and Q could be found out. Since Q can take any value from 0 to 5 hence the sum cannot be uniquely determined.
For Q. 60 one could have equated the 2 functions and found out the discriminant of the resulting quadratic equation. If the 2 curves were never to intersect then discriminant must be less than 0.This happens for integer values of a are from 1 to 9.Hence a takes 9 values here.
Q. 63 was a sitter. The values of the sum of roots can be uniquely determined here to be 1.5.A must attempt question.
In Q. 65 the value of x=(y+120)/5.Clearly, y should be a multiple of 5 and for negative values of y from -5 to –115 the value of x will be a positive integer.-5 to -115 are 23 in number and hence the answer.
A quicker way to tackle Q.67 could have been to directly calculate the value of sum from 55/2 to 5/64, which comes out to be around 45.234. If one notices the terms to be subtracted one can observe that their sum is very negligible and it can never exceed 1.Hence, after subtracting such a small value our answer must be close to 45 and hence the answer.
Q. 70 was again a sitter and one just needed to check that the roots of the given equations were 0 and 2 which could be checked on the given answer options. A must attempt.
In Q. 76 the common factors of A and B which are factors of C are (1 and 2), (1 & 5), (1& w), (x0 to x7). Hence total 2 x 2 x 2 x 8=64
In Q. 77 one could have deduced a+2o=2 and m=4, where a, o and m denote apple, orange and mango respectively. Hence 3a+m+6o=10,but the list price is 12.Hence discount =12-10.5/12=12.5%.
Q. 78 could also have been done easily. Most students won't be reading this problem because it looks intimidating, whereas it was one of the easiest questions in the section. Just note that a cut triangle would result in 4 triangles. Suppose one goes on creating four triangles from each of the triangles in the subsequent process. In such a case one will get 64,256 and 1024 as the number of metal sheet at some moment during the process. Hence we can eliminate these options.
This translates to an attempt of about 10 questions. For a realistic accuracy of 75%, a score of 11-12 could have been achieved.
SECTION III: LOGICAL REASONING AND DATA INTERPRETATION
LR/DI section in this paper was of relatively high difficulty level compared to the previous mock. The best way to tackle this type of a DI paper would have been to spend a few minutes in the beginning scanning through the sets and then selecting which ones to do.
| Set 1 | This set could have given you a familiar sight but there was something here with a difference. First question says maximum number of bottles of Z manufactured, which can only be possible if the for all the 30 days 5 bottles are manufactured and if that is done, to have 60 bottles unsold, 3 bottles per day must be sold for 30 days. For the next question, one should realize that in order to maximise the number of days on which 5 bottles were manufactured, the number of days on which 3 bottles of Z were sold has to be maximised. The next question deals with minimizing the same for X and hence the number of days on which 1 bottle of X was sold has to be maximised. This set was a bit too lengthy for a 1 marker set. One could have attempted 2-3 questions from here and left the set. |
| Set 2 | This set on logic was a definite sitter. A rare one for this paper. It could have been easily deduced by the given four statements that the couple are (A, E), (H, D) and the rest 2 could toggle. The directions for question 88 and 89 ensure that G and C are couple and so are F and B and the possible anniversary dates of each couple could also be deduced. One could have attempted all 4 questions from this set in not more than 8 minutes. |
| Set 3 | This set could have enticed one for sure since it had a pictorial representation of the tournament proceedings. One may feel the set to be on the easier side but once into the set one may be lost amidst the mother data. Each question here had to be tackled with the extra condition given in each of the problems. One of the most important deductions was the fact that irrespective of N losing or winning any of the match, P must have played 3rd round at least as the number of wins for P is 2 more than that of N. The questions were not at all difficult to handle given that one understands the possible formations. If one were clear with the idea to proceed, one could have attempted this set and solved all the questions.
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| Set 4 | This is one of the sets that should have been attempted. It is worth noting here that the number of days for which Tata Steel's share witnessed an increase was one more than the number of days on which it witnessed a decline. Hence in 5 consecutive days, there were 3 increments and 2 decrements but if the price had been higher than 527.5 on 3rd February, then there would have been 3 consecutive increments and 2 consecutive decrements. Therefore, the share price on 3rd February was lower than the price on 2nd Feb and even lower than 527.Also, since the price of Modi Steel increased on 4 days and decreased on 1 day, therefore the share price would have increased on 1st Feb because there is already a decrease from 4th to 5th Feb. On the above basis one could tabulate the offer prices of companies from 1st Feb to 6th Feb. After this each of the questions could have been tackled easily. |
| Set 5 | One could be bamboozled not to find any thing other than percentages in this set, but that is the beauty of this set. The best way to tackle such questions is to treat the percentages as ratios and then establish equations as we can do here with respect to the gross salary and net salary relation. Once the idea strikes that there is nothing left in these 2 problems, one must attempt set though. |
| Set 6 (Q.101 to 105) | This is another set one could have attempted. A pretty straightforward set. It would have been wise to first go for questions 101,104 and 105 as these could have been done pretty quickly. For e.g. in 105 it is not known as to how many students would be there next year for that paper, because by then, the new batch would be appearing for the paper also. It is not clear whether the new batch strength of ME would be 90 or not.102 and 104 are also easy just that you have to check out or each of the departments separately and then compare. One should have done all the questions from this set. |
| Set 7 (Q.106 to 110) | This set of questions was seemingly new for all. One could have better left this set or at the best attempted question 110, 109 and 107 in that order. For, 110 the sequence will be 1,2,3,4,5,6. So 21 steps. For 109,7 throws are needed with top face as 4,5,6,3,4,5,6 in that order. Distance moved will thus be 4+5+6-3+4+5+6=27.In 107, the maximum distance covered by A after 6 consecutive throws will be achieved if the number appearing on the top face of the dice is 1,2,6,3,4,5 in that order. One could have avoided 106 and 108 to save time. |
| Set 8 (Q.111 to 113) | A set that one could have comfortably understood in terms of the presented data. It is advisable to quickly scan through the questions in a set and start from the easiest one and here Question 113 was the easiest of the lot and should have been picked up right away. Average weight of students in terms number of boys B in the class is B/20 + 45.Clearly, B has to be a multiple of 20 and hence we can find out all the possible values of Bs and number of girls and hence the ratio. One would have required a great deal of patience to handle the other 2 questions of this set but they were definitely doable. One could have attempted 2 questions from this set. |
| Set 9 | One should have left this set alone as one could have in all probabilities failed to understand and clarify the statements given in the mother data. Here one needed to check what happened after round 1 and after round 2 and formulate a table containing the various cases for the given differences and also the ranks. Then, one could have gone about attempting the questions. But as said earlier, understanding the statements and formulating the right table could get difficult here, one could be better off leaving this set after a few minutes of inspection. |
| Set 10 | A fairly easy read set and a type. One could have attempted at least questions 118 and 120 here and even 119, if one had time for the same. Revenue increased or decreased in each year. However, only the absolute values were known and also a final increase or decrease of 5% was mentioned. This was the basis for handling all the questions of this set depending upon the conditions for that particular question. |
This translates into an attempt of about 12 questions. For a realistic accuracy of 75%, a score of 14 could have been achieved.
All the best!!
Career Launcher Team
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Answer and Explanation corrections:
Question No. 62 was not considering while evaluating your score.
Ideally Q. No. 62 option (3) should be read as:
Option (3) 3 (312 - 213) - 9
Explanation: Total number of ways in which 13 balls can be put into three different boxes = 313.
Total number of ways in which 13 balls can be put into three different boxes such that each box contains at least 1 ball = Total ways – (exactly two empty boxes) – (exactly one empty box) = 313 – [3] – [3C1 (213 – 2)] = 313 – 3 – 3 (213) – 6
= 3 (312 - 213) - 9
For Questions 72 and 73:
Read the list five lines as:
We have 6 solutions. Hence option (3) and (1) are the correct choices for Q. 72 and 73 respectively.
We can go on to prove that for minimum greater than 11 we cannot have a valid set.
The total 6 sets being {10, 11,19, 21}, {10, 11, 13, 27}, {10, 13, 17, 21}, {11, 12, 17, 21}, {11, 13, 16, 21}, {11, 12, 13, 25}.
Question No.104 was not considering while evaluating your score.
Ideally Q. No. 104 option (4) should be read as:
Option (4) 99
Explanation: Total students who applied for re-evaluation = 12 + 10 + 14 + 20 + 14 = 70
Number of students who passed after re-evaluation = 0.5 x 70 = 35
Total number of students who should appear next year = failed + absentees
= (124 – 35) + (2 + 1 + 6 + 1) = 89+ 10 = 99
Q. 108: Correct answer is (4) in place of (3).
Explanation: At least 4 throws are required in order to make the distance between the two brothers ‘zero’.
The following steps be:
| | Person Name | Number | Steps |
| Initial Throw | B | 1 | 14 |
| Throw 1 | A | 5 | 9 |
| Throw 2 | B | 2 | 11 |
| Throw 3 | A | 6 | 5 |
| Throw 4 | B | 1 | 0 |
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