MOCK CAT – 6 (2007)
Overview:
When the going gets tough, the tough get going! Persistence is the key --- in love, war and Mock 6!!!
Over all a real tough paper, which would have shaken you, all ends up. Almost midway en route to the journey to CAT 2007 and what better reason to work on the weak areas, which were revealed along with the end of this test!!
In this paper, Logical Reasoning based Data Interpretation and Verbal Ability sections were above average, and Quantitative Ability section was tough.
The information the cover page made available to you was:
| There were 90 questions in all, distributed over three sections. Each section had 30 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:
| Section | Topic | Total Qs | Suggested Time (in min.) | Possible number of attempts | Possible Score* |
| I | Logical Reasoning and Data Interpretation | 30 | 50 | 14 | 40 |
| II | Verbal Ability | 30 | 50 | 11 - 13 | 38 |
| III | Quantitative Ability | 30 | 50 | 12 | 36 |
| | Total | 90 | 150 | 37-39 | 114 |
*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: LOGICAL REASONING AND DATA INTERPRETATION
LR/DI section in this paper was of above average difficulty level but there were sets where one could have capitalized and utilized upon the opportunity to score. Again as always making the smart choice of doable sets is of utmost importance. Also, attempting few questions from many sets is also not a bad option always. Lets see the sets one by one.
| Set 1 | This was one of the sets, which were easy to crack but demanded patience. From Dream Team 7, it could have been deduced that P12 must be from Chicago Bulls. From Dream Team 6, it could have been deduced that P15 belonged to Utah Jazz and that P14 belonged to Chicago Bulls. Dream Team 1 could have led to P1 belonging to UTAH Jazz and Dream Team 4 could have ensured P10 belonging to Chicago Bulls. From Dream Team 2, P11 belongs to Chicago Bulls and P13 belongs to Utah Jazz. From Dream Team 3, P5 belongs to Utah Jazz and so on. One can finally find all the belongings. La Lakers: P2, P3, P4, P6, P8, P9, P16 Then all the questions could have been answered one by one. |
| | |
| Set 2 | This set was pretty simple to read and understand. One should have done the first 3 questions of this set without much effort and left this set to attempt others. The main thing to keep in mind while solving this set of questions was the start and stop operations of the machine and the bit manipulations associated with them. Last 2 questions could have been left untouched on account of opacity and length. |
| Set 3 | This was again a set, which could have been done after getting the hang of it. One could have left the last question and attempted all the others. One needed to pay heed to each statement in the mother data and then could have played with the various combinations so as to obtain different possibilities of total amount paid by each girl and the different combinations of eatables. Q. 12 and Q. 14 were very easy, which did not even require one to do any calculation. This set could have asked for time investment but returns could have been fruitful. |
| Set 4 | This was one set that could have been avoided. A tabular combination of weights, maximum and minimum possible members belonging to a party and parliament and state 1, 2 and 3 could have been established. Then depending upon the individual questions one could have gone on to solve each of them individually. This set was a test of your analytical as well as logical ability to the extremes. It would have been wise to do around 2 questions from this set and find an escape route. All in all a tough set. |
| Set 5 | This set was again an easy read and therefore a must read. One could have done the first 3 questions from this set and left the set. One could have taken realistic values like 23,17,26,25 or 22,24,18,27 and 21 and 22 were done. One could have seen that for the second smallest value more than 23, other values could not have been chosen so as to satisfy the sum as expected. Last 2 questions could have been attempted only if there was ample time and no hurry. |
| Set 6 (Q.26 to 30) | This set was more into numbers and those who are not very comfortable dealing with maximum or minimum value question would have been better off not attempting this set. However, the concept was not alien and only the theme was different. If at all one were to attempt this set, a table revealing the Fresh participation roll over participants and total participants should have been formulated. Q.26, Q.29 and Q. 30 should have been attempted and other should have been skipped or avoided. |
This translates into an attempt of about 14 questions. For a realistic accuracy of 75%, a score of 40 could have been achieved.
SECTION II: VERBAL ABILITY
This was a Verbal ability section that really tested your
You really needed to display your question selection skills and solve them with a cool, logical mind. You may have had to read each question once, twice or even thrice. But you may have found out that many questions revealed answers when you persisted a second or third time.
Following is an analysis of the question types:
| Q 31-35 Para-jumbles | These 5 questions ranged from medium to difficult to very difficult. Q 31 and Q 33 could be called as the medium ones. Q32, 34 and 35 were difficult. You could have attempted 3 Qs here getting 2 correct. |
| Q 41-45 Para-completion | These were a mix of Easy-Medium-Difficult types. Q 43 and 45 were easy. Q41, 42 and 44 were medium to difficult. You could have again attempted 3qs getting 2 correct. |
| Q 51,52 Analogies | Q 51 was medium whereas Q 52 was very difficult. You may have got 1 correct out of 1 attempt. |
| Q 53,54,55 Grammar-Errors | These were tricky but a must attempt chunk. You could have got 2 correct out of 3 attempts. |
| Reading Comprehension | |
| Passage 1- Qs 36-40 | The passage was moderate to read. Questions were a mix of medium to difficult ones. Q 36,37 and 40 were moderate and Q 37 and 38 were difficult. A doable passage in which you could have got 3 correct out of 5 attempts. |
| Passage 2-Qs 46-50 | A difficult passage to read and questions were on the difficult side. This passage could have been left. |
| Passage 3-Qs 56-60 | A MUST ATTEMPT PASSAGE- because it was short and readable. Q 56, 57 AND 59 were easy to medium. Q 58 AND Q. 60 were difficult. |
This translates to an attempt of about 11-13 questions. A SCORE of 32-34 is possible considering that many of you would have spent more time on the English section –seeing that QA and DI were tough.
SECTION III: QUANTITATIVE ABILITY
The quantitative ability section of this Mock CAT was yet again high on the difficulty index. One could have sorted out the relatively easier ones and then went on to solve them.
Q. 64 could have been attempted. The lengths of EH, BC and perpendicular distance between BC and EH could have been found out and then the area of the trapezium could have been found. The area of the annular ring was the difference between the area enclosed by the larger circle and that of the smaller circle. Now, the ratio could have been equated to the given ratio and found out the desired ratio.
For Q. 65 one could have just seen that the last digit from the product of 13 and 17 is 1, hence the desired prime numbers should be such that their product ends in 7,which is possible only when their last digits are (1,7), (3,9)(9,3) or (7,1). The product should be different and hence one should have taken care that repetition must not occur and thus found out 55 different products.
In Q.67 one could have seen that there were 12 items (10 beads and 2 pendants), out of which 2 pendants were identical. Thus, the number of ways of forming a necklace =(11! /2) x (1/2) =11! /4
Q.69 again was a must attempt. One could have found out the length of DE or CF, which would have meant that angle ECF is known. Now CP = 4 and BC = 4, so PB could have been found by using cosine law.
For Q.72 one could have calculated the horizontal distance and vertical distance between the two points and then applied Pythagoras theorem to get the distance as 6(7) 1/2 or if some one is very comfortable with fixing coordinates, one can do that and directly calculate the distance between the 2 points.
Q.74 again could have been done by observing the fact that if the 3 digit number ranged from 100 to 109 then the resulting 2 digit numbers could have proceeded like 81,72,63.54 and so on. Hence the 3-digit number must have been a multiple of 9 and so it must have been 108.Only option (2) was a factor of 108.
In Q.76 apart from the pure geometrical way as given in the solution booklet if one could have fixed the coordinates of P, O and Q which was very easy, one could have easily known the area of the triangle without much manipulation.
Q. 77 was a must attempt and one could have observed that the numbers were in the form a2 x b2 x c x d or a2 x b2 x c3 for a, b, c, d primes. 22 x 32 has to be there and to minimize the LCM; the numbers must have been minimum. So, the numbers should have been 23 x 32 x 52 and 22 x 33 x 72 their LCM was 23 x33 x 5 2 x 72
Q. 80 was a relatively problem wherein one could have converted the base 8 numbers into base 10 and added them and just compared to the RHS which was already in base 10.One could have found P=2 and Q=3 and so, P+Q=5
In Q. 85, it was worth noting that one had to multiply all the perfect squares from 49 to 289 and check the powers of 22 and 3.The power of 22 was 11 and that of 3 was 8. Hence, the highest power of 12 was 8 in the product.
For Q. 90, the best way was to take a value of ‘a’ and then find corresponding value of b and then substitute both in the required expression. A =1 would have led to b=1/(3)1/2 and the expression becomes 20+8(3)1/2
This translates to an attempt of about 12 questions. For a realistic accuracy of 75%, a score of 36 could have been achieved.
All the best!!
Career Launcher Team
…………………………………………………………………………
Answer Key and Explanations Corrections:
Q. 65 Explanation: Read the case I and case IV as:
Case I: If units digit of N is 3, then units digit of M will be 9.
Then, N = 13, 23, 43, 53, 73 or 83 and M = 19, 29, 59, 79, 89
So, number of distinct products = 6 × 5 = 30
Case IV: If unit digit of N is 9, then unit digit of M will be 3.
Then N = 19, 29, 59, 79 or 89 and M = 13, 23, 43, 53, 73 or 83
So number of distinct products = 5 × 6 = 30
Total number of distinct products = 55.
Q. 85. Correct answer is (4) in place of (5).
Explanation: Only perfect squares have odd number of factors
P = 49 × 64 × 81 × 100 × 121 × 144 × 169 × 196 × 225 × 256 × 289
Where N is neither a multiple of 2 nor a multiple of 3.
Here highest power of 3 is 8, and highest power of 2 is more than 12. So the highest power of 12 is 8.
Q.86 Correct answer is (2) in place of (5).
……………………………………………………………………………………………..