Mock CAT-10 Analysis

Overview:

The pattern has been adapted keeping in mind the likely possibility of CAT 2007.

In this paper, Logical Reasoning based Data Interpretation and Quantitative Ability sections were above average and Verbal Ability section was of average difficulty level, having few questions in each section that could be managed easily.

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	8	22
II	Verbal Ability	30	50	16-18	33-35
III	Quantitative Ability	30	50	10	30
	Total	90	150	34	85

*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

One should have looked at the questions, which were easy and placed randomly in the paper.

To begin with, the choice of the first set should have been made on the basis of comfort or familiarity of the set. The decision of leaving a set should be made within 5 minutes on the set unless you are very clear of what you are doing and the concept had been cracked.

Set 1 (Q.1 to 3)	This set could have been solved by considering the relative difference of the ages with respect to any one child. One could have listed down the ages of all the children as on any fixed date and then could have handled all the 3 questions without any hassles. In the third question one could have considered the answer options and taken the age of the youngest child as x and expressed the other 2 in terms of x. The sum of all the three should have been a multiple of 3.It was a must attempt set.
Set 2 (Q.4 to 5)	These were one of the easiest questions in the set. One needed to find the missing link between the starting statement and the conclusion. It could have been seen that in 4 only 1 or 2 could have been possible which could have been further solved to know that such values did not exist below 50. Hence, the answer was option 2. Similarly in 5 one should have realised that Poornima’s husband’s brother’s wife was to be considered.
Set 3 (Q.6 to 10)	This was a calculation driven set and one needed to be very careful here. A slight mistake and one could be misled to the wrong answer. One could have looked at attempting 1 or 2 questions only and then left this set for other doable sets.
Set 4 (Q.11 to 15)	This again could have intimidated you, but after some pencil pushing one could have realised that first 2 questions of this set could have been easily tackled. One just needed to check for the LCM of the denominators of the elements used in the respective compound row and also check for the common factor in the numerator and hence found out the minimum amount of each element used to form various compounds. This knowledge was sufficient to crack the first 2 problems and then one could have skipped this set.
Set 5 (Q.16 to 20)	This set was on familiar lines and hence one could have been tempted to go for it, but considering the standalone nature of all the questions, there was a possibility of one getting lost amidst so much data. Again, one could have looked at attempting 1 or 2 questions and left the set.
Set 6 (Q.21 to 25)	This was an open set wherein after a certain stage of deductions one would need question conditions one by one to make new possibilities. This question is seemingly easy but when one starts doing the same it is possible that one gets lots of arrangements and hence loses the patience to continue with the set and mark some wrong answer in haste. One should have rather selected one or two easy ones and went about doing them and then skipped the set.

Set 7 (Q.26 to 30)

One could have looked at doing a few questions from this set as well if none at all. On the basis of the given table one could have filled up some of the spaces in the second table. Now on the basis of each question one could have checked the tables and come up with new deductions. One should have definitely attempted questions 27 and 28.

Hence, about 8-9 questions from this section could have been attempted.

Assuming 75%+ accuracy, a score of 22+ was easily possible.

SECTION II: VERBAL ABILITY

Q 31-35 RC Passage- Huxley	A passage, which was moderately challenging to read. The questions required good inference skills. You could have got 2 questions correct out of 4attempts.
Q 36-40-Poem	A tough proposition. It was not easy to understand the poem and answer the questions. You could have left this.
Q 41-45 RC Passage- Sustainability	A passage, which tested your reading comprehension skills. The passage was reasonably tough to read and the questions challenged your inference and further application skills. If attempted you could have got 2 questions correct out of 5.
Q 46-48 Para-completion	A set of easy – moderate Para-completion questions with a twist. You had to find the opening statement. You could have easily got 2 questions correct out of 3 attempts.
Q 49-52 Critical Reasoning	Q 49 and 50 were doable whereas questions 51 and 52 were on the tougher side. You could have got 2 correct out of 2 attempts.
Q 53-57 Cloze	This was a moderately difficult cloze set. You could have got 2 questions correct out of 5 attempts.
Q 58-60	A different question type in which you had to rearrange paragraphs. You may have left this because it was time consuming.

Overall, an attempt of 16-18 Qs could have yielded a score of 33-35.

SECTION III: QUANTITATIVE ABILITY

This section had a good mix of easy questions spread evenly across the section. One should have identified them and attempted maximum from them in order to push the score higher.

In Q62, the combined weights could have been represented by a + b, a + c, a + d, b + c … and so on and then each of these values could have been found out individually.

In Q. 65, if one observed carefully, you could see that the difference of any two consecutive terms was in increasing powers of 5.This was enough to know the next term in the series.

In Q. 68, for minimum value one’s intention should have been to take max x and also keeping in mind the integers being distinct. The values of x should have been 15,1,4 respectively giving the sum as 20.For maximizing one should have taken maximum value for z, so the values should have been 1,2 and 72 respectively for x, y and z so that the sum was 75.Hence, the difference was 55.

Although Q. 69 looked dreadful to read, it could have been solved easily. The key was just to read through the problem. If the number of sheep in the year 2006 with Ramu were n, then one could have easily established the relation n²-97n-3500=0 and solved the quadratic to get to n=125.

In Q. 72 one needed to express the equation as y=x+1/y=>4x=4(y-1/y). Now putting the extreme values of y from the given range to get the range of 4x as 3.33<=4x<=8.4.Hence option (4) was correct.

Q. 73 and Q.74 could have been very easily cracked if one could have realised that CF=2AB=2 units and in turn GJ=2CF=4 units. The area of ABKLMN could have been calculated as sum of the regular hexagon JKLMNG and the trapezium ABJG. Also, for calculating AM one could have calculated the horizontal distance between A and M and also the vertical distance and then used the Pythagoras theorem.

In Q.76 one could have found more than one pair of (x, y) satisfying the condition out of which max value for xy could have been 32.Now, one needed to find the highest power of 6 in 32! => the highest power of 3 in 32! =[32/3]+[32/9] +[32/27]=10+3+1=14.

Q.77 was a must attempt and one needed to consider the 8 - units digit of each multiplied term to reach the conclusion. This could have been done by using the cyclicity funda

Q.78 was a must attempt. One could have put values and realised that only for N=1 the expression is satisfied because |2N-1|>N for N>1 or N<=0.

Q.84 was interesting and easy. Initially, the sum is odd and at any stage a value of 2 is incremented to the sum, so after every stage an odd sum will be left. Hence, all the terms can never be same.

In Q.86, one could have saved the time by calculating the first 3 terms approximately as the higher terms are almost negligible in magnitude. Hence 2/3 could have been the closest value.

Q.89 was on fibonacci series. and one could have easily calculated the values of f(4),f(5),f(6),f(7),f(8) and found out the value of the expression as 13/2

Hence, about 10-11 questions from this section could have been attempted.

Assuming 75%+ accuracy, a score of 30+ was easily possible.

All the best!!

Career Launcher Team