Mock-6 Analysis

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 (Q.1 to 5)	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. Chicago Bulls: P7, P10, P11, P12, P14 La Lakers: P2, P3, P4, P6, P8, P9, P16 Utah Jazz: P1, P5, P13, P15 Then all the questions could have been answered one by one.
Set 2 (Q.6 to 10)	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 (Q.11 to 15)	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 (Q.16 to 20)	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 (Q.21 to 25)	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 Reading ability, Comprehension ability, Concentration, ability to think through logically when confronted with close choices and ability to understand the theme, tone and scope of whatever was being said.

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).

……………………………………………………………………………………………..

NMAT Notification

NMAT Exam for entry into the PG Program of NMIMS

Date of Exam: December 30, 2007
Time: 9:00 AM
Sale of forms begin: September 17, 2007
Cost of form: Rs.1500
Last date of reciept of form: September 30, 2007

For detailed itenary of NMAT 2007 please click here.

JMET NOTIFICATION IS OUT

JMET 2008 will be conducted on Sunday, December 09, 2007 from 10:30 AM to 12:30 PM.

Applications are to be made online.
Start Date: September 03 2007
End Date: October 03 2007

Announcement of Results of JMET 2008 : Friday, January 4, 2008

For detailed list & description of Institutes participating in JMET 2008 and their profiles please click here.

SPJAIN, NIRMA, FORE, K.J.Somaiya, IMDR

1.SPJAIN, Mumbai

Website: http://www.spjimr.org

Important Dates:
Last Date of Online Application: 30 November 2007

2.NIRMA

Cost of Form: Rs. 1,100

Website: http://www.imnu.ac.in

Important Dates:
Start of Form Sale: 25 August 2007
Last Date of Form Receipt: 30 November 2007

3.FORE

Cost of Form: Rs. 1,300 (for PGDM) and Rs. 1,350 (for PGDM-IBM)

Website: http://www.fsm.ac.in

Important Dates:
Start of Form Sale: 13 August 2007
End of Form Sale: 14 November 2007
Last Date of Form Receipt: 17 November 2007

4.K.J.Somaiya PGDM

Website: http://simsr.somaiya.edu

Important Dates:
End of Form Sale & Last Date of Form Receipt: 15 December 2007

5.IMDR, Pune

Cost of Form: Rs. 1,000

Website: http://www.imdr.edu

Important Dates:
Start of Form Sale: 14 August 2007
Last Date of Form Receipt: 30 November 2007

Mock CAT-5 Analysis

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 3^1/2.So, tan of the required angle = b/z= (3^1/2 - 1)/ 3^1/2.

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 (Q.81 to 85)	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 (Q.86 to 89)	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 (Q.90 to 94)	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.
Set 4 (Q.95 to 98)	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 (Q.99 to 100)	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 (Q.114 to 117)	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 (Q.118 to 120)	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

…………………………………………………………………………………………………..

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

……………………………………………………………………………………………………………….