Cl Mock Cat 9 Key 2008

Mock CAT – 9

Answers and Explanations 1

4

2

5

3

1

4

5

5

3

6

2

7

1

8

1

9

2

11

4

12

2

13

4

14

4

15

5

16

1,4,5

17

4

18

2,3

21

3

22

1

23

4

24

5

25

3

26

2

27

1

28

1

29

4

31

2

32

4

33

3

34

1

35

3

36

3,5

37

3,5

38

2,3,4

39

3,4

41

5

42

4

43

5

44

3

45

1

46

5

47

4

48

4

49

2

10

2

19 2,3,4,5 20 1,2,4 30

4

40 1,2,4 50

3

51

2

52

3

53

1

54

5

55

4

56

4,5

57

5

58 1,3,4,5 59

2

60

3,5

61

1

62

4

63

1

64

3

65

5

66

5

67

3

68

2

69

1

70

5

71

2

72

2

73

3

74

4

75

2

76

2,4

77

4,5

78

1,2,5

79

1,2

80

1,3

MY PERFORMANCE Total Time Taken Total Correct Incorrect Net Questions (Min) Attempts Attempts Attempts Score Logical Reasoning based Data Interpretation Verbal Ability Reading Comprehension Quantitative Ability TOTAL

Section A Section B Section A Section B Section A Section B Section A Section B

15 5 15 5 15 5 15 5 80

150

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MCT-0013/08 009

1

For questions 1 to 4: If Mr. Mathew would have taught each of the given ten classes on exactly three different days of the week ‘W’, then the total number of students he would have taught in week ‘W’ would be (1 + 2 + 3 +.....+ 9 + 10) × 3 = 165. Total number of students taught by Mr. Mathew on the six days of the week ‘W’ = 18 + 12 + 23 + 19 + 32 + 25 = 129.

3. 1

Class C is not taught by him on two consecutive days of the week ‘W’.

4. 5

It cannot be uniquely determined which class was taught by him on three consecutive days of the week.

For questions 5 to 7: 5. 3

Difference = 165 – 129 = 36. Therefore, the total number of students in the two classes that were

Let the total revenue generated by “Pristine Careers” in the year 2007 be ‘R’. Total revenue generated through “Media Interaction” = (0.15 × 0. 20 + 0.25 × 0.40 + 0.10 × 0.65 + 0.20 × 0.55) × R = 0.305R Required Percentage = 30.5%.

 36  not taught by him is   = 12.  3 

So, the two classes that were not taught by him could be (B and J); (C and I); (D and H) and (E and G)

6. 2

On Tuesday the total number of students taught by him is 12.

Option (1): It is not true as the total revenue generated through “Media Interaction” in 2007 is more than the total revenue generated through ‘Publishing’ in 2007.

This is possible in two cases: Case I: He taught the classes A, B, C and F Case II: He taught the classes A, B, D and E

Option (2): It is true as the total revenue generated through ‘Consulting’ and not by “Media Interaction” in 2007 is 17.5% of the total revenue generated through ‘Others’ in 2007.

From the two cases stated above, we can definitely conclude that he taught the classes A and B.

Option (3): It is not true as the total revenue generated through “Media Interaction” is 200.5% more than the revenue generated through ‘Coaching’ in 2007.

On Friday the total number of students taught by him is 32.

Option (4): It is not true as the total revenue generated through ‘Others’ and not by “Media Interaction” in 2007 is 36% of the total revenue generated through ‘Publishing’ in 200. Hence, option (2) is the correct choice.

This is possible in two cases: Case III: He taught the classes E, H, I and J. Case IV: He taught the classes F, G, I and J. From the two cases stated above, we can definitely conclude that he taught the classes I and J.

7.1

The total revenue contributed through the role played by “Media Interaction” is maximum in “Others”.

Also, he taught either class E or G.

For questions 8 to 11:

Therefore, the two classes that Mr. Mathew did not teach are D and H. 1. 4

Number of movies NOT Liked

Number of movies Liked

Shefali

6

2

The classes taught by him on Friday is F, G, I and J. Hence, option (5) is the correct choice.

Shreyas

3

5

Nitin

4

4

For questions 3 and 4: Given that on Saturday of week ‘W’, Mr. Mathew did not teach class C but taught class J.

Richa

5

3

Devendra

7

1

Pankaj

0

8

Since, he did not teach class J, which means that he definitely taught class I as on saturday Mr. Mathew taught 25 students an only combination possible is (C, F, G, I).

Abhishek

2

6

Ashraf

1

7

Priyanka

4

4

Hence, he definitely taught class E on each of the three days namely Monday, Wednesday and Thursday as he has not taught class E on Friday, Saturday and Tuesday.

Amit

5

3

Total

37

43

2. 5

8. 1 The possible combinations of the classes taught by him on Thursday is (E, I, C and B) or (E, A, C and J). Therefore, he definitely taught the class C on Thursday. The only possible combination of the classes taught by him on Monday is A, B, E and J.

Monday Tuesday

Wednesday

Thursday

Friday

Saturday

A, B, E and J

(E, C, A (E, G, J and and J) or A) or (E, G, I (E, C, I and B) and B)

F, G, I and J

C, I, F and G

2

A, B, C and F

Aggregate number of students who liked the movies P, Q, R, S, T, U and V is 10 + 9 + 1 + 4 + 6 + 3 + 8 = 41 So, the number of students who liked the movie W = 43 – 41 = 2. Therefore, the number of students who did not like the movie W = 10 – 2 = 8.

009

For questions 9 to 11: Since, every student liked the movie P and Devendra liked only one movie, therefore Devendra definitely liked P. Nine students liked the movie Q and hence everyone except Devendra liked the movie P. Shefali liked only 2 movies and they have to be P and Q. Hence, every student except both Devendra and Shefali liked the movie V. Following similar logic, we can get the list of the students who did not like each of the given eight movies.

Further, as the number of buyers must be integers therefore from the definition of ESE, we conclude that a and c must be multiples of 8 and 14 respectively. The sets of all the possible values that integers that a, b, c and d can take are as follows:

a = {48,56,64,...,296} b = {27,28,29} c = {42,56,70,...,294} d = {26,27,28} 12. 2

As the Fridge Balance for variety “E” of the pastries is zero, all the pastries produced on 25th July 2008, must have been sold on the same day. Hence (2) is the correct answer.

13. 4

Let S A ,SB,SC,SD and SE denote the number of pastries of varieties A, B, C, D and E respectively that were sold on 25th July, 2008. So, we can write: SA = 50 × 100 – a SB = b × 100 + 180 SC = 25 × 100 – c SD = d × 100 – 560 SE = 4500 The number of pastries sold would have been maximum when a = 48, b = 29, c = 42 and d = 28 accordingly, the maximum values of SA, SB, SC, SD and SE are 4952, 3080, 2458, 2240 and 4500 respectively. Hence, the maximum number of all the pastries, sold could have been 17230. Hence (4) is the correct answer.

14. 4

The Fridge Balance (F) and the number of retail shop buyers are related to each other by the following relation:

The following is the list of all the students who did not like each of the given eight movies P–0 Q – 1 – Devendra V – 2 – Devendra and Shefali R – 9 – Devendra, Shefali, Amit, Richa, Nitin, Priyanka, Shreyas, Ashraf, Abhishek S – 6 – Devendra, Shefali, Amit, Richa, Nitin, Priyanka T – 4 – Devendra, Shefali, Amit, Richa U – 7 – Devendra, Shefali, Amit, Richa, Nitin, Priyanka, Shreyas W – 8 – Devendra, Shefali, Amit, Richa, Nitin, Priyanka, Shreyas, Abhishek 9. 2 10. 2 11. 4

Out of the given eight movies, there is only one movie, i.e. T, which is not liked by Richa but is liked by Nitin.

For questions 12 to 15: On any given day, none of the pastries produced, get wasted. Hence all the pastries produced in a day either get sold on the same day or are kept in the fridge for sale on the next day. Hence, for any two consecutive days, (N – 1) th and (N) th, we must have:

 (number of pastries produced)Nth Day     + (number of pastries left unsold)(N – 1)th Day     (number of pastries sold)Nth Day   =  + (numberof pastries left unsold)Nth Day    Or,

(number of pastries produced)Nth Day − (number of pastries sold)Nth Day (Fridge Balance)

ESE =

Absolute value of Fridge Balance Number of RetailShopBuyers

The sets of all possible values that integers a and c can take are as follows:

a = {48,56,64,...,296} c = {42,56,70,...,294} Accordingly,

RA =

a = {1, 2, 3, 4, 5, 6… 37} 8

RB =

180 = 15 12

RC =

c = {1, 2, 3, 4… 21} 14

560 = 35 16 As the Fridge Balance of “E” variety of the pastries is not known, we cannot comment on the possible values(s) of RE. Option (4) is definitely invalid. RD =

Nth Day

Further, we are given that a, b, c and d are integers and they satisfy: 25 < d < b < 30 < c < a < 300

009

3

15. 5

The number of pastries that were left unsold on the two consecutive days, are listed below:

17. 4

Difference between the marks obtained by Anup in section C and D =

Variety of Pastry A

24th July 2008 x

25th July 2008 x+a

B

x

x - 180

C

x

x+c

D

x

x +560

E

x

x

18. 2,3 Only Case II and Case III are possible, hence the marks obtained by Rohan can be either 900 or 1200. 19. 2,3,4,5 Case I: Marks obtained by Tarun in section B

=

We are given that: (x + a) + (x – 180) + (x + c)+ (x + 560) + x = 870 or 5x + (a + c) = 490 As (a + c ) has the minimum possible value, it must be 90, which gives x = 80. Hence, option (5) is the correct choice.

A 17r 50

B 9r 50

C r 5

D 7r 25

Tarun

8t 25

9t 50

t 5

3t 10

Anup

a 2

a 10

3a 20

a 4

The marks obtained by each of the three candidates in sections A, B, C and D should be divisible by 2, 3, 5 and 6 respectively.

=

=

3 3 (a + 2t ) = (600 + 200 ) = 120. 20 20 Option (4): It can be true when the aggregate marks obtained by Tarun and Anup in the recruitment test is 700 and 600 respectively. Hence, options (1), (2) and (4) are the correct choices. is

If we take 'r' = 100, then the marks obtained by Rohan in section D is 28, which is not possible. 21. 3

This is the only option that continues with the theme of the last sentence regarding the private economy and mentions the role of ‘compensatory fiscal policy’ mentioned in the passage. The keyword ‘governor’ in this option can also be linked to the same keyword in the last sentence. Option 1 is wrong because what happened in the seventh edition would follow after what is mentioned in option 3. Also both option 1 and option 2 appear to be beginnings of paragraphs rather than ending sentences. Both options 1 and 2 as well as Option 5 refer to ‘earlier edition’ while the passage refers to the first edition. Option 4 mentions ‘new economics’ which is not mentioned anywhere in the passage.

22. 1

The passage is about the ‘single critic’ and option 1 continues with the theme of the individual critic. Also the third sentence mentions that the critic ‘can only interpret an author in the light of his own age’ and sentence 1 continues with ‘the individual critic therefore can only make a contribution to a portrait which…must remain unfinished’. Option 2 is not correct because the passage has a somewhat negative tone about the limited ability of an individual critic and does not mention any ‘justification’ of the critic’s role. Option 3 is not correct because the passage makes no mention of the ‘reader’. Option 4 is not correct because it talks specifically about ‘psychoanalytical’ critic while the passage talks about the critic in general. Option 5 is not correct because it talks about the critic stimulating the reader which is nowhere a part of the original passage.

Similarly, the minimum possible value of 't' is 100 and 't' will always be a multiple of 100. Also by the same logic the minimum possible value of 'a' is 600 and 'a' will always be a multiple of 600.

For questions 17 to 19: Given that the aggregate marks obtained by Rohan, Tarun and Anup is 1900 and the marks obtained by Rohan is not less than the marks obtained by Anup. The following are the cases that are possible. Case I: Rohan = 600, Tarun = 700 and Anup = 600 Case II: Rohan = 900, Tarun = 400 and Anup = 600 Case III: Rohan = 1200, Tarun = 100 and Anup = 600

4

9t 9 × 100 = = 18 50 50

20. 1,2,4 Option (1): It can be true when Tarun has 300 marks and Anup has 2400 marks in the recruitment test. Option (2): It can be true when the marks obtained by Rohan and Tarun in the recruitment test is 300 and 100 respectively. Option (3): It cannot be true as the minimum sum of the marks obtained by Anup and Tarun in sections C and D respectively

If we take 'r' = 50, then the marks obtained by Rohan in section A is 17, which is not possible.

16. 1,4, 5 Minimum possible aggregate marks obtained by Rohan, Tarun and Anup are 300 + 100 + 600 = 1000. Option (4) can be possible if the marks obtained by Rohan, Tarun and Anup are 600, 100 and 600 respectively. Option (5) can be possible if the marks obtained by Rohan, Tarun and Anup are 300, 200 and 600 respectively.

9t 9 × 400 = = 72 50 50

Case III: Marks obtained by Tarun in section B

It is clear that 'r' has to be a multiple of 50.

Therefore, the minimum possible value of 'r' is 300 and 'r' will always be a multiple of 300.

9t 9 × 700 = = 126 50 50

Case II: Marks obtained by Tarun in section B

For questions 16 to 20: Let the total marks obtained by Rohan, Tarun and Anup in the test be 'r','t' and 'a' respectively.

Rohan

a 3a a − = = 60 4 20 10

009

23. 4

The last sentence of the passage says ‘Still, it was not art, it was religious art’; the art had a purpose’. Option 4 continues with the theme of ‘religious art’, by mentioning ‘cathedrals’. Options 3 and 5 are close but do not bring out the connection with ‘religious art’ as well as option 4 does. Option 2 is not correct because the passage gives an explanation for the purpose behind cave art. Option 1 is also not relevant to the passage because it does not explicitly link to the religious purpose cited in the passage.

24. 5

The central idea of the passage is conveyed in ‘relentless purpose’ in sentence 2 and ‘He determined…to wrest back the ‘ashes’…and to put Bradman in a reasonable…place. Option 5 carries on with the idea of ‘relentless purpose’ and also his determination to achieve the two objectives cited above. 2 is close but only touches upon one of the two objectives mentioned in the last sentence. Option 1 and 4 do not touch upon any of the main ideas. Option 3 is too general and does not touch upon the specific objectives mentioned in the passage.

25. 3

26. 2

27. 1

D is the opening sentence since it introduces the subject of study and what it aims at. So options 2, 4 and 5 are safely eliminated. The area of the study is the “relationships at the cellular level”. This idea is being carried forward in A where it mentions “cellular microenvironment” and “cellular functions”. C further talks about the “microenvironment” which was mentioned in A. The role of complex sugars is being questioned in C. E directly follows C and A since it mentions “the areas which are probed”. Also E talks about dissecting the “complex sugars” which were mentioned in C. B finally concludes by stating the use of such a study. It says the “understanding can be used...”. This understanding comes from the “observed changes” which were talked about in E. So, DACEB forms the correct sequence. E is the opening statement because it introduces “Authorship” as a subject. The rest of the statements give details about authorship. In B, the phrase “this issue” refers to “authorship” mentioned in E. This makes EB a mandatory pair. B also talks about the “development of modern concepts of authority” and “late medieval period as a crucial moment”. This idea is elaborated in A where “changing status” of “late medieval” writers is mentioned. A also cites “more authorial identity”. D introduces the name of a book “designed for students”. C mentions the work of the author of the book Jacqueline saying that she provides “detailed analysis for students”. This makes DC a mandatory pair. So the correct sequence is EBADC. ‘A’ cannot be the opening statement because it starts with the phrase ”in the field”, but the field hasn’t been mentioned. B, C start with pronoun “this” but it is not clear what the pronoun refers to. E says “on the other hand” which suggests that it is comparing something with mainstream finance, and therefore the subject needs to be introduced. D is the opening statement, it starts with the development of the subject. Options 2, 4 and 5 are ruled out. B follows D, since it talks about the “advancements” and D had talked about the ‘development’ of the subject. C refers to the “direction” of these advancements. Therefore, BC forms a mandatory pair. A follows C, it talks about what scholars in the field believe. E concludes by contrasting mainstream finance with mathematical finance.

28. 1

Statement E forms the opening statement since it introduces the subject of the collapse of “Union of Socialist Soviet Republic” (giving the full name of the country first). So options 2, 4 and 5 are ruled out. The next statement is A which describes the scene after world war II and introduces ‘cold war’. Next comes B because B takes the idea of cold war further and names Mikhail Gorbachev (gives full name first). Statement D evidently follows B, it says that the ‘end of the cold war’ isn’t the only event ‘Gorbachev’ is associated with (gives only the last name). Statement C follows D, it tells with which other event ‘he’(using pronoun for Gorbachev) was involved - fall of communism. BDC is a mandatory sequence which is only present in option 1.

29. 4

Option 4 best supports the view expressed in the passage. It presents a very close picture of the existence of life on Mars.

30. 4

Aromatherapy may have a number of uses and the given paragraph/argument assumes that a cure or healing is the main attraction.

31. 2

Since ‘… agonising’ is juxtaposed with ‘..self congratulation’, it should be preceded by ‘introspective’. ‘Cognitive’ is not an appropriate option here because it refers to the act of perception of some external being or idea. ‘Perceptive’ is ruled out on similar grounds and ‘volatile’ and ‘random’ are out of context.

32. 4

Since this particular statement is in contrast with the French people’s usual mood, ‘triumph’ cannot be preceded by ‘apologetic’. ‘Reluctant’ and ‘Deliberative’ do not make any sense with ‘triumph’ and ‘significant’ cannot express the extreme feelings in this case.

33. 3

The other four words do not fit in the context because they have a negative tone.

34. 1

‘Redemptive’ is the right option here because it refers to the act of redeeming which goes with ‘slick’ and ‘warm’; whereas ‘tedious’ and ‘dry’ are inappropriate. ‘Amenable’ and ‘listless’ are negative words and inappropriate here.

35. 3

‘Staggering’ is the most appropriate word in the given context. Staggering creates the necessary impact in the context as compared to massive.

36. 3,5 The belief that all that exists around us came by chance weakens the need for creating a God. Also option 5 disproves the paragraph. 37. 3,5 The passage is talking about the efficacy of continuity. Options 3 and 5 best support the argument. 38. 2,3,4 The incorrect statement is ‘A’. In ‘A’, ‘probing’ should be followed by an article rather than a preposition; therefore it should be ‘probing the.. .’ rather than ‘probing about ..’. 39. 3,4 In ‘A’ it should be ‘...confronted by’ rather than ‘..with’. In ‘B’ it should be ‘... this is...’ instead of ‘these are’, since the referred subject is ‘ the show’. Hence, options 3 and 4 are correct. 40. 1,2,4 In C it should be ‘an increase ...’ since it is being referred to the first time. The other statements are correct.

009

5

41. 5

You have to choose the statement that is incorrect. The third line of the passage denies the statement in option 5 that Orwell would do it for the sake of producing a great work of art. Hence option 5 is false and the correct answer thereof.

42. 4

Option 4 is the correct answer as it cannot be concluded from the passage that Orwell had an exaggerated notion of himself. The other statements can be concluded from the passage.

43. 5

The passage seems to be exploring the purpose behind Orwell's writing and more specifically - the purpose behind writing 'Animal Farm' The purpose is not to criticize Animal Farm. Hence option 1 is ruled out. The other options are too general or off the central theme.

52. 3

The answer here is directly linked through the main idea of the passage given in para 3. ‘Contestedness’ involves those mediating practices between “conflict” on one hand and ‘consensus’ on the other.’ ‘Neither the essential agreement of ‘cultural consensus’ nor the absolute contradiction of ‘class conflict’ pervades the notion of ‘contested identity’.

53. 1

Refer para 6 ‘Modern Macedonia thereby ‘personifies’ contested identities, with the culinary symbol ‘Salade Macedoine’ consisting of numerous sliced fruits and vegetables.

54. 5

Refer para 3, ‘Contested identity’ …requires a semiotic analysis by the anthropologist, in which icons, indices, and symbols constitute the interpretive field.’ Also in the same para ‘Again, as Appadurai noted in a keen semiotic perception, the hyphen in ‘nation-state’ can now ‘serve less an icon of conjunction, than an index of disjunction’. Therefore, both the anthropologist researching socio-cultural symbols (point 1) and the linguist looking for meaning in words that would help explain a certain context would find semiotic perception a useful trait.

55. 4

It is clearly mentioned in the 5th paragraph the Macedonia is characterised as an ambiguous region between the East and the West - this ambiguous nature makes its identity contested.

44. 3

It can be discerned from line 11 of the paragraph that Animal farm attempts to clarify in the minds of readers, what Orwell felt Russia had become. It was also intended to be a satire on dictatorship in general. A holistic reading of the passage leads to option 3 as the best among the choices. Option 3 is best in terms of tone and theme.

45. 1

Option 1 can be concluded from the last part of the passage where the ‘moral’ purpose behind Orwell’s work is connected to the form traditionally associated with the moral-the animal fable.

46. 5

The lesson finds mention in the first Para. Read and summarize it to reach the answer and eliminate the rest.

47. 4

Refer to para 3 line 4.The term railroad orientation means the same as that towards movie business i.e., product orientation where as transportation orientation and entertainment orientation match with each other which means customer orientation. In the question statement the author criticizes product orientation in contrast to customer orientation and also considers the same as a reason for the failure of the railroad industry. Option 3 is talking of the same points. But it is stating the reason of its success where as option 4 highlights the reasons of failure of a business.

57. 5

48. 4

A company that thinks customer wise would not receive the fate of New England Co. as it has been mentioned in the passage as a failed concern on account of its product orientation.

59. 2

49. 2

The given rhetoric in the context of the passage sets a prelude to the fact that the companies that grow but are not customer oriented may decline tomorrow, their glamorous growth notwithstanding. Hence the organizations that failed due to lack of customer orientation are being indicated here. Option 2 is best.

56. 4,5 Statement 4 has not been mentioned anywhere in the passage. Statement 5 is contradicted in the last line of the second paragraph.. Rest of the options figure in the passage. Refer to the 3 rd Para. Here ‘man’ is used for the ‘brain’ and ‘artifact’ for the ‘tool of language and writing’ produced by the brain itself as per the passage. This is a metaphorical allusion.

58. 1,3,4,5 Only option 2 is correct according to the passage. Refer to Para 3 line 1; Option 1 is untrue because the passage says that Neuroscience is beginning to do this. It has not already as mentioned in option 1. The others options do not match with what is given in the passage. In the context of the passage, ‘man’ is used for the ‘brain’ and ‘artifact’ for the ‘tool of language and writing’ produced by the brain itself. Hence option 2 is correct.

60. 3,5 In the context of the passage, pedestrian means prosaic/dull.

50. 3

Deja vu means sudden recall; delirium means ecstasy; equanimity means composure;

51. 2

The idea of the hyphen in ‘nation-state’ serving as an icon of disjuncture happens when, a state is formed, usually through under some extraordinary circumstance and is made up of people who share a tradition of an earlier existence of the ‘nation’ they hail from. These leads to some of the problems exemplified through the case of Macedonia. Only in case of the formation of Yugoslavia is there no clear instance of the existence of a previous idea of Yugoslavia. Hence this is the only option that would not qualify as a relevant example.

6

61. 1

1

Given that F(n – 1) =

(2 − F (n))

For n = 2: F(1) =

1

( 2 − F ( 2 ))

and F(1) = 2.

⇒ F(2) =

3 , 2

Similarily, one can find the values of F(3), F(4), F(5) as

4 5 6 , and respectively. 3 4 5

n +1 n From this we can say that every term except [F(1)], of the series [F(1)] + [F (2)] +…………+ [F(50)] is equal to 1 as for ‘n’ > 0, F(n) lies between 1 and 2. Therefore, [F(1)] + [F(2)] +……+ [F(50)] = 51. Hence, option (1) is the correct choice. ⇒ F (n ) =

009

62. 4

The following are the two cases that are possible. Case I: P = Q. There are 9 such ordered pairs (P, Q).

1 1 1 1 1 1 1 1     =  1 − + − + − + .....∞  +  1 − 2 + 2 − 2 + .....∞  = 2 2 2 3 3 4 2 2 3    

66. 5 Case II: P = 4, 6 or 8 and Q = 4, 6 or 8. There are 6 such ordered pairs (P, Q). Therefore, there are 9 + 6 = 15 such ordered pairs (P, Q). Hence, option (4) is the correct choice. 63. 1

Note: Every number has a cyclicity of 4, i.e. any number raised to power 4n + 1 viz. 5, 9, 13, 17….. will have the same unit’s digit as the original number itself. Hence, the number ‘a’ has the same unit’s digit as a5.

This means that he has taken out 3 rotten apples and 1 apple that is not rotten. The one apple that is not rotten could be taken out on any of the four days.

Similarly, b and c have the same unit’s digit as b5 and c5 respectively. It implies that a + b + c would have same unit’s digit as a5 + b5 + c5.

Case I: The apple that is not rotten is taken out on the first day. 9 6 5 4 Probability = × × × 15 14 13 12

As a, b and c are distinct digits their maximum possible sum is 24(9 + 8 + 7). Hence, a + b + c = 9 or 19. If a + b + c = 9, then the maximum possible value of a5 + b5 + c5 would be 32769.

Case II: The apple that is not rotten is taken out on the second day.

This is possible when a, b and c = (8, 1, 0) not necessarily in that particular order.

6 9 5 4 × × × Probability = 15 14 13 12 Case III: The apple that is not rotten is taken out on the third day. Probability =

Since, 91849 is much greater than that, hence a + b + c = 19. Sum of digits at the odd places starting from the left =2+5+1=8 Sum of the digits at the even places starting from the left = a + b + c = 19

6 5 9 4 × × × 15 14 13 12

Therefore, the remainder when 2a5b1c is divided by 11 will be when 8 – 19 = –11 is divided by 11.

Case IV: The apple that is not rotten is taken out on the fourth day.

Probability =

Hence, the remainder will be 11 – 11 = 0.

6 5 4 9 × × × 15 14 13 12

Hence, option (5) is the correct choice.

6 5 4  12  9 Required Probability = 4 ×  × × × =  15 14 13 12  91

64. 3

67. 3

Given that f(x) = x3 – x2 (3 + a) + x( 2 + 3a) – 2a > 0 ⇒ x3 – x2a + 2x – 2a + 3ax – 3x2 > 0 ⇒ x2(x – a) + 2(x – a) – 3x(x – a) > 0 ⇒ (x2 – 3x + 2)(x – a) > 0 ⇒ (x – 1)(x – 2)(x – a) > 0 Since, ‘a’ is an odd prime number, therefore ‘a’ is not less than 3. Therefore, the range of values of ‘x’ for which f(x) > 0 is 1 < x < 2 or x > a. Hence, option (3) is the correct choice.

68. 2

Let, Mr. Ganguly had a total of N toffees.

Perimeter of the base of cuboid = 64 cm Length of the base + Width of the base = 32 cm Length of edge of every cube = 2 cm Therefore, the total number of cubes along the length and the

32 = 16. 2 Therefore, the number of cubes along the length and the width of the cuboid can be (1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9) and (8, 8) not necessarily in that particular order. Also, in the cuboid (Number of cubes along the length) × (Number of cubes along the breadth) × (Number of cubes along the height) = 78 The only possibility = 3 × 13 × 2 = 78 Hence, the number of cubes along the height of the cuboid is 2. Hence, option (3) is the correct choice. width of the cuboid =

65. 5

The given series is S=

5 12 × 22

+

11 22 × 32

+

19 32 × 4 2

+

29 42 × 52

+ ..... + ∞

 1× 2 + 1 + 2   2 × 3 + 2 + 3   3 × 4 + 3 + 4  = 2 + +  + .....∞  1 × 22   22 × 3 2   3 2 × 4 2  2 2  1 1 3 2 − 22  1   2 −1 = + + + ..... + ∞  +  2 2 + 2 + .....∞  2   1 2 2 3 3 4 × × 3 ×2  ×   2 ×1 

009

Given that a5 + b5 + c5 = 91849. Therefore, the unit’s digit of a + b + c would also be 9.

Toffees got by the first player = 50 +

N − 50 N + 450 = 10 10

Toffees got by the second player

9N − 1450 9N + 8550 = 100 100 The same holds true for the toffees got by the subsequent players in the team. As each player gets an equal number of toffees, therefore = 100 +

N + 450 9N + 8550 ⇒ N = 4050 = 10 100 ⇒ Each player gets

N + 450 9N + 8550 = = 450 toffees 10 100

4050 + 1 = 9 + 1 = 10 450 Hence, option (2) is the correct choice. ⇒ Number of players in his team =

7

69. 1

The total number of silver coins has to be multiple of three i.e, the total number of coins in five out of the six boxes should be a multiple of three.

73. 3

a B re akfa st

We see that three boxes (containing 31, 19 and 16) has 3n + 1 number of coins and one box containing 20 coins is of the form 3n + 2.

b

Hence, the total number of gold coins has to be 20 as the coins in the other five boxes sum up to a multiple of three. Hence, option (1) is the correct choice.

U

x =z 3 a + b + c + d = 700 d+a=x a+b=y a=z

It is given that:

T

3 0°

Q

R O

1 un it

Join the points O and T.

...(ii) ...(iii) ...(iv) ...(v)

y =z ...(vi) 2 From (i), (iii) and (v) d = 2a From (iv), (v) and (vi) b=a ⇒ 700 = a + a + c + 2a As c = 200, a = 125 ⇒ z = a = 125 Hence, statement A alone is sufficient to answer the question.

2 2+ 3 1 +1= unit ⇒ QR = 3 3 3

∴ PT = PQ − QT =

...(i)

Using statement A:

Now, here ∠TOQ = 30° and ∠TQO = 60°

2+ 3 1 1+ 3 − = 3 3 3

PT = 3 +1 TQ Hence, option (5) is the correct choice. ∴

71. 2

d

z = (x ∩ y) a = Number of members who have neither breakfast nor dinner. b = Number of members who have only breakfast. d = Number of members who have only dinner. c = Number of members who have both breakfast and dinner.

P

∴ QT =

c

Let x = Number of members who do not have breakfast. y = Number of member who do not have dinner.

For questions 70 and 71: 70. 5

D in n er

Using statement B: a + b = 100 Hence, y = 100. From (vi), z = 50 Hence, statement B alone is sufficient to answer the question. Hence, option (3) is the correct choice.

We can write here 2

1  2+ 3  × PU PT2 = PR × PU ⇒  1 +  = 3 3 

2 3 Radius of semi circle, OR = 1 unit ⇒ PU =

2 PU = OR 3 Hence, option (2) is the correct choice.

74. 4

The following table lists down all the possibilities when the total number of candies with all the kids is an integer, which is less than 100.

Therefore,

72. 2

Let, a man and a woman can complete the work alone in ‘M’ and ‘W’ days respectively. Let the group initially have ‘m’ men and ‘n’ women.

⇒

m n 1 + = M W 120

...(i)

(m + 1) + (n – 1) =

1 ⇒ ...(ii) M W 96 Assume that after two women have been replaced by two men, the group can complete the same work in ‘p’ days. ⇒

(m + 2 ) + (n – 2 ) = 1 M

W

p

Now, (ii) – (i) = (iii) – (ii) ⇒

...(iii) 1 1 1 1 ⇒ p = 80 = – – 96 120 p 96

Hence, option (2) is the correct choice.

8

Himanshu

Prachi

Case I

Veena

1

1

1

Case II

2

4

16

Case III

3

9

81

Case IV

4

16

256

Case V

5

25

625

Using statement A: Even if the number of candies with Himanshu is a perfect square, he can have 1 or 4 candies. Therefore, statement (A) alone is not sufficient to answer the question. Using statement B: Possible cases are Case II and Case IV. Therefore, statement (B) alone is not sufficient to answer the question. Combining statements A and B together: If each kid has an even number of candies, then the only case that is possible is Case IV. Hence, option (4) is the correct choice.

009

75. 2

Using statement A: Given that a × b = 60. The values of ‘b’ and ‘a’ can be (2, 30), (3, 20), (4,15), (5,12), (6,10). Therefore, ‘b’ can be either odd or even. Hence, statement (A) alone is not sufficient to answer the question.

Since, AF is the angular bisector of ∠ DAE, therefore

AD DF = AE FE

⇒ DF = 6 units, therefore DE = DF + FE = 6 + 3 2 = 3 (2 + 2)

Using statement B: Given that b × c = 12. The values of ‘c’ and ‘b’ can be (1, 12), (2, 6) and (3, 4) in that particular order as and c < b. So, in each of the cases ‘b’ is an even number. Hence, statement (B) alone is sufficient to answer the question. Hence, option (2) is the correct choice.

76. 2,4 Given that

x 2 = y 7

Since A.M

≥ G.M

Since, AE = DE ⇒

(

)

⇒ X = 6 1 + 2 units.

Also,

AF2

=

AE2

+ FE2

((

⇒ AF2 = 3 2 + 2

))

2

+ 18 = 72 + 36 2

A

79. 1,2 D

N

M

P

Q

x x z z y y ⇒  + + + + + ≥6 y z y x x z

O

3

B

x x z z y y x y ≠ , therefore ⇒  + + + + +  > 6. y x y z y x x z Join ON to form an isosceles triangle ∆ONB. Since, ABOD is a parallelogram, we have AB = DO. Therefore, DM = 2 units as OM = 3 units.

Out of the given options only option (2) and option (4) is greater than 6.

 x x z z y y  61 For z = 1,  + + + + +  = . y z y x x z 7

OP = OB2 − BP2 = 32 − 22 = 5 units

 x x z z y y  94 . For z = 3,  + + + + +  = y z y x x z 7

Area of ∆ONB =

Hence, option (2) and option (4) are the correct choices.

Now QN = OP.

1 × 4 × 5 = 2 5 square units 2

Therefore, the area of the ∆ONM =

77. 4,5 When Rohan encounters the first gate, he can select any 2 out of the five fingers of his right hand. It implies he has 5C2 ( i.e. 10) number of choices. Since two doors can have at most 1 finger common. He is left with 9 choices at the second door. Similarly at the 3rd door he is left with 8 choices. Total number of ways to open all 5 doors = 10 × 9 × 8 × 7 × 6 = 30240 It can also be denoted as 10C5 × 5!. Hence, option (4) and (5).

A

B

F G

Area of the quadrilateral ANMD = Area of ABOD – Area of NMBO 1   3 5 = 5 5 – 2 5 + × 3 5 = square units 2 2   Also, the length of BD can not be equal to 5 units because it has to be greater than OD. Hence, options (1) and (2) are the correct choices.

80. 1, 3

1 1 1 ...(ii) + + =1 a b c The only possible set of values of a, b and c that satisfy both the equations is 4, 2 and 4 not necessarily in that particular order. Therefore, the absolute difference between a and b could be either 0 or 2. Hence, options (1) and (3) are the correct choices. Also,

E D

1 × 3 × 5 square units 2

Let the number of bananas with Moti, Sumit and Manky is a, b and c respectively. Then, a + b + c = 10 ...(i)

78. 1, 2,5

C

Given that FE =3 2 units. Let, AD = X units, therefore AC = X 2 units. Therefore, AE =

009

X = 3 (2 + 2) 2

Therefore, AF = 6 2 + 2 units. Hence, options (1), (2) and (5) are correct.

x x z z y y 1  + + + + +   y z y x x z  ≥  x × x × z × z × y × y 6   6 y z y x x z

Since,

X DF =  X  3 2    2

⇒

X 2

9

Space for rough work

10

009

Space for rough work

009

11