Cl Mock Cat 7 Key 2008

Mock CAT – 7

Answers and Explanations 1

3

2

5

3

3

4

1

5

2

6

1

7

2

8

4

9

1

10

5

11

3

12

5

13

4

14

3

15

2

16

2

17

3

18

1

19

5

20

1

21

2

22

2

23

2

24

1

25

3

26

4

27

2

28

3

29

1

30

5

31

3

32

1

33

3

34

5

35

4

36

4

37

2

38

3

39

5

40

1

41

4

42

2

43

5

44

3

45

1

46

3

47

1

48

5

49

2

50

4

51

1

52

4

53

5

54

5

55

1

56

5

57

4

58

4

59

1

60

3

61

3

62

5

63

2

64

3

65

1

66

4

67

5

68

5

69

2

70

4

71

3

72

1

73

5

74

2

75

2

MY PERFORMANCE Total Time Taken Total Correct Incorrect Net Questions (Min) Attempts Attempts Attempts Score Language Comprehension Section I and English Usage Logical Reasoning based Section II Data Interpretation Quantitative Ability TOTAL

Section III

25 25 25 75

150

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MCT-0011/08 007

1

1. 3

2. 5

3. 3

4. 1

2

The main idea of the three panels is given in the third panel which says ‘I somehow wish I had been in Auschwitz. With my parents, so I could really know what they lived through. I guess it’s some kind of guilt about having had an easier life than they did. Option (3) completes the paragraph best with the speaker talking about reconstructing the reality of the Holocaust and feeling inadequate to do so because he was not present at the Holocaust with his parents. The first panel also talks about the speaker having nightmares about SS men coming into the class and dragging Jewish kids away and this would tie in with the ‘darkest dreams’ mentioned in option (3). (1) is not correct because (1) would seem to imply that the speaker is somehow guilty about the fact that Richieu, rather than he, was a victim of the Holocaust. But in the first panel the speaker clearly says ‘I never felt guilty about Richieu.’ (2) is not correct because it would seem to imply that the speaker’s experience of Auschwitz was from the perspective of a distant observer reading about it in a history book. This is not true since the speaker’s parents were themselves in Auschwitz. (4) and (5) are statements that talk about the problem representing the reality through comics and should logically follow after (3). The paragraph begins with the mouse wearing glasses talking about the victims of the Holocaust being blamed because they are no longer living. He goes on to talk about the fact that even though many books have been written about the Holocaust, this has not changed people (and presumably their attitudes towards the victims of the Holocaust). Then he goes on to say that since the victims who died can never tell their side of the story, ‘so maybe it’s better not to have any more stories’. Option (5) ties in best with this last statement because it seems to imply that words or stories about the Holocaust can never really give voice to the victims who cannot speak for themselves and hence become ‘like an unnecessary stain on silence and nothingness’. Option (1) is close but on closer inspection you will see that it actually fits in better with the second panel, which talks about how the different books written about the Holocaust have failed to change people. Sentence (2) reads like a statement about a book written by the listener. Therefore it should ideally come before the first and second panels and cannot in any case complete the paragraph. (3) also refers to the statement made in the second panel about the ‘newer, bigger Holocaust’ and does not therefore tie in with the rest of the passage. Sentence (4) about ‘life equals winning so death equals losing’ is also a statement that fits in best with the first panel ‘Yes, life always takes the side of life”. In C, the second speaker says "Now that you are becoming successful…" and in E the first speaker says, "No matter, what I accomplish…'carrying on with the idea of being successful mentioned in the previous panel. The two therefore form a mandatory pair. Choices then get reduced to options 3, 4 and 5. A & C can be quickly identified as a mandatory pair because A ends with the second speaker saying '…its hard for me to remember' and C begins with 'Mainly I remember…' The choices are then reduced to options 3 and 4 where this mandatory pair exists. B and A also form a mandatory pair because in B the discussion is about exposing the second speaker's father to ridicule and in A the first speaker says "even so every boy when he's little, looks up to his father", thus continuing with the same theme. D & A form a mandatory pair because in D the speaker says “I saved only so I can have a little something for my old age” and in A he continues with “So, now I have my old age…”. A & B form a mandatory pair because in A the speaker says,

“…and look what I have,” and in B he goes on to describe the things he has (tank with oxygen, weak heart and diabetes). C and E form a mandatory pair because in C, the speaker urges the other person to come and stay with him and his wife and when the other person refuses to do so in E he goes on to say “So how have I to live Artie…” and goes on to speak about not being able to live in a retiring home. 5. 2

Refer to the last sentence of the last para: “Maus rewrites the cultural norm. and invents a new discursive space to address the questions and…the transmission of these conflicts from one generation to the next.” This is an interpretive role since it gives a new cultural interpretation to the Holocaust. 1 is not correct because there is no mention of ‘release’ of feelings happening through the comic book medium. 3 is incorrect because there is no mention in the passage that Maus borrows upon the literary genre of Melancholia. 4 is incorrect because while Maus does subvert the traditional role of the comic book as being representative of popular culture, that is not the only or the principal role that it performs in the context of depicting the Holocaust. 5 is also incorrect because it refers to a particular style that Maus uses and does not indicate the principal role that it plays.

6. 1

Refer to the second paragraph: “Both within these genres and modes there are both high and low forms…Spiegelman by utilizing the comic book as the textual medium of a story of the Holocaust succeeds in breaking the ‘taboo’ or ‘ritualized fixity’of confronting the Holocaust. Also refer to the third paragraph where it is mentioned that ‘The reduction of the players to cats, mice…pointing up…the reduction and simplification present in many “responses” to the Holocaust as well. In this way, Spiegelman literalizes the call for petits recits so prevalent in postmodern discourse today.’ So it can be inferred that the term petits recits would refer to a small, simplified narrative that would form part of many theoretical standpoints about an issue. 2 is not correct because as per the passage a petits recits does not use a grand, unified approach to legitimize a particular version of the truth. 3 talks about rejecting definitions that claim to have discovered absolute ‘truths’ which is not mentioned anywhere in the passage. 4 talks about ‘a small piece of poetry’ which cannot be inferred either directly or indirectly from a reading of the passage. 5 talks about the narrative style that is specifically used by Maus and there is no reference in the passage that the same necessarily applies to petits recits as a genre.

7. 2

Refer to the second paragraph. ‘There would at first sight seem to be an inalterable cultural hierarchy of forms, media, genres…’ that form part of “appropriate” forms of literary representation’. ‘My view is that Spiegelman, precisely by utilizing the “comic-book” as the textual medium of a story of the Holocaust, succeeds in breaking the “taboo” or “ritualized fixity” of confronting the Holocaust.’ This would imply that Maus does not form part of the cultural hierarchy of forms, media and genre that form appropriate forms of literary representation of the Holocaust.

8. 4

The Holocaust. Refer to the second para ‘The philosopher Berel Lang has argued in his book Act and Idea of the Nazi Genocide that there are only certain appropriate and ethically responsible ways of representing the Shoah. In this respect, many critics have said that the Holocaust requires an “elevated” genre, that it is the stuff of “high” literature and should not be “desecrated” by allowing low genres to communicate the destruction of the European Jews.’ The Hebrew word ‘Shoah’ then obviously refers to the Holocaust.

007

9. 1

10. 5

The main statement talks about the Ansoff’s shift in views from rigidity to flexibility in the context of planning. ‘1’ will follow the idea logically because it talks about giving up the search for ‘prescriptions’ and adopting a flexible approach. ‘2’ is exactly opposite since it talks about ‘Rigid controls’. And ‘3’ is completely irrelevant in the given context because it lays stress on ‘cost’. ‘A’ is irrelevant because the main statement talks about the ‘underlying reason’ that is ‘integration’, besides the immediate problem related to ‘wiring up the huge aircraft, brought on by the use of incompatible software’. Also the ‘wiring problem’ was in the past and was resolved. ‘2’ can logically follow the main statement because it reflects towards the state of affairs pertaining o the lack of integration. ‘3’ talks about ‘gaining traction with the customers’ which means ‘getting toehold or grip’. This statement is definitely irrelevant to the main idea because the latter refers to the lack of integration within the company.

11. 3

‘A’ will logically follow the main statement because the main statement says that the whole energy sector is riddled with subsidies therefore if existing forms of renewable energy is subsidized there is nothing wrong; it just levels the playing field. ‘2’ will also follow logically because it talks about all the areas in which energy is cheap. ‘3’ is a parallel argument because the context is not of energy but of food. And this also talks about the situation below the surface.

12. 5

According to para 2 “Something-In-It” visibly take on flesh, fill out the hollows of uncertainty,” which in the context of the passage implies that a sham is slowly converted into an apparently convincing stance by the defenders with the help of arguments. Thus option 5 is the appropriate one. Options 1 and 2 are incorrect because the former talks about magic and the latter takes the argument into an extreme which makes it irrelevant. Options 3 is not specific to sham. Option 4 is completely irrelevant.

13. 4

14. 3

15. 2

007

Option 1 is absolutely nonsensical in the given context. Option 2 is factually incorrect. Option 3 is incorrect. Option 4 gives the exact answer to the question. Option 5 gives the credit to astrology for ‘enlivening the mood of the participants’ which is a far-fetched opinion. Option 1 is incorrect because it negates the view of the author, who believes that ‘the something’ is an absolute sham however hard one tries to justify it. Option 2 talks about the conversion of sham into truth which is again incorrect according to the author; a sham will remain a sham and any amount of justification cannot convert it into a truth. Option 4 is incorrect because the whole passage deals with the ‘sham’ and its defenders and since the author deeply believes that a sham will remain a sham, it can be logically concluded that as soon as the as the false justifications (it can be referred to as a veil of pretense) are proved false the reality will emerge. (‘sham’ -a spurious imitation; fraud or hoax). Option 5 is incorrect because the author would not agree with the statement that ‘the something’ is a reality. Option 3 can be inferred from the passage. 1 is incorrect because it says that ‘illogical argumentation’ drives the half hearted fellows into defending the position taken; rather these fellows use the illogical argumentation as a tool to defend…2 is correct because it brings out the theme of the passage through which author conveys that any sham (something which may be half- truth or misconception or widely believed but not proved). 3 is incorrect because it

does not represent the complete theme of the passage. 4 is just a fact expressed in the passage because it does not highlight the real concern of the author. 5 is factually incorrect because it talks about ‘elixir’ whereas author refers to the ‘drug of incredulity’ which has a negative connotation. 16. 2

The passage takes a relook or a fresh look at Art Noveau-not just Art Noveau exhibitions. Hence option 2 is best.

17. 3

See para 7 where the writer castigates the book for its undecipherable technical jargon.

18. 1

Art Nouveau is not a deprecation of art but rather needs to be reassessed. Hence option 1 is best.

19. 5

Refer to the last para where option 5 is clearly mentioned.

20. 1

DBAEC Sentences A, B and E cannot be the obvious starters. D is the opening sentence of the paragraph where accommodation theory is defined. The adaptations mentioned in D are further exemplified in B. A further discusses the topic and cites an example when accommodation tends to occur. E further discusses a situation when accommodation has been observed and C complete the discussion

21. 2

EADCB E is a general stand alone statement that introduces Edmund Husserl’s phenomenology and attempts to define intentionality. A discusses the first and second part of his work, D and C talk about his other work. B concludes by talking about the consequences of his work.

22. 2

Helotry means slavery and is opposed to sovereignty which has connotations of political freedom. Incredulous is the opposite of credulous which is synonymous to naïve, susceptible. Banausic and Mechanical are synonyms. Mordacious means truculent. Words in option 4 have no specific relationship. A shrew is a woman.

23. 2

The question has a relation: The word stationary is given and the meaning furnished is that of its homonym stationery. Similarly, in Option 2, the meaning furnished is of the homonym of complaisant - complacent. In options 1 and 3, the correct meanings of the given words are furnished. Affect means ‘to have an influence on’ and effect means to ‘bring about or create’.

24. 1

Hobson’s choice means a choice that appears to be a free choice but is really no choice as there is no genuine alternative. In this case the Judge is considering which one of two courses to choose. Therefore the right idiom to use in this context would have been ‘Justice Robinson was on the horns of a dilemma in the last case…”. The other idioms are all appropriately placed.

25. 3

When people say that you can’t squeeze blood out of a turnip, it means that you cannot get something from a person, especially money, that they don’t have. The same cannot be inferred about Dover Foods, Inc. which Robert takes over. The other idioms are all placed appropriately.

3

For questions 26 to 30: Let the distance covered by A on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday be denoted by MON, TUE, WED, THU, FRI, SAT and SUN respectively. As per the additional information 1: MON > WED and MON > TUE As per the additional information 2: THU < SUN and THU < SAT As per the additional information 3: THU + TUE < MON + WED As per the additional information 4: WED + SUN < TUE + SAT As per the additional information 5: THU > MON

29. 1

Case 1: SAT > FRI > SUN > THU > MON > WED > TUE Case 2: SAT > SUN > FRI > THU > MON > WED > TUE Therefore, on four days namely Thursday, Monday, Wednesday and Tuesday, B definitely covered less distance than what he/she covered on Friday. 30. 5

Consider additional information 3 and 5: Since THU > MON and THU + TUE < MON + WED, therefore we can conclude that WED > TUE.

If the distance covered by A on Friday is 33 kms, then the distance covered by A on Saturday, Sunday, Thursday, Monday, Wednesday and Tuesday is 43, 38, 27, 25, 21 and 18 kms respectively.

Case 2: SAT > SUN > THU > MON > WED > FRI >TUE In this case the difference between the distance covered by C on Monday and Friday is 19 – 14 = 5 kms. Hence, the required difference can be 6 kms or 5 kms. Hence, option (5) is the correct choice. For questions 31 to 35: The following table provides information about the number of units of work completed on each of the twelve given days.

Therefore, the difference between the distance covered by A on Saturday and Monday is 43 – 25 = 18 kms. 27. 2

No information is given about the distance covered by any of these persons on Friday. If the distance covered by a person on Friday is not the least among all the days, then we can definitely say that the distance covered by the person will be least on Tuesday. For each of the persons we will assume that the distance covered by them is least on Friday. If the above assumption is true, the data given in the following table must also hold true.

A

B

C

D

Saturday

43

24

28

36

Sunday

38

18

23

32

Thursday

33

17

21

31

Monday

27

10

19

30

Wednesday

25

9

17

26

Tuesday

21

7

14

25

Friday

18

5

13

22

On checking the consistency of the data in the table with the additional information given in the question, the above data is possible for only C. Therefore, for three persons namely A, B and D, the distance covered by each one of them is definitely the least on Tuesday. 28. 3

4

If we consider the distances covered by C on the seven days of the week, the following cases are possible. Case 1: SAT > SUN > THU > MON > WED > TUE > FRI In this case the difference between the distance covered by C on Monday and Friday is 19 – 13 = 6 kms.

Consider additional information 4: Since we have derived that WED > TUE and WED + SUN < TUE + SAT, therefore SAT > SUN Therefore, SAT > SUN > THU > MON > WED > TUE The above holds true for each of the mentioned six persons. 26. 4

If we consider the distances covered by B on the seven days of the week, the following cases are possible.

On the

Rohan

Deepak

First day

10

12

Tripti Sonal Tarun Total 5

20

18

65

Second day

12

20

10

5

2

49

Third day

2

18

10

15

24

69

Fourth day

16

2

25

10

4

57

Fifth day

20

32

30

15

2

99

Sixth day

8

36

16

15

45

120

Seventh day

72

20

12

20

5

129

Eighth day

8

16

32

25

12

93

Ninth day

28

16

0

15

3

62

Tenth day

4

8

0

10

5

27

Eleventh day

4

0

0

0

40

44

Twelfth day

16

0

0

0

80

96

31. 3

On the sixth day maximum number of units of work, i.e. 129 was completed by all the five friends.

32. 1

Average units of work completed per day by all the mentioned

910 = 75.83 12 On Day 3, the total units of work completed is 69. Hence, out of the options given Day 3 is the only day on in which the total units of work completed is less than the average units of work completed per day by all the mentioned persons. Hence, option (1) is the correct choice. persons is

If we consider the distances covered by D on the seven days of the week, the only possibility is that the distance covered by D on Friday is less than the distance covered by him/her on Monday but more than the distance covered by him/her on Wednesday. Therefore, SAT > SUN > THU > MON > FRI > WED > TUE Therefore, the distance covered by D on Friday is 26 kms.

007

For questions 33 and 34: By comparing the number of units completed by each of the five persons on each of the twelve days we get the following conclusion. Rohan is more efficient than both Tripti and Sonal. Deepak is more efficient than both Rohan and Tripti Tripti is more efficient than none of the mentioned persons Sonal is more efficient than Tarun Tarun is more efficient than both Tripti and Rohan. 33. 3

Out of the mentioned persons Tarun is more efficient than Rohan.

34. 5

There are three persons namely Rohan, Deepak and Tarun who are more efficient than exactly two persons.

The only possibility is that the number of black shirts of brands Lacoste and Dockers gifted to Larry by his mentioned friends is 35 and 31 respectively. Therefore, the number of black shirts of brand Dockers gifted to Larry by Simar is 13. 38. 3

The following table lists down the range of the total number of white shirts of each brand gifted to Larry by his mentioned friends.

Number of white shirts

35. 4

36.4

37. 2

On seven days namely the second, third, fourth, fifth, sixth, seventh and ninth days the number of units of work completed by Sonal is more than at least one and at most two out of the mentioned persons. Minimum possible value of C = 7 + 3 = 10 Maximum possible value of C = 14 + 8 = 22 Minimum possible value of L = 7 + 10 = 17 Maximum possible value of L = 18 + 18 = 36 It is given that L > 3 × C, therefore the possible values of C are 10 and 11. Hence, option (4) is the correct choice.

Diesel

Lacoste

Dockers

Ravneet

3-5

3-8

13 - 18

3-9

Heena

3-8

3-5

3 - 12

3-4

Sarah

3-8

3-6

15 - 18

3-4

Total

9 - 21

9 - 19

31 - 48

9 - 17

Since, the maximum possible number of white shirts of brand Caterpillar gifted to Larry is less than the minimum possible number of white shirts of brand Lacoste gifted to Larry, therefore the number of white shirts of brand Caterpillar gifted to Larry is definitely less than the number of white shirts of brand Lacoste gifted to Larry. 39. 5

The following table lists down the range of the total number of black shirts of brands Lacoste and Dockers gifted to Larry by his mentioned friends.

Number of black shirts Lacoste

Caterpillar

The following table lists down the range of the number of white and black shirts of each brand gifted by Anjana to Larry

Caterpillar Black 10 - 25

Diesel

3

18 - 25 9 - 18 7 - 10

Dockers

Anjana

7 - 10

7 - 22

Urvashi

7 - 18

11 - 25

Simar

7

13 - 25

Total

21 - 35

31 - 72

Lacoste

Dockers

White Black White Black White Black 3-6

White

7 - 22 11 - 18

68 - 127 Given that the number of shirts bought by Larry is same as the total number of shirts gifted to him by Anjana. Therefore, at least 750 – 2 × 127 = 496 shirts are there with Larry that are neither bought by him nor gifted to him by Anjana.

Given that out of the black shirts gifted to Larry, the number of shirts of brand Lacoste is 4 more than the number of shirts of brand Dockers. 40. 1

The following table lists down the range of the total number of shirts of each brand gifted to Larry by his mentioned friends.

Number of Shirts Caterpillar

Diesel

Lacoste

Dockers

Black

White

Black

White

Black

White

Black

Urvashi

7 - 20

11 - 18

15 - 25

3-6

7 - 18

10 - 18

11 - 25

3 - 11

Simar

10 - 25

3

7 - 18

15 - 18

7

3-4

13 - 25

13 - 18

Total

17 - 45 14 - 21 31 - 56

22 - 43 18 - 24 40 - 67

14 - 25 13 - 22 27 - 47

White

24 - 50 16 - 29 40 - 79

Given that the total number of shirts of each of the mentioned brands gifted to Larry by Urvashi and Simar is the same, which means that the answer should be a multiple of 4. We can conclude from the table that the range of the total number of shirts of each brand gifted to Larry is 40 – 47. So, the range of the total number of shirts gifted to Larry will be (40 × 4 – 47 × 4) = (160 – 188) Only, option (1) lies within the permissible range. Hence, option (1) is the correct choice.

007

5

For questions 41 to 45: The number of times each of the given bookies placed a bet on each of the given six players is provided in the table given below. A

B

C

D

E

F

Kar im

4

3

1

1

3

0

Jos e ph

1

4

2

1

1

3

Har vinde r

2

2

2

2

2

2

Suk hde v

2

2

4

3

0

1

Tony

1

2

1

2

2

4

Car los

4

2

2

1

1

2

Total number of centuries scored in the cricket series = 18. Since, each of the given players has scored a century thrice in the cricket series it is obvious that Karim is not the bookie such that the players on whom he placed his bet in every match always scored a century. For the same reason Joseph, Sukhdev, Tony and Carlos cannot be the bookie such that the players on whom he placed his bet in every match always scored a century. Hence, Harvinder is the bookie such that the players on whom he placed his bet in every match always scored a century 41. 4

Since, we know Harvinder is the bookie such that the players on whom he placed his bet in every match always scored a century; we can uniquely determine the two players who scored a century each in each of the given six matches. As we are given that for every player who scored a century in a given match, there is at least one bookie who had placed a bet on him in that particular match. Therefore, the list of all the possible players who could have been the third player to have scored a century in each of the given six matches is as follows: Match 1: A / C / D / E Match 2: A / B / F / E Match 3: B / F Match 4: A / C Match 5: A / B / D / F Match 6: F / E / C

42. 2

Required Answer is Match 3.

43. 5

Given that A did not score a century in Match 4 and E did not score a century in Match 6, which means the third player to score a century in Match 4 and Match 6 is C and F respectively. Therefore, the third player to score a century in Match 3 has to be B or E. Hence, (A, D and B) or (A, D and E) scored a century in Match 3. Hence, option (5) is the correct choice.

For questions 46 to 50: Given that the price tags on all the packs labeled as “Diamond Pack” and “Pearl Pack”, summed to Rs.640000 and Rs.450000 respectively. Therefore, out of the 25 packs, the number of packs labeled as “Diamond

640000 =8 80000 Similarly, out of the 25 packs, the number of packs labeled as “Pearl

Pack” is

450000 =9 50000 Therefore, out of the 25 packs, the number of packs labeled as “Mixed Pack” is 25 – 9 – 8 = 8. Total number of rings with Mr. Gold Smith 2 × 25(In Packs) + 25(Single) = 75 Pack” is

From additional information II: Total number of diamond rings with Mr. Gold Smith = 15 × 2 = 30 Total number of pearl rings with Mr. Gold Smith = 75 – 30 = 45. Also, the number of diamond rings for gentlemen and ladies’ is 15 each. From additional information I: Given that out of the 25 rings that remained single, the number of gentlemen’s diamond rings is 5. Therefore, in the 25 packs made by Mr. Gold Smith there are 10 gentlemen’s diamond rings. 46. 3

Total number of diamond rings in the 25 packs made by Mr. Gold Smith 8 × 2(Diamond Packs) + 8 × 1(Mixed Packs) = 24. Number of ladies’ diamond rings that remained single = 30 – 5 – 24 = 1.

47. 1

Out of the 25 packs made by Mr. Gold Smith, the number of packs labeled as “Mixed Pack” = 8.

48. 5

We have already calculated that in the 25 packs made by Mr. Gold Smith there are 10 gentlemen’s diamond rings Therefore, in the 25 packs made by Mr. Gold Smith the number of gentlemen’s pearl rings is 25 – 10 = 15. Out of the 25 packs, we know that 9 are labeled as “Pearl Pack” and 8 are labeled as “Mixed Pack”. Therefore, we can say that out of the 15 gentlemen’s pearl rings in the 25 packs made by Mr. Gold Smith, 9 are in packs labeled as “Pearl Pack” and 6 are in packs labeled as “Mixed Pack”. Therefore, 6 gentlemen’s pearl rings were packed with ladies’ pearl rings.

49. 2 For questions 44 and 45: Comparing the players on whom Carlos placed his bet with the players on whom Harvinder placed his bet, we can conclude that there are a maximum of three matches such that in each of these three matches, the player on whom Carlos placed his bet always scored a century. These three matches will be Match 3, Match 4 and Match 6. 44. 3

The maximum possible value of “X” is 3.

45. 1

Given that the value of “X” = 1 and B scored a century in Match 3, therefore Match 3 is the match such that the player on whom Carlos placed his bet in Match 3 scored a century. Therefore, A cannot be the player who scored a century in Match 4 as then there will be two such matches such that the player on whom Carlos placed his bet always scored a century.

6

50. 4

Total number of pearl rings in the 25 rings that remained single = 25 – 5 (gentlemen’s diamond rings) – 1 (ladies’ diamond rings) = 19. Assuming these 19 rings are all gentlemen’s pearl rings. Number of gentlemen’s pearl rings in the 25 packs made by Mr. Gold Smith = 15 Therefore, maximum possible number of gentlemen’s pearl rings = 19 + 15 = 34.  30  Required percentage =   × 100 = 40%  45 + 30 

007

54. 5

51. 1 B

P

C

Let the original cost price of milk per litre = Rs.y Total cost price of the solution = Rs.((26 – x) × y) Total selling price of the solution = Rs.(26 × 1.1 × y) Profit percentage =

⇒ x = 6 litres Total cost price of the solution when 6 litres of water is added to 30 litres of pure milk = Rs.(30 × y) Total selling price of the solution when 6 litres of water is added to 30 litres of pure milk = Rs.(36 × y)

M

R

D

A

Profit percentage =  

Q

Since, the radii of the circular arcs are 5 units, therefore AD = 5 + 5 = 10 units. Drop a perpendicular from point B on AD meeting AD at the point R.

∴ AR = 5 – ⇒ BR =

2

6 = 5 – 3 = 2 units 2

The HCF of the two given numbers is 1011. Hence, all the factors of 1011 would be common to both the numbers 3011 and 2013. 1011 = 211 × 511 ⇒ Total number of factors = 12 × 12 = 144

56. 5

Using statement A: The possible sets of values of ‘x’ and ‘y’ such that the product of the unit’s digit of ‘x’ and ‘y’ is 4 in that order are (6, 4); (2, 2); (8, 3); (7, 2); (4, 1); (9, 6) and (8, 8). Hence, statement A alone is not sufficient to answer the question.

2

6 − 2 = 4 2 units

1   1 [6 + 10] × 4 2 + 10 × 5 − 2 π × 52  2  

Using statement B: No definite information is given above the value of ‘x’. Hence, statement B alone is not sufficient to answer the question.

= 32 2 + 50 − 12.5π sq. units. 52. 4

53. 5

It is given that x3 – ax2 + bx – c = 22 Given that T1 + T2 + T3 = 25 ⇒ a + b + c + 22 = 25 ⇒ a+b+c=3 The number of such equations will be equal to the number of non negative integral solutions of the equation a + b + c = 3 Required number of solutions = 3+3–1C3–1 = 10 EA2 = 82 = 64 and ED × (ED + EC) = 4 × 16 = 64 ∴ EA2 = ED × (ED + EC) Since, DB = BC, therefore keeping B as the center, a circle can be drawn that passes through the points C, D and A. C

D

B

A

57. 4

Using statement A: It is given that length of two of the sides of the triangle is 10 m each. The length of the third side may or may not be equal to 10 m. This means the triangle can either be equilateral or isosceles which means that there will be different values for the expression (N2 + 2A) as the values of ‘A’ and ‘N’ would be different in all such possible triangles. Hence, statement A alone is not sufficient to answer the question.

As height = 5 3 m

∴ BD = BC = AB = Radius of the circle

1 × base × 5 3 2 ⇒ Base = 10 m Hence, statement B alone is not sufficient to answer the question as we do not have the data for the other two sides of the triangle. 25 3 =

Now, in ∆ECA EA2 + AC2 = EC2 ⇒ AC2 = 192

⇒ AC = 8 3 units AC 8 3 = = 4 3 units 2 2

Perimeter of the quadrilateral ∆BDE = 12 + 8 3

(

)

= 4 2 3 + 3 units

007

Combining both the statements together: Possible sets of values of ‘x’ and ‘y’ such that the remainder when the unit’s digit of ‘x’ is divided by the unit’s digit of ‘y’ is 2 in that order are (6, 4); (8, 3) and (2, 7). So, even after combining both the statements together we are not able to answer the question. Hence, option (5) is the correct choice.

Using statement B: It gives information about the area and the height of the triangle,

E

∴ AB =

36 – 30  × 100 = 20% 30 

55. 1

Area of the figure ABCDQ =

(26 × 1.1 × y ) – ((26 – x ) × y ) 43 = 100 ((26 – x ) × y )

Combining both the statements together: One becomes sure that triangle is an equilateral. In this triangle keeping every side as base a square can be inscribe which gives N = 3 and A = 25. So the value of expression N2 + 2A = 59. Hence, option (4) is the correct choice.

7

58. 4

Since, 83 is a prime number, therefore the age of one out of the 3 persons A, C and E has to be equal to 83 years.

62. 5

Using statement A: Given that Age of E > Age of C. Case I: E = 83, C = 4, B = 2, (A, D) = (2, 4) not necessarily in this order. Case II: A = 83, E = 3, C = 2, B = 1 and A = 6. Similarly, there will be more cases possible. Hence, statement A alone is not sufficient to answer the question.

(0 , 6 )

59. 1

5 C

A

(0 , 0 )

If A (0, 0); B (6, 0) and C (0, 8)….triangle ABC is a right triangle. Hence the circle passing through ABC has a diameter of 10 cm. There two possible circle which will touch the given 3 unit circles and their radii will be Case I: (5 – 1) = 4 cm(circle touching unit circles externally) Case II: (5 + 1) = 6 cm(circle touching unit circles internally) Hence, a unique radius cannot be determined. 63. 2

One of the possible combinations out of the choices.

So, dimension of the rectangle is (4 × 4) and the minimum numbers of blocks required is 4.

E

64. 3

G Q

P

F

Using statement A: As the number of steps is odd, Sanjay could not reach the red light, which is only 2 steps (even number of steps) away from Sanjay. Sanjay should have reached the blue light. Hence, statement A alone is sufficient to answer the question.

H

D

A

I

C B

Using statement B: As the net movement is three steps to the left side, he will reach the blue light. Hence, statement B alone is sufficient to answer the question. Hence, option (3) is the correct choice.

J

∠PDQ = 360° – ∠EDC − ∠GDC = 120° Now, since DP = DQ, therefore DP = DQ = 1 cm. ∴ ∠DPQ = ∠DQP = 30°

Therefore, PQ = 2 × PD × cos 30° = 61. 3

It can be easily concluded that ABCD is a rhombus. Now, if AO + BO = 20 units, then AC + BD = 40 units.

1 × AC × BD 2 Area of ABCD will be maximum when AC = BD.

We know that area of rhombus =

1 × 20 × 20 2 = 200 square units.

Hence, the maximum possible area =

8

x (8 , 0 )

Using statement A: aaabbb = aaa × 1000 + bbb, where ‘aaa’ and ‘bbb’ are three-digit numbers. The number can also be written as 37 × 3 [a × 1000 + b] = 37 × 3 [a00b] If a00b is a prime number, then number of factors of aaabbb = 2 × 2 × 2 = 8. Hence, statement A alone is sufficient to answer the question. Using statement B: No unique values of ‘a’ and ‘b’ can be found out Hence, statement B alone is not sufficient to answer the question. Hence, option (1) is the correct choice.

60. 3

O

4

6

Using statement B: Given that the age of 4 friends is the same. Case I: A = 83, B = C = D = E = 3 Case II: C = 83, B = A = D = E = 3 Case III: E = 83, B = C = D = A = 3 Hence, statement B alone is not sufficient to answer the question. Combining both the statement together: The only sets of values of A, B, C, D and E that satisfies both the statements is 3, 3, 3, 3 and 83 respectively. Therefore, the age of C is 3 years. Hence, option (4) is the correct choice.

B

65. 1

3 cm

Total distance travelled by the train to cross the tunnel = Half the length of tunnel + Length of the train

1 (200 ) + 150 = 250m 2 We know that, when time taken is equal, then speed is directly proportional to the distance travelled =

⇒

Speed of train 250 5 = = Speed of Schumi 200 4

007

⇒ =

Relative speed while moving in same direction Relative speed while travelling in opposite direction

Speed of train – Speed of Schumi Speed of train + Speed of Schumi

⇒

5–4 1 = = 5+4 9 ⇒

66. 4

Time taken while moving in same direction 9 = Time taken while moving in opposite direction 1

Initially total volume of mixture = 100 litres Water : Milk = 4 : 1 ⇒ Water = 80 litres and Milk = 20 litres Operation

Water(in litres)

Milk(in litres)

1

72

28

2

50.4

49.6

3

25.2

74.8

After 3 operations  Water   25.2  1  Milk  =  74.8  > 3    

252.a – a2  a2  = a −   252 252  

⇒ a2 should be divisible by 252 As 252 = 7 × 22 × 32 ⇒ a2 must contain 71 × 22 × 32 ⇒ a must contain 71 × 21 × 31 ⇒ Possible values a are a = 7 × 21 × 31 = 42 a = 7 × 22 × 31 = 84 a = 7 × 21 × 32 = 126 Sides of rectangle could be: (42, 210)(84, 168)(126, 126) ⇒ So, minimum possible area of the part of the rectangular sheet of

paper that each of the 252 persons get =

pn +1 + pn = 2n − 5

70. 4 71. 3

If graduates are denoted by G1, G2, G3, G4, G5, G6 and G7, then we have to make groups of 3 engineering graduates such that every pair of group has exactly one graduate common. (G1, G2, G3)(G1, G4, G5)(G2, G4, G6)(G3, G5, G6) (G2, G5, G7)(G1, G6, G7)(G4, G3, G7) There are seven such groups.

72. 1

Given that when f(x) is divided by (x – 1), (x – 2)……. and (x – 51), it leaves remainders 1, 2, 3, 4…. and 51 respectively. Therefore, F(x) = a (x – 1) (x – 2)……(x – 51) + x …(i) where a is any constant. Now putting x = 52 in the above equation we get, f (52) = a × 51 × 50 ×……..× 1 + 52 ⇒ f (52) = a × 51! + 52. ...(ii) Also, f(0) = a × (–1)51 × 51! + 0 Adding equations (i) and (ii), we get f (52) + f (0) = 52.

73. 5

Let Vikram and Ravinder has a total of Rs.x and Rs.y each. Now as per the statements in the question, we can write,

⇒ pn + 2 + pn +1 = 2 (n + 1) − 5 ⇒ pn + 2 = 2 (n + 1) − 5 − pn +1 = 2n − 3 − (2n − 5 − pn ) ∴ pn + 2 = pn + 2 Now, p(n + 2)+ 2 = pn + 2 + 2 = pn + 2 + 2

⇒ pn + 4 = pn + 4 ∴ pn + 2k = pn + 2k Here ‘k’ is a natural number.

∴ p1+ 2×1002 = p1 + 2 × 1002 = −5 + 2004 = 1999 68 . 5

Date 18th Nov 2007

Bank Balance 4000

19th Nov 2007

2000

20th Nov 2007

–2000

21st Nov 2007

–4000

22nd Nov 2007

–2000

23rd Nov 2007

2000

24th Nov 2007

4000

25th Nov 2007

2000

th

26 Nov 2007

–2000

Thus, we see that Bank Balance repeats after a cycle of 6 days The number of days from 18 th November 2007 to 16th November 2008 is = 365 days (As 2008 is a leap year) The remainder when 365 is divided by 6 is 5 Therefore, his bank balance on 16th November 2008 = –Rs.2000

007

42 × 210 = 35 cm2 252

69. 2

 1 ⇒ In the fourth operation required ratio will be less than   . 3

67. 5

For questions 69 and 70: If ‘a’ and ‘b’ are the length and breadth of the rectangular sheet of paper then 2a + 2b = 504 and a + b = 252. ⇒ Area of the rectangular sheet = a × b = a(252 – a) It is given that area is exactly divisible by 252

 1  1  4  x +  7  y = 68 ⇒ 7x + 4y = 68 × 28      1  1   x +   y = 75 ⇒ 4x + 7y = 75 × 28 7 4 Dividing (i) by (ii)

...(i)

...(ii)

x 16 = y 23 ⇒ y has to be multiple of 23 and 28 both hence no solution possible. Hence option (5) is the correct choice.

9

74. 2

There are only three possibilities: 75. 2

X

Y

Before Meeting Red

Green

After Meeting Case I

Blue

Red

Case II

Green

Blue

Case III

Green

Red

As the three cases are mutually exclusive, the required probability is given by: P = P(case I ) + P(case II ) + P(case III) 3  3  2  1 P=   ×   +  ×  + 5 4 5  4 Hence (2) is correct.

10

 2 3 7  5  ×  4  = 10    

P  x  First installment =   + (P) ×   4  100  Remaining principle amount after the first installment P 3P  = P –  = 4 4    P   3P   X  Second installment =   +   ×    4   4   100   P   2P   X  Similarly, Third installment =   +   ×    4   4   100  P P  X  Fourth installment =   +   ×    4   4   100  Total amount paid in the four installments  X   5P  =P+   ×   = Rs.300  100   2  Out of the choices only option (2) satisfies, and hence is the correct choice.

007