Answers And Explanations

Proctored - Mock CAT 8

Answers and Explanations 1

c

2

c

3

b

4

a

5

c

6

b

7

d

8

a

9

d

10

a

11

b

12

a

13

d

14

a

15

d

16

b

17

c

18

b

19

a

20

c

21

a

22

d

23

d

24

b

25

b

26

d

27

d

28

a

29

b

30

d

31

b

32

c

33

c

34

a

35

c

36

c

37

c

38

b

39

d

40

a

41

b

42

b

43

d

44

b

45

c

46

a

47

b

48

c

49

d

50

d

51

a

52

d

53

b

54

a

55

b

56

d

57

a

58

c

59

b

60

c

d

62

c

63

a

64

a

65

c

66

b

61

Page

1

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Proctored - Mock CAT 8

For questions 1 to 5: Let the maximum marks which can be obtained in the subjects P, Q, R and S be 40p, 50q, 60r and 80s respectively. From the given information we can conclude that:

For questions 8 and 9:

15p ≥ 20q, 20p ≥ 10q, 25p ≥ 15q, 10p ≥ 18q, 25p ≥ 32q and30p ≥ 12q ⇒ p ≥ 1.8q

...(i)

Ye ar Total am ount (in 000's Rs .) of claim s paid

1991

1992

12

7 + (18 – 12) = 13

Ye ar Total am ount (in 000's Rs .) of claim s paid

1993

15p ≥ 8r, 20p ≥ 16r, 25p ≥ 20r, 10p ≥ 24r, 25p ≥ 10r and30p ≥ 20r ⇒ p ≥ 2.4r

...(ii)

15p ≤ 25s, 20p ≤ 15s, 25p ≤ 20s, 10p ≤ 30s, 25p ≤ 10s and30p ≤ 35s ⇒ p ≤ 0.4s

...(iii)

1994 7 + (45 – 27) 2 + (27 – 18) + + (23 – 15) + (15 – 7) = 19 (5 – 2) = 36

Ye ar

1. c

⇒ 20q = 8r ⇒

q 2 = . r 5

Re quired percentage = 2. c

3. b

50q 50 2 100 × 100 = × × 100 = . 60r 60 5 3

The marks obtained by F in subject P is 30p. ⇒ p needs to be maximized. Marks obtained by C in subject S is 20s. Given that 20s = 20 or s = 1.

5. c

Given that X = 20p – 10q. For the minimum value of X, p should be the minimum and the minimum value of p is 1.8q (from (i)). ⇒ The minimum value of X is 20(1.8q) – 10q = 26q. So, the students A, B, C, D and F have definitely obtained lesser marks than 'X' in subject Q. Given that the value of 'X' is 91.

7 2 Maximum possible marks that can be obtained by A, B, C, D, E and F in subject Q are 70, 35, 52.5, 63, 112 and 42 respectively. So, none of B, C, D and F can obtain more marks than 65 in subject Q. ⇒ 91 ≥ 26q ⇒ q ≤

For questions 6 to 9: 6. b

7. d

Page

The total amount (in Rs.) of claims paid in the year 1997 with respect to the accidents that occurred in the year 1992 was 43000 – 39000 = Rs. 4000. The total amount (in 000' Rs.) of the claims paid during the year 1998 is 39 + (17 – 11) + (43 – 37) + (33 – 29) + (33 – 31) + (17 – 12) + (45 – 43) + (87 – 71) = 80.

2

1995

Ye ar

1997 11 + (71 – 67) + Total am ount (in (43 – 39) + (12 000's Rs .) of claim s – 9) + (31 – 25) paid + (29 – 19) + (37 – 21) = 54

In subject P, the students who have obtained not less than 40% of the maximum marks that can be obtained are B, C, E and F. There is no other subject in which there are atleast 3 students who have passed.

From (iii), p ≤ 0.4s or p ≤ 0.4 (as s = 1) ∴ The maximum value of p can be 0.4. ⇒ The required answer is 30 × 0.4 = 12. 4. a

1996 21 + (67 – 15 + (53 – 45) + 53) + (39 – Total am ount (in (29 – 23) + (7 – 29) + (9 – 7) 000's Rs .) of claim s + (25 – 17) + 5) + (17 – 7) paid = 41 (19 – 15) = 59

The marks obtained by A in subjects Q and R are 20q and 8r respectively.

1998

80

8. a

The total amount (in Rs. ) of claims paid during the year 1999 = 80000 + (80000 – 54000) = 106000.

9. d

In the year 1996, the total amount of claims paid was more than that in the year in 1997.

For questions 10 to 13: From additional information (I), we can conclude that the type of pen which was not purchased by A and C could be (Q, P), (R, Q), (R, S), (S, T), (S, P) and (Q, T) in the same order. Similarly, from additional information (II), we can conclude that the type of pen which was not purchased by B and D could be (Q, P), (R, Q), (R, S), (S, T), (S, P) and (Q, T) in the same order. The following table lists down the type of pen which was not purchased by each of the people. Cases People

I

II

III IV

V

VI VII VIII IX

X

A

Q

Q

R

R

R

S

S

S

Q

Q

XI XII S

R

B

R

S

S

S

Q

Q

R

R

R

S

Q

Q

C

P

P

Q

Q

S

T

T

P

T

T

P

S

D

S

T

T

P

P

P

Q

Q

S

P

T

T

E

T

R

P

T

T

R

P

T

P

R

R

P

10. a

It could be either Q or S.

11. b

The four types of pens purchased by D could be (P, Q, R and T); (P, Q, R and S); (P, R, S and T) and (Q, R, S and T). Therefore, the amount (in Rs.) spent by D on purchasing the four pens could be 110 or 120.

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12. a

13. d

14. a

15. d

Given that the total amount (in Rs.) spent by C in purchasing the four pens was 110. It means that the four pens purchased by C could be (P, R, S and T) or (P, Q, R and T). In the table, it corresponds to the cases III, IV, V or XII. In all the four cases, E and D are the people who have not purchased the pens of types P or T and hence, have spent the maximum amount (in Rs.) in purchasing the four pens.

As AB and CD are two-digit numbers, their sum EFG cannot be more than 198. So E has to be equal to 1. So, conclusion III can be derived. Given that AB is a prime number, therefore B has to be equal to 3 as E = 1 and no two-digit prime number can end with 2, 4, 5 or 6. Therefore, A could be 2 or 4 or 5. Since, the maximum value that C can take is 6, therefore A cannot be equal to 2 as EFG is a three-digit number. Also, AB cannot be equal to 53, because then C has to be equal to 5 (since, F = 0) and this is not possible. So, AB = 43, CD = 62 and EFG = 105. Hence, conclusion II can be derived. As per the information given in the question, we can conclude

0 ⋅ 2a + 0 ⋅ 25b + 0 ⋅ 4c = 0.3 a+b+c

⇒ 0 ⋅ 1c = 0 ⋅ 1a + 0 ⋅ 05b ⇒ 2c = 2a + b b 2 So, the only conclusion that can be drawn is c > a. ⇒c =a+

17. c

The roots of the equation x2 – Ax + B = 0 are 1 and m. So, 1 + m = A and m = B. The roots of the equation x2 – Bx + C = 0 are 1 and n. So, 1 + n = B and n = C Hence, m = n + 1 If C = 2, then n = 2 and m = 3. Hence, conclusion (I) can be derived but conclusion (II) cannot be derived. If 1, ‘n’ and ‘m’ (in that order) form a G.P., then, n2 = m = n + 1. This is not possible for any interger ‘n’. Hence, conclusion (III) can also be derived. Clearly, ∠AED = 180° – ∠DEC = 125°. In quadrilateral ABDE, the sum of opposite angles

∠ABD and ∠AED = (55 + 125)° = 180°. So, the points A, B, D and E are concyclic. Let, ∠ACB = x° So, ∠EBD = x° as EB = EC. By the exterior angle property, ∠DAC = ∠DAE = (80 – x)° But, ∠EBD = ∠DAE (Points A, B, D and E are concyclic) So, x = 80° – x

Page

A

3

(80 – x)° E 55 °

Given that C spent the maximum amount (in Rs.) in purchaisng the four pens. It means that any of the eight cases I, II, VI, VII, VIII, IX, X and XI could be possible. Therefore, either B or E could not have purchased the pen of type R and hence could have spent the minimum amount (in Rs.) in purchasing the four pens.

that

16. b

Therefore, x = ∠EBD = 40° .

55 °

x°

x°

80 °

B

D

C

Hence, ∠DAC = ∠DCA and thus, AD = DC. Hence, conclusion (I) can be derived. Also, ∠ABE = ∠ADE = 55° – x = 15° as the points A, B, D and E are concyclic. Hence, conclusion (II) can be derived.

∠BED = ∠BAD = 45° as the points A, B, D and E are concyclic. Hence, conclusion (III) can be derived. For questions 18 to 22: From Statements 1 and 2: The number of players in rounds 1, 2 and 3 must be 6, 5 and 4 respectively. In round 1, the possible points received by the 6 players are 29, 23, 19, 17, 13 and 11. In rounds 2 and 3, the possible points received are 19, 17, 13, 11 and 7. Hence, the players with 11 and 7 points must have got eliminated in round 1 and round 2 respectively. None of A, D or F got eliminated in round 1, as the table gives some points for them in further rounds. Neither, it was C as he got 13 points in Round 1 Hence, either B or E got eliminated in round 1. From Statement 5: E got higher points than B in Round 1. So, B must be the one, who got eliminated in Round 1 with 11 points. Either A or D received the highest points in Round 1 as A received more points than E. From Statement 4: The player who received the highest points in round 1, got eliminated in round 3. D did not get eliminated in round 3 as his/her points were more than that of F. So, A got the highest points of 29 in round 1 and 7 points in round 3. (A’s points in round 3 must have been 7 because F had 11 points and he was not eliminated). Since, D received 17 points in round 3, E must have received 17 points in round 1 and only E could get an aggregate score of 24 i.e. (17 + 7) in the first two rounds. So, E got eliminated in round 2. The table at this stage looks like:

Maximum Points 30

Round

A

B

C

D

E

F

13

19

17

23

1

29

11

20

2

11

X

20

3

7

X

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7 17

X

11

Proctored - Mock CAT 8

conducts oneself. ‘Demeanor’ means behaviour toward others. ‘Convalescence’ means recovery of health and strength after illness. ‘Recuperation’ is its synonym. ‘Anathema’ refers to someone or something who is intensely disliked or loathed. ‘Bete noire’ refers to a person or thing strongly detested or avoided.

Since, C could not receive 13 points in round 3, he must have received 19 points in round 3. Since, C and D could not receive 19 points in round 2, F must have received 19 points in round 2. ⇒ D received 13 points and C received 17 points in round 2. The final table looks like:

Maximum Points 30

Round

A

B

C

D

E

F 23

1

29

11

13

19

17

20

2

11

X

17

13

7

19

20

3

7

X

19

17

X

11

Total

-

-

-

49

49

-

53

18. b

From the table it’s clear that B got the least points i.e. 11 in round 1 and hence, got eliminated.

19. a

From the table it’s clear that A received the highest points i.e. 29 in round 1.

20. c

From the table it’s clear that C received 17 points in round 2.

21. a

From the table it’s clear that F received the third highest points i.e. 11, in Round 3. C got the highest while D got the second highest.

22. d

F’s aggregate points of 53 was the highest and hence he was declared as the “Champion”.

23. d

In the passage, the author uses all the elements mentioned in the answer options. The author with reference to Freud’s analogy of the iceberg uses analogy. The tone of the author is conversational especially in the lines “You probably are beginning to sense that both the id and the superego make life rough for the ego. Remember that Freud considered personality like…” The author uses a question at the start of the fourth paragraph.

24. b

Option (a) does not answer why the id makes life rough for the ego. Option (c) is incorrect as the ego is partly subconscious and partly conscious like the iceberg partly submerged in water. Option (d) is distorted. It is anxiety that triggers the mechanisms of repression and not the reverse. Refer to the last line of the second last para “The anxiety alerts the ego to resolve the conflict by means of defense mechanisms.” Option (b) is correct. The id is made up of subconscious impulses that can be inferred as ‘wanting pleasure even if it causes harm’. The superego places moral restrictions on this. Thus the life of the ego becomes rough.

25. b

The ‘id’ represents the unconscious impulses which are often dangerous for a person and have to be restricted by the ego. The ego uses reason and interacts with reality and even distorts it. The superego is the moral department. Option (a) represents an unrestricted impulse which can cause harm to the person. Hence it would most probably belong to the ‘id’. Option (c) is a conclusion based on reality. Hence it would belong to the ego. Option (d) belongs to the moral departmentthe superego. But option (b) seems to have been generated by the conscience or the superego as it deals with right/ wrong.

26. d

Page

The relationship between the words in the given pair is that of antonyms. ‘Opulence’ means wealth or affluence. ‘Pauperism’ means poverty. Option (d) exhibits a similar relationship, ‘Clemency’ means disposition to be merciful. ‘Mercilessness’ is its opposite. ‘Deportment’ means the manner in which one

4

27. d

Statement 1 may seem correct but it has a spelling mistake. The noun ‘gallows’ should replace ‘gallow’. Statement 1 should read as, “Who, by and large, are the men whom the gallows swallow?” ‘Gallows’ is an instrument of execution consisting of a wooden frame from which a condemned person is executed by hanging. Its plural is gallows or gallowses. Statement 4 has two subject-verb-agreement errors. The correct sentence should read as ‘...-it is these people who are the morning meal of the macabre executioner’ as the sentence takes various people into consideration.

28. a

The second part of the sentence holds the key to solving this question. Unattributed works are investigated. So, anything that is unattributed will be investigated for its origin and source. Hence, we cannot use ‘authenticity’ and ‘truthfulness’ in the first blank. ‘Plausibility’ means the quality of appearing worthy of belief; it is not fit for the first blank. So, only option (a) can give the clear meaning of the sentence.

29. b

Option (a) is incorrect as the author is discussing the responsibility of writers like Benjamin but the author does not pronounce any judgement on him. Option (c) is again an inference beyond the scope of the passage as the author seems to have just broached the topic of ‘writers’ responsibility’ and he has not made any judgement on them. Option (d) cannot be inferred from the passage as Benjamin was aware that his writings had an effect, but we do not know whether he was fully aware of the effect of his writings on the populace. Option (b) can be inferred as the possibility is open that Benjamin’s writings could have influenced the course/ trajectory of the red army. In fact the author seems to believe that the writings and the trajectory of the red army could be connected.

30. d

All the statements given as answer options are untrue. In fact, towards the end of the passage the author mentions “But a difficult question remains. Does any kind of fuse or trail lead from his words to their deeds? If so, it would mark a striking instance of the general problem : how responsible is a thinker for the fate of his/ her ideas? Such questions were hardly foreign to Benjamin.” These lines clearly show that Benjamin, (an anarchist, according to the author) was not unaware of the effect of his writings. The passage nowhere states that the anarchists were aware of one another; it states that the anarchists worked on their ‘infernal machine’ without knowing of one another. (c) cannot be inferred from the passage, as the idea of ‘fighting against evil rulers’ has not been hinted at in the passage.

31. b

Option (a) is part of the purpose but it leaves out the influence of writers who give a spark to the dormant tensions. Option (c) is discussed to a very small extent in the fourth para. The passage is concerned about the responsibility of writers and not the futility of talking about them. This makes (d) incorrect. The primary theme of the passage can be broadly described as ‘ Discussing the fact that the inner tensions in a population are just waiting for a spark to blow up into a violent movement - and that the responsibility of writers who fuel this needs to be examined’. Hence option (b) emerges as an option that is closest to the primary theme of the passage.

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32. c

The given argument highlights that the request for the protest march satisfied all the mandatory requirements and the government showed apathy by not responding to the request. In option (c), it is mentioned that the landowner got the map prepared from a government-approved architect, it shows that he complied with the norms. No response from the government shows its apathy; this equates it with the reasoning given in the argument. Options (a) and (b) are inconsistent with the reasoning given in the argument as they do not highlight any instance of compliance or apathy. Option (d) is incorrect because the criterion for compliance is not fulfilled- the student knowing his subject well does not imply that he will pass the examination.

33. c

Option (a) is erroneous. The adjective ‘at least’ has been wrongly written as ‘atleast’. Also there is no word such as ‘markmanship’. It is incorrect and it should be ‘marksmanship’ which means skill in shooting. Moreover, the preposition ‘of’ has been wrongly placed after killing. ‘Killing Jackson’ is the needed noun phrase, they had to make sure of it and execute with the requisite carefulness. The verb, ‘make sure’ is often used with a noun to form a phrase. This error is also seen in option (b). Option (b) has another imprecision: The adverbial expression ‘so much as’ is distorted as ‘so much so’. ‘So much as’ is used as an intensive to indicate something unexpected; even. The sentence means to say that none of them had even seen a proper revolver. Using the expression ‘so much so’ would impart a vague meaning to the given sentence, making it look awkward. ‘So much so’ means to such a great degree. It is often followed by a clause beginning with that. In option (d), ‘marksmanships’ is incorrect.

34. a

35. c

36. c

Page

The question statement clearly talks about wasteful expenditure. So, the first blank should be filled with a word that can mean profuse or to bestow someone with something. For this ‘lavish’ is the best word. Moreover, the speaker intends to communicate that something is done in memory of Mahatma Gandhi. So ‘commemorative’ is the most appropriate word to communicate the meaning of the sentence. Now coming to option (b), extravagant already means something that exceeds limits. So, ‘extravagant excess’ will bring redundancy error in the sentence. Options (c) and (d) are incorrect as the second blank cannot be filled with the word ‘remembrance’ or ‘honour’. The tone of the paragraph shows a dichotomy. The key to solving the question lies with the second blank, which should convey the opposite meaning to the sense of contentment displayed by America. ‘Fastidious’ (reflecting a meticulous, sensitive, or demanding attitude) is the apt choice for the second blank. Also, ‘dysfunctional’ (not functioning normally) is the correct word for the first blank. Option (a) is incorrect as ‘complacent’ should not be used for the second blank. Similarly, ‘sedated’ in option (d) cannot fit the context. Option (b) is incorrect as crotchety (Capriciously stubborn or eccentric; perverse) does not fit in the second blank. In option (a), the relative clause modifies the noun ‘town’, which gives a wrong impression that the town was usually hated by the locals although its architects won a major Welsh housing award in 2006. Option (b) also has the same error. In addition to this, in option (b) ‘bottle bank’ has been incorrectly written as ‘bottles bank’. Option (d) has two errors- (i) a comma is placed before ‘that’; here that is being treated as a restrictive clause and should not have a comma before it. (ii) ‘it’s architects’ is incorrect; the correct usage here would be ‘its’ as the possessive form is required here. (iii) the expression ‘most often usually’ is a case of redundancy. It should be either ‘most often’ or ‘usually’.

5

37. c

Option (a) can be inferred as a hurdle in doing justice to the promise of justice. Refer to the discussion in the opening para “Nevertheless, how is it possible to do justice to the promise of justice? Especially when this very promise brings with it all the suffocating memories of the disappointments, and worse, that attend its history.” Option (b) also is a hurdle as seen from the lines “One obvious way to do justice to the promise of justice is to realize its promise, yet this realization might be to betray both promise and justice. For a promise of justice can also be threat and its realization the wreaking of vengeance.” Option (c) is not a hurdle but a factor which revives the need to do justice to the promise of justice. Refer to the line in the first para ,“The prefigurations……….revive the force of the promise, in spite of everything.”

38. b

Refer to the lines in the last para “The tension between regarding the ‘promise of justice’ in terms of either divine vengeance, on the one hand, or divine grace, mercy or forgiveness, on the other, is not only an urgent legal and political issue but also an intrinsic articulation of the promise of justice itself.” This leads to option (a). Option (c) can be inferred from the lines in the second last para “Derrida confirms his consistently held view that justice can only be promised; to realize it is to betray its promise to calculation or vengeance.” The 4th line in the last paragraph clearly states that the character of the justice promised is inseparable from the form of its realization. This clearly contradicts option (b).

39. d

‘Eschatology’ is a branch of theology concerned with the final events in the history of the world or of humankind. None of the options come close to the meaning of eschatology.

40. a

The implicit premise of GP’s argument is that they are on course. Satya, through his argument, has rejected this implicit premise and this has led to a different conclusion. Hence, option (a) is the correct answer. Option (b) is not the correct answer, as accepting the truth of GP’s premises does not imply that GP’s argument is invalid. Option (c) is not the correct answer as Satya offers the reason for rejecting GP’s implicit premises. Option (d) is incorrect as Satya is not noncommittal. He, in fact, feels that they need to stop.

41. b

It is given in the question statement that the job of a counselor requires no educational qualification. But this does not mean that Mr. Padhai is not an educated person. So, in order to weaken the reasoning one has to offer information that can clarify the ambiguity of the question statement. Option (a) is not the correct choice, as working at Career Launcher for 10 years does not weaken the reasoning.

42. b

‘Lesion’ as such is a damage resulting to body parts from a disease. So option (b) is correct . The term does not connote time-early/advanced symptoms. Hence options (a) and (d) which respectively speak about ‘early symptoms’ and ‘advanced signs’ are inappropriate. Option (c) is incorrect as the term does not imply dead body parts.

43. d

Option (a) is explicit in the 3rd last line of the last para of the passage. Option (b) can be discerned from the lines in the 4th para, “The medical gaze required a new concept of pathology as well as a shift in the place of death in the medical field.” Option (c) can be inferred from the discussion in the last para. But option (d) cannot be inferred. Definitely the ultrasound has added a new dimension to the medical gaze, but we cannot infer that the stethoscope is redundant; the author clearly mentions, “The invention of the stethoscope was strategic in this process”.

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44. b

45. c

Since we are looking at the author’s purpose in quoting the comment by Barbara Duden, ‘we are overwhelmed with fetuses.’, in the overall development of the passage, we need to look at the parts of the passage before and after the comment. The comment is used to highlight the blinding effect created by the images of the fetuses. So in the context of the development of the passage the comment is used not only to highlight the new medical stage but to draw attention to the implications of the creation of the images of fetuses. This makes option (b) score over option (c). Option (a) does not qualify as the primary intent of the author. Option (d) is not discussed in the passage. Although, the author raises a question, “What are its political and ethical implications”, with reference to the images of fetuses but he does not pass a judgement on it. For the 1st step either of the ants has 4 choices of moving: east(E), west (W), north (N) or south (S) and for the subsequent steps, each of the ants has 2 choices (Taking a left or right turn). ⇒ Total number of different paths each ant can take = 4 × 2 × 2 = 16 ⇒ Total number of pairs of distinct paths =

46. a

16

49. d

A

T

F

N

B

O

C

D

E

OF = OA = OE = Radius = 3 cm. FN = AT = DE; FD = NE = 1 cm. ON = OE – NE = 2 cm.

∴ FN = 32 − 22 = 5 cm ∴ TO =

(3 )2 − (

5

)

2

= 2 cm

∴ Area of the rectangle ABCD = (2 + 3) × 8 = 40 cm2. 50. d

16 × 15 C2 = = 120 . 2

Given that 2

Final distance between the two ants will be maximum, if one of the ants moves maximum possible distance westward and the other ant moves maximum possible distance eastward. ( 3 Moving in the opposite directions will maximise the distance between the two ants, and since, 3rd step should be either eastward or westward, the first step should also be eastward or westward) One of the cases when the distance between the 2 ants will be maximum is as follows: N

1 1 1 1  logx  + + + + ....infinitely many terms  = 2  2 5 9 14  Now, let

1 1 1 1 + + + + ....infinitely many terms = P 2 5 9 14

⇒

1 1 1 1 + + + + ....infinitely many terms = P ...(i) 1× 2 1× 5 3 × 3 2 × 7 Dividing both sides by 2, we get

1 1 1 1 P + + + + .... infinitely many terms = ...(ii) 1× 4 2 × 5 3 × 6 4 × 7 2   1  1 1  1 1  1 −  +  −  +  −   1  4   2 5   3 6   ⇒  3  1 1 +  −  + ...infinitely many terms   4 7    

A n t B (–1 0,2) E

W

Let us do the construction as shown in the figure below.

A n t A (6,–4 )

=

S

47. b

48. c

3P  1 1 = 1+ +  2  2 3

Required distance = (6 – (–10))2 + (2 – (–4))2 = 292 cm.

⇒

Let us assume f(0) = K, where ‘K’ is a constant. Then, f(0 + y) = f(0.y) = f(0) = K and f(x + 0) = f(x.0) = f(0) = K.

11 9 Therefore, from the given equation, we get

This proves that the function is a constant function. So, the value of f(– 25) = f(–5) = 5. So, f(– 25) + f(25) = 10.

11  11  2 9 =x ⇒x= 9.  

The digit at the ten’s place of 62, 63, 64, 65 and 66 are 3, 1, 9, 7 and 5 respectively. Also, the digit at the ten’s place of 67 is 3. ⇒ The cyclicity of the ten’s digit of 6N is 5. (N ≥ 2) The remainder when 117 is divided by 5 is 1. ⇒ The digit at the ten’s place of the given number N will be 6 same as the digit at ten’s place of 6 which is 5.

Page

P 2

6

⇒P =

2

51. a

(xy) × (pq) – (yx) × (qp) = (10x + y) (10p + q) – (10y + x) (10q + p) = 99(xp – yq) Since 99(xp – yq) < 100 (as per the given condition) ∴ (xy) × (pq) – (yx) × (qp) = 99 ⇒ (yx) × (qp) = (xy) × (pq) – 99 Now since the product (xy) × (pq) ends with 8 and 3, therefore (yx) × (qp) must end with 8 and 4.

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52. d

Let the distance from point ‘X’ to school be D. Let Ram leaves the cycle at a distance f × D from point ‘X’. (f is a fraction of the total distance from point ‘X’ to the point A) Then, (fD) + (1 – f)D = (fD) + (1 – f)D . 15 4 5 12

∴

55. b

5 Solving for f, we get f = . 9

56. d

∴ 8 + 8xt2 = 15 ⇒ xt2 =

4 1 = . 8 2

AO = tan60° EO Let EO = x units

7 . x

Similarly, 5 + 5xt1 = 15 ⇒ xt1 = 2 Total time taken by Vicky = 5t1 =

In ∆AOE,

10 . x

⇒ The required ratio = 7 : 10. 57. a

N × M = (52p – 1) (5q + 1)2 (5r + 1)2 = (52p – 1) (52q + 1 + 2.5q) (52r + 1 + 2.5r) (52p – 1) when divided by 6 will yield a remainder of ‘0’. ⇒ (N × M) when divided by 6, remainder is 0.

58. c

Total number of balls transferred by Amit and Bineet to the box B:

Hence, AO =

3 x units . Also, OD = x units. In ∆ADB ,

n(7 ∪ 27) = n(7) + n(27) – n(7 ∩ 27) 1000  1000  1000  = + −   7   27   189  = 142 + 37 – 5 = 174 Finally, Ishu transferred back all the balls to the box A which are numbered multiple of 37. ⇒ The number of balls transferred by Ishu

)

1+ 3 x AD BD = = . tan60° 3 In ∆ADC ,

)

AD = 3 1+ 3 x tan30°

BC = BD + DC =

7 . 8

Total time taken by Sanjeev = 8t 2 =

∠A = 90°, ∠B = 60° and ∠C= 30°,

(

1 2

1 = 15. 2 Let Sanjeev and Vicky take t2 and t1 min for 1 step respectively.

∠A: ∠B : ∠C = 3:2:1 So,

DC =

Let Amit takes 1 step in t minute and speed of escalator be x steps per minute.

Number of steps in escalator = 10 + 10.

Total number of ways in which A can be eliminated = 4

(

)

If b = 1: 10 = 2 × 5, possible values of 'a' are from 9 to 18.

⇒ xt =

2. A gets 2 votes and both B and C get 1 vote each, this can also happen in 2 ways: a. A votes for B, B votes for C and C votes for A. b. A votes for C, C votes for B and B votes for A.

54. a

(

⇒ 10 + 10xt = 30 – 30xt

Number of ways in which A can be eliminated: 1. A gets 3 votes or in other words when B, C and D all vote against A, this can happen in 2 ways (A votes for either B or C)

Probability of A getting eliminated =

)

If b = 3: 103 = 23 × 53. So, the value of 'a' should be such that a! has at least three multiples of 5. So, the value of 'a' should be greater than 14. Possible values of 'a' are 15 and 16. So, total number of pairs, (a, b) = 10 + 8 + 2 = 20.

Solving for the average speed, we get average speed = 6.75 km/hr. Since, they take an hour to reach the school, D = 6.75 km. ‘A’ has 2 options to vote against either B or C. Similarly B and C also have 2 options each to vote against. D has only 1 option ‘A’. So, total number of ways participants can vote against each other = 2 × 2 × 2 = 8.

(

If b = 2: 102 = 22 × 52. So, the value of 'a' should be such that a! has at least two multiples of 5. So, the value of 'a' should be greater than 9. Possible values of 'a' are from 10 to 17.

5 So, Ram leaves the cycle at a distance   D from point ‘X’. 9 The average speed of both Ram and Shyam is the same, since they reach the school simultaneously.

53. b

OF 3x 3 3 = = BC 4 1 + 3 x 4 1+ 3 . 3

= n(7 ∩ 37) + n(27 ∩ 37) – n(7 ∩ 27 ∩ 37)  1000   1000   1000  = + – =4  7 × 37   27 × 37   7 × 27 × 37 

4(1 + 3 )x . 3

⇒ The final count of balls in the box B = 174 – 4 = 170.

Also ∠OAF = 60°. So in ∆AOF , OF = AO × tan60° = 3x

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59. b

Let the six committees be C1, C2, C3, C4, C5 and C6. Let one of the committees ‘C1’ have ‘n’ members. Each member must be in exactly one of the other 5 committees. So there must be exactly 5 members (say A, B, C , D and E) in committee ‘C1’ (so n = 5). Now, let committee ‘C2’ have member ‘A’ in it. There must be four more members (say F, G ,H and I) in C2 who are also members of other 4 commiittees C3, C4, C5 and C6. Subsequently we’ll get 3, 2 and 1 new member in C3, C4 and C5. The whole arrangement is given as Committees

62. c

Since, one angle is 65°, one of the other angles has to be 115o, since, a pair of opposite sides are parallel. Let the other two angles be x° and (180 – x)°. If x < 65, then (180 – x)° > 115º and if x > 65, then (180 – x)° < 115º. Since, we are looking for the largest possible angle of the trapezium, xº, 65º, 115º, (180 – xº) are in A.P. ⇒ 65° – x = 115 – 65º, ⇒ x = 15º ⇒ Largest possible angle = 180º – 15º = 165º.

63. a

As t1 = 1

∴ t101 = 1013 – 993 + t100 = 1013 – 993 + 1003 – 983 + t99 = 1013 – 993 + 1003 – 983 + 993 – 973 + .... + 43 – 23 + 33 – 13 + (23 – 0) + (t1) = 1013 + 1003 – 13 – 0 + t1 = 1013 + 1003 = 2030301.

Members

C1

A B

C

D

C2

A

F G

H

I

C3

B

F

J

K

L

C4

C G

J

M

N

C5

D H

K

M

O

C6

E

L

N

O

I

E

E

A

64. a

So the total number of members will be 5 +4+3+2+1 = 15. 60. c

F

Given that 11[y] + 23{y} = 250 ...(i) Now 0 < {y} < 1 So, 0 < 23{y} < 23. Comparing the above with (i) 227 <11[y]< 250 ...(ii) As [y] is always an integer the only possible values of [y] in (ii) are 21 and 22. (this is because only multiples of 11 between ‘227 and 250’ are 231 and 242) when [y] = 21,

{y} =

O B

Since, AD is the median of the ∆ABC We have, ...(i) (Apollonius theorem) AB2 + AC2 = 2(AD2 + BD2) In ∆CBE : CF is a median, ...(ii) So, CB2 + CE2 = 2(CF2 + FE2) 3 AB = 4 cm, AC = 6 cm, AD = 3 cm, from (i) 16 + 36 = 2(9 + BD2)

250-231 19 = 23 23

Subsequently y = [y] + {y} = 21 +

⇒ BD2 = 17 ⇒ BD =

Also, when [y] = 22,

68 + CE2 = 2(25 + 16) ⇒ CE = 14 cm. 65. c

The number of people in the office, originally, was N. Since, 5 more people joined the office the number of telephone lines would increase by (N) + (N + 1) + ( N + 2) + (N + 3) + (N + 4) = 75. So, N = 13. Hence, the number of people in the office now is 18. The number of telephone lines = 18C2 = 153.

66. b

Since, |x| ≤ 100, y ≤ 100

8 8 Subsequently y = [y] + {y} = 22 + or y = 22 . 23 23 So there are exactly two possible solutions to the equation, y = 21 61. d

19 8 and y = 22 . 23 23

Common Ratio ‘2’: PQR = 124 and 248 Common Ratio ‘3’: PQR = 139

'3' : PQR = 469 2 The number formed by reversing the digits of these numbers are also in GP. Product of all such three-digit numbers = 124 × 248 × 139 × 469 × 421 × 842. Unit's digit of the above product is 4.

Common Ratio

17 cm

∴ BC = 2 17 cm From equation (ii):

19 19 or y = 21 . 23 23

250 – 242 8 {y} = = . 23 23

C

D

⇒ – 100 ≤ x ≤ 100 As, y is a square of a natural number, the possible values y can take are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100, i.e. a total of 10 values. For each value of y, x can take two possible values (one positive and one negative) except when y = 100, where x can take only one value i.e. ‘99’. ⇒ There are a total of (10 × 2– 1) = 19 solutions.

(Since, 931 and 964 are greater than 900 hence they are not considered in the product)

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