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Page 1 Online Mock CAT 9 - Unproctored

1 1 2 5 3 4 4 2 5 4 6 3 7 4 8 2 9 1 10 4

11 1 12 3 13 1 14 4 15 2 16 1 17 4 18 2 19 5 20 3

21 4 22 5 23 4 24 2 25 5 26 4 27 5 28 5 29 4 30 3

31 1 32 3 33 1 34 2 35 2 36 5 37 2 38 1 39 3 40 5

41 5 42 4 43 1 44 3 45 3 46 3 47 5 48 3 49 2 50 5

51 2 52 2 53 3 54 2 55 1 56 1 57 3 58 3 59 5 60 4

61 5 62 2 63 2 64 3 65 4 66 5 67 1 68 4 69 3 70 4

71 5 72 3 73 4 74 1 75 2 76 1 77 1 78 3 79 3 80 1

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For questions 1 to 5:

Selectors

GradesP Q R S T Total

A 0 5 3 2 3 13

B 3 0 2 3 3 11

C 4 2 1 4 0 11

D 3 0 3 1 3 10

E 1 4 2 1 2 10

Total 11 11 11 11 11 55

This table gives you the break-up of the grades given by the selectors.In the second table there are two grades that are disguised as X andY The total number of Bs, Cs, Ds and Es in the second table is 11, 11,10 and 10 respectively.Therefore, the grades disguised as X and Y are both As.

1. 1 The grade disguised as Y is A.

2. 5 Both P3 and P6 have the least number of points.Hence, option (5) is the correct choice.

3. 4 Since both the grades that are disguised as X and Y are Aand both these grades are given by S, therefore S did not givegrade A to P10.Therefore, the three As that are given to P10 are by Q, Rand T.Therefore, either P or S could have given C grade to P10.

4. 2 Given that the grade given to P3 by S is X

P3 got grade A by S and hence not by Q.Q doesnt give grades B or D to any one. Grades secured byP3 are E, E, D, X(A) and B, of which grades A, B and D havenot been given by Q and hence Q must have given grade E toP3.

5. 4 Since, Q gave grade A to each of P2, P3, P5, P7 and P11,therefore he necessarily gave grade C to P10 and P4.Hence, Q gave the grade E to P9 as that is the only grade heis left with.

For questions 6 to 10:

For questions 6 and 7:

Given that In Lab 2 as well as Lab 3, the number of bottles of acids in

the category XL as a percentage of the total number of bottles ofacids in the respective laboratories is not more than 1%.Total number of bottles in these laboratories = 2400.

6. 3 A maximum of 6 bottles each in Lab 2 and Lab 3 can be in thecategory XL.Also, a maximum of 688 bottles in Lab 4 can be in the categoryXL.

Total number of XL bottles302400 720

100= =

So, in Lab 1, the number of bottles of acid in the categoryXL cannot be less than 720 6 6 688 = 20

Required percentage20 100 4.

500= =

7. 4 Let the number of the bottles of acids in Lab 1 and Lab 4 thatare in the category XL be 2x and 7x respectively.Therefore, the number of bottles of acids in Lab 1 and Lab 4that are not in the category XL will be

500 2x and 688 7x respectively.Also, let the number of bottles of acids in Lab 2 and Lab 3 thatare in category XL be Y.Therefore, Y + 2x + 7x = 720 and Y cannot be more than6 + 6 = 12.So, the value of x can be 80 or 79.Required difference 500 2x (688 7x) = 5x 188If x is equal to 80, then the required difference is 212 and if xis equal to 79, then the required difference is 207.

For questions 8 and 9:Given that all the bottles containing one or the other of the three acidsnamely Sulphuric, Nitirc and Nitrous are in one or the other of the threecategories S, M and L.

Total number of bottles containing one or the other of three acidsnamely Sulphuric, Nitric and Nitrous acid is384 + 367 + 402 = 1153Total number of bottles of acids in the three categoriesS, M and L = 360 + 600 + 240 = 1200.Difference = 1200 1153 = 47.

a + b + c = 47

8. 2 Out of the options given only option (2) can be a possible ratioof a: b: c.

9. 1 Maximum possible value of a could be 47.So, in order to minimize the number of bottles of Benzoic acidin Lab 2 that do not belong to any of the three categories S, Mand L we need to maximize the number of bottles of Benzoicacid that belong to one or the other of three categoriesS, M and L.So, required answer is 123 47 = 76.

10. 4 Average number of bottles per variety of the acids is 400.Number of bottles of Sulphuric, Nitric and Salicylic acid is lessthan the average number of bottles per variety of acid.

For questions 11 to 15:Given that the total number of stocks distributed byMr. Moneywaala is X.

So, X14440

Therefore, X 5760 .Also, Q got 18.75% of the X stocks distributed, which means the

number of stocks got by Q is3X

.16

Now, 3X 14416

(as no one got less then 144 stock.) or, X 768

So, all multiples of 16 from 760 to 5760 (both inclusive) can be apossible value of X.Total multiples of 16 (in this range)

5760 7681 360 48 1 313.

16 16 = + = + =

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11. 1 There are 313 possible values of X.

12. 3 For option (1):424 cannot be the total for Q, R and T as then X will become424 + 184 + 144 = 752. But X 760.For option (2):1220 cannot be the total for Q, R and T as then X will become1220 + 184 + 144 = 1548. But 1548 is not divisible by 16.Similarly for option (4): 3824 cannot be the total.For option (3):Here X = 600 + 184 + 144 = 928928 is a multiple of 16 and hence this is possible.

For questions 13 to 15:Let, the total number of stocks with the son who got stocks of type Aand B both be Y.Therefore, the number of stocks with sons who got stocks of type

A only 4Y15

= Also, the number of stocks with sons who got stocks of type B only= 4YTotal number of stocks distributed by Mr. Moneywaala = X

4Y 79YX Y 4Y15 15

= + + =

Since, each son received atleast 144 stocks, therefore

4Y 28815

4y(as15 is the total number of stock with two sons who got

stock A only.)

79YY 1080 and X 5688.15

=

Since, X 5760, therefore the only possible value of X is 5688 (asonly 5688 is divisible by 79 in the range) and the corresponding valueof Y is 1080.The sons who received only stock A received 144 stocks each.The sons who received only stock B received 184 and 4136 stocksrespectively.

13. 1

14. 4

15. 2

For questions 16 to 20:From Table 2: Number of movies seen once by Jane= 55 + 18 + 19 + 33 + 13 = 138.

From Table 3: Number of movies of duration 150 minutes that havebeen seen once by Jane = 138 (32 + 33 + 19 + 29) = 25.

From Table 1: Number of movies of duration 150 minutes seen by Jane= 19 + 29 + 39 + 67 + 25 = 179.

From Table 3: Number of movies of duration 150 minutes that havebeen seen four times by Jane = 179 (25 + 21 + 53 + 39) = 41.

From Table 2: Number of movies seen four times by Jane= 18 + 59 + 23 + 31 + 39 = 170.

From Table 3: Number of movies of duration 90 minutes that have beenseen four times by Jane = 170 (43 + 39 + 41 + 39) = 8.

Similarly, the number of movies of duration 90 minutes that have beenseen exactly twice by Jane is 73 and the number of movies of duration180 minutes that have been seen exactly twice by Jane is 13.With the help of the above drawn information table 4 can be completed.The values in bold are the values that have been uniquely determined.

Once Twice Thrice Four times Five times

60 minutes 32 39 * 43 *

90 minutes 33 73 29 8 60

120 minutes 19 40 * 39 *

150 minutes 25 21 53 41 39

180 minutes 29 13 * 39 *

16. 1 Five out of the eleven deleted values can be uniquelydetermined.

17. 4 The number of movies of duration 90 minutes that have beenseen atleast thrice by Jane = 29 + 8 + 60 = 97.

For questions 18 and 19:Let the number of movies of duration 60 minutes and 120 minutes thathave been seen thrice by Jane be denoted by x and y respectively.

Table 3:

Once Twice Thrice Four times Five times60 minutes 32 39 x 43 68 x90 minutes 33 73 29 8 60

120 minutes 19 40 y 39 103 y150 minutes 25 21 53 41 39180 minutes 29 13 132 (x + y) 39 (x + y) 42

The number of movies of duration 60 minutes seen thrice by Jane ismaximum possible.Maximum possible value of x is 68.

18. 2 Number of movies of duration 60 minutes that have been seenmore than thrice by Jane = 43 + 68 x= 43 + 68 68 = 43.

19. 5 Number of movies of duration 120 minutes that have beenseen five times by Jane is 103 yTo minimize 103 y, we need to maximize the value of y.

Number of movies of duration 180 minutes seen thrice byJane = 132 68 y = 64 y.Maximum possible value of y is 64.

Therefore, at least 103 64 = 39 movies of duration 120minutes have been seen five times by Jane.

20. 3 The number of 180 minutes movie that is seen five times byJane is (x + y) 42. To maximize it, we need to maximize bothx and y.Maximum value of x = 68 and y = 64.Hence, maximum number of 180 minutes movies that havebeen seen five times by Jane is (68 + 64) 42 or 90.

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For questions 21 to 25:For convenience, let us identify Avni, Bindu, Charu, Devi, Ela, Fatima,Gauri and Hemlata as A, B, C, D, E, F, G and H respectively. Representingthe conditions pictorially,

Rotten Smirched Tarnished

A BE

(G H) and (D F)

F B/C

It is given that at least 2 students must go to a village. That means, themaximum number of students in any village could be 4.

21. 4 B and G go to Rotten village. H, who always accompanies Galso goes to Rotten village. Had F been in Smirched village, Bshould also have been in Smirched village. But B is in someother village, thus F is not in Smirched village.But, A and F are together. Hence, they are not in Smirchedvillage. If A and F go to Rotten village, it would make 6 studentsfor this village (including Ela), which is not possible.Hence, A and F have to go to Tarnished village.Only option (4), i.e. Charu, Devi and Ela go to the same village(Smirched) could be true.

22. 5 G and H go to Tarnished village. No other student goes to thisvillage. C, who cannot go to Rotten vil lage, must go to Smirchedvillage. Thus, multiple cases arise.Suppose that F goes to Smirched village. Then, B should alsoaccompany her. Since B is in Smirched village (if A goes torotten, B and E should also be in rotten), A should also be inthe same village. Now, this village has 4 students A, B, C andF. It cannot have more than that and Tarnished village has onlyG and H. Hence, the rest of the two students (D, E) must go toRotten village.

Now suppose that F doesn t go to Smirched village. Then, Fshould go to Rotten village. D, which cannot be with F, goes toSmirched village. The remaining students A, B, E all can go toRotten village.

1. (Rotten F, A, B, E / Smirched C, D / Tarnished

G, H)2. (Rotten D, E / Smirched A, B, C, F / Tarnished G, H) are two of the many possibilities.Other possibilitiesare:Rotten - F, B, E/Smirched - C, D, A / Tarnished - G, HRotten - F, B/Smirched -C, D, A, E / Tarnished - G, HRotten - F, E/Smirched - C, D, A, B / Tarnished - G, HThus, A, B, D, E and F is the exhaustive list of studentswho may visit Rotten village.

23. 4 4 students in Tarnished means 2 students each in Rotten andSmirched.In Tarnished village, B and D are there. Since D and F cannotbe together, F is not in Tarnished village. Further, if F were tobe there in Smirched, B would also have been there. Since Bis in some other village, F is not in Smirched either. Only villageleft for F is Rotten. Similarly, A is not in Rotten. C is definitelynot in Rotten as per the given conditions. Now, G and H haveto be together, so they can be in Smirched or Tarnished. Thetwo persons in Smirched/Tarnished have to be A and Ctogether. Thus, the only student left is E, who is with F inRotten village.From the options, it can be seen that only (4) cannot happen.

24. 2 From option (2),

R S T

A DA in Rotten would imply B and E also in Rotten.F cannot be in Tarnished because D is there. Also, since B isnot in Smirched, F cannot be in Smirched. Thus, only villageleft for F is Rotten. By now, Rotten village has 4 students viz.A, B, E and F. Thus, other two villages must be having 2students each. G and H, who have to be together, go to

Smirched. The only student left is C, who goes to Tarnished.Thus, entire assignment of students for these villages canbe determined.We can find that all the other options will not help us touniquely identify the villages to which each of the studentsgo.

25. 5 F and H go to Smirched. B also goes to Smirched with Fwhereas G accompanies H. Since B doesn t go to Rotten, Acannot go to Rotten. There are already 4 students forSmirched, so A goes to Tarnished. C cannot go to Rotten soC goes to Tarnished. Remaining 2 students D and E can go toRotten only. Only option (5) is not possible.

26. 4 x + y + 31 = 2xy 62 = 4xy 2(x + y) 63 = 1 + 4xy 2(x + y) = (2x 1)(2y 1)

(2x 1)(2y 1) = 63 1 xy = 32 1 = 32Maximum possible integral power of 3 in 32! is

2 332 32 323 3 3

+ + = 10 + 3 + 1 = 14.

Note: We have taken only the maximum possible value ofxy, which should hence give the maximum possible integralpower of 6 in (xy)!Maximum possible integral power of 6 in (xy)! = 14.

27. 5 Total number of possible ways in which 3 distinct numberscan be chosen = 20C3 = 1140.Favourable number of ways is as follows: (6, 2, 3); (8, 2, 4);(10, 2, 5); (12, 2, 6); (12, 3, 4); (14, 2, 7); (15, 3, 5); (16, 2, 8);(18, 2, 9); (18, 3, 6); (20, 2, 10) and (20, 4, 5).Therefore, the favourable number of ways in which thenumbers can be chosen is 12.

Required probability12 1

.1140 95

=

28. 5 7! = 5040

k k k 1[7!] [5040] [5040 (5040) ]= =76 76 76

k 1[1260 (5040) ]= .19

Remainder when 5040 (k -1) is divided by 19:

(k 1) (k-1)5040 [(5035 + 5) ] = .19 19

5035 is a multiple of 19.So the net remainder is when 5 (k 1) is divided by 19.(k 1) = 2 2 + 2 3 + 2 4 ++2 9 + 2 10 =4(1 + 2 + 2 2 + 2 3 ++ 27 + 2 8)= 4m, where m =511.

4m m m5 625 (627 2) = =19 19 19

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m 5112 2 = = 19 19

7 56[(2) (512) ] = 19

56[(128) (513 1) ]= .

19Now 513 is a multiple of 19. 128, when divided by 19, leavesa remainder of 14.Therefore, when 5 4m is divided by 19, the remainder = 5.

Remainder when 1260 is divided by 19 = 6. The remainderwhen 1260 (5040) k 1 is divided by 19 is when 6 5 = 30 isdivided by 19 i.e. 11.Since we have cancelled a common factor of 4 from thenumerator and denominator, the net remainder = 11 4 = 44.

29. 4 Distance remaining after the first day

1= 110 1 .

4x

Distance remaining after the second day

1 1= 110 1 1

4x 3y

1 1 3Or, 110 1 1 = 110 1

4x 3y xy

3y + 4x = 37.Possible integer values of x and y that satisfy the aboveequation are (x = 1 and y = 11); (x = 4 andy = 7) and (x = 7 and y = 3).Minimum possible product of x and y = 11.Maximum possible aggregate distance covered on the first

two days 3=110 = 30 km11

Minimum possible distance left to be covered by Gypsa afterthe first two days = 110 30 = 80 km.

30. 3 A D

CFEB

B E

Let the length of B F and EF be x cm and y cmrespectively.Also, B E = BE = 1 cm.In triangle ABF : AB 2 + BF 2 = AF 2

2 2 2(y 1) 4 (4 x) + + = +2(y 1) x 8x2 + = + ... (i)

In triangle EB F : EB 2 + B F2 = EF 2y2 = 1 + x 2 ...(ii)

Solving (i) and (ii) we get that =8 17

x = and y15 15

Area of the shaded region = Area of the squareABCD (Area of the triangle ABE + 2 Area of the triangleAEF)

= + = 21 1 17 14216 1 4 2 4 cm

2 2 15 15

31. 1 2 x + yz = 2 + ( x + yz ) = (1 + x + y + z ) + ( x + yz )

or 2 x + yz = (y + 1)(z + 1)Similarly other denominators can be expressed as factorsand the given expression can be rewritten as:

( )( ) ( )( ) ( )( )1 1 1+ +

y +1 z +1 x +1 z +1 x +1 y +1

=( ) ( ) ( )

( )( )( )x +1 + y +1 + z +1

x +1 y +1 z +1

=

( )( )( )( )

x + y + z + 3

x +1 y +1 z +1

= ( )( )( )4

x +1 y +1 z +1

we are given that x + y + z = 1or (x + 1) + (y + 1) + (z + 1) = 4The maximum value of the product (x + 1)(y + 1)(z + 1) occurswhen

( x +1) = (y + 1) = (z + 1) =43

so the minimum value of the given expression is 34 27 .

1643

= Alternative:

Since there is symmetry(any two of x, y, z can be interchangedwithout changing the condition or the given expression) in the

question put x = y = z =13 for an extreme case.

32. 3 If a + b = 9, then possible pairs of values of a and b notnecessarily in that order is (1,8); (2,7); (3,6) and (4,5)

Case 1: (a, b) is (1, 8): a 243 + b 243 (ab) 243Using cyclicity the units digit should be = 1 + 2 2 = 1.For the below mentioned values of a and b, the value of(ab) 243 is greater than the value of (a 243 + b 243 ).

Case 2: (a, b) is (2, 7): (ab) 243 (a 243 + b 243 ) = 4 (8 + 3)= 3.

Case 3: (a, b) is (3, 6): (ab) 243 (a 243 + b 243 ) = 2 (7 + 6)= 9.

Case 4: (a, b) is (4, 5): (ab) 243 (a 243 + b 243 ) = 0 (4 + 5)= 1.

33. 1 Assume that f(0) = K.If a = 0 and b = 0, then f(0 + 0 + c) = f(0. 0. c) = f(0)= K f(c) = KIf a = 0 and c = 0, then f(0 + b + 0) = f(0. b. 0) = f(0)= K f(b) = KSimilarly if b = 0 and c = 0, then f(a) = f(0) = K

f is a constant function for all values of a, b, c

= 1f(1001)3

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34. 2 Product of the first n terms

2 26! 7! 8! 9! 10! (n 4)!

1 ...............3! 5! 7! 9! 11! [n (n 1) ]!

+

First five terms are all greater than one and considering3!5! = 6! their product should be equal to 8!

The 6th term is1

11

7th term is 112.13The product 11.12.13 is clearly less than 8!

The 8th term will be1

13.14.15The product 11.12.13.13.14.15 is clearly more than 8!(as 8! = 12.3.4.5.7.8)Hence, product of the first 8 terms

6! 7! 8! 9! 10! 11! 12!1 1.

3! 5! 7! 9! 11! 13! 15!= 3x x 12 x 3 + > >Also, x + 12 > 3x 2x 12 x 6 < + > > 8

(k 1)x 8 k 1x

Also, PB + BC > PC + > < +8x 8 kx k 1x

< < +8 81 k 1x x

...(ii)

Using (i) and (ii) we get that <

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61. 5 Option 1 is incorrect as it cannot be inferred from the passagethat infant damnation means rejection of infant procreation .Option 2 again is exaggerated in its reference to superiority .Option 3 cannot be inferred as the parson was opposed tothe earlier ruler-King George. Also the dress and hat showedhis sense of dignity and not really his commitment to the ruler.Option 4 outlines the expectations from the Parson s role andnot his desires.

62. 2 The statement in the context of the passage means to becompletely thorough and well versed with everything andhave the qualities of the great people, before one starts topreach or guide anyone. The option which comes closest tothis sense is option 2. Option 1 is incorrect as it does notconvey the intended meaning. Option 3 is narrow as it talksonly about commanding . Options 4 and 5 are part of theanswer but are milder in tone. The sense of the statement isstrong and completely thorough fits the tone of the statementaptly.

63. 2 Dazzling means amazingly impressive. The word agrees withthe tone of the sentence. Glaring is incorrect as it meansobvious. Burnished means glossy and does not convey themeaning intended by the sentence. Lambent means flickeringwith a glow. Homely means unattractive.

64. 3 Option 3 is correct here as mingling would result in adiscourse , which means a conversation. Grandiloquence ,which means a pompous speech, would not fit in here.Monologue is out of context with the theme of the sentence.Sermon , which means a religious discourse delivered as apart of the church service or a lengthy and tedious speech,doesn t fit the context. Explanation is out of context.

65. 4 Powerful means having or exerting great power or forceand is the best answer in the context as the second part ofthe sentence talks about reliance on mathematical models.Firm means rigid which is not true in the context. Robust means vigorous and is not applicable in the context. Capable implies competency, the sentence implies something more thanmere competency.

66. 5 Imperative means absolutely necessary or required;unavoidable. The sentence has a ring of finality which is bestconveyed by this option making it the correct answer. Desirous means to express desire; it is a weak option. Optional conveysa sense of not being necessary, hence it is incorrect. Unlimited means limitless which is not appropriate in the context.Discretionary conveys a sense of flexibility, which again isnot in the purview of the sentence.

67. 1 It is clear from the passage that the deep rooted primitiveideas and mindset of people prevent discussions from takingplace. People mistake predicting troubles for causingtroubles . Hence statement A represents an obstacle. StatementB is not the correct representation of an obstacle. The author,in the 1st paragraph, states that One is that by the very orderof things such evils are not demonstrable until they haveoccured... ; however, the author does not mention ability orinclination on part of the people. Statement C is distorted.Deep rooted in the passage refers to obstacles and notevil.

68. 4 The author, in the first line of the passage, states the functionof statesmanship, which he says is to provide againstpreventable evils . Preventable evils here refer to the futureproblems as becomes clear by reading the subsequentparagraphs. The author, through a major portion of thepassage, emphasizes on the relevance of discussing the

future problems and how such discussions face variousproblems. Option 1 becomes too narrow as regards thisprimary purpose. Option 2 is distorted-it is obstacles and notevils that are deep rooted. Option 3 is one of the obstaclesand not the primary purpose. Option 5 is not clearly delineatedin the passage. Option 4 stands out as an option which isclose to the primary purpose and the structure of the contentin the passage.

69. 3 Option 1 does not take into account the example which coversa major part of the passage. Also, rather than providingcounter-arguments, the author is discussing the rigid beliefsof people. Option 2 is too narrow. The author does argue hiscase for some time. In fact, in the end he comes back to it afteran example. Option 4 is incorrect as the opinions/argumentsat the start of the passage do not look like a hypothesis,counter-arguments are nil; the author in the initial half arguesto establish his viewpoint. Option 3 is correct as the first halfof the passage establishes the author s viewpoint and thesecond half shows, through an example, that the viewpointsare relevant.

70. 4 D follows the first sentence as it voices the general perceptionabout gangsters. B would follow D as it continues thenarrative. C continues the discussion by bringing in the Frenchperspective; A continues the discussion about gangsters inFrench films. Thus the correct sequence of sentences isDBCA.

71. 5 Sentence D would follow the first sentence; the clue lies inthe words these social critics . B would follow D as ithighlights the wit mentioned in D. Now the confusion mayarise while choosing between sentence C and A to follow B.If we look at the sixth sentence, we would be able to chooseC over A as the sixth sentence talks of reminds the State of itsevident failures , these words would go with the irrationalityof the state mentioned in A. Thus, A would immediately precedethe sixth sentence. The correct sequence would be DBCA.

72. 3 BD is a mandatory pair since only B has a component of abuilding an office. D refers to the building. Only option 3 hasthis mandatory pair. Statement 1 is talking about the wall as aperfect subject for art . C is talking of exceptions. So theexamples B, D and A are most probably the exceptions.Statement 6 is conclusive. The only concrete clue here is themandatory pair BD. This leads us to CBDA.

73. 4 The best statement which continues the idea/question instatement 1 is C. C continues the theme. A, B and D seem tobe part of a specific description of events. DB would qualifyas a mandatory pair- since the fan s booing , logically wouldbe a reaction to Davis behavior. The comment in A completesthe event being described in B.

74. 1 From the argument we can infer that Pawan interprets Shivku sstatement to be applicable to anyone . This leads us to option(1). Option (2) mentions most modes of world travel , whichare not discussed in the argument. Option (3) cannot be inferredfrom the argument; in fact, Pawan s statement undermines it.Option (4) is an incorrect inference. Option (5) again cannotbe inferred from the argument.

75. 2 According to the argument there are only two ways for Sudipto travel to the office-either by car or by bus number 411.Since the argument states that bus number 411 is not runningin the current week, we can infer that Sudip used his car tocome to the office. The method of reasoning followed by theargument to arrive at the conclusion is highlighted by option(2). Option (1) is incorrect as the argument is not bringing to

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light any exception. Option (3) is incorrect as the argument isnot talking of any possibility but it highlights one of the certaintythat has occurred. Option (4) does not underline the methodof reasoning as is followed by the argument.

76. 1 From the argument, it is clear that Anup suggests that readinghistory would help one not to repeat the mistakes made in thepast. However, Lachiramka doubts this on the ground that thehistorical facts might help to prove a point but at the same timethey might aid proving something that is contrary to the point.Thus, Lachiramka s concern is aptly expressed by option (1).

77. 1 The argument clearly reasons since drinking is a health hazard,the advertisements that promote it should be banned. Tosupport this conclusion, we must provide a statement thatwould talk against advertisements featuring unhealthyproducts. Option (1) aptly does this. Option (2) to an extentweaken the argument. Options (3) and (4) are genericsuggestions that lie beyond the scope of this argument. Option(5) is wayward as it brings in an extraneous element, whichlies beyond the theme of the argument.

78. 3 Caret refers to the proof-reading marks placed below a lineof text to indicate a proposed insertion or correction. Carat refers to a measure of fineness in gold. Perspicacious means

having penetrating mental discernment. In this context it wouldmean shrewd. Perspicuous means something that is plain tothe understanding especially because of clarity of expressionand presentation. In this context delusion , which means afalse belief strongly held in spite of invalidating evidence,especially as a symptom of mental illness would fit. Decry means to condemn or depreciate by proclamation. It fits thecontext. Descry means to catch sight of or to discover. Epitaph means a commemorative inscription on a tomb. Epithet meansa term used to characterize a person or thing. It fits in here.

79. 3 In the first sentence, troupe is correct. Troupe refers to agroup of theatrical performers. Troop refers to a generalcollection of people or things. In the second sentence, wriggle would fit as it means to extricate oneself. Writhe means totwist so as to distort. It also means to suffer keenly. In the thirdsentence, biannual would fit as it means occurring twice ayear. Biennial means occurring every two years. In the fourthsentence assassin would fit; it means a person who murdersan important political figure either for hire or from fanaticalmotives. In the last sentence, sleep a wink is the correctexpression to use.

80. 1 Calender is a machine that smoothens or glazes paper orcloth by pressing it between plates or passing it throughrollers. Calendar means a system of determining the beginning,length, and divisions of a year and for arranging the year intodays, weeks, and months. In the second sentence, callous would fit as it means showing no sympathy for others. Callus refers to the thickening of the skin. In the third sentence,epigram would fit in; the clue lies in the words though notpoetic per se . Epigram , here would refer to the terse andparadoxical statement of Oscar Wilde. Epigraph refers to anengraved inscription. It also means a quotation set at thebeginning of a literary work or one of its divisions to suggestits theme. Callow means young and inexperienced; immature.Shallow means lacking depth of character, intellect, ormeaning; superficial. Here shallow would fit. In the lastsentence enthralled , which means to hold spellbound wouldfit.

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