mock 8 sol

8/13/2019 mock 8 sol

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Online Mock CAT 8 - Unproctored

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

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

21 2 22 4 23 3 24 4 25 3 26 4 27 4 28 5 29 2 30 1

31 1 32 3 33 1 34 2 35 5 36 4 37 2 38 4 39 5 40 3

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

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

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

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1. 3 Since, W was not the third product that was launched,therefore it was not launched in May or March.Also, X and W were launched in months that have namesstarting from the same alphabet; therefore X and W arelaunched in January and June not necessarily in that order.

Using statement A:Given that Y was not the second or the last product that waslaunched, which means that Y was not launched in the monthsof March and August.

So, Y was launched in May.It is given that Z was not launched in the last month, i.e.August, so we can conclude that Z was launched in themonth of March.Hence, statement A alone is sufficient to answer the question.

Using statement B:Given that K was not the third or the last product that waslaunched, which means that K was not launched in the monthsof May and August.So, K was launched in March.We also, know that Z was not launched in August, so we canconclude that Z was launched in the month of May.Hence, statement B alone is also sufficient to answer thequestion.Therefore, option (3) is the correct choice.

2. 2 Let the number of dogs, cats and rabbits be ‘d’, ‘c’ and ‘r’respectively.It is also given that d + c + r = 12 and .The possible sets of values of ‘d’, ‘c’ and ‘r’ not necessarily inthat particular order are (2, 4, 6); (2, 5, 5); (3, 4, 5); (3, 3, 6)and (4, 4, 4).

Using statement A:The number of cats and dogs with Carol could be(5, 5) or (3, 3) or (4, 4).The number of rabbits when the number of dogs is5, 3 and 4 is 2, 6 and 4 respectively.It is also given that the number of rabbits is not more than thenumber of dogs, so the number of rabbits could be either 2 or4.

Hence, statement A alone is not sufficient to answer thequestion.

Using Statement B:The number of cats and dogs with Carol could be(5, 2) or (6, 3) in this particular order.The number of rabbits with Carol when there are5 cats is 5 and the number of rabbits with Carol when thereare 6 cats is 3.Since the number of rabbits with Carol is less than the numberof cats, therefore the number of rabbits with Carol is 3.Hence, statement B alone is sufficient to answer the question.

3. 1 Using statement A:Given that the angular distance covered by the line segmentAB in every 5 seconds is 615°.Therefore, the angular distance covered by the line segmentAB in 30 seconds = 615°× 6 = 3690°= 10 × 360°+ 90°.Therefore, we can conclude that at time t = 30 seconds theline segment AB will lie in the second quadrant.Hence, statement A alone is sufficient to answer the question.

Using statement B:Given that the angular distance covered by the line segmentAB in every 10 seconds is 640°.Therefore, the angular distance covered by the line segmentAB in 30 seconds = 640°× 3 = 1920°= 5 × 360°+ 120°.Therefore, we cannot conclude that at time t = 30 secondsthe line segment AB will lie in which of the four quadrants.Hence, statement B alone is not sufficient to answer thequestion.

Hence, option (1) is the correct choice.

4. 1 From the given function we get the values of f(x) as

2 3 4, , for x = 2, 3 ,4

4 8 16∞

Let S = f(1) + f(2) + f(3) + .... .

⇒ 1 2 3 4S = + + + + + infinite terms

2 4 8 16 ...(i)

⇒S 1 2 3

= + + + + infinite terms2 4 8 16

...(ii)

Subtracting equation (ii) from (i), we get

S 1 1 1 1= + + + + + infinite terms

2 2 4 8 16

1

2 = 1

11 –

2

=

⇒ S = 2.

5. 1 The sum of digits of the four-digit number can be even in thefollowing cases.

Case I: All the four digit selected are oddFour digits out of five odd digits (1, 3, 5, 7, 9) can be selectedin 5C4 = 5 ways.These 4 digits can arrange among themselves in 4! ways

Total ways = 5 × 4! = 120 ways.

Case II: All the four digit selected are non-oddThere can be two sub cases:(a) All four digits are (2, 4, 6, 8)The number of ways 4 digit number can be formed= 4! = 24 ways.(b) One of the digit is zero and the three digits are selected outof (2, 4, 6, 8) in 4C3 ways.A four-digit number in which one-digit is zero can be formed in3 × 3 × 2 × 1 ways.Total ways = 4C3 × 3 × 3 × 2 × 1 = 72 ways.

Case III: Two digits are non-odd and the other twodigits are odd.There can be two sub cases:

(a) Two odd digits are out of (1, 3, 5, 7, 9) and two even digitsare out of (2, 4, 6, 8)These four digits can be arranged in 4! ways.Total ways = 5C2 × 4C2 × 4! = 1440

(b) Two digits are out of (1, 3, 5, 7, 9), one digit is zero and onedigit is out of (2, 4, 6, 8)Total ways = 5C2 × 4C1 × 3 × 3 × 2 × 1 = 720Total number of all such four-digit numbers= 120 + 24 + 72 + 1440 + 720 = 2376.

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6. 5 Factorizing the numbers:120 = 23 × 31 × 51

1320 = 120 × 11 = 23 × 31 × 51 × 11

⇒ X definitely has to be a multiple of 11 but it cannot be a

perfect square as X cannot contain (112)1680 = 120 × 2 × 7 = 24 × 31 × 51 × 71

⇒ Y should be multiple of 24 × 71 but it cannot be a perfect

square as Y cannot contain 72.1800 = 120 × 3 × 5 = 23 × 32 × 52.

Since the LCM of 120 and Z is 1800, therefore Z can be aperfect square. One of the possible values of Z can be 32 × 52.But it is not necessary that Z has to be perfect square as itcan be 2 × 32 × 52 also.Hence, option (5) is the correct choice.

7. 2 The given equation is ab + 2 = 2a + b + 600. ⇒ ab – 2a – b + 2 = 600 ⇒ (a – 1)(b – 2) = 600

Now, it is given that ‘a’ is an odd number and ‘b’ is an evennumber which implies that both (a – 1) and (b – 2) are evennumbers.Therefore, the possible pairs of values of (a – 1) and (b – 2)that satisfy the given equation are (2, 300); (4, 150); (6, 100);(10, 60) (12, 50). (20, 30); (30, 20); (50, 12); (60, 10); (100, 6);(150, 4); and (300, 2).Therefore, there are 12 solutions for the given equation.


8. 1

Join EP such that EP || DC. Also join EC

Now, consider ∆ ADB:

⇒ BD2 + AB2 = 82 + 42 = 16 + 64 = 80.

2 {AE2 + BE2} = 2{42 + 2(2 6) } = 80.

∴ 2(BE2 + AE2) = BD2 + AB2.By Apollonius Theorem, BE is the median, hence E is the mid-point of AD and P is the mid point of BC.Now,

1area ( CEB) area (ABCD)

2∆ =

1area ( CEP) area (ABCD)

4⇒ ∆ =

In ∆ =BC

CEF,CF5

1 1area( CEF) area( CEB) (ABCD)

5 10⇒ ∆ = ∆ =

area ( DEC + CEF) = area(CDEF)∆ ∆

7area (CDEF) area (ABCD)

20⇒ =

1(As area ( CDE) area ( ABCD))

4∆ =

9. 1 Let the earlier cost price of the item = Rs.100

⇒ Earlier marked price = Rs.105.

On that day, 30% discount is offered on Rs. 3 × 105 = Rs.315Thus, new selling price = Rs.220.50New Profit percentage = 120.50.

10. 5 Given that a5 + b5 + c5 = 91849.Therefore, the unit’s digit of a + b + c would also be 9.

Note:Any number raised to power (4n + 1) i.e. 5, 9, 13, 17….. willhave the same unit’s digit as the original number itself.

Hence, the number ‘a’ has the same unit’s digit as a5.

Similarly, b and c have the same unit ’s digit as b5 and c5

respectively.

It implies that (a + b + c) would have same unit’s digit as(a5 + b5 + c5).

As a, b, and c are distinct digits their maximum possible sum is24 i.e. (9 + 8 + 7).

Hence, a + b + c = 9 or 19.

If a + b + c = 9, then the maximum possible value ofa5 + b5 + c5 would be 32769.

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

Since, 91849 is much greater than that, hence a + b + c = 19.Sum of digits at the odd places of (2a5b1c) starting from theleft = 2 + 5 + 1 = 8

Sum of the digits at the even places of (2a5b1c) starting from

the left = a + b + c = 19Therefore, the remainder when 2a5b1c is divided by 11 willbe the same as when |8 – 19| = 11 is divided by 11.

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


11. 4 There are three possibilities.Case I:

A

B CD

30 °

A’

60 DAC DA'C 120° < ∠ + ∠ < ∠ °

(as DA 'C 30 and 30 DAC 90 )∠ = ° °< ∠ < °

Note: Line segment A’D is the bisector of ∠ ∠BA ' C, BAC, a nd

side BC. A’D is also perpendicular to BC(why?).

Case II:

A’

B CD

30°

A

30 DAC DA'C 60° < ∠ + ∠ < ∠ °

(as DA 'C 30 and 0 DAC 30 )∠ = ° °< ∠ < °

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Case III:

A

B C

A’

30°

D

30 DAC DA'C 120° < ∠ + ∠ < °

(as DA 'C 30 and 0 DAC 90 )∠ = ° °< ∠ < °

Hence, the aggregate measure of ∠ + ∠DAC DA'C cannot be

equal to 120° (as according to all three cases).

For questions 12 and 13:If we assume (1, 7) as the first block, (1, 1, 7, 7) as the second blockand so on till mth block, then T5001 will be in the mth block.

There are 2 terms in the first block, 4 terms in the second block and so

on.⇒ 2 + 4 + 6 + 8 + ..... + 2m ≥ 5001 ⇒ 2(1 + 2 + ..... + m) ≥ 5001

⇒ ( )2m m 15001

2

+≥ ⇒ m(m + 1) ≥ 5001

Least possible value of ‘m’ that satisfy the above is m = 71.

⇒ T5001 lies in 71st block.Total number of terms in the first 70 blocks = 70 × 71 = 4970 terms.In the 71st block, there will 71 one’s and 71 seven’s.Therefore, the sum of first 5001 terms of the series= (8 + 16 +.... 70 terms) + 1 × 31 = 8(1 + 2 + ..... + 70) + 31

70 718 31

2

×= × + = 19911

12. 3

Alternative method:Note: All the numbers in this sequence are odd and hence the sum

of 5001 terms has to be an odd number greater then 5001,hence option (3).

13. 5 All the terms from T4971 to T5041 are ‘1’ and all the terms from

T5042 to T5112 are ‘7’.

⇒ T5040 + T5042 + T5044 + T5046 = 1 + 7 + 7 + 7 = 22.

14. 2 To find the maximum possible value of( )

1

yz we need to find

the minimum (positive) possible value of the product of y andz.

y = 4x + 41 – x will be an integer only at x = 0 and x = 1 (wherex is an integer)The value of ‘y’ at x = 1 is 5 and at x = 0 is 5 as well.The value of ‘z’ at x = 1 is 3 and at x = 0 is 8.Minimum product = 5 × 3 = 15Therefore, the maximum possible value of the expression

( ) ( )= =

×1 1 1

yz 5 3 15


15. 5

A

5

F

3

CBxE

Let, the length of BE be ‘x’ cm.

= − =2 2BC AC AB 4 cm

Using the angle bisector theorem in ABC,∆

we get thatAB BE 3 x

x 6 cmAC CE 5 x 4

= ⇒ = ⇒ =+

2 2AE AB BE 45 cm 3 5 cm.⇒ = + = =

16. 2 When z is of the form 9x or 9x + 1 or 9x + 8 then Cz willcontain a multiple 9.From 72 to 136 there are 8 numbers each of the form 9x and9x + 1, and 7 numbers each of the form 9x + 8.So, the number of sets with a multiple of nine = 8 × 2 + 7 × 1 =

16 + 7 = 23 sets.

17. 1 From 10 to 100 there are 7 perfect square.One set will contain 100 and each of the remaining perfectsquares will be contained in 3 sets each.So in total there are 6 × 3 + 1 × 1 = 19 sets.

18. 5

A B

410 x – 4 10

230x – 2 30

If x is the width of the river then speed at which they travel is

proportional to the distances they travel.Hence from the diagram, we get

410 x 410 230

x 410 410 x 230

− +=

− + −Solving the quadratic equation, we get x = 1000 m.

For questions 19 to 23:Let the total marks obtained by Rohan, Tarun, and Anup in the test be'r', 't', and 'a' respectively.

A B C D

Rohan17r

50

9r

50

r

5

7r

25

Tarun8t

25

9t

50

t

5

3t

10

Anupa

2

a

10

3a

20

a

4

The marks obtained by each of the three candidates in sections A, B,C, and D should be divisible by 2, 3, 5, and 6 respectively.It is clear that 'r' has to be a multiple of 50.If we take 'r' = 50, then the marks obtained by Rohan in section A is 17,which is not possible.If we take 'r' = 100, then the marks obtained by Rohan in section D is28, which is not possible.

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Therefore, the minimum possible value of 'r' is 300 and 'r' will alwaysbe a multiple of 300.Similarly, the minimum possible value of 't' is 100 and 't' will always bea multiple of 100.Also by the same logic the minimum possible value of 'a' is 600 and 'a'will always be a multiple of 600.

19. 1 Minimum possible aggregate marks obtained by Rohan, Tarun,and Anup is 300 + 100 + 600 = 1000.So, option (1) is the correct choice.

For questions 20 to 22:It is given that the aggregate marks obtained by Rohan, Tarun, andAnup is 1900 and the marks obtained by Rohan is not less than themarks obtained by Anup.The following cases are possible:-Case I: Rohan = 600, Tarun = 700, and Anup = 600Case II: Rohan = 900, Tarun = 400, and Anup = 600Case III: Rohan = 1200, Tarun = 100, and Anup = 600

20. 3 Difference between the marks obtained by Anup in section C

and Da 3a a

604 20 10

= − = =

21. 2 Only Cases II and III are possible, hence the marks

obtained by Rohan can be either 900 or 1200.

22. 4 Case I:Marks obtained by Tarun in section B

9t 9 700126

50 50

×= = =

Case II:Marks obtained by Tarun in section B

9t 9 40072

50 50

×= = =

Case III:Marks obtained by Tarun in section B

9t 9 10018

50 50

×= = =

23. 3 Option (1): It can be true when Tarun has 300 marks and Anuphas 2400 marks in the recruitment test.Option (2): It can be true when Tarun has 5100 marks andRohan has 4800 marks in recruitment test.Option (3): It cannot be true as the minimum sum of the marksobtained by Anup in section C and Tarun in section D is

( ) ( )3 3

a 2t 600 200 120.20 20

+ = + =

Option (4): It can be true when Tarun has 3600 marks andAnup has 4200 marks in the recruitment test.Hence, statement in options (3) is definitely false.

For questions 24 to 26:

24. 4 Let’s start from the box B6, as there has to be one death boxadjacent to it and there is only one box available i.e. C5. So,C5 is a death box. Now C5 is also adjacent to D6 and B5; asC5 satisfies the death box number for D6 and B5; so otherboxes adjacent to D6 and B5 cannot be death boxes; so E5and C4 are life boxes.

Now for E6, F5 has to be a death box.For box B4 there has to be two death boxes, adjacent to it.One is in C5, the other one has to be in C3 as it is the only boxleft adjacent to B4.Now for box F4, one death box is in F5, other one must be inbox E3 as it is the only box left. So all the boxes adjacent tobox D4 are opened and there are only three death boxesamong them. So, the box D4 shows 3 when opened. Hence,the following table can be concluded on the basis ofinformation given in the question:

A B C D E F G H I

1 * * 2 2 1 1

2 2 4 * 4 * 2

3 2 * 4 * 2

4 2 2 3 2 2 2 1 1

5 1 * 1 1 * 2 *

6 1 1 1 1 1 2

7 1

8 1 *

9 1

25. 3 From the table shown above, the ‘death’ boxes adjacent tothe box G6 are F5 and H5.

26. 4 From the condition given in the question following cases arepossible.

Case 1:

H I

7 1 1

8 * 1

9 1 1

Case 2:

H I

7 2 2

8 * *

9 2 2

Case 3:

H I

7 1 1

8 * 2

9 2 *

Case 4:

H I

7 2 2

8 * *

9 3 *

Hence the required sums in case 1, case 2, case 3, and case4 are 5, 8, 6, and 7 respectively. Hence option (4) is thecorrect choice.

27. 4 Total weight ( in ‘000 tonnes) of the crops produced in thestates of HR, UP, PB and TN = 18 + 11 + 14 + 13 = 56.Total weight (in ‘000 tonnes) of the crops W, P, B, M produced= 18 + 25 + 9 + 17 = 69.Assuming that the excess weight (in ‘000 tonnes) is all that ofcrop P which will be equal to 69 – 56 = 13.Maximum weight (in ‘000 tonnes) of crop P produced in thestate GJ = 13

Required percentage = × =13 100 81.2516

28. 5 The possible combinations of states where the crop PO andO can be grown are (WB, AP, GJ), (WB, TN, KR), (WB, MR,KR), (MP, AP, PB), (TN, MR, MP), and (UP, KR, MP).So, HR is definitely not the state where the crop PO is grown.

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For questions 29 and 30:Minimum weight (in’000 tonnes) of the crop amongst the ten states isproduced in the state WB, i.e. 7.

Lets say ‘x’ number of crops are produced in equal quantities in thegiven states and the value of ‘x’ is maximum possible.

For this the aggregate weight of these ‘x’ crops produced in Indiashould not be more than 70.

The value of ‘x’ cannot be more than 5 as no matter which 6 crops weselect their total weight will exceed 70.

There can be six different cobinations of 5 crops such that their totalweight does not exceed 70.

These 6 combinations are as follows:1. B, J, R, PO, M2. B, J, R, PO, W3. B, J, R, PO, O4. B, J, R, M, W5. B, J, R, M, O6. B, J, R, W, O

If we take one more crop, then the total weight will cross70, therefore maximum possible value of ‘x’ is 5.

29. 2

30. 1 The aggregate weight (in ‘000 tonnes) of the crops R, M, B, J

and PO produced in the state66

MP 6.610

= =

So, maximum possible weight (in ‘000 tonnes) of the crop Wproduced in the state MP = 9 – 6.6 = 2.4.

31. 1 As per the information given in the question, the followingtable can be drawn.

Amar Bhuvan Chirag Dhruv Elan

A × × ×

B × × ×C × × ×

D ×

E × × ×

Given that the code assigned to Amar is D.This implies that code assigned to Elan is B and code assignedto Dhruv is E.Also, the code assigned to Chirag is A.Therefore, the code assigned to Bhuvan is C.

32. 3 When “Jersey 1” begins the chain of passes, some of theplayers repeatedly get the ball, making a loop while othersnever receive the ball as illustrated in the figure below:

13

4

7

10

1

If “Jersey 6” is not included in the loop, he will not get anychance to pass the ball to any of his team members andhence the minimum number of passes that he can make is 0.When he is included in the loop, he gets to pass the ball backevery fifth time. Given that “Jersey 6” is in the loop, the minimumnumber of passes that he can make, out of the given 1001 is

200 and the maximum number of passes that he can make is201(that is when he makes the first pass.)

3

9

12

15

6

For questions 33 to 36:Let the marks obtained by Aseem, Ayaan, and Samantha in all thementioned six subjects be ‘x’, ‘y’, and ‘z’.

33. 1 Given that z : y : x = 1 : 2 : 5z = k, y = 2k and x = 5k

Marks obtained by Ayaan in Social Studies = 10% of y =y

10

Marks obtained by Samantha in Social Studies

= 20% of z =z

5

Marks obtained by Aseem in Commerce = 16% of

x =4x

25

Required percentage

y

10 100z 4x

5 25

= × +

5y100

10z 8x

= × +

10k100 20%

10k 40k

= × = +

34. 2 Marks obtained by Aseem, Ayaan, and Samantha in English

arex 6y 4z

, , and10 25 25

respectively.

Marks lost by Aseem in English = 40% ofx x

10 25=

Similarly, marks lost by Ayaan and Samantha in English is

3y 4zand

50 125respectively.

As per the question,

x 3y 4z

25 50 125= = 10x 15y 8z⇒ = =

Marks lost by Ayaan in Science due to reason R

36 1 1 9yy

100 4 10 1000

= × × × =

Required Percentage9 y

1001000 x

= × ×

= × × =

9 10100 0.6.

1000 15

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For questions 35 and 36:

Marks obtained by Aseem in Maths =3x

25

Marks obtained by Ayaan in Social Studies =y

10

Marks obtained by Ayaan in English =6y

25

Marks obtained by Samantha in Hindi =z

10

Now as per the information given3x y 6y z

and25 10 25 10

≥ ≥

35. 5 Given that6z

144,25

= therefore z = 24 × 25.

Number of marks obtained by Aseem in Science9x

50=

Since,3x y 6y z

and25 10 25 10

≥ ≥ x 25

z 72⇒ ≥

Therefore, the marks obtained by Aseem in Science is at least

9 25 124 25 37

50 72 2× × × =

36. 4 Marks lost by Ayaan in English due to reason

Q36 1 24

y100 4 100

= × ×

Therefore,36 1 24

y 45100 4 100

× × × =

125 50y

3

×⇒ =

Marks lost by Samantha in Maths due to reason

z 1 1 zT

5 5 5 125= × × =

Also, y 5

z 12≥

Maximum possible value of ‘z’ = 5000Maximum possible number of marks lost by Samantha in Mathsdue to reason T is 40.

For questions 37 to 40:There are 4 non-vegetarian items and only 2 non-vegetarians D & E, itimplies that both(D, E) have exactly two non-veg items (from statement2)According to information no (1) the meals taken by C, D, E could bea. Biscuit and burgerb. Rice and rotis (this has to go with C as only vegetarian combination)c. Continental and Chinesed. Continental and chicken.

Toast and Fruit can go to A only (these should go to a vegetarian)andnow Dosa & Upma combination has to go to B only. As E doesn’t likechicken it should go to D, this implies that Continental-Chinesecombination should go to E and Biscuit-Burger combination to DHence the meal arrangement will be:

Breakfast Lunch Dinner

A Toast Fruits Pizza

B Milk Dosa Upma

C Salad Rice Rotis

D Biscuit Burger Chicken

E Eggs Continental Chinese

37. 2 38. 4 39. 5 40. 3

41. 4 From the first paragraph, one can infer that ambivalence orhesitation as also self consciousness- which facilitated self-criticism in the earlier days is lacking nowadays. So option(1), (2) and (3) are the characteristics that lack in modernromantic ideology. Option 4 seems to be the trend of themodern times. In fact it is this positive belief and absorption-which is opposite of ambivalence, that is criticized by McGann.Option 4 is the best available choice.

42. 5 Refer to the last paragraph. From the 2nd sentence it can beinferred that McGann does not support extreme viewpoints.Option 2 is rejected in the same sentence by McGann. Option3 is incorrect. Infact, McGann advocates recognizingparaochial ideological formulations. Option 4 is the author’s

view and not of McGann’s. Option 5 is in line with McGann’sapproach of skepticism or circumspection. Option 5 is adoptedby Heine and McGann supports his work.

43. 2 The author is trying to show that the uncritical acceptance ofthe presumptions of Romanticism results in problems as theideology becomes debased and this debased ideology startspermeating or informing other branches like poetry. Option 1 isincorrect. Option 3 is not the main argument of the author.Option 4 is incorrect as the author is not only challenging thedebased Romantic ideology – but showing that the reasonbehind this is ‘ uncritical acceptance’ of it. Also he is talkingabout the problems resulting from it. Option 5 is general andexaggerated: the author does not call the ideology- pristine.

44. 2 Option 2 can be inferred from the 2nd para which talks of

‘antagonistic tendencies’ in poetry developing an ‘allergicreaction’ to ‘intellectual self-consciousness’. Options 1 and 4are not indicated in the passage. In fact the debased Romanticideology is seen to inform American Poetry rather than beingrejected by it. Option 3 is also not seen to be happening. The‘antagonistic tendencies’ favour acceptance rather thanmoving into constructive criticism. Option 5 is not completelysubstantiated by the passage.

45. 2 Heine, in the last para is in line with McGann’s approach. He isquestioning the immediate relevance of far-removed texts orthe source texts which were considered the main influencesin German Romanticism. So like McGann, he is generatingskepticism which would lead to criticism. Options 1 and 3differ from the purport of the last paragraph. Options 4 and 5miss the point and are exaggerated.

46. 3 In sentence B we would use ‘send word’ to indicate that amessage is being sent. In D the correct expression to be usedis ‘give ear to’.

47. 5 All the sentences are grammatically correct. The only confusionmight arise while looking at the phrase, ‘similar twins’. At firstglance, it might appear to be incorrect however, the expressionis correct and means ‘monozygotic twins’.

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48. 4 In A, i t should be ‘shortcomings ’ as we are refering to‘shortcomings ’ in various fields. In E ‘between’ should be usedinstead of ‘among’ as the countries-France, Italy, Austria andGermany are distinct entities which are seen as points thatdefine the boundaries of Switzerland.

49. 5 Options 1 and 2 are entirely the views of “Wallace” and notthe author. Options 3 and 4 cannot be validated by the passage.

50. 4 The question is asking the reason for the ‘numerous variations’

in species and not just the reason for variation. Option 1 isdefinitely a reason for the variation but it does not explain thereason for the numerous variations. Option 2 is too specifican answer. Option 3 cannot be inferred from the passage.The passage nowhere states that species cannot adapt to just one or two conditions. Option 5 talks about the ‘pressureof natural condition’. But unless the conditions “change” andthis change is “diverse” there would be no pressure createdto adapt and develop “numerous” variations. Option 4 answersthe question. The chief reason for the numerous variations isthat the changes in natural conditions which create pressureon species to adapt are “diverse”. Hence the variations arenumerous.

51. 3 Refer to the 2nd paragraph. Options 1, 2 and 5 representpossibilities which depend on other conditions or assumptions.

Option 4 is not indicated in the passage. Option 3 is a definiteoutcome of the intensified conditions.

52. 2 The paragraph ends on the completion of an idea that everydaylife is inextricably bound together by the music. Choices (1),(3) and (5) seem like a repetition of this idea. Choice (4) talksabout ‘cultural revolution’ which seems disjoint as per thetheme of the paragraph. The paragraph talks about the socialcontext of music-making in Tibetan culture. Choice (2) is bestin line with the theme which is indicated at the start – disruptionin cultural life due to the ban.

53. 1 (3), (4) and (5) talk about education, awareness andintroduction of modern science which seem like abrupt ideasafter the paragraph. (2) is disjointed from the flow of ideas inthe paragraph. (1) continues the idea of ostracization-by talking

about repatriation.

54. 3 The paragraph has introduced Johnston Central Library. Afterthis, choice (3) is the best to continue the paragraph-the mainclue being the less intimidating part! Choices (2) and (4) seemto be good choices but are also subtly intimidating. ‘Humblerand friendlier’ go well with the tone of choice (3) rather thanchoice (2). Choice (1) refers to the overall ambience of alibrary in general whereas the paragraph specifically mentionsthe ambience of the Johnston Central library. Therefore, itdoes not continue the flow of thought as per the last lines ofthe paragraph.

55. 4 Option 1 cannot be inferred as the meaning of Lenin’sstatements or the “single” directive is interpreted by the author

as “making organizations parallel”. Option 1 is Cardan’s viewwhich is being refuted by the author. Option 2 cannot beconcluded as the author mentions that the ‘Taylor system’was not introduced during “Lenin’s lifetime”. It could havebeen introduced later. Option 3 is not indicated or implied in thepassage. Option 5 is again Cardan ’s view which is beingrefuted by the author in the second last para. Option 4 issomething to which the author would agree as he refutesCardan’s argument by interpreting the meaning of Lenin’sstatement as “Organizations should be parallel”.

56. 1 Option 1 is correct. Statement A is what the author mainly triesto indicate as the flaw in Cardan’s reasoning. The terms“parallel organizations” and “discipline” etc. carried a differentmeaning when seen in a particular context: when the statewas four months old. The terms when taken for their literalmeaning lead to Cardan’s incorrect interpretation. StatementC is also correct as factual data is used by the author to delinkcause and effect. For example, the subjective-objectiveconnection is delinked by the author in the 2nd last para bypointing to where capitalism originated. B and D are only

referred to and neither discussed nor described by the author.

57. 3 The author is totally denouncing or dismissing Cardan’s views.He calls them a lie. The tone is extreme. The author does notappear frustrated. ‘Argumentative’ is the method used by theauthor and not his attitude. ‘Questioning’ is a mild tone whearesthe author’s tone is extreme here.The author is not ‘regretful’of anything here- regret is normally felt for one ’s own wrongactions. ‘Dismissive’ – is appropriate here as the author istotally dismissing Cardan’s views.

58. 5 The argument clearly states that 'community pressure isexerted' on the people who win the jackpot throughout theKyakya Island community. It then goes on to illustrate thedifferences in the behaviour of the urbanites and people fromthe rural community while spending their earnings from the

jackpots. The question asks us to provide a reason for thedifference in the behaviour of both the peoples. Option (1) toan extent, weakens the argument. Options (2), (3) and (4) donot provide any plausible reason for the difference betweenthe behaviour of the people from the two communities. Option(5) conforms to the premise of community pressure beingexerted on the jackpot winners. It can be safely concludedfrom the option that the community pressure exerted on therural population is more than their urban counterparts, whichmakes the people in the countryside to share their 'goodfortune' with their neighbours.

59. 2 The argument speaks of Nishant's predicament on the matterof making his grandmother's will public or not. The argumentfurther provides the conclusion that after due considerationNishant decided against making the will public. The question

asks us to provide an reason which governed Nishant'sdecision. Option (1) is not the answer as it does not affectNishant's decision on the will. Option (2) is the correct answer-choice as it aptly governs Nishant's decision not to make thewill public. In case Nishant would have decided to make thewill public, Rishi would have wasted the money, whicheventually would have benefited no one. However, by notmaking the will public, Nishant empowered his mother whowould use the money to benefit herself and others. Thus, thisacted as a reason for Nishant to decide on the will as hewanted it to benefit the maximum number of people, harmingno one. Option (3) is a generic statement, which does notaffects the argument. Option (4) is incorrect as Nishant violatesthe promise by not making the will public. Option (5) suggestsof a decision in favour of the choice that 'harms no one andbenefits some' over the one that 'harms some and benefitsnone'. However, the argument does not, in any manner,suggests that had the money in the will gone to Rishi, it wouldhave harmed anyone.

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60. 1 The argument restricts the rest-house owners to only twoways for increasing profits- either by adding more rooms orby enhancing the already existing infrastructure. Then, theargument also rules out the expansion of rest house capacity.This leaves the owners with only one way- 'improving whatis already there'. To complete the argument, a reason must beprovided for the guest- house owners' inability to increasetheir profits. Option (1) provides an apt reason as it statesthat the rest houses are in their best shape and offer luxury tothe extent beyond which even the wealthy customers would

not pay. Thus, it rules out maximizing profitability through anyimprovement to the already existing structures. In option (2)‘Increasing the occupancy’ is not one of the two given waysof increasing profitability. Option (3) does not address theissue of profitability. Option (4) is a way-ward assumptionthat lies beyond the scope of the argument. Option (5) doesnot address the key issue in the argument and lies beyond thetheme of the argument.

61. 3 Option (1) states that the step may help the Taj Hotel to increaseits profits. However, it doesn’t address the core issue.Option (2) is not tackling the real problem. The issue is toincrease the profitability. Maintaining something at a lessercost does not necessarily guarantee profits. It may just happenthat the mini bars, which have a higher maintenance cost,might generate higher profits.

Option (3) is the correct option as it states clearly thatmaintaining mini-bars is a loss making activity.Option (4) does not take profitability into account.Option (5) states that the Oberai Hotel’s profitability is morebut we cannot deduce whether the profitability comes fromempty refrigerators.

62. 1 The argument clearly states the incompetence of Gautam asa detective. To weaken the claim in the argument, we wouldhave to provide information that proves Gautam to be acompetent detective. Option (1) fulfills this purpose, hence itbecomes the correct answer-choice. Option (2) is a far fetchedassumption. Options (3) and (4) strengthen the argument.Option (5) lies beyond the scope of information given in theargument as it talks of a situation which is distinctively different,hence it has no impact on the argument.

63. 2 ‘Phlegmatically unraveling’ as well as the ‘harsh’ treatment ofthe film make it explicit that the unraveling is ‘non-sentimental’and ‘stoic’. Option 1 is incorrect. Option 3 becomes excessiveas a harsh/stoic treatment does not mean ‘cruel undertones’.Option 4 is talking about the intention behind ‘unraveling’. Option5 is talking about the effect of the unraveling on the audience.

64. 2 Except for option (2), all the other options count as remarkablefeatures of the movie ‘Samson & Delilah’.

65. 5 Option 1 generalizes the meaning of real characters from thespecific context. By real, what is meant is ‘that which is closeto the actual real life situation’ and not ‘those without overtemotion’. Option 2 cannot be inferred as nothing has beenmentioned about the experience of the characters. By ‘painful’the author is referring to his and the viewers’ experience ofthe movie. Also ‘painful’ with reference to acting would meana role which demands effort and not literal pain. Option 4again distorts things. The critics choked on superlatives andnot on pain and emotion. Option 5 can be clearly inferred asthe author emphasizes the ‘real characters’ and also theireffect on the audience and the critics.

66. 2 ‘Circumspect’ refers to one who is cautious; it is difficult tocheat or ‘gull’ such a person. In option 1 ‘loquacious ’ refers toone who is very talkative or ‘garrulous’. The pair does notdemonstrate a relationship similar to the stem pair. In option(2), ‘vigilant’ refers to one who is alert or watchful; it is difficultto ‘ambush’ such a person. The relationship demonstrated inthe pair is similar to the one in the stem pair. In option (3), theword pair doesn’t exhibit any relationship - ‘Arabesque’ meansan ornate design of intertwined floral, foliate, and geometricfigures. ‘Emphasize’ means to given emphasis to. In option

(4), ‘dilemmatic’ refers to a situation where one requires tomake a choice between options that are or seem equallyunfavourable or mutually exclusive. In option (5) ‘taciturn’ refersto one who is habitually reserved; ‘goad’ means to provokesomeone to take some kind of action, usually in anger. Therelationship exhibited in the stem-pair is not evident here.

67. 5 ‘Tourniquet’ checks the flow of blood. Similarly, ‘dam’ checksthe flow of water. Bucket does not control a ‘well’. Similarly,the relationship between ‘bridge’:’river’, ‘antiseptic’ : ‘surgery’and ‘pressure’ : ‘air’ do not exhibit the relationship demonstratedby the question statement.

68. 4 ‘Relapse’ means to regress after partial recovery from illness.‘Convalescence ’ means to return to health and strength after

illness. The relationship in this pair is of antonyms. In option(1), ‘Exhibitionism’ refers to the act or practice of deliberatelybehaving so as to attract attention. ‘Suspiciousness’ is notrelated to ‘Exhibitionism’. In option (2), ‘Moral’ and ‘Principle’ donot demonstrate a relationship similar to the one in the stem-pair. In option (3), ‘Dissonance’ and ‘Euphony’ do not have thesame relationship as demonstrated in the stem-pair. Although,they have an antonymous relationship, ‘Dissonance’ does notconvey a sense of ‘regression’ as is conveyed by ‘Relapse’ inthe stem-pair. In option (4) ‘Recidivism’, which means a habitualrelapse in crime has an antonymous relationship to‘Rehabilitation’; one of the meanings of ‘Rehabilitation’ is “torestore to useful and constructive activity ”. Hence, (4)becomes the correct answer choice. In option (5), ‘Prolificacy’and ‘Completion’ do not have the same relationship as thestem pair.

69. 2 The tone of the question suggests that something is beingsacrificed for something else. Also, the usage of ‘unfortunately’in the beginning, suggests something good being sacrificed.Following this logic, options (4) and (5) can be ruled out.Option (1) contains synonyms. Options (4) and (5) do not fitthe context. Option (3) cannot be the answer as‘responsibilities’ will not, unfortunately, crowd out ‘privileges’.Hence the answer is (2).

70. 1 ‘Anomie’ means a lack of purpose or identity; it also meansrootlessness. ‘Siddle’ means to move sideways, especially ina sly or stealthy manner. From the first sentence we cangauge the gloomy mood. Options (2), (3) and (4) do not carrygloominess. ‘Ominousness’ is not the right-fit for the first blank.The author wants to highlight the gloomy mood, ‘ominousness’,which means menacing or threatening is too extreme for that.Also, distracted makes no sense in the sentence. Hence the

answer is (1).

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