Mock Cat 5 Sol

Mock CAT – 5

Answers and Explanations 1 11 21 31 41 51 61 71 81 91 101 111

2 1 4 2 1 3 1 1 3 4 4 3

2 12 22 32 42 52 62 72 82 92 102 112

2 4 4 1 4 2 3 3 4 1 4 2

3 13 23 33 43 53 63 73 83 93 103 113

3 1 2 1 1 1 3 1 1 4 2 2

4 14 24 34 44 54 64 74 84 94 104 114

3 2 2 3 4 2 4 3 3 4 4 2

5 15 25 35 45 55 65 75 85 95 105 115

4 3 3 4 2 4 2 3 2 2 4 4

6 16 26 36 46 56 66 76 86 96 106 116

3 4 4 2 1 4 3 2 1 1 2 1

7 17 27 37 47 57 67 77 87 97 107 117

4 2 2 1 3 1 3 4 2 3 3 4

8 18 28 38 48 58 68 78 88 98 108 118

2 2 1 1 2 2 1 2 1 1 3 4

9 19 29 39 49 59 69 79 89 99 109 119

3 3 4 4 4 4 2 2 3 1 1 1

10 20 30 40 50 60 70 80 90 100 110 120

4 4 2 4 4 2 3 4 3 2 2 4

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

30 10 30 10 30 10 120

150

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MCT-0005/07 555

1

1. 2

We try to reconcile differences. Hence irreconcilable differences is the most appropriate in the context. Also irretrievably and irreconcilable go well with the tone of the sentence and give parallel emphasis. Essential/contradictory/ opposite differences do not create the necessary emphasis.

2. 2

There is no need for a contrast between the two parts of the statement. In fact the second part of the sentence seems to complement the first. Hence ‘reigning and manager’ – which is complementary and positive is the appropriate choice.

3. 3

The correct sentence is “… were left to our discretion.”

4. 3

The sentence is incorrect, breeding requires an adjective “…good breeding “ as it is preceded by ‘the standard’, thus the specific standard needs to be defined.

5. 4

The author talks about this in the fourth paragraph at the beginning – “The philosophes had also attacked the Church because it blocked human reason. The Romantics attacked the Enlightenment because it blocked the free play of the emotions and creativity. The philosophe had turned man into a soulless, thinking machine — a robot.” This proves that the Romantics who came after the philosophes were fighting against the over dependence on reason and harking back to emotions. This makes choice 4 correct.

15. 3

D & B introduce the subject matter through a report. A introduces the subject who is being discussed. C concludes by presenting Olmert’s point of view.

16. 4

D is the opener stating a fact about Madonna’s accomplishments. A goes on to describe the reactions of the crowds. C closes by explaining the reason for the mania making option ‘4’ correct. BC is a mandatory pair as C explains ‘the long lines’ in B.

17. 2 B introduces the subject about Delta’s fortune. Also it has a general idea - ‘frustrations of airline travel’. BD forms a mandatory pair the frustrations being the pricing. C expands the topic further; A explains the reason for the frustration, making option 2 correct. 18. 2

The 2nd sentence is incorrect, the article preceding ‘…great Sanctuary’ should be ‘the’. The 4th sentence is incorrect, there is an article missing, ‘simple crown’ needs to be defined, preceded by either ‘the’ or an ‘a’.

19. 3

Sentence 1 is incorrect as ‘who’ has been used incorrectly for inanimate objects ‘that’ is the correct pronoun to be used. Sentence 2 is incorrect, the correct subject-verb-agreement, should be ready-to-wear clothing ‘is’, making option 3 correct.

6. 3

The author mentions this at the end of the third paragraph. “The Romantics were conscious of their unique destiny. In fact, it was self-consciousness which appears as one of the key elements of Romanticism itself” making choice 3 correct.

20. 4

Sentence 1 is incorrect ‘the’ should precede ‘greatest naval…’, In sentence 2 the correct phrase is ‘to build up’, in Sentence 3 ‘for acquiring …’ should be ‘in acquiring’.

7. 4

The author mentions in the second paragraph – “To speak of a Romantic era is to identify a period in which certain ideas and attitudes arose, gained currency and in most areas of intellectual endeavor, became dominant. That is, they became the dominant mode of expression. Which tells us something else about the Romantics: expression was perhaps everything to them”. None of 1, 2 or 3 talk about this.

21. 4

Sentence 4 is incorrect. Error of subject-verb-agreement, should be ‘humans in their’ not its, the subject is humans.

22. 4

The précis in the question is talking about the ‘first abacus’, the fourth option elaborates this point and we have a complete description of the original abacus.

23. 2

The answer lies in the line ‘… warmest passions are directed’, making option 2 the most appropriate answer. 1 & 3 talk of wish / repressed wish. 4 does not go into the cause of dreams.

24. 2

Issue forth: Come out of a place.

25. 3

Fall away - to withdraw support or allegiance.

26. 4

Walked through - To perform (a play, for example) in a perfunctory fashion, as at a first rehearsal is the best choice. It is better than ‘lounged about’ which means ‘to laze around’. ‘Pulled over’ is used with vehicles.

27. 2

The author mentions this in the first paragraph, where he talks about how it opposed Victorian architecture which emphasized on ornamentation. This makes choice 2 correct.

28. 1

The author mentions this in the fourth paragraph, and talks about how the Modernists were going to create simpler designs which would be cheap and therefore be suitable for the working classes. This makes choice 1 correct.

29. 4

The author mentions this in the fourth paragraph – “modernist architects in the early twentieth century acted as much out of disdain as empathy for the lives of the middle-class people meant to crowd into the “machines for living.” This makes choice 4 correct. ‘Disdain’ - which means considering others beneath oneself is synonymous with condescension. ‘Antipathy’ means ‘aversion’. ‘Empathy’ is closer to acceptance than benevolence.

8. 2

The author mentions this in the fourth paragraph – “The philosophe had turned man into a soulless, thinking machine — a robot,” Making choice 2 correct.

9. 3

Statements 1 & 2 are clearly Facts as they deal with pieces of information open to verification. Statement 3 conveys an opinion that involves occurrences of the past. Statement 4 is an instance of an inference, as it a conclusion drawn about the unknown on the basis of the known.

10. 4

Statements 1 & 4 express opinions making them Judgements. Statement 2 is stating a verifiable piece of information making it a Fact. Statement 3 is an inference - a conclusion about the unknown - ‘contributors’, based on the known - ‘that Wikipedia’ is the biggest encyclopedia.

11. 1

All four statements are expressions conveying opinions making option 1 correct.

12. 4

Statement 1 is clearly a fact. Statement 2 is an opinion and Statements 3 & 4 are facts which can be verified.

13. 1

C is the obvious opener. It introduces the subject matter and the works of Paulo Freire. ‘A’ goes on to elaborate the subject, also, forms a mandatory pair with ‘A’. BD is better after CA since it seems an expansion of Freire’s methodology.

14.2

The opener is B, sets the tone of the passage, D goes on to elaborate the same. ‘The industry’ in D refers to the ‘Scent business’ in B. C adds further specific information to the topic. A discusses the advances made by Givaudan in the field, making BDCA the correct option.

2

555

30. 2

31. 2

The author mentions this in the seventh paragraph - “Nearly all our suburbs—tracts of Georgian revivals, Cape Cod bungalows, faux adobes—evoke the past rather than the Modernists’ future,” writes Glazer”, making choice 2 correct.

P

41. 1

O H3

The author mentions this in the third paragraph – “Bioaccumulation of most major toxins insures that those of us on the top of the food chain will absorb the full brunt of an overdeveloped industrial ethos”, making choice 2 correct.

D A

H1

H2 B

C

Q

32. 1

The author mentions in the sixth paragraph – “….that we in the developed North have lost nearly all conception of our daily bread as the staff of life”, making choice 1 correct.

Let the radii of H1, H2 and H3 be R1, R2 and R3 respectively. Volume (H2 ) 1 = Volume (H1) 8

33. 1

The author uses this example in the last paragraph, while he talks about how strict adherence to orthodox commitments to a particular diet might go against responsible ecological food choices. The purpose of the author is to show the irony of the situation rather than just mock the situation. This makes choice 1 correct.

2 3 πR2 R 1 3 ⇒ = ⇒ R2 = 1 2 3 8 2 πR 3 1

34. 3

35. 4

36. 2

37. 1

38. 1

Consider the triangle OAB.

The author mentions this is the second last paragraph, and begins the last paragraph praising the anarchists for the way they treat food and says ”…anarchists just love to eat”. This makes Choice 3 correct.

O R3

The author makes this comment at the end of the passage – “Moral orthodoxy, whether in religious or dietary conviction, shuts down more studied and complex understandings of the world around us”, which is closest to choice 4.

R1

R2

OC2 = AO2 – AC2 2 ⇒ (R3 + R2)2 = R12 − R22 = R1 −

⇒ R3 +

Options (1), (2) and (3) are synonymous. They refer to cheaters and ways of deception. Apparition is the odd word which means ‘ghost’ or a ‘ghostlike form’.

R12 3 2 = R1 4 4

R1 3 R = 2 2 1

 3 − 1 ⇒ R3 =   R1  2  3

⇒

Choices (2), (3) and (4) represents fear of amphibians, insects and reptiles respectively. But (1) is fear of ‘pain’ – hence the odd one out.

40. 4

B

C

R2

The paragraph refers to the popularity of Grimm’s Fairy Tales all over the world. It starts from general information regarding its overall impact and gradually refers to its impact in specific regions. It ends with the dedication of the Japanese to the Grimm’s. So, reference to the United States is the logical extension to the passage.

Choices (1), (2) and (3) are tombs whereas Sacellum is a Monument within a church – a small chapel.

9 0°

A

The theme of the paragraph is the Egyptian Kings, the Egyptian people and the concept of immortality. The paragraph emphasizes the fact that the people willingly participated in immortalizing their King, even though they had no hope of a berth in the tomb. So choice (2) which specifically refers to the people - ‘hoi polloi’ and suggests ‘immortality by proxy’ continues the idea left off in the last line. Choices (1) & (4) could come one step later since they are general. Choice (3) is vague.

39. 4

R1

D

3

 3 − 1 Volume (H3 )  R3  3 3 −5 =  =  =  Volume (H1)  R1  2 4  

Alternative Method: The ratio of radii of H2 & H1 is 1 : 2 and from the figure radius ‘r3’ of H3 is less than r2. So, the ratio of volume of H3 and H1 is less than 1 : 8. Hence, option (2) is right. 42. 4

[x + 3 ] < 5 −5 < [x + 3] < 5 If we take x = 2, [x + 3] = 5 and if we take x = –8 [x + 3] = –5. Therefore all integer value of ‘x’ which are greater than –8 and less than 2 satisfy the given inequality. Therefore 9 integer values of x satisfy the inequality.

555

3

43. 1

and (1 + b 4 + a5 ) ≥ 3 3 b 4a5 ...(iii) Multiplying the three inequalities, we get: A

X

(1 + a + c3)(1 + b2 + c3)(1 + b4 + a5) ≥ 273 (a.c 3 ) × (b2c 3 ) × (b4a5 )

P

G

6

 1  1 3 = 273 (abc)6 = 27   = 27   9 9

Q G

∴ Minimum possible value of the product is

R

Let, ∠XRQ be ‘θ’ ∴ ∠XAQ = 180 – θ [XAQR is a cyclic quadrilateral] ∴ ∠PAQ = 180° – ∠XAQ = 180° – (180° – θ) = θ ∴ ∆XPR ~ ∆QPA

⇒

46. 1

XP PR RX = = QP PA AQ

XP PR XA + AP PR = ⇒ = QP PA QP AP Let AP = x

⇒ (x − 6)(x + 8) = 0 ∴ AP = 6 units  XP  ⇒ RX = AQ   = 4 5 units  QP  2

2

∆ XPR 9XAQR + ∆ APQ 4 = = 1 ∆ APQ ∆ APQ [3 Area of ∆XPR = 9XAQR + ∆QPA]

3 ∆XPR 4

Now, we can observe that XR2 + XP2 = PR2 \ DXPR is a right angled triangle.

1 1 (RX )( XP ) = × 4 5 × 8 = 16 5 sq. units 2 2

⇒ XAQR = 44. 4

45. 2

Akhil

No. of phases

4

5

Age (in yrs)

36

18

Therefore,

x 20 = (a + r) (b –r)

Also, (x +60) = 40 (a + r) (b + r)

As a, b and c all are positive numbers, we can apply AM ≥ GM property.

Given that b = 5r

1+ a + c3 ≥ 3

3

1× a × c3

Bunty Chameli

Disha

Eku

x

8

8

9

72

24

For questions 48 and 49: Let the distance between point Q and point P be ‘x’ m. Let the speed of A, B and the river be a, b and r respectively.

Obviously all prime numbers till 20 can be elements of ‘N’. ‘1’ can also occur along with those prime numbers. There are 8 prime numbers till 20. Even if we add ‘4’, ‘6’ and ‘9’ in this list, that will not violate any condition. So there can be a maximum of 12 elements in ‘N’. They are 1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 17 and 19.

So,

4

3 × 16 5 = 12 5 sq. units 4

Name

Since Disha is the eldest, she would do least work per phase. Let us assume that the work done per phase by Disha = 1. Then, work done per phase by Akhil = 2 Bunty = 4 Chameli = 8 Eku = 3 Total work done by all = 4 × 2 + 5 × 4 + 8x + 8 × 1 + 8 × 3 = 60 + 8x Chameli did 40% of the work. ⇒ Others did 60% of the work ⇒ 60% of the work = 60 ⇒ 40% of the work = 40 = 8x ⇒ x=5 Hence, Chameli worked on 5 phases of the project.

XP  4 8 Ratio of areas of ∆XPR to ∆QPA =   =4 = 1  QP   

∴ 9 XAQR =

1 . 3

Let, x – 2y + w = 3k ...(i) Therefore y + 2z – 2x = 2k ...(ii) and z – w + y = k …(iii) Adding (i), (ii) and (iii) we get 3z – x = 6k (i) – 2(ii) + 2(iii) = 5x – 2y – 2z – w = 3k – 4k + 2k = k Required Ratio =(5x – 2y – 2z – w): (3z – x) = k: 6k = 1:6.

47. 3

⇒ x 2 + 2x − 48 = 0

∴ ∆XPR =

2

... (i) ...(ii)

Dividing equation (i) by (ii) we get: 2x (b + r) = (x + 60) (b – r)

Therefore,

...(iii)

2x = 1.5 ⇒ x = 180 m (x + 60)

Distance between points Q and R = 180 + 20 = 200 m.

⇒ (1 + a + c 3 ) ≥ 3 3 a.c 3

...(i)

Putting value of x = 180 in (i) we get:

Similarly, (1 + b2 + c 3 ) ≥ 3 3 b2c 3

...(ii)

a+r =9 4r ⇒ a = 35r = 7b

(a + r) = 9. (b – r)

⇒

555

48. 2

Ratio of speeds of A and B = 7 : 1

49. 4

When B turned back, at that instant A was ‘y’ m from point P. Distance traveled by A = 180 – y.

In ∆RNP : tan θ =

Distance traveled by B = (180–y) 9

53. 1

Time taken by A to reach the point R, from this instant =

( y + 20) (a + r)

Time taken by B to reach the point R, from the point where he turned back = (180 – y) . (9b + 9r) Therefore

(y + 20) (180 – y) = . (a + r) (9b + 9r)

Perfect square less than 200 are 1, 4, 9………196. Perfect cubes less than 200 are 1, 8, 27, 64 and 125. So, ‘P’ will contain the following 21 elements: I. (1 × 1), (4 × 1), (9 × 1)………(196 × 1) ⇒ 14 elements II. (1 × 8), (4 × 8), (9 × 8) and (16 × 8) ⇒ 4 elements III. (1 × 27) and (4 × 27) ⇒ 2 elements IV. (1 × 125) ⇒ 1 element In order to have exactly 12 factors, a number has to be expressed in the form of a2 × b3, where ‘a’ and ‘b’ are prime numbers. Out of the above, only 2 elements can be expressed in the form of a2 x b3, where ‘a’ and ‘b’ are prime numbers. They are (9 × 8) and (4 × 27).

( y + 20) (a + r) 2 = = (180 – y ) (9b + 9r) 3

Or,

Therefore y = 60 m. So, the required probability is 50. 4

There are two ways of writing 50 as a sum of two perfect squares, namely 52 + 52 and 72 + 12. So there are 12 integral solutions of the equation x2 + y2 = 50. x = ±5, x = ±1, x = ± 7,

51. 3

54. 2

y = ±5 ⇒ 4 solutions. y = ±7 ⇒ 4 solutions. y = ±1 ⇒ 4 solutions.

LCM = 3 × 11 × 41 × 43. The two numbers are four-digit numbers. Thus following cases arise: HCF 1st Number 2nd Number 33 × 41

33 × 43

66

41

41 × 33

41 × 43

410

43

43 × 33

43 × 41

344

55. 4

θ

30° = 15° 2 In ∆A3A8A4,∠A3QA4=∠A3A8A4 + ∠A12A4A8=15° +60°= 75° Angle subtended by A12A9 at the center = 90°

∴ ∠A12A4A9 =

Q

Similarly ∠A1A9A4 =

Since ∠PSQ = 45°, SQ = H m

∠RQS = ∠QRT = 60° RS x + H = = tan60° = 3 SQ H

90° = 45° 2

90° = 45° 2 In ∆A1A9A4, A1PA4 = ∠A12A4A9 + ∠A1A9A4 = 45° + 45° = 90° Required Ratio = 90° : 75° = 6 : 5

Let PQ be the building and RS be the unbroken part of the tree. Let the angle of elevation of the broken point of the tree from the top most point of the building be ' θ ' . Let PQ = H m and RN = x m

x = 3 −1 H

Since the circle is divided into 12 equal parts, each part will subtend the same angle at the center.

∴ ∠A3A8A4 =

P

Hm

⇒

...(iv) ...(v)

120° = 60° 2 Angle subtended by A3A4 at the center = 30°

4 5° S

5(ii) − 3(i) : 14x − 9z = −17 5(iii) − 2(i) : 19z − 4x = 87

∴ ∠A12A4A8 =

6 0°

N

...(i) ...(ii) ...(iii)

360° = 30° 12 Angle subtended by A12A8 at the center = 4 × 30° = 120°

T

xm

2x + 5y + 3z = 4 4x + 3y = −1 2y + 5z = 19

The angle subtended by any part =

Hence option (3) can never be possible.

R

2 . 21

Solving (iv) and (v) we get that z = 5 and x = 2 Putting the value of x = 2 in (ii) we get that y = –3 Therefore x + y + z = 2 – 3 + 5 = 4.

Difference

33

52. 2

2 3 +1

 2  θ = tan−1    3 + 1

  (a + r) Since (b – r) = 9   

Now they meet at the point R

x = 3 − 1 ⇒ tan θ = H

56. 4

Assume PQR is greater than RQP, then P is greater than R. After checking units digit of both the numbers, we get that (6 + R – P) = P. Therefore R is an even number. So it can be either 2 or 4. If R is 2 then P is 4 and the difference will be in three digits, which is not possible. So, R is 4 and P is 5. Now, 6 + Q –1 – Q = P therefore Q can take, any value from either 0, 1, 2 or 3. So the value of ‘P + Q + R’ cannot be determined.

555

5

Time taken from the instant pipe B is opened till the tank is 57. 1

7 25  3 x 2 − 2(p − 1)x +  p2 + p − >0 3 2  4 7 25  3 ⇒ x 2 − 2(p − 1)x + (p − 1)2 +  p2 + p − − (p − 1)2 > 0 3 2  4

completely filled =

Total time taken to completely fill the tank

13 27  2  1 p− ⇒ [x − (p − 1)] +  − p2 + >0 3 2   4

[x − (p − 1)]2

= 2+3+ 62. 3

is always positive for any real values of x and p.

13 27  −3p2 + 52p − 162  1 Now  − p2 + p − = 3 2  12  4 The least integral value of p for which

−3p2 + 52p − 162 is 12 63. 3

59. 4

Water initially in the glass = 100 ml In each sip, quantity of liquid inside the glass is reducing by 10 ml. Hence, total number of times the drunkard sipped it = 9. Net quantity sipped = 9 × 20 = 180 ml. Out of 180 ml, 100 ml was water. Thus, 80 ml of alcohol was sipped.

Let the other root be β. As α is one of the roots,

or α2 − 3α + 2 = 0

⇒ α = 2 or 1 Now,

α+β = ⇒β=

8 7 × ml 10 9

1+ α 2

1 − α −1 or 0. = 2 2

3 or 1. 2 Maximum possible value of the sum of roots = 1.5 ⇒ (α + β) is either

Similarly, water in the glass after 4th sip

8 7 6 5 × × × = 33.33 ml 10 9 8 7 Net quantity in the glass after 4th sip was taken and 10 ml alcohol was added = 100 – 4 × 10 = 60 ml Remaining (60 – 33.33) = 26.66 ml must be alcohol. Ratio after the 4th sip = 26.66 : 33.33 = 4 : 5

Three balls can be selected from the 13 balls in 13 C3 ways. These 3 balls can be put into three boxes in 3! ways. Now the remaining 10 balls can be put into these 3 boxes in 310 ways. Therefore total number of ways in which 13 balls can be put into three different boxes such that each box contains at

2(α )2 − (1 + α )α + 2(1 − α ) = 0

Water in the glass after first sip = 80 ml In the glass, 10 ml of alcohol was added. Thus, water and alcohol should be in the ratio 80 : 10, even when the sip of 20 ml is taken. ⇒ Water in the glass after second sip = 100 ×

45 80 3 = = 11 hrs 7 7 7

least 1 ball = 13 C3 ×310 × 3!

greater than zero is 5. 58. 2

(3x5x4x3) 45 hrs. = 4(12 + 15 − 20) 7

= 100 ×

60. 2

x3

64. 4

= p ´ 42 + p ´ 4 ´ 42 + 62 = 16 p + 4 p 52

ax2

+ + 3x + 11 f(x) = and g(x) = x3 – x2 + ax + 10 The points of intersections of the two curves are the roots of the equation f(x) – g(x) = 0. i.e. (x3 + ax2 + 3x + 11) – (x3 – x2 + ax + 10) = 0 ...(i) or (a + 1)x2 + (3 – a)x + 1 = 0 If the two curves do not intersect each other then the determinant of equation (i) must be negative.

Surface area of the cone P

= π × 42 + π × 4 × 42 + 52 = π(16 + 4 41) Surface area of the part of the cone Q which is left out = 4π( 52 + 41) Required Ratio =

– 4 × 1 × (a + 1) < 0 ⇒ ∆ = (3 – or a2 – 10a + 5 < 0 a)2

⇒ 5−2 5
Initially Pipe A is opened: Time taken till the tank is half filled with water = 2 hrs. Then Pipe A and C both are open: Time taken from the instant pipe C is opened till pipe B is opened =

(4x3) = 3 hrs. 4(4 − 3)

Since the height of the cone P is a positive integer and the surface area is maximum, therefore the height of cone P is 5 units and radius is equal to 4 units. Surface area of the cone Q

65. 2

4 + 41 4 + 41 π(16 + 4 41) = = 4π( 52 + 41) 52 + 41 2 13 + 41

5x = 120 + y

y 5 For ‘x’ to be an integer ‘y’ must be a multiple of 5. Since the values that ‘x’ assumes have opposite signs as compared to the corresponding values of ‘y’, ‘y’ will have to be negative. Or, x = 24 +

So all multiples of 5 from – 5 to –115 (both inclusive) will give positive integral values of ‘x’ Therefore number of integral solutions = 23.

Pipe A, B and C are open:

6

555

66. 3

Let ‘x’ be the cost price of the article.

Half of the class = 100 students ⇒ a + b + c = 100 and d = f + 10 Here, a + d + e + g = 80 b + d + f + g = 75 c + e + f + g = 60 Adding all of the equations, ⇒ (a + b + c) + 2(d + e + f) + 3g = 215 ⇒ 100 + 2(d + e + f) + 3g = 215 Let (d + e + f) = S ⇒ 2S + 3g = 115, where S is greater than or equal to 10.

p  p   Here x – x  1 −   1 + 100  = 21  100   

⇒ x×

p2 1002

= 21

... (i)

 p2  Also (x – 21)  1 − = 2058  1002   

p2

⇒ ( x – 21) – x ×

⇒ x − 21 − 21 + ⇒ x + 21×

2

100

+ 21×

p2 1002

= 2058

(because d = f + 10). 68. 1

21p2 = 2058 100

p2 1002

= 2100

...(ii)

Solving (i) and (ii) we get that 'p' ≈ 10

67. 3

Students who enrolled for at least one of the three activities = (a + b + c) + (d + e + f) + g = 100 + S + g Thus, we have to minimise the above value. If 2S + 3g = 115, 2S + 2g = 115 –g

⇒ (S + g) =

55 45 35 5 5 15 55 + + + ........ + - ...... 2 4 8 64 128 256 4096

115 – g 2

\

115 – g 2 To minimise 100 + (S+g), we have to maximise ‘g’ so that the expression in the right hand side is minimum. Thus, 2S + 3g = 115 and S ≥ 10

S 55 10 10 10 10 10 10 55 ....... – = - ....... + 2 2 4 8 64 128 256 4096 8192

⇒ maximum value of g = 31, when S would be equal to 11. Therefore, minimum value = 100 + 31 + 11 = 142

Let S =

⇒ 100 + (S+g) = 100 +

S 55 45 15 5 5 45 55 ........ = + + ...... + + 2 4 8 64 128 256 4096 8192 Subtracting (ii) from (i), we get

S é 55 55 ù 1 ù é1 1 – 10 ê + + ...... = + 2 êë 2 8192 úû 4096 úû ë4 8 é1 æ 1 öù ê 4 ´ èç1 - 11 ø÷ ú S 55 ´ 4097 2 ú – 10 ê Þ = 2 8192 ê æ1 - 1 ö ú êë èç 2 ø÷ ûú

Þ

69. 2

x2 – 2x = 0

70. 3

⇒ x(x − 2) = 0 ∴ x = 0 and x = 2 Checking with the options: Option (3): y = –(|x–1|–1) If we put x = 0, we get y = –(|0 – 1|–1) = 0 If we put x = 2, we get y = –(|2 – 1| –1) = 0 ∴ y = –(|x –1|–1) has identical roots as that of the equation x2 – 2x = 0.

S 55 ´ 4097 10 ´ 2047 = 2 8192 4096

S 184395 = 2 8192

\S =

Students enrolled for exactly 2 activities = S. If 2S + 3g = 115, maximum value of S would be 56, when g = 1.

184395 » 45 4096

71. 1 A

For questions 68 and 69: S w im m ing (8 0)

S katin g (7 5) a

d

b Q

g f

e c

D a ncin g (6 0)

D

C

O

P

S

E M B

Let, the radius of the smaller circle with center at P be r. ∴ OP2 = (R – PS)2 = (R – r)2 ⇒ OM2 + MP2 = OP2 ⇒ OM2 = (R – r)2 – r2 Let the side of the square OCDE be ‘a’ units ⇒ R2 = OD2 = a2 + a2 ⇒ a =

555

R 2

7

R ⇒ OM = OE + PQ = +r 2

 R  +r ⇒ (R – r)2 – r2 = OM2 =   2  ⇒

R2

R2 + 2Rr + r 2 – 2Rr = 2

⇒ 2r2 + 2r

−2 ⇒ r= ⇒ r=

2

( 2R + 2R ) – R2 = 0   2 2 ( 2 + 1)R +  8 ( 2 + 1) + 8  R  

R 2 2 4 

 4

(4 + 2 2 )



−2 2

(

 2 +1  

)

R  2 2 + 4 – 2 − 1  2  = 0.14 R

∴ r=

72. 3

In all possible sets, two numbers are prime.

73. 1

5 sets

74. 3

Since, 9xy + 18(xy 2 + y) + 27(x 2 y 2 + 1) = 19y 2

⇒

9x 18(xy 2 + y) 27(x 2 y 2 + 1) + + = 19 y y2 y2

⇒

  9x 1 1  + 18  x +  + 27  x 2 +  = 19 2  y y  y  

  1  1 1  19 ⇒ ( x )   + 2  x +  + 3  x2 + =  y y     y 2  9  2

   1  19 1 1 ⇒ 3  x +  + 2  x +  − 5 (x )  = y y     y 9 To minimise the value of x + of

For questions 72 and 73: Highest two digit prime number is 97. So the maximum possible sum of these four number is 97 – 36 = 61. The other possible sums are 89 – 36 = 53 and 83 – 36 = 47. The minimum number we can have is 10 and maximum cannot be more than 28 .Why? 61 – (28 + 10) = 23 = (11 + 12). If we have the maximum number greater than 28 then we cannot get the other two middle numbers as distinct numbers greater than 10. Also, minimum cannot be more than 14. Why? Suppose we start with 15 and take 4 consecutive integers {15, 16, 17, 18}. The sum will be more than 61, which is not possible. Starting with 10 and 28 as extreme numbers, we have one combination {10, 11, 12, 28} which leads to a sum of 61. But this combination violates the second condition that sum taken two at a time should yield a prime number in three occasions. Next we try with 10 and 27 as extreme numbers. We get the following combination. {10, x, y, 27} Now, (61 – 37) = 24 24 we can get in (11, 13) which satisfies the condition for sum taken two at a time and three at a time. Hence, one solution is {10, 11, 13, 27}. Basically, we have to take care of the number to be added to the sum of extreme numbers so as to get a prime. Next we see that the 2nd condition is valid only for {10, x, y, 21}. Here 61 – 31 = 30 10,12,16 cannot be middle numbers in this case. So we can have only (11, 19) and (13, 17) as the middle numbers if the extreme numbers are 10 and 21 leading to two different cases. Next we can see that other possible combinations of extreme numbers are 11 and 21. Thus we get the following combination: {11, a, b, 21} Here, (61 – 32) = 29. Here 11, 15, 21 cannot be middle numbers. So, we can have only (12, 17) and (13, 16) as middle numbers in this case. We have 6 solutions. Hence option (3) and (1) are the correct choices for Q.72 and Q.73 respectively. We can go on to prove that for minimum greater than 11 we cannot have a valid set. The total 6 sets being {10, 11, 19, 21}, {10, 11, 13, 27}, {10, 13, 17, 21}, {11, 12, 17, 21}, {11, 13, 16, 21} {11, 12, 13, 25}.

8

...(i)

1 we need to maximise the value y

x . y

Applying A.M. ≥ G.M. , maximum possible value of

1 1 x +  4 y

x is y

2

Putting this maximum value of

x in equation (i), we get y

2

 7 1 1  19  x +  + 2 x +  = 4 y y 9  Let, x +

⇒

1 =a y

7 2 19 a + 2a − =0 4 9

⇒ 63a2 + 72a − 76 = 0 ⇒ 63a2 + 114a − 42a − 76 = 0 ⇒ (3a − 2)(21a + 38) = 0

 1 2 We get  x +  = , which is the minimum possible value of y 3   1  x + . y 

∴ Minimum possible value of

1 1 1 2 1 x +  = × = 4 y 4 3 6

555

75. 3

z: [4] = 1 choice ∴ Total number of such multiples = 2 × 2 × 2 × 8 = 64.

R E G Q

D

77. 4

C

S F P A N

B

Percent discount =

Let QR intersect DC at E and SP intersect DA at F. Also let PQ extended meet DC at G. Area of pentagon DEQPF = [ ∆AGD − ∆APF − ∆QGE]

There is 1 equilateral triangle initially. Whenever we pick any triangle and cut out one equilateral triangle, we are getting 4 equilateral triangles as a result. Thus, each such cutting increases our count from 1 to 4, that means an increment of 3. Therefore, the number of triangular metal sheets at any moment during the process has to be of the form (1 + 3n), where ‘n’ is any integer. Here, only option (2) is not of that form and is our answer.

79. 2

Here,

2 4 cm and AG = cm 3 3

1 2 2 cm ×2× = 2 3 3 Drop a perpendicular PN from P to AB. Let AN = x cm.

(12–10.5) × 100 = 12.5% 12

78. 2

In ∆AGD, AD = 2 cm. Therefore, DG =

Let the cost of 1 apple be ‘a’, cost of 1 mango be ‘m’ and the cost of 1 orange be ‘o’ Therefore, (2a + 3m + 4o) – (a + 2m + 2o) = 6 ...(i) And (3a + 3m + 5o) – (a + 2m + o) = 8 ...(ii) 2 x (ii) – (i): 3a + m + 6o = 10. To make a profit of 5 % one should sell 3 apples, 1 mango and 6 oranges at Rs. 1.05 × 10 = Rs. 10.5

Area of ∆AGD =

A B A

1 Then AP = 2x cm and PN = 3x cm ⇒ x = cm 8

× 2 3 1

1 Obviously PF = cm 4 3

A B A ×

Area of ∆APF =

A B A A B A × A B A ×

1 1 1 1 cm2 × × = 2 4 4 3 32 3

A B A × × A B A × ×

4 1  In ∆QGE, QG = AG − AP − PQ = − 2 1 +  8 3  =

∴ EQ =

(

3 × 16 − 9 3

Area of ∆QGE =

16 − 9 3 cm 4 3

) = 16 − 9 4

4 3

3

cm

 16 − 9 3  1 × 3 ×   2 4  

2

3  16 − 9 3   2   1  ∆AGD − ∆APF − ∆QGE =    − − 2 4  3   32 3   

=

76. 2

For, the digit at the tens place of the product, B + 3A = Base + A or 2A + B = Base ...(i) [In case of carry over 2, hundred digit ‘4’ is not possible] For the digit at the hundreds place of the product, 3A + 3B, = Base + Base + 4 + 1(carry over from the tens place) ...(ii) ⇒ 2B + A + 1 = Base + 4 from (i) and (ii) we get 2B + A + 1 = 2A + B + 4 ⇒ B=A+3 For thousands place of the product: 2(carry over from the hundreds place) + 3A + 2B = Base + Base + A 2

864 3 − 1434  717  2 2 =  27 −  cm = 1.15 cm 32 3 16 3  

2 + 3A + 2B = 2 (2A + B ) + A

⇒ A =1 ⇒ B = A + 3 = 1+ 3 = 4 Hence, Base = 2A + B = 2 × 1 + 4 = 6 [In case of other bases 5, 7, 8 or 9, this product is not possible]

A = 10 × w 9 × x 7 × y 3 = 2 × 5 × w 9 × x7 × y3 s

B = 50 × w × x13 × z 4 C = A × B = 22 × 53 × w10 × x 20 × y 3 × z 4

Since we need to find the common multiples of A and B such that they are factors of C, so the choices for the different powers of 2, 5, w, x, y and z available are as follows Powers of 2: [1 or 2] = 2 choices 5: [2 or 3] = 2 choices w: [9 or 10] = 2 choices x: [13, 14, ...20] = 8 choices y: [3] = 1 choice

555

9

82. 4 A

80. 4

Q

P

That means, (106 – 60) = 46 bottles out of the total bottles of Z, which were manufactured in February were left unsold. Since we have to maximize the number of days on which 5 bottles of Z were manufactured, we will maximize the number of days on which 3 bottles of Z were sold. So maximum possible number of days on which 3 bottles of Z were sold = 30.

R D G B

C

S

In-radius of the ∆ABC =

Therefore number of bottles of Z that were manufactured in February = 46 + 3(30) = 136. Let the number of days on which 5 and 4 bottles of Z were manufactured be ‘a’ and ‘b’ respectively. Therefore the number of days on which 2 bottles of Z were manufactured = (30 – a – b). 136 = 5a + 4b + 2(30 – a – b) or 3a + 2b = 76. Maximum possible value of a is 24 when b = 2. Therefore maximum possible number of days on which 5 bottles of Z were manufactured = 24. In the month of March, (166 – 106) = 60 bottles of Z out of the bottles of Z manufactured in the month of March were left unsold. So, maximum possible number of days on which 5 bottles of Z were manufactured in the month of March = 30, when 3 bottles of Z were sold on each of the 30 days in March. Therefore, maximum possible number of days on which 5 bottles of Z were manufactured across all the three months = 30 + 24 + 30 = 84.

2 3 = 1 cm 2 3

Circumradius of the ∆ABC =

2 3 = 2 cm 3

Join CG and extend it to intersect the side AB at D. Area of ∆ADC =

1 1 3 3 3 cm2 ∆ABC = × × (2 3)2 cm2 = 2 2 4 2

AP = 1 cm ∴ PD = ( 3 − 1) cm CR = 2 cm ∴ AR = 2( 3 − 1) cm ∴ Area of ∆APR =

1 3( 3 − 1) cm2 × AP × AR × sin60° = 2 2

∴ Area of ∆CRG =

1 × CR × CG × sin30° = 1 cm2 2

∴ Area of ∆PDG =

1 ( 3 − 1) 2 cm × PD × DG = 2 2

83. 1

∴ Area of ∆PRG = Area of [ ∆ADC − ∆APR − ∆CRG − ∆PDG] =

81. 3

Number of bottles of Z left unsold at the end of January = 60. Maximum possible number of days on which 5 bottles of Z were manufactured in January is 30, when the numbers of days on which 3 bottles of Z were sold = 30.

3 3 − 4 2 3 3  3( 3 − 1) ( 3 − 1)  − + 1+  =   cm 2 2 2 2    

Number of bottles of Z left unsold at the end of January = 60. If maximum possible number of bottles were manufactured, then maximum possible number of bottles should be sold. That means that on each of the 30 days in the month of January, 5 bottles were manufactured and 3 bottles were sold. That means 5 units of ‘Y’ can not be manufactured and 3 units of Y can bot be sold on any day of January. In the month of January, let, the number of days on which 4 bottles of Y were manufactured be x, therefore the number of days on which 2 bottles of Y were manufactured = 30 – x. Let the number of days on which 1 bottle of Y was sold be ‘y’, therefore the number of days on which 2 bottles of Y were sold is (30 – y). Obviously ‘x’ and ‘y’ are integers less than or equal to 30.

Total number of bottles of X manufactured in all the three months = 144 + Total number of bottles of X sold in all the three months. Let the number of days on which 5 and 4 bottles of X were manufactured in all the three months be p and q respectively. Or, 5p + 4q + 2(90 – p – q) = 144 + 90 (to maximise the number of bottles of X sold in all the three months) 3p + 2q = 54. Minimum value of p will be 0, when the value of q is 27. Therefore minimum possible number of days on which 5 bottles of X were sold in all the three months = 0.

84. 3

Ratio of number of days on which 2, 4 and 5 bottles of Y were manufactured in the month of February is 1:2:3, therefore the number of days on which 2, 4 and 5 bottles of Y were manufactured is 5, 10 and 15 respectively. Number of bottles of Y manufactured in the month of February = 2(5) + 4(10) + 5(15) = 10 + 40 + 75 = 125. Let the number of days on which 1, 2 and 3 bottles of Y were sold in the month of February be d, e and f respectively. Therefore, 125 – d – 2(30 – d – f) – 3f = 110 – 53 Or, f – d = 8 Or, f = 8 + d. Therefore on at least 8 days 3 bottles of Y were sold. So, minimum value of ‘f’ is 8, and in that case d = 0. Number of days on which 2 bottles of Y were sold in February = 30 – 8 = 22.

53 = 4x + 2(30 – x) – y – 2(30 – y) = 2x + y. Maximum possible value of y = 29, because if y = 30 then the value of x is not an integer.

Minimum possible value of x =

10

53 − 29 = 12 . 2

555

85. 2

To maximize the difference between the number of days on which 2 bottles of Y were manufactured in January and March we need to maximize the number of days on which 2 bottles of Y were manufactured in January and minimize the number of days on which 2 bottles of Y were manufactured in March. January: Let the number of days on which 2 and 4 bottles of Y were manufactured be ‘g’ and ‘h’ respectively. So, the number of days on which 5 bottles of Y were manufactured = (30 – g – h). 2g + 4h + 5(30 – g – h) = 53 + number of bottles of Y sold in January. Or, 3g + h = 97 – number of bottles of Y sold in January. In order to maximize the value of ‘g’ we need to minimize the number of bottles of Y sold in January. Minimum number of bottles of Y sold in January will be when 1 bottle of Y is sold on all 30 days. Or, 3g + h = 67. Maximum possible value of g will be 22, when the value of h = 1. March: Number of bottles unsold at the end of February = 110. Number of bottles unsold at the end of March = 140. Total number of bottles of Y manufactured in March = 140 – 110 + Number of bottles of Y sold in March. Minimum value of days on which 2 bottles of Y manufactured can be zero. One such case is when 4 bottles of Y were manufactured on all days in March and 3 bottles of Y were sold on all days in March. Therefore, maximum possible difference between the days on which 2 bottles of Y were sold in January and March is (22 – 0) = 22.

For questions 86 to 89: From statement 1, E’s anniversary date can be 14th or 24th and that of H can be 7th or 12th in that order. From statement 2, the anniversary date of D can be 7th May or 12th December and correspondingly C’s anniversary date can be 12th or 24th. From statement 3, the anniversary date of A-E can be 24th May or 14th May or 14th March. From above conclusions and statement 4, we can conclude that A’s marriage anniversary is on 14th March and D’s marriage anniversary is on 7th May. Now C’s anniversary date comes out to be 12th. For both D and H, the marriage anniversary date comes out to be 7th May. So, they must form a couple.

Husband

A

B

C

D

Wife

E

F/G

G/F

H

Anniversary date

14

24

12

7

Month

March Oct/Dec Dec/Oct May

86. 1

24th March cannot be the anniversary date for any of the given couples.

87. 2

Only option (2) is a possible combination.

555

For questions 88 and 89: The possible anniversary dates of F can be 24th October or 24th December or 12th October or 12th December. Among these only 24th October and 24th December gives the maximum value, in other cases sum is not maximum. But 24th December = 24 + 12 = 36 is a perfect square. So F’s anniversary is on 24th October. As B has his anniversary date on 24th so F is the wife of B. 88. 1

B is the husband of F.

89. 3

G’s marriage anniversary is on 12th December.

90. 3

N beats W and never meets L. That means N looses in the second round. P wins two more matches than N. That means P wins in the third round and looses the final. So P is the runner-up of the tournament.

91. 4

P played in the final, that means he has won at least 3 matches. So N has won at least one match. T is from group A and plays the final match. So he represented group A in the third round. Therefore T, V and X are from group A. Both T and V won their first round matches in group A, so N can not play with them. Z wins at least one match so he cannot play with N in the first round. P plays final so he also cannot play with N in the first round. J, M and S also cannot play with N from given conditions (1) and (7). If J, M, N and S are in the same group then N and S must win their first round matches and meet in the second round, which is not possible. So N may have beaten any one among K, L, Q, R, U, W and Y in the first round.

92. 1

Please note that ‘N’ appears in every option. Assume N reaches the third round, and then he plays with L. So P can not play his third round match with N. So option (3) is ruled out. Therefore P, N and S are in the same group. By checking options, we observe the following: If P plays the first round match with S, then he plays second round with N. Thus S and N should be the first two opponents. Only option (1) holds true.

93. 4

N wins round 2 Match against W that means N reaches third round, then he plays with L, who is from group D. So N is from group C and other players of groups C are S, W and Q. So Q cannot be in the group B. Therefore option (2) is ruled out. Assume L, U, J and M are from group D and R, T, V and X are from group A, then K, P, Y and Z are in group B, which is possible. Assume P, T, U and R are from group A and L, R, K and Y from group D, then J, M, V and X are in group B, which is also possible. Hence option (4) is the answer.

94. 4

If P is either from group A or B then P definitely plays round 3 match 1. This is because even if N fails to win any match, P must win at least 2 matches (refer statement 6). If P is either from group C or D, then we cannot definitely comment about round 3 match 1 players.

11

For questions 95 to 98: 1.

The number of days for which Tata Steel’s share witnessed an increase was one more than the number of days on which it witnessed a decrease. Therefore in 5 consecutive days, there were 3 increments and 2 decrements. But if the price would have been higher than 527.5 on 3rd February, then there would have been 3 consecutive increments and 2 consecutive decrements. Therefore, the share price on 3rd February was lower than the price on 2nd February and even lower than Rs. 527.

2.

Also, since the price of Modi steel increased on 4 days and decreased on 1 day, therefore the share price would have increased on 1st February because there is already a decrease from 4th to 5th February.

Based on 1 and 2 above, the following is the offer price of the companies from 1st February to 6th February.

Sl.No

Nam e of the Com pany

1 2 3 4 5

Offe r pr ice of the Com panie s Indus tr y

Gr oup

Tata Steel

Steel

Tata

594

595

596

594

595

593

JK Steel

Steel

JK

592

590

588

589

590

591

Essar Steel

Steel

Essar

591

592

593

594

595

596

Modi Steel

Steel

Modi

596

597

598

599

600

598

Nippon Steel

Steel

Nippon

598

596

597

598

599

597

1/2/07

2/2/07

3/2/07

4/2/07

5/2/07

6th Fe br uar y 2007

Based on the above table, all the questions can be answered. 95. 2

Modi Steel had the highest offer price of Rs.598 as on February 6th.

96. 1

Tata Steel, JK Steel and Nippon Steel each had a price difference of Rs. 1 per share on 6th February as compared to 1st February.

97. 3

As on 5th February, the highest price offers were from Modi Steel and Nippon Steel and therefore these two companies dropped out. That means the next highest bid was Rs. 595 which indicates a tie between Essar Steel and Tata Steel.

98. 1

As on 4th February, only two companies had an offer price that is higher than Rs.595, which happened to be Modi Steel and Nippon Steel. Therefore the remaining 3 companies were not eligible for further participation.

For questions 99 and 100:

For questions 101 and 105:

For A, B, C, D and E respectively, ratio of gross salary = 4 : 8 : 10 : 3 : 5 ratio of tax paid = 4 : 16 : 17 : 3 : 10 ratio of net salary = 36 : 64 : 83 : 27 : 40

101. 4 Pass percentage of the whole batch

Since (gross salary – tax) = net salary, we can form the following 5 equations: 4x – 4y = 36z 8x – 16y = 64z 10x – 17y = 83z 3x – 3y = 27z 5x – 10y = 40z Solving them, x : y : z = 10 : 1 : 1 99. 1

100. 2

A’s tax = 4y = 2000 ⇒ y = 500 ⇒ x = 5000 ⇒ gross salary of E = 5x = Rs. 25,000 Tax paid by B = 16y Net salary of D = 27z ⇒ ratio = 16y : 27z = 16:27

=

63 + 70 + 38 + 54 + 61 × 100 90 + 90 + 60 + 90 + 90

=

286 × 100 = 68.1% 420

102. 4 Maximum number of additional students passed from ECE is less than 70% of 12 ≈ 8 Maximum number of additional students passed from CSE is less than 70% of 10 = 6 Maximum number of additional students passed from IT is less than 70% of 14 ≈ 9 Maximum number of additional students passed from ME is less than 70% of 20 ≈ 13 Maximum number of additional students passed from EE is less than 70% of 14 ≈ 9 Passed students in ECE = 63 + 8 = 71 Passed students in CSE = 70 + 6 = 76 Passed students in IT = 38 + 9 = 47 Passed students in ME = 54 + 13 = 67 Passed students in EE = 61 + 9 = 70 Ratio Ratio Ratio Ratio Ratio

12

of passed students to failed students for ECE = 71 : 19 of passed students to failed students for CSE = 76 : 12 of passed students to failed students for IT = 47 : 12 of passed students to failed students for ME = 67 : 17 of passed students to failed students for EE = 70 : 19

555

Clearly ratio is the maximum for CSE in this case. But if we assume that no additional student passed from CSE, then the ratio is maximum for ME department. So correct answer is option (4).

108. 3

I. The number that appeared on the top face of the dice when A threw the dice is 6, then the distance between the brothers becomes = 14 – 6 = 8 steps. II. The number that appeared on the top face of the dice when B threw the dice is 2, then the distance between the brothers becomes = 8 + 2 = 10 steps. III. The number that appeared on the top face of the dice when A threw the dice is 5, then the distance between the brothers becomes = 10 + (6 – 5) = 11 steps. IV. The number that appeared on the top face of the dice when B threw the dice is 1, then the distance between the brothers becomes = 11 – (6 – 1) = 6 steps. V. The number that appeared on the top face of the dice when A threw the dice is 6, then the distance between the brothers becomes = 6 – 6 = 0.

103. 2 Failed students in ECE = 27 – 6 = 21 Failed students in CSE = 18 – 5 = 13 Failed students in IT = 21 – 7 = 14 Failed students in ME = 30 – 10 = 20 Failed students in EE = 28 – 7 = 21

Failed percentage in ECE =

21 × 100 = 23.3% 90

Failed percentage in CSE =

13 × 100 = 14.4% 90

Failed percentage in IT =

14 × 100 = 23.3% 60

Failed percentage in ME =

20 × 100 = 22.2% 90

21 × 100 = 23.3% 90 Minimum percentage of failed students is from CSE.

109. 1 This can be achieved in seven throws. If the number appearing on the top face of the dice in the seven throws is 4, 5, 6, 3, 4, 5 and 6 in that particular order. Distance moved forward after these seven throws = 4 + 5 +

Failed percentage in EE =

104. 4 Total students who applied for re-evaluation = 12 + 10 + 14 + 20 + 14 = 70 Number of student who passed after re-evaluation = 0.5 × 70 = 35 Total number of students who should appear next year = failed + absentees = 35 + (2 + 1 + 6 + 1) = 45 105. 4 It is not known how many students will be there next year for this paper, as then new batch will be appearing for the paper also. It is not clear whether the new batch strength of ME will be 90 or not. 106. 2 Maximum possible distance between the two brothers is when one brother moves maximum possible distance in 4 r o u n d s and the other brother moves minimum possible distance in those 4 rounds. Maximum possible distance that a brother can move in 4 rounds is equal to 18 steps. This is possible when the number appearing on the top face of the dice is 3, 4, 5 and 6 in the four rounds in that particular order. Therefore the distance moved by the brother in first, second, third and a fourth round is 3, 4, 5 and 6 steps respectively. Minimum possible distance that a brother can move in 4 rounds from start is equal to 0 steps. This is possible when the number appearing on the top face of the dice is 4, 2, 4 and 2 respectively or 5, 1, 5 and 1 respectively in those four rounds in that particular order. Therefore maximum possible difference between the brothers after 4 rounds = 18 steps. 107. 3 Maximum possible distance covered by A after 6 consecutive throws in which the number appearing on the top face of the dice are distinct and 6 appears in the third throw is 15 steps. This can be achieved if the number appearing on the top face of the dice in these 6 consecutive rounds is 1, 2, 6, 3, 4 and 5 in that order. Therefore the distance covered in these 6 rounds = 1 + 2 + 6 – 3 + 4 + 5 = 15 steps.

555

At least 5 throws are required in order to make the distance between the two brothers ‘zero’. Initial distance between the two brothers = 14 steps.

6 – 3 + 4 + 5 + 6 = 27 steps. 110. 2 The sequence will be 1, 2, 3, 4, 5, 6. So the answer is 21 steps. 111. 3

Let the total number of boys in the class be ‘x’, therefore the total number of girls in the class will be equal to (200 – x). Average weight of the students in the class 55x + 45 (200 − x )

 x  =  + 45  20  Minimum possible average weight of the students in the class

=

200

 20  is   + 45 = 46. Therefore there are 20 boys and 180 girls  20  in the class in this case. Let the number of boys in section A, B, C and D be a, b, c and d respectively. Therefore, 70a + 60b + 55c + 40(20 – a – b – c) = 55 × 20 Or, 6a + 4b + 3c = 60 Maximum value of c is 14 when a = 1 and b = 3.

Let the number of girls in section A, B, C and D be p, q, r and s respectively. Therefore, 50p + 55q + 45r + 40(180 – p – q – r) = 180 × 45 Or, 10p + 15q + 5r = 180 × 5 Or, 2p + 3q + r = 180. Maximum value of r is 175 when p = q = 1. Therefore at most there could be 14 + 175 = 189 students in section C. 112. 2 Since we need to find the maximum possible ratio of the number of boys to the number of the girls in section D, we need to maximise the number of boys and minimise the number of girls in section D and hence overall. So, we have to assume that there are 180 boys and 20 girls in the whole class. Let the number of boys in section A, B, C and D be a, b, c and d respectively. Therefore, 70a + 60b + 55c + 40(180 – a – b – c) = 180 × 55 Or, 6a + 4b + 3c = 540.  2b  c Or, a = 90 –   –    3  2

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Since we need to maximise the value of d, b = 3 and c = 2. For b = 3 and c = 2, the value of a = 90 – 2 – 1 = 87. Therefore maximum possible value of d = 180 – (87 + 3 + 2) = 180 – 92 = 88. Let the number of girls in section A, B, C and D be p, q, r and s respectively. Therefore, 50p + 55q + 45r + 40(20 – p – q – r) = 20 × 45 Or, 2p + 3q + r = 20. Maximum possible value of (p + q + r) is when r = 15, p = q = 1. Therefore, s = 20 – (15 + 1 + 1) = 3.

After round 2: After round 2, the sum of absolute differences in ranks of friends after round 1 and present ranks is 10 and three friends have same absolute differences in ranks. So the possible combinations of differences can be (2,2,2,4,0), (3,3,3,1,0) or (1,1,1,4,3) There is no possible arrangement of ranks with combinations (2,2,2,4,0) and (3,3,3,1,0). With combination (1,1,1,4,3) the possible ranks can be as follows:

Rank After Round 1

Maximum possible ratio of the number of boys to the number of the girls in section D = 88:3 113. 2 Let the total number of boys in the class be ‘x’, therefore the total number of girls in the class will be equal to (200 – x). Average weight of the students in the class 55x + 45 (200 − x )

 x  = =  + 45 200  20  The following table gives the possible number of boys and girls in the class.

Boys

20

40

60

80

100

120

140

160

180

Girls

180

160

140

120

100

80

60

40

20

Ratio

1:9

1:4

3:7

2:3

1:1

3:2

7:3

4:1

9:1

Rank after Round 2 Differences (1,1,1,4,3) Case I

Case II

Case III

A

2

1

3

5

B

4

3

5

3

C

5

2

1

1

D

1

5

4

2

E

3

4

2

4

114. 2 Possible initial ranks of A can be 1, 2 and 5. So, among the given numbers, only 1 and 2 are possible. 115. 4

Maximum sum of initial rank of C and rank secured after each of the 2 rounds can be = 4 + 5 + 2 = 11

116. 1 A cannot have 4 as initial rank or rank after any round. Therefore except for 4:3 and 3:1, every other ratio could be a possible ratio of the number of boys to the number of girls in the class.

117. 4 Only A, B and D can possibly secure rank 5 after round 2. For questions 118 to 120: Revenue increased or decreased in each year. However, only the absolute value of the increase/decrease is given.

For questions 114 to 117: After round 1: A’s rank is higher than B’s rank with both of them having even numbered rank. So A’s rank is 2 and B’s rank is 4. C, D and E can occupy ranks 1, 3 and 5 not necessarily in that order. The difference between the ranks of D & E; and E & C is same, with D having highest rank among the given three. So, after round 1, D’s rank is 1, E’s rank is 3 and C’s rank is 5.

1990 to 1991

Year

Change in Revenues (in Rs. Lakhs) 20

1991 to 1992

30

1992 to 1993

10

1993 to 1994

40

Also, it is given that the final increase or decrease was of 5%. The sum of absolute differences between the ranks of friends after round 1 and the initial ranks is 8 and two different pairs of friends has same absolute difference in ranks. So the possible combinations of differences can be (0, 0, 4, 2, 2) , (0, 0, 3, 3, 2) and (1, 1, 3, 3, 0) So the possible initial ranks can be as tabulated below:

Differences (0,0,2,2,4) (1,1,3,3,0)

Names

Rank after Round 1 Case I Case II Case I Case II A

2

2

5

1

2

B

4

4

1

5

4

CASES Revenues Change in revenue from in 1990 1990 to 1991 to 1992 to 1993 to 1991 1992 1993 1994 1 800 +20 –30 +10 +40

Revenues in 1994 840

C

3

1

4

2

5

2

D

5

3

2

4

1

3

800

–20

+30

–10

+40

840

E

1

5

3

3

3

4

800

–20

+30

–10

–40

760

Initial Rank There is no combination of initial ranks possible with difference in rank as (0,0,3,3,2)

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118. 4 It is given in the statement that revenues in 1990 were Rs. 8 crores or Rs. 800 lakhs. Thus, taking the 5% increase or decrease, it can be said that the revenues in 1994 could be Rs. 760 lakhs or Rs. 840 lakhs. Thus, the change has been of Rs. 40 lakhs. 40 lakhs can be possible in only these 4 circumstances.

800

+20

–30

+10

–40

760

Only the cases (2) and (4) are in accordance with our conclusion. Checking the options, it can be seen that in option (4) whenever there was a decrease in the revenue from 1990 to 1994, the revenue from 1993 to 1994 has also decreased necessarily.

555

119. 1 According to the statement, there was a decrease in revenues from 1992 to 1993. Here, considering the conclusion, it can be seen that revenue between 16 to 20 crores is a big figure. Infact, the maximum possible revenue in 1990 could be when there has been a consistent increase of 100 lakhs from 1990 to 1994. The revenue would be 20 crore in that case. Therefore, as per the conclusion, there could be only two cases: Revenues in 1990 1600

1990 to 1991 20

1991 to 1992 30

1992 to 1993 –10

1993 to 1994 40

Revenues in 1994 1680

2000

–20

–30

–10

–40

1900

The conclusion would be true only when the information in the options restrict to one (or both) of the cases above. Along with the statement, if you consider option (1), you can see that there is a consistent drop in the revenues and hence, the final revenue lies between 16 crore and 20 crore.

555

120. 4 From the statement, the rise from 1990 to 1992 is just 1.25%. From 1990 to 1992, the increase could be possible in only two cases: + 20 + 30 = + 50 ⇒ 1.25% = 50 ⇒ 100% = 4000 –20+30 = +10 ⇒ 1.25% = 10 ⇒ 100% = 800 But revenues of Rs. 4000 lakhs is not possible, because then the final change in revenue can never be 5%. 5% of Rs. 40 crores is Rs. 2 crores. Even if we consider successive increase of Rs. 20 lakhs, Rs. 30 lakhs. Rs.10 lakhs and Rs. 40 lakhs (from 1990 to 1994), we can only get an overall increase of Rs. 1 crore. Thus, the revenues in 1990 have to be Rs. 800 lakhs and option (1) is inconsistent with our main statement. To get an overall increase/decrease of 5%, there should be a decrease of 10 lakhs in 1993 and increase/decrease of 40 lakhs in 1994 (as illustrated in the explanation of quesiton 118). Thus, option (2) is superfluous. The information in this option is implicit in the main statement. Similarly, it can be seen that option (3) is inconsistent. Option (4), if true, would lead you to the stated conclusion.

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