e-prac-cat # 13 solutions

8/8/2019 E-Prac-CAT # 13 Solutions

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e-Prac-CAT # 13ANALYSIS

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Solutions

1. The first number in the given range that is divisible by 23 is 805 and the last one is 1196.Common difference = 23Tn = a + (n - 1) dWhere a = first termn = number of termsd = common difference

Tn = nth

termSo, 1196= 805 + (n 1) 23Or n = 18Therefore, sum of all the 18 terms by the formula: Sn = n/2 [2a + (n 1) d]= 18/2 [(2 805) + (18 1) 23] = 18009 Ans. (1)

2. The distance AC = 55 5 = 275 kmSimilarly, BC=150 kmSince, D is the midpoint of BC,BD = 75 kmThe distance from A to D can be obtained from the median theorem:-AB2+AC2=2(BD2+AD2)Hence, AD=228.4 or approximately 228 km. Ans. (1)

3. Solution: G(x) = x2 + nx 24 = 0x(x + n) = 24x can take the values 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24

Thus, n can take the corresponding values: 23, -23, 10, -10, 5, -5, 2, -2, -2, 2, -5, 5, -10, 10, -23, 23But, -10 < n 10Hence, n cannot be equal to 23, 23 or 10Hence, total 10 values are possible.Ans. (2)

4. Any of the 4 colours can be chosen for the 1st stripe. Any of the remaining 3 colours can be chosen for the 2nd stripe. The 3rd stripe can again becoloured in 3 ways (excluding the colour of the 2nd stripe but including the colour of the 1st) and so on up to the 6th stripe, each of which can becoloured in 3 ways.Thus, number of ways the flag can be coloured is 4 3 5 = 12 81Ans. (4)

5. A D = (A B)(C D) /(B C) = 4B E = (B C)(D E)/(C D) = 10C F = (C D)(E F)/(D E) =18Hence, the ratio is4 : 10 : 18Or 2 : 5 : 9Ans. (1)

For Q.6 & Q.7:If the boy was born on the 31st of December and they met on 1st January, then according to the statement in the question, 2 days back, i.e., on 30thDecember (say 2007), the boy was 10 years old. Now, on 31st December 2007, he turns 11. Today, i.e., on 1 st January 2008 he is 11 and this yearon his birthday he will turn 12, i.e., on 31 st December 2008. Therefore, next year on 31st December 2009, he will turn 13.

6. Ans. (4)7. Ans. (4)

8. Let A = a, B = ar, C = ar2 and D = ar3, where r is the common ratio.Then, the equation becomesX2a2r3 + a2r2 + a2r4 = a2r3

r2 + (X2 1)r + 1=0As r is a real number, so, Discriminant = (X2 1)2 4 0(X2 3) (X2 + 1) 0Or X 3 or X 3 Ans. (4)

9. Given equation (15x2 +8y2)/22 = xy can be written as,15x2 22xy + 8y2 = (3x 2y)(5x 4y) = 0 Case I:y = 3/2xThere are 33 such pairs (2, 3) (4, 6)(66, 99) Case II:y = 5/4xThere are 19 such pairs (4, 5) (8, 10)(76, 95) Ans. (3)


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10. Let the increase in yoga fee be x, then according to question,Yoga teacher earns the amount = (40 + x)(600 5x) = 24000 + 400x 5x 2 = 5[6400 (x 40)2]The above expression would have maximum value when x 40 = 0 or x = 40 and the maximum value is 32000. Ans. (3)

11. We have,x + [y] + {z} = 3.8 (1)[x] + {y} + z = 3.2 (2){x} + y + [z] = 2.2 (3)

Adding the above three equations we get,2(x + y + z) = 9.2 (Since, x, y and z are positive real numbers)Or, x + y + z = 4.6 (4)Subtracting equation (1) from equation (4) we get,{y} + [z] = 0.8{y} = 0.8 and [z] = 0,Subtracting equation (2) from equation (4) we get,{x} + [y] = 1.4{x} = 0.4 and [y] = 1,Subtracting equation (3) from equation (4) we get,[x] + {z} = 2.4[x] = 2 and {z} = 0.4So, x = 2.4, y = 1.8 and z = 0.4Hence, [(x2 + y2 + z2)] = 9Ans. (3)

12. From the figure, the trapezium ABFC has HJ as the line joining mid-points of the non- parallel sides. G and I are the points of bisection of the twoheights of trapezium and lie on HJ.Now, AD = 2 cmApplying Pythagoras theorem in triangle ACD,CD = 2 cmIn similar triangles ACD and AGH,(AG/AD) = (HG/CD)Since, AG = 1 cm, so, HG = 1 cmAlso by symmetry, HG = IJ,And given that AB = GI = 4 cmHG +GI+ IJ = 1 + 4 + 1 = 6 cm Ans. (3)

13. The intersection of the 2 equations y = 8 - |x| and y= |x| - 8 is:(0,8)

(0,-8)

(-8,0) (8,0)

A B4

H G

C D E F

JI

22


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One of the sides would be 82. The sides would be inclined at 45 with the diagonal as it is a square field.

Hence, the perpendicular dropped from the centre of the field to one of the sides of the square would be 4 2.The radius of the largest circle will be 42.Thus, the area that the cow can graze upon is p(42) 2

So, the area of the remaining field where there is still grass = (8 2) 2 32p = 128 - 32pAns. (4)

14. Since, total 260 votes are cast in each round, let the number of votes cast for L.K. Advani be x and for Manmohan Singh be 260 x.Now the number of votes for L.K. Advani in the second round is (1.25)x. Then the number of votes obtained by Manmohan Singh = 260 1.25xAs per the given conditions,2(260 - 2x) = 2.5 x - 2603 260 = 6.5xx = 120The votes cast for Manmohan Singh are 140 and 110 in the first and the second round respectively.Hence, the percentage decrease in the number of votes = (30/140) 100 = 21.4%. Ans. (3)

15. A number when divided by 100 leaves the same remainder as the quotient only when the number is of the form abab. Also if the last 2 digits aredivisible by 4 and 3, only then the number is divisible by 12. So the possible numbers are 1212, 2424, 3636, 4848, 6060, 7272, 8484, and 9696 i.e. atotal of 8 numbers.

Ans. (4)

16. (A) xy(x3 + y3) = xy(1 3xy) = 1/3 (3xy (1 - 3xy))Since, 3xy + (1 3xy) = 1 is a constant value, so the maximum value of 3xy (1 - 3xy) occurs when 3xy = 1 3xyOr, xy = 1/6On substituting this value of xy in (A),q = 1/12(B) If n is positive, then [n/2] + [n/4] can be at the most 3n/4. Thus, no solution for positive value of n exists. For non-negative n, n = 0 is a trivialsolution. For n < 5, LHS > RHS. For 5 n < 0, we have 3 solutions, n = -2, -3, -5. Thus, p = 4. So, p q = 1/3. Ans. (3)

For Q.17 & Q.18:By construction, we can observe the following:Since, ABC = 90, hence, AD is a diameter of length 2 units.Now, since, BGFE is a square, hence, GBO = 45Hence, GBA = 45Thus, GBA = ADB = 45 (Alternate Segment Theorem)

17. In s BDP and BGO,PBD = GBOBPD = BOGHence, they are similar.PD : OG = BP : BO = 1 : 1But, BO can be of any length.Ans. (4)

18. IfACB = 67.5 then BAC=22.5 and hence AC is angle bisector ofBAD.BP/BA = 1/2Since BP = 1.Hence, BA=2unitsAD = 2Hence, by angle bisector theorem, AB : AD = BC : CD = 1/2. Ans. (3)

42 4

2

(0,8)

(0,-8)

(-8,0) (8,0)8


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19. q(x) =1/(x21) and p(x)= q(x 1)Hence, p(x) = 1/[(x - 1)2 - 1] = 1/(x22x)Now, if p(x) and q(x) intersect, then p(x) = q(x)Hence, 1/(x21) = 1/(x22x)Or, x = 1/2Hence, the two functions intersect only once. Ans. (4)

20. Odds against P = 7:4Thus, probability of P occurring = 4/7 + 4 = 4/11

Odds in favour of Q = 8:7Probability of Q occurring = 8/(8+7) = 8/15Therefore, the probability that at least one of the events will happen

= - - -

= -

= -

=

1 14

111

8

15

17

11

7

15

149

165116

165

[( )( )]

[ ]x

Ans. (3)

21. The given condition suggests that a lies between the roots.

So, D 0 and if f(x) is the given equation, f(a) < 0

Or, 4(2 1 8 1 02a a(a+ - + ) )

8[ 1 2 02a a+ + / ]

Now, f(a) < 02a2 2(2a + 1)a + a(a + 1) < 0-a2 - a < 0Or, a2 + a > 0So, a > 0 and a < -1 Ans. (4)

22. D = where A picks up BC = Where C got off

ID = m

I C B

D C

BC = kCD = x

If we take the ratios of the speeds of A and C,Speed of A during the distance DB/ Speed of C during the distance BC = 50/10(because when C got off at point C and A went back for B, they all reached the destination at the same time, so the time is constant)

(2x + k)/k =50/10x/k = 2/1(IC + CD)/ID = 50/10(2x + m)/m = 5/1x/m = 2/1x = 2k = 2mk = m = x/2x + k + m = 120k = 30 kmm = 30 kmx = 60 kmTotal distance = IC + CD + DB = m + x + x + x + k = 240 kmTime required = 240/50 = 4.8 hours Ans. (3)

23. In his way to reach his destination the person adopted two types of walking patterns.In the first type he took 2 steps forward and 1 step backward, which he did for a number of times and the other one in which he took 3 stepsforward and 1 backward, he did for b number of times.Since, he takes a minute for each step, the total time taken to reach the destination becomes3a + 4(b 1) + 3 = 34 minutes (as in the last three minutes, he reaches his destination by taking 3 steps forward)The total number of steps taken in the backward direction are a + (b 1) = 9Therefore, a = 5 and b = 4So, after 3a minutes or 15 minutes, the man changed his walking pattern. Ans. (1)


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24. 4 cubes above x-axis and 4 below x-axis are possible.From the figure, it is clear that no other cube is possible except the cubes shown. Ans. (4)

25. Among the 60 people, 2/3rd of males and 3/7th of females were not feeling well. This means that we have to divide 60 in 2 parts such that we get onedivisible by 7 and the other by 3.This can be done in 2 different ways:-(a) Males = 39, Females = 21

Number of times each step is performed is[(1/3) 39] [21 4/7] = 156(b) Males = 18, Females = 42[(1/3) 18] [42 4/7] = 144Hence either (1) or (2) is possible.Ans. (4)

26. In the equation |x|3 > 2x2 - |x|Case I:Taking x > 0, the equation becomes,x3 - 2x2 + x > 0or x(x - 1)2 > 0(x - 1)2 is already greater than zero. Hence, x must be greater than zero.Also, x can neither be 1.Hence, x > 0 and x 1Case II:Taking x < 0, the equation becomes,

x3 > 2x2 + xor x (x + 1)2 < 0(x+1)2 is already greater than zero, hence, x < 0Also, x can neither be 1.Hence, x > 0 and x 1Combining both the cases we getx(,) {1} {1} {0}Ans. (2)

27. In all, let there be 100 students in the school.b = 10p = 0y = 40c + y + z = 80Hence, c + z = 40Also, a + b + c + x + y + z + p = 100a + 10 + 80 + x + 0 = 100a + x = 10

Since, c + z = 40And, a + x = 10Hence, to maximize b + x + z,a = 0 and x =10 c = 0 and z = 40So, a maximum of 40 + 10 + 10 = 60 students are possible admirers of History. Ans. (4)

28. (1 + x)a (1 x)b = (1 + aC1x + aC2x2 ) ( 1 bC1x + bC2x2..)Coefficient of x = aC1 bC1 = 3Therefore, a b = 3 (1)Coefficient of x2 = aC2 + bC2 aC1 bC1 = 6Or, a(a 1) + b(b 1) 2ab = 12a2 + b2 2ab (a + b) = 12(a b)2 (a + b) = 129 + 12 = a + bOr, a + b = 21 (2)From (1) and (2), a = 12 and b = 9. Ans. (3)

29. The more relevant thing here is not the number of strikes but the number of time intervals. In other words, if a clock has to strike 4, there are 3 timeintervals between the 4 strikes (the 1st strike happens at the 0th sec). So, in 7 seconds, the pendulum had 3 time intervals. To strike 11, there have tobe 10 time intervals which will take (10 7)/3 = 70/3 seconds. Ans. (3)

30. (P 1)! is not divisible by P when P is Prime. There are 10 prime numbers in this range. Also P = 4 satisfies this condition hence total numbers are11. Ans. (3)

31. Ans. (4). In the fourth paragraph, the author states that every man has the right to occupation and uses this to prove further arguments. Clearly, thisis one of his assumptions. This same line can lead us to the assumption that all men are born equal. When the author states that to live every manmust need land or material on which to cultivate, it becomes clear that he is aware of only these ways of earning the livelihood. Option (3) is evidentin the paragraph again through the line that every man has the right to occupation. Option (4) can be chosen as the incorrect one as this is not an

a x b

p

y z

c

Mathematics

Historyscience


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assumption, but a conclusion that the author draws on the above assumptions. Option (2) can be easily said to be an assumption as the authoragrees that this is a premise on which he draws conclusions.

32. Ans. (4). In the fifth paragraph, the author defines an usufructuary which is a function of utilizing the land by ensuring that the land is not harmed inany way. This definition is best phrased in option (4). Option (1) is far-fetched as it says that an usufructuary excludes the function of a proprietor.This is not true as it can be seen that the proprietor is acting from the background by making the usufructuary pay him. Option (2) is nowherementioned in the passage. Option (3) is partially correct but misses out the imporatnt fact that the usufructuary must ensure that the land is notgetting harmed in any way.

33. Ans. (3). Duranton says "Possession is a matter of fact, not of right." By this he means that this is an agreement in which the tiller of the land canwork on it till the owner deems it fit. This is a situation that can be called as a matter of fact, as "this is something that is". No rights get accrued to thetenant is the meaning of the other part. Option (3) is the correct answer. Option (1) is vague. (2) is missing out the 'rights' part of the statement.Option (4) is in contrast to the passage in a sense that the passage doesn't say that the tiller doesn't make profit out of the activity.

34. Ans. (4). The author mentions in the first paragraph that this is a vain distinction that has been adopted in order to justify the abuse of the landwherein the abuse has not been defined. The author goes on to mention in the passage that abuse may be defined as use in the matter of property.If we ask the question that why has this justification been attempted, then we would get to know that the only utility for this purpose is that userestricts the occupation of land to a proprietor and gives him authority over land. But this authority should not be over-exercised and should stay in alimit, not in the law's eyes but in a way that it does not hamper any proprietor's practice. Simply speaking, a proprietor should not encroach any otherproprietor's domain.

35. Ans. (3). Jus in re and jus ad rem are defined as 'right to it' and the 'right in it'. We can understand this with the example provided in the passage.When one marries, one gets the legal right to the other person and the marriage. But when one gets engaged, there is no legal right involved exceptthat there is a right to the engagement itself. Thus, the closest of the options is option (3) where a signed agreement holds a legal right whereas, a

promise is oral and hence lacks the legal enforcement. If we were to compare two rights or two duties, we might get closer to the answer.

36. Ans. (4). In option (1), secure is used with the meaning to make certain ensure. In option (2), it means to bring about effect. Option (3) uses it tomean to obtain. In option (4), secure is used to mean something that is not likely to fail or give way.

37. Ans. (2). Option (2) is incorrect because the correct usage would have been He played ten .... Played down means to belittle. Option (1) usesplay to mean To assume the role of act as. In option (3), it means to perform or put into effect, especially as a jest or deception. Option (4)implies that play means here to emphasize or publicize. In the last option, it means freedom or occasion for action.

38. Ans. (4). In option (1), the usage means 'to devote, apply, or direct'. In option (2), it means 'to have'. Option (3) implies 'to abandon relinquish'. Inoption (4), throw means 'to make more accessible, especially suddenly or dramatically'.

39. Ans. (4). We will just take a look at all the idioms used here so that the answer becomes self-evident.Set out: to start a journeySet forth: to express in words

Set forward: to begin a journeySet to: to begin working energeticallySet at: to attack or assailSet apart: to make noticeableSet aside: to discard or rejectSet in: to insertSet off: to direct attention to by contrast accentuate.

40. Ans. (2). Again, let's see what every idiom means:Put by: to save for later usePut away: to renounce discardPut down: to attributePut on: to addPut in: to introduce, as in conversation interposePut off: to persuade to delay further actionPut through: to cause to undergo

41. Ans. (1). Let on: 'to allow to be known'

Let go: 'to cease to employ dismiss'Let up on: 'to be or become lenient with'Let (someone) in on: to reveal (a secret) toLet alone: not to mention much lessLet up: to slow down diminishLet loose: to allow to escape

42. Ans. (1). In a previous paragraph, author speaks about the importance of archetypes. In the next, he gives examples. It is evident that this is done inorder to cement any conclusion that the author is trying to draw from the paragraph. Thus, option (1) is the best answer. Option (2) is incorrect as inthe fifth paragraph, the author is not concerned with the origin of the archetypes. Option (3) is unlikely to be the answer as it is incorrect. The authordoes not intend to contrast the two known archetypes, instead he wants to use the two contrasting archetypes to prove his point. Option (4) which


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says that the examples help build the stage for the introduction of Carl Jung's theories can be said to be less than true as this job is accomplished ina better manner by paragraph six.

43. Ans. (4). In the passage, the author first of all introduces what is the meaning of Carl Jung's views. He then provides some examples, like the'Persona' and the 'Shadow'. He supposes the hypothesis to be true as he does not venture to provide any analytical dissection of Jung's theories. Hethen goes on to say that one should worry in the least about the origins of the archetypes and instead focuses the discussion on the importance ofthe archetypes.

44. Ans. (2). The author is completely confident of what Carl Jung said in his theories. This is evident through the paragraph in which he says that wecan completely believe Carl Jung and move on to study considering him to be correct. The author is not believing Jung blindly because he has givenreasons for the confidence that he shows in the theories. Diffidence meaning lack of self-confidence is irrelevant here. The author is neither criticalof nor is trying to justify Jung's theories. He is only giving his opinion about what can be learnt from them.

45. Ans. (4). One look at paragraph nine would reveal the correct answer to be option (4). Options (2) and (3) are both incorrect as they attributearchetypes to only one of the two: either conscience or anti-conscience. This is incorrect as can be seen in the passage. Option (1) is false as itassumes that both conscience and anti-conscience are reflected through same archetypes.

46. Ans. (4). Paragraph four of the passage states that the primitive nature is bent on destroying the better, civilized human nature. This can lead us tothe conclusion that the author would surely agree that human beings are getting better with the development of the conscience. He speaks nothingabout the evolution in general. Option (2) is false as the archetypes are present everywhere and in everyone's dreams as is stated in the opening ofthe passage. Thus, they cannot represent any single person's character. Option (3) makes the mistake of quoting 'will not degrade' for the anti-conscience. The author makes no such claims.

47. Ans. (2). The civilian customs are looked down upon by a military officer and hence the right word to describe the civilian behaviour is "effeminate".Effeminate implies behaviour that is worthy of women. Downtrodden meaning 'wrongfully subjugated and oppressed' is incongruous here. 'Largerthan life' is something that is considered not real. This is not the correct phrasing of the intended meaning. Even hyperbole is not apt as it implies thatthe officer considers the civilian behaviour to be more than what it needs to be. He considers it 'unworthy something degraded' and not 'anythingmore'.

48. Ans. (4). The analogy of a good habitation, well adorned but lacking in a good roof clearly shows that an institution that lacks the army support isgoing to suffer some disadvantages. Hence, we can negate options of a positive connotation like synchronized (option (1)), arranged (option (3)).'Kept' (option (2)) is not fitting here. Option (4) is the best fit.

49. Ans. (2). The author in this line and the lines coming up next is revering the fact that army leads a man to a life of complete loyalty to the country andso on. The author is praising the army and hence cannot use words like negated or hyped. Option (4) may seem a fine enough option but it does notprovide justice to the flow of the paragraph and seems awkward. 'Doubled' is the best fit as it justifies that army is the best place to look for a loyaland God-fearing man.

50. Ans. (3). Sentence A has a missing 'the' before French constitution. In sentence B, immigrant should replace emigrant as the people are coming into

the country. Sentence D needs a 'that' before 'the legal citizens'. In Sentence E, 'hardly won' which means 'lost' isn't appropriate and should bereplaced with 'hard won' that implies a hard fought victory.

51. Ans. (3). Instead of can you be in sentence A, it should be you can be and 'two first' in sentence B should be replaced with 'first two'. Sentence Eshould utilize preferable to instead of preferable than.

52. Ans. (4). In A, we should use the construction 'No sooner had' instead of No sooner has. Instead of when in C, it should be than as the correctusage should be 'No sooner than'. In D, it should be he was much annoyed instead of they were very annoyed.

53. Ans. (1). The author has stated that he has the tendency of following streaks and betting against it mostly on the fourth chance after a streak ofthree. He does this because he has shown that the probability of the win on the fourth hand is meager as compared to the probability of the win forothers. Option (2) introduces a new fact which is unknown to us and hence, can be avoided. Option (3) is a way to consciously lose money as it asksthe gambler to bet on the least possible option. Option (4) asks the gambler to progressively bet which is one thing that the author has asked not todo.

54. Ans. (4). The author advises the reader to stay away from progressive betting in order to ensure that one does not get lost in the flow and spendmore than she/he can afford. Anyone who bets progressively while he is losing is making a mistake as the outcome cannot be influenced by theamount of money bet and hence, the probability of winning/losing doesn't necessarily change. Options (1), (2) and (3) make an assumption that aplayer losing is also a streak and hence can be discarded.

55. Ans. (4). The author in the passage advises the readers to be mindful of streaks and employ their own styles for winning at Baccarat. Theparagraph, on the other hand, says that every single hand is unique in its probability or that a hand is independent of the results of the previoushand, and thus, any strategy that asks the reader to build a style upon streaks is incorrect. Thus, the two thoughts are contrasting in nature.

56. Ans. (1). Clearly, the paragraph tries to draw a conclusion from what has been discussed in the passage. It agrees with the passage on most lines,like betting on the banker is boring, pattern chasing can instill fun in an otherwise simple game and so on. Thus, the two thoughts are completelyaligned with each other.


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57. Ans. (4). The paragraph agrees with the passage on the lines that the Baccarat game is simple and the best way to play it is to have fun. But itdisagrees in a major sense that there are no effective ways to improve your performance at the game. Whereas, the passage totally speaks aboutthe streaks that can lead a person to winning in the game. Thus, there is partial disagreement between the two. The paragraph does draw aconclusion but the conclusion drawn is not based on the passage and hence, the two options (1) and (2) can be rejected.

58. Ans. (3). A major link here is that A must precede E i.e. A must be anywhere before E. The reason for this is that A introduces the hypnosis conceptand E takes it forward. Thus, options (1) and (4) can be negated. Another major link is BA as the 'then' used in A is answering a situation which isposited in B. No other sentence provides a situation that can be answered with a then, and since A doesn't start the paragraph (evident from options)

it has to follow B. In option (2), it can be seen that D cannot precede E as the situation that D speaks of is explained only in C. Thus, option (3) is theonly plausible one.

59. Ans. (2). If we take a look at the ending of the previous question, we can easily conclude that the start for this question has to be D. Even withoutthis, we can stumble across key-links like DA and AC. The 'it' in A refers to a table and this finds mention in D. The 'seeming' phenomenon visible inA is carried forward in C. Thus, only option (2) satisfies the key-links.

60. Ans. (4). The toughest of the three questions, this one should be solved keeping in mind the above questions and solving through the options. Oneseeming key-link is that E has to be the start where we say that 'there's nothing to prove'. This seems to be a reply to the last question where we ask'how to prove...' Sentence A with the 'this' must follow E. The 'this' in A refers to the hypothesis referred to in the previous question. CB is anotherkey-link with its established preposterousness and the definition of the two. This leaves out option (4) as the correct answer.

For 61 to 65:From I4, I5, I6, I8 and I11, the following table can be prepared:

Table - I

Position on bench from left Left-most 2 3 4 5 6 Right-most

Name of the person DGender M F F F M M M

State of belonging Bengal Tamil Nadu Gujarat

Intelligence position (1

being most intelligent)7 1

Marital Status Married

From I7, the above table becomes:Table II


Name of the person D C

Gender M F F F M M M

State of belonging Bengal Tamil Nadu Gujarat

Intelligence position (1being most intelligent)

7 1 2

Marital Status Married Unmarried Married

From I7, C's husband is from Uttar Pradesh and the husband of the lady from Tamil Nadu is from Maharashtra. So, D and the person from Gujaratare unmarried males. From I9, the person from Madhya Pradesh is the most intelligent, so she is the female sitting on the 3 rd position from the left.

Table III


Name of the person D C


State of belonging Bengal Tamil NaduMadhya

PradeshGujarat


being most intelligent)7 1 2

Marital Status Unmarried Married Unmarried Married Married Married Unmarried

From I12, F is a female from Tamil Nadu and D is the 6 th most intelligent person. From I9, B is an unmarried man so he must be from Gujarat.Combining this with I13, the following table can be prepared:

Table IV


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Name of the person D F E C A/G G/A B


State of belonging Bengal Tamil NaduMadhya

PradeshDelhi

Maharashtra/U

ttar Pradesh

Maharashtra/U

ttar PradeshGujarat


being most intelligent)6 7 1 2 3/4 4/3 5

Marital Status Unmarried Married Unmarried Married Married Married Unmarried

61. Ans. (4)

62. Ans. (2)

63. From Table-IV, either A and E or E and G are sitting next to C. Ans. (4)

64. From Table-IV, it is clear that the person from Delhi or Gujarat can be sitting next to the person from Uttar Pradesh but the person from Maharashtrais definitely sitting next to the person from Uttar Pradesh. Ans. (1)

65. From Table-IV, either C-A and F-G are married couples or C-G and F-A are married couples. Ans. (4)

66. This equation has five roots as degree of equation is five. Using statement (A) alone we can find four roots as irrational roots occur in conjugate

pairs. So roots of equation will be 5 2 , - 5 2 , but we cannot say anything about the fifth root. Hence, statement (A) alone is notsufficient. Statement (B) alone is also not sufficient. Using both the statements we can find fifth root as sum of the roots = 4/3 and we already knowthe four roots. Once all roots are known equation can be found out. Ans. (3)

67. Let x = the number of Rs.20 tickets sold by Yogi and y = the number of Rs.10 tickets sold by Sachin. Then, Sachin sold 2x (Rs.20 tickets) + y (Rs.10tickets). Yogi sold x, (Rs.20 tickets) + 3y (Rs.10 tickets). Statement (B) implies: x + 3y = 35 but we cannot say anything about the number of ticketssold by Sachin. Hence Statement (A) is insufficient. Statement (B) implies: 70 = x + 2x + y + 3y and 1000 = 20(x + 2x) + 10 (y + 3y). We have nowtwo equations that can be solved simultaneously to get value of the tickets sold by Sachin. Ans. (2)

68. If the number of toffees with them initially is x, x + 4 and x + 8. Statement (A) is insufficient since we cannot determine a unique value of x. Statement(B) implies that after distributing, the number of toffees with them is: x 2, x and x + 4 respectively. Therefore, x 2 = (x + 4)(x 2). Hence, x = 4.Ans. (2)

69. From statement (A), the amount of the bill was between Rs.15 and Rs.50, which means that, say, for a bill of Rs.20, the tip would have been Rs.4,which is 20%. However, for a bill of Rs.19, the tip would be Rs.2, which is 10.5% < 15%. So, statement (A) alone is not sufficient. From statement(B), the tip was calculated at Rs.8, which means that the bill was at most Rs.49.99, and the tip was at least 16% of the bill. Thus, statement (B) aloneis sufficient, but statement (A) alone is not sufficient to answer the question. Ans. (2)

For Q.70 to Q.73:Male players of all three clubs (19-21)

Table 1

X-Changer EAOL INFY- the Infinite

Byron Gregory Oman

Pete Marshall

Male players of all three clubs (17-19):Table 2


Clive Ibrahim Amin Qrar

Don Ken Ritwik

AbouTaleb Janser

Female players of all three clubs (19-21):Table 3


Eiona Leizani Sue

Female players of all three clubs (17-19):Table 4


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Hasim Natalie Tania

Frida Margerret

70. Since any player of age group (19-21) of EAOL does not want to play in the team with any player of X-Changer of age group (17-19), so, from Table3 & 4, Leizani does not want to play with either Hasim or Frida.Option (4) shows the legal team. Ans. (4)

71. From the conditions given in the question, Janser does not want to play with Leizani or Natalie so they cannot be in the team. The remaining playersare Janser, Gregory, Ken, Margerret and Ibrahim Amin. The team must include one female player so Margerret must be included in the team. Thus,the team comprises Janser, Gregory, Ken and Margerret. Ans. (2)

72. For this question we have to check out the options.Option (1):Sue, Tania, Don, Oman, RitwikGiven that no male player of age group (17-19), who plays for X-Changer wants to play with any female player who plays for INFY the Infinite,hence, Don cannot play with Sue. So, this team is illegal.Option (2):Byron, Abou Taleb, Oman, Ritwik, TaniaIn this team, because of the condition that no male player of age group (17-19) who plays for X-Changer wants to play with any female player whoplays for INFY the Infinite, Abou Taleb cannot play with Tania.So this team is also illegal.Option (3):Byron, Oman, Ritwik, Hasim, TaniaThis team does not contradict any conditions and follows all rules, so this team is legal.

Option (4):Byron, Oman, Abou Taleb, Hasim, TaniaAgain, Abou Taleb cannot play with Tania in a team and hence, this team is also illegal. Ans. (3)

73. Check option (1)Byron, Janser, Pete Marshall, Abou Taleb, Leizani, NatalieIn this team there is only one player of INFY - the Infinite. But condition given in the question is that there must be 2 - 2 players of each club.So this team is illegal.Option (2)Gregory, Janser, Pete Marshall, Eiona, Leizani, TaniaIn this team there is only one player of X-Changer, but condition given in the question is that there must be 2 - 2 players of each club.So this team is also illegal.Option (3)Byron, Gregory, Oman, Frida, Sue, TaniaIn this team there is only one player of EAOL The condition given in the question is that there must be 2 - 2 players of each club.So this team is also illegal.Option (4)

Eiona, Frida, Ken, Leizani, Ritwik, SueThis team does not contradict any of the given conditions or rules. So this is the right option. Ans. (4)

For Q.74 to Q.77:We have the purchases as follows:3 items (D + 2B) = Rs.40 + Rs.180 = Rs.2202 items (A + C) = Rs.1804 items (E + 2D + B) = Rs.45 + Rs.80 + Rs.90 = Rs.215The combinations of possible purchases and the corresponding points earned in each case are:Rs.220 4 = Rs.880 (12 1000) (120 100) = 0 pointsRs.180 5 = Rs.900 (10 1000) (100 100) = 0 pointsRs.215 4 = Rs.860 (16 1000) (140 100) = 2000 points2(Rs.220 + Rs.180) + Rs.180 = Rs.980 (12 1000) (20 100) = 10000 points2(Rs.220 + Rs.215) = Rs. 890 (14 1000) (110 100) = 3000 points2(Rs.215 + Rs.180) + Rs.180 = Rs.970 (14 1000) (30 100) = 11000 points

74. Thus the maximum items to be purchased to maximise the points is 14. Ans. (2)

75. The number of items of B and D are 2 and 4 respectively. Ans. (4)

76. The maximum possible points are 11000. Ans. (3)

77. Minimum possible unspent amount when he spent Rs.980 is Rs.20. Ans. (2)

For Q.78 to Q.80:78. The table shows the program written by all three players in the game, which has a memory attainder of 6 MB.


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S.No.

P1 P2 P3 P1

1 1 1 1 3

2 1 1 3 1

3 1 1 2 2

4 1 2 3

5 1 3 2

6 2 1 3

7 2 3 1

8 3 2 1

9 3 1 2

10 2 2 2

11 3 3

Total possible cases of winning of any player

All players play the game intelligently and in an unbiased manner. So, if P 1 wants to win, he will always write a program of 1 MB. So, the onlypossible cases are:

S.No.

P1 P2 P3 P1

1 1 1 1 3

2 1 1 3 13 1 1 2 2

4 1 2 3

5 1 3 2


Since, P2 is also an intelligent player, so he will understand that he cannot win. Therefore, there are equal chances for him writing a program of anyof the given sizes. Also, from the above table if P2 writes a program of size 1 MB then P1 always win.Therefore, the required chances of P1 winning are 1/3 (chances of writing a program of size 1 MB by P 2.). Ans. (4)

79. The table shows the program written by all three players in the game with a memory attainder of 7 MB:

11 3 3

S.No.

P1 P2 P3 P1 P2

1 1 1 1 1 3

2 1 1 1 3 1

3 1 1 1 2 2

4 1 1 2 3

5 1 1 3 2

6 1 2 1 3

7 1 2 3 1

8 2 1 1 3

9 2 1 3 1

10 1 2 2 2

11 2 1 2 2

12 1 3 3


All players play the game intelligently and in an unbiased manner. So, if P 1 wants to win, he will always write a program of 1 MB or 2 MB. So, theonly possible cases are


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S.No.

P1 P2 P3 P1 P2

1 1 1 1 1 3

2 1 1 1 3 1

3 1 1 1 2 2

4 1 1 2 3

5 1 1 3 2

6 1 2 1 3

7 1 2 3 18 2 1 1 3

9 2 1 3 1

10 1 2 2 2

11 2 1 2 2

12 1 3 3

15 2 2 3

16 2 3 2


Then the chances of P2 winning:

Case I:P1 writes a program of 1 MBChances of P1 writing a program of 1 MB = 1/2P2 plays intelligently so he always writes a program only of 1 MB to win.So, the chances of P2 writing a program of 1 MB = 1Now, since P3 is also an intelligent player, so, he will understand that he cannot win.

Therefore, it is equally likely that he being an unbiased player can write the program of any of the given sizes.Now, for P2 to win, P3 must write a program of size 1 MB with a probability of winning: 1/3.So, the overall chances of winning of P2 = 1/2 1 1/3 = 1/6

Case II:P1 writes a program of 2 MBChances of P1 writing a program of 2 MB = 1/2Now, P2 is an intelligent player, so, he knows that he cannot win anymore.On combining both the cases the chances of winning of P2 = 1/6Ans. (2)

80. The table shows the program written by all three players in the game, which has a memory attainder of 6 MB.

S.No.

P1 P2 P3 P1

1 1 1 1 3

2 1 1 3 13 1 1 2 2

4 1 2 3

5 1 3 2

6 2 1 3

7 2 3 1

8 3 2 1

9 3 1 2

10 2 2 2

11 3 3


All players play the game intelligently and in an unbiased manner. So, if P 1 wants to win, he will always write a program of 1 MB. So, the onlypossible cases are:

S.No.

P1 P2 P3 P1

1 1 1 1 3

2 1 1 3 1

3 1 1 2 2

4 1 2 3

5 1 3 2


Now, since P2 is also an intelligent player, so, he will understand that he cannot win. Therefore, there are equal chances that P 2 writes the programof any of the given sizes. Also, from the above table, if P2 writes a program of size 1 MB then P1 always wins.


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Therefore, the required chances of P1 winning are 1/3 and the required chances of P3 winning = 1 - 1/3 = 2/3 Ans. (4)

For Q.81 to Q.85:From the information in the question, following table can be prepared:

Table I

D a t e

G h a n a t o

Nigeria (via

T o g o )

G h a n a t o

Nigeria (via

S u d a n )

1 6850 7250

2 6650 7675

3 8200 6875

4 6100 6500

5 6275 6000

6 6150 6825

7 5850 6450

8 5800 6375

9 6925 5950

10 7300 5925

11 7300 5875

12 7000 5975

13 6650 6275

14 6400 6600

15 6000 5 4 0 0

16 6480 5750

17 6725 5425

18 7050 5475

19 7225 5850

20 6800 6175

81. From Table I, distance between Ghana and Nigeria is minimum on 15th of November 2008 when the Professor goes via Sudan. The distance in thiscase is 5400 light-years. Ans. (1)

82. From Table I, there are a total of 15 days when the distance traveled could be less than 6600 light-years. Ans. (4)

83. From the information in the question, following table can be prepared:

Table II


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Date Ghana to Sudan to Nigeria Ghana to Togo to NigeriaDifference in distance

between 2 routes

1 7250 6850 400

2 7675 6650 1025

3 6875 8200 -1325

4 6500 6100 400

5 6000 6275 -2756 6825 6150 675

7 6450 5850 600

8 6375 5800 575

9 5950 6925 -975

10 5925 7300 -1375

11 5875 7300 -1425

12 5975 7000 -1025

13 6275 6650 -375

14 6600 6400 200

15 5400 6000 -600

16 5750 6480 -730

17 5425 6725 -1300

18 5475 7050 -1575

19 5850 7225 -1375

20 6175 6800 -625

The difference is least on the 14 th of November, i.e., 200 light-years. Ans. (3)

84. If the cost increases by 12%, the new cost would be:Table III

Date Ghana to Sudan to Nigeria Ghana to Togo to NigeriaDifference in distance

between 2 routes

1 7250 7672 -422

2 7675 7448 227

3 6875 9184 -2309

4 6500 6832 -332

5 6000 7028 -1028

6 6825 6888 -63

7 6450 6552 -102

8 6375 6496 -121

9 5950 7756 -1806

10 5925 8176 -2251

11 5875 7300 -1425

12 5975 7000 -1025

13 6275 6650 -375

14 6600 6400 200

15 5400 6000 -600

16 5750 6480 -730

17 5425 6725 -1300

18 5475 7050 -1575

19 5850 7225 -1375

20 6175 6800 -625

Hence, observing Table III, we conclude that on 2nd and 14th November he would still prefer the route via Togo. Ans. (2)

85. On observing Table II, there are 13 days on which travelling via Sudan is economical as compared to travelling via Togo. Ans. (4)

For Q.86 to Q.90:86. Total sales of all the six companies taken together = Rs.64 billion or Rs.64000 million

Average selling price (in million Rs.) per ton = 64000/3500 = Rs.18.3 million. Ans. (2)87. Quantity of butter sold by Amul in the year 2050 = (15 3500)/100 = 525 tons


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Total expenditure incurred (in million Rs.) by Amul = 10.54 525 = 5533.5Total sales achieved by Amul = (11 64000)/100 = Rs.7040 millionAmul s profits in the year 2050 = 7040 5533.5 = Rs.1504.5 millionSo, approximate profit percentage = (1504.5 100)/5533.5 = 27.2% Ans. (1)

88. Marked price = Selling price + DiscountQuantity of the butter sold by Kera = (21 3500)/100 = 735 tonsTotal sales of Kera = (29 64000)/100 = Rs.18560 millionNow, per ton selling price of Kera's butter = 18560 / 735 = Rs.25.25 million

Discount = 10% of 25.25 = Rs.2.525 millionSo, marked price = 25.25 + 2.525 = Rs.27.775 million or Rs.27.8 (approx.). Ans. (3)

89. Quantity of the butter sold by Nestle = (18 3500)/100 = 630 tonsTotal sales of Nestle = (17 64000)/100 = Rs.10880 millionNow, per ton selling price of Nestle's butter = 10880 / 630 = Rs.17.27 millionSo, cost price per ton of Nestle's butter = 0.8 17.27 = Rs.13.82 millionQuantity of butter sold by Gayatri Dairy in the year 2050 = (16 3500)/100 = 560 tonsTotal sales of Gayatri Dairy = (13 64000)/100 = Rs.8320 millionNow, per ton selling price of Gayatri Dairy's butter = 8320 / 560 = Rs.14.86 millionSo, cost price per ton of Gayatri Dairy's butter = 0.75 14.86 = Rs.11.15 millionTherefore, the required approximate ratio = 15 / 11 Ans. (4)

90. Selling price per ton of butter of the six companies:

Company

Quantity of Butter sold (in

tons)

Total Sales (in million

Rs.)

Selling price (in

million Rs.) per ton

Amul 52 5 7040 13.4

Nestle 630 10880 17.3

Sanchi 560 8960 16

Kera 735 18560 25.25

Mother Dairy 490 10240 20.9

Gayatri Dairy 560 8320 14.85

So, from the table we can observe that Kera has the highest selling price per ton. Ans. (4)

e-prac-cat # 13 solutions

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