CAT 2005


Section - 1


Sub–Section I-A : Number of questions = 10


Note: Questions 1 to 10 carry one mark each.


Directions for questions 1 to 5: Answer the questions independently of each other.



Ans .

(1)


    Explanation : x=(\(16^{3}\)+\(17^3\)+\(18^{3}\)+\(19^{3}\)) is an even number Therefore, 2 divides x. (\(a^{3}\)+\(b^{3}\)) = (a + b)(\(a^{2}\) –ab + \(b^{2}\)) ⇒ a + b always divides \(a^{3}\) + \(b^{3}\). Therefore, \(16^{3}\) + \(19^{3}\) is divisible by 35. 183 + 173 is divisible by 35. Thus, x is divisible by 70. Hence, option (1) is the correct choice.





Ans .

(3)


    Explanation : CAT 2005
    D gets emptied first, it gets emptied in 20 minutes. Hence, option (3) is the correct answer.





Ans .

(4)


    Explanation : CAT 2005
    Shaded area = 2 × (area of sector ADC – area of ∆ADC) =2* [\(\frac{π}{4}\) * \(1^2\) - \(\frac{1}{2}\) *1*1] = \(\frac{π}{2}\) - 1





Ans .

(4)


    Explanation : CAT 2005
    Let r be the radius of the two circular tracks
    ∴The rectangle has dimensions 4r × 2r
    A covers a distance of 2r + 2r + 4r + 4r = 12 r
    B covers a distance of 2πr + 2πr = 4πr
    Time taken by both of them is same.
    Therefore, \(\frac{4πr} {S_{B}}\) = \(\frac{12r} {S_{A}} \) : \(S_{B}\) = \(\frac{π}{3}\) * \(S_{A}\)
    Therefore, Required percentage = \(\frac{S_{A}-S_{B}}{S_{A}}\) * 100
    = \(\frac{π-3}{3}\) * 100 = 4.72%





Ans .

(1)


    Explanation : Let there be m boys and n girls
    \( ^{n}C_{2} \) = 45 : \(\frac{n(n-1)}{2} \) = n(n-1) = 90 : n=10
    \( ^{m}C_{2} \) = 190 : \(\frac{m(m-1)}{2} \) = m(m-1) =380 : m=20
    Numbers of games between one boy and one girl = \( ^{10}C_{1} \) * \(^{20}C_{1} \) = 10*20 = 200



Directions for questions 6 to 7: Answer the questions on the basis of the information given below. Ram and Shyam run a race between points A and B, 5 km apart, Ram starts at 9 a.m from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed, Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.



Ans .

(2)


    Explanation :
    CAT 2005
    It is clear that Ram and Shyam shall meet each other between C & B, sometime after 10:00 AM. At 10:00 AM they are moving as shown below:
    CAT 2005
    From now, time taken to meet = \(\frac{2.5}{10+5}\) * 60min = 10min
    So, they meet each other at 10:10 AM.





Ans .

(2)


    Explanation : It is clear from the diagram that at 10:30; Shyam overtakes Ram.



Directions for questions 8 to 10: Answer the questions independently of each other.



Ans .

(4)


    Explanation : R = \(\frac{30^{65}-29^{65}}{30^{64}+29^{64}}\)
    R = (30-29) * \( \left(\frac{^{65}C_{0}30^{64}+^{65}C_{1}30^{63}*29+^{65}C_{2}30^{62}*29^{2}+.....^{65}C_{65}29^{64}}{30^{64}+29^{64}}\right)\)
    R = 1 * \(\left(\frac{1+ ^{65}C_{1}30^{63}*29+^{65}C_{2}30^{62}*29^{2}}{30^{64} + 29^{64}}\right)\)
    R = (1+ +ve quantity)
    R>1





Ans .

(4)


    Explanation : Case I:Chords on same side of the centre.
    CAT 2005
    \(OB^{2}\) = \(OA^{2}\) – \(AB^{2}\) = \(20^{2}\) - \(16^{2}\) = 144
    OB = 12
    \(OD^{2}\) = \(20^{2}\) - \(12^{2}\) = 400 – 144 = 256
    OD = 16
    BD = 4 cm Case II: Chords on opposite side of the centre.
    CAT 2005
    AB = 32cm
    CB = 24cm
    OP= \(\sqrt{AO^{2}-AP^{2}}\) = \( \sqrt{20^{2}-16^{2}}\)
    OP=12cm
    OQ= \(\sqrt{OC^{2}-CQ^{2}}\) = \( \sqrt{20^{2}-12^{2}}\)
    OQ=16cm
    Distance = PQ = 12 + 16 = 28 cm.





Ans .

(3)


    Explanation :
    \(y^{2}\) = \(x^{2}\)
    2\(x^{2}\) - 2kx + \(k^{2}\) -1 =0
    D=0
    4\(k^{2}\)=8\(k^{2}\)-8
    4\(k^{2}\)=8
    \(k^{2}\)=2
    k= \(\pm\sqrt{2}\).
    k= +\(\sqrt{2}\) gives the equation 2\(x^{2}\)-2\(\sqrt{2}\)x+1=0;
    its root is \(\frac{-b}{2a}\)=+\(\frac{1}{2}\),
    k= -\(\sqrt{2}\) gives the equation 2\(x^{2}\)+2\(\sqrt{2}\)x+1=0;
    its root is \(-\frac{1}{\sqrt{2}}\) this root is -ve, will reject k=-\(\sqrt{2}\).
    Only answer is k = + \(\sqrt{2}\).




Sub–Section I-B : Number of questions = 20


Note: Questions 11 to 30 carry two marks each.




Ans .

(4)


    Explanation :
    If p = 1! = 1, then
    p + 2 = 3 when divided by 2! will give a remainder of 1. If p = 1! + 2 × 2! = 5, then
    p + 2 = 7 when divided by 3! will give a remainder of 1.
    Hence, p = 1! + (2 × 2!) + (3 × 3!) + ... + (10 × 10!) when
    divided by 11! leaves a remainder 1.





Ans .

(1)


    Explanation :
    CAT 2005
    equation of line ≡ x + y = 41. If the (x, y) co-ordinates of the points are integer, their sum shall also be integers so that x + y = k (k, a variable) as we have to exclude points lying on the boundary of triangle; k can take all values from 1 to 40 only. k = 0 is also rejected as at k = 0 will give the point A; which can’t be taken. Now, x + y = k, (k = 1, 2, 3, ...,40) with k = 40; x + y = 40; taking integral solutions. We get points (1, 39), (2, 38); (3, 37); ..., (39, 1) i.e. 39 points x + y = 40 will be satisfied by 39 points. Similarly, x + y = 39 is satisfied by 38 points. x + y = 38 by 37 points. x + y = 3 by 2 points. x + y = 2 is satisfied by 1 point. x + y = 1 is not satisfied by any point. So, the total no. of all such points is:39 + 38 + 37 + 36 + ... + 3 + 2 + 1 = \(\frac{39*40}{2}\) = 780 points.





Ans .

(2)


    Explanation :
    Let A = abc. Then, B = cba.
    Given, B > A ⇒ c > a
    As B – A = (100c + 10b + a) – (100a + 10b + c)
    ⇒ B – A = 100 (c – a) + (a – c)
    ⇒ B – A = 99 (c – a). Also, (B – A) is divisible by 7. But, 99 is not divisible by 7 (no factor like 7 or \(7^{2}\)). Therefore, (c – a) must be divisible by 7 {i.e., (c – a) must be 7,\(7^{2}\) , etc.}. Since c and a are single digits, value of (c – a) must be 7. The possible values of (c, a) {with c > a} are (9, 2) and (8, 1). Thus, we can write A as:
    A : abc ≡ 1b8 or 2b9 As b can take values from 0 to 9, the smallest and largest possible value of A are:
    \(A_{min}\)=108 and \(A_{max}\)=299
    Only option (b) satisfies this. Hence, (2) is the correct option.





Ans .

(3)


    Explanation :
    \(a_{1}\)=1, \(a_{n+1}\)-3\(a_{n}\)+2=4
    \(a_{n}\)=3\(a_{n}\)+4n-2
    when n=2 then \(a_{2}\)=3+4-2=5
    when n=3 the \(a_{3}\)=3*5+4*2-2=21
    from the options, we get an idea that \(a_{n}\) can be expressed in a combination of some power of 3 & some multiple of 100.
    (1) \(3^{99}\) – 200; tells us that \(a_{n}\) could be: \(3^{n-1}\) – 2 × n; but it does not fit \(a_{1}\) or \(a_{2}\) or \(a_{3}\)
    (2)\(3^{99}\) + 200 ; tells us that \(a_{n}\) could be: \(3^{n-1}\) + 2 × n;again, not valid for \(a_{1}\), \(a_{2}\) etc.
    (3) \(3^{100}\) – 200; tells \(3_{n}\) – 2n: valid for all \(a_{1}\), \(a_{2}\), \(a_{3}\).
    (4) \(3^{100}\) + 200; tells \(3_{n}\) + 2n: again not valid.so, (3) is the correct answer.





Ans .

(2)


    Explanation :
    CAT 2005
    (i) Counting from the LMD-end:
    CAT 2005
    We have 1, 2, 3, 4 & 5 to be filled in these blocks. Odd nos. (1, 3, 5) to be filled in at odd positions. Other places are to be filled by even numbers (2 or 4) Let’s count, how many such numbers are there with 2 at the unit’s digit
    CAT 2005
    Odd numbers can be filled in 3P2 = 6 way. The remaining two places are to be filled by 2 numbers (one odd number left out of 1, 3, 5 & one even i.e. 4) in = 2 ways. So, there are 6 × 2 = 12 number with 2 at the rightmost place. Similarly; there are 12 such numbers with 4 at the rightmost digits. The sum of rightmost digits in all such number = 12(2 + 4) = 72
    (ii) Now counting from the RMD-end. Let’s place 1 at the units place and check, how many numbers are possible with (1, 3) at the odd positions:
    CAT 2005
    Number of such cases = 2 × 2 = 4 ways
    CAT 2005
    Here again number of ways = 2 × 2 = 4 ways So, there are 4 + 4 = 8 numbers, in which (1, 3) are at odd positions. Similarly there are 8 numbers in which (1, 5) are at odd positions. So, in all there are 16 numbers where 1 is at unit’s place. Similarly there are 16 numbers with 3 at unit’s place and 16 more with 5 at unit’s place. Summing up all the odd unit’s digits = 16(1 + 3 + 5) = 144 From (i) and (ii) we can now, sum up all (even or odd) numbers at units place = 72 +144 = 216. Hence answer is (2)





Ans .

(1)


    Explanation : \( \left(30^{4}\right)^{680}\)
    . Hence, the right most non-zero digit is 1.





Ans .

(2)


    Explanation :
    CAT 2005
    Drawn figure since it have not to be within distance of 1 m so it will go along APQD, which is the path of minimum distance.
    AP=\(\frac{90}{360}\)*2π*1=\(\frac{π}{2}\)
    Also,AP=QD=\(\frac{π}{2}\)
    So the minimum distance =AP+PQ+QD=\(\frac{π}{2}\)+1+\(\frac{π}{2}\)=1+π.





Ans .

(4)


    Explanation :
    P=\(\log_{x}\left({\frac{x}{y}}\right)\) + \(\log_{y}\left({\frac{y}{x}}\right)\)
    = \(\log_{x}{x}\)- \(\log_{x}{y}\)+\(\log_{y}{y}\)-\(\log_{y}{x}\)=2-\(\log_{x}{y}\)-\(\log_{y}{x}\)
    Let t=\(\log_{x}{y}\)
    P=2-\(|frac{1}{t}\)-t= \(-\left[\sqrt{t}-\frac{1}{\sqrt{t}}\right]^{2}\)
    Which can never be positive. Out of given options, it cannot assume the value of +1.





Ans .

(4)


    Explanation : It is given that 10 < n < 1000.
    Let n be a two digit number.
    Then, n = 10a + b ⇒ \(p_{n}\) = ab, \(s_{n}\) = a + b
    Then, ab + a + b = 10a + b ⇒ ab = 9a ⇒ b = 9
    ∴There are 9 such numbers 19, 29, 39, …, 99.
    Now, let n be a three digit number.
    ⇒ n = 100a + 10b + c ⇒ \(p_{n}\) = abc, \(s_{n}\)= a + b + c
    Then, abc + a + b + c = 100a + 10b + c
    ⇒ abc = 99a + 9b
    ⇒ bc = 99 + 9\(\frac{b}{a}\)
    But the maximum value of bc = 81 (when both b & c are 9) and RHS is more than 99. Hence, no such number is possible. Hence, there are 9 such integers.





Ans .

(3)


    Explanation :
    CAT 2005





Ans .

(3)


    Explanation :
    \(\mid x+y \mid + \mid x-y \mid \)=4
    Replacing “+x” by “– x” & “+y” by “–y” everywhere in the curve, we again get the same equation. ⇒ Curve is symmetric in the 4-quadrants of X–Y plane. In I-quadrant (x, y > 0)
    \(\mid x+y \mid + \mid x-y \mid \)=4
    (x+y)+(y-x)=4;y>x
    (x+y)-(y-x)=4;y< x
    y=2; y>x
    x=2;y < x
    The graph looks like below
    CAT 2005
    Area of quadrant = \(2^{2}\) =4
    \(\mid x+y \mid + \mid x-y \mid \) =4 is
    4 × (area of I-quadrant) = 4 × 4 = 16.





Ans .

(2)


    Explanation :
    CAT 2005
    AE = 1 cm, BE = 2 cm & NL = 1 cm, ML = 2 cm
    HL=OE= \(\frac{1}{2}\)
    DL=DH+HL
    DL=DH+ \(\frac{1}{2}\)
    OB = AO = radius = 1.5
    \(DO^{2}\) = \(OL^{2}\) + \(DL^{2}\)
    \(\left(\frac{3}{2}\right)^{2}\) = \(\left(\frac{1}{2}\right)^{2}\) + \(\left( DH + \frac{1}{2}\right)^{2}\) : DH= \(\sqrt{2}\) - \(\frac{1}{2}\) = \(\frac{2\sqrt{2} -1}{2} \)





Ans .

(1)


    Explanation :
    CAT 2005
    Here \(\angle ACB \) = \(\theta\) +[180-(2 \(\theta\) + \(\alpha\)] = 180 - ( \(\theta\) + \(\alpha\) )
    So here we can say that triangle BCD and triangle ABC will be similar.
    ∆BCD ~ ∆BAC Hence, from the property of similar triangles
    \(\frac{AB}{12}\) = \(\frac{12}{9}\) : AB = 16
    \(\frac{AC}{12}\) = \(\frac{12}{9}\) : AC = 8
    Therefore, AD = 7
    \(S_{ADC}\) = 8+7+6 = 21
    \(S_{BDC}\) = 27
    Hence r = \(\frac{21}{7}\) = \(\frac{7}{9}\)





Ans .

(1)


    Explanation :
    CAT 2005
    Here cos30 = \(\frac{a}{2r}\)
    a= r \(\sqrt{3}\)
    Here the side of equilateral triangle is r \(\sqrt{3}\)
    From the diagram cos120 = \(\frac{x^{2}+x^{2}-a^{2}}{2x^{2}} \)
    \(a^{2}\) = \(3x^{2}\)
    x=r
    Hence the circumference will be 2r(1+ \(\sqrt{3}\))





Ans .

(1)


    Explanation :The 100th and 1000th position values will be only 1. Different possibilities of unit and tens digits are (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3), (7, 9), (9, 1) and (9, 7). Hence, there are 9 elements in S.





Ans .

(3)


    Explanation : From the given equation, it is obvious that 2 < x < 3. Option (c) satisfies the condition.
    Alternative solution:
    \(\sqrt{4+\sqrt{4-x}}\)
    \(x^{2}\) = 4 + \(\sqrt{4-x}\)
    \(x^{2}\) -4 = \(\sqrt{4-x}\)
    Now putting the values from options, we find only option (3) satisfies the condition.





Ans .

(4)


    Explanation :
    g(x + 1) + g(x –1) = g(x)
    g(x+ 2) + g(x) = g(x + 1)
    Adding these two equations, we get
    g(x + 2) + g(x – 1) = 0
    ⇒ g(x + 3) + g(x) = 0
    ⇒ g(x + 4) + g(x + 1) = 0
    ⇒ g(x + 5) + g(x + 2) = 0
    ⇒ g(x + 6) + g(x + 3) = 0
    ⇒ g(x + 6) – g(x) = 0





Ans .

(4)


    Explanation :
    There are two equations to be formed 40 m + 50 f = 1000
    250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
    850 m + 800 f = A
    m and f are the number of males and females A is amount paid by the employer.
    Then, the possible values of f = 8, 9, 10, 11, 12
    If f = 8, m = 15.
    If f = 9, 10, 11 then m will not be an integer while f = 12, then m will be 10.
    By putting f = 8 and m = 15, A = 18800. When f = 12 and m = 10, then A = 18100
    Therefore, the number of males will be 10





Ans .

(3)


    Explanation :
    Frenchmen : \(F_{1}\) , \(F_{2}\) ,\(F_{3}\)
    Englishmen : \(E_{1}\) , \(E_{2}\) ,\(E_{3}\)
    Let \(E_{1}\) knows French
    I round of calls :
    CAT 2005
    In the 6th call \(E_{1}\) knows all the secrets . Similarly after 9th call , everybody knows all the secrets .





Ans .

(2)


    Explanation :Let the rectangle has m and n tiles along its length and breadth respectively. The number of white tiles W = 2m + 2(n – 2) = 2 (m + n – 2) And the number of Red tiles = R = mn – 2 (m + n – 2) Given W = R ⇒ 4 (m + n – 2) = mn ⇒ mn – 4m – 4n = – 8 ⇒ (m – 4) (n – 4) = 8 As m & n are integers so (m – 4) & (n – 4) are both integers. The possibilities are (m – 4, n – 4) ≡ (1, 8) or (2, 4) giving, (m, n) as (5, 12) or (6, 8) so the edges can have 5, 12, 6 or 8 tiles.




Section - II


Sub–Section II-A : Number of questions = 10


Note: Questions 31 to 40 carry one mark each.


Directions for questions 31 to 34: The passage given below is followed by a set of four questions. Choose the best answer to each question. A game of strategy, as currently conceived in game theory, is a situation in which two or more “players” make choices among available alternatives (moves). The totality of choices determines the outcomes of the game, and it is assumed that the rank order of preferences for the outcomes is different for different players. Thus the “interests” of the players are generally in conflict. Whether these interests are diametrically opposed or only partially opposed depends on the type of game. Psychologically, most interesting situations arise when the interests of the players are partly coincident and partly opposed, because then one can postulate not only a conflict among the players but also inner conflicts within the players. Each is torn between a tendency to cooperate, so as to promote the common interests, and a tendency to compete, so as to enhance his own individual interests. Internal conflicts are always psychologically interesting. What we vaguely call “interesting” psychology is in very great measure the psychology of inner conflict. Inner conflict is also held to be an important component of serious literature as distinguished from less serious genres. The classical tragedy, as well as the serious novel reveals the inner conflict of central figures. The superficial adventure story on the other hand, depicts only external conflict; that is, the threats to the person with whom the reader (or viewer) identifies stem in these stories exclusively from external obstacles and from the adversaries who create them. On the most primitive level this sort of external conflict is psychologically empty. In the fisticuffs between the protagonists of good and evil, no psychological problems are involved or, at any rate, none are depicted in juvenile representations of conflict. The detective story, the “adult” analogue of a juvenile adventure tale, has at times been described as a glorification of intellectualized conflict. However, a great deal of the interest in the plots of these stories is sustained by withholding the unraveling of a solution to a problem. The effort of solving the problem is in itself not a conflict if the adversary (the unknown criminal) remains passive, like Nature, whose secrets the scientist supposedly unravels by deduction. If the adversary actively puts obstacles in the detective’s path toward the solution, there is genuine conflict. But the conflict is psychologically interesting only to the extent that it contains irrational components such as a tactical error on the criminal’s part or the detective’s insight into some psychological quirk of the criminal or something of this sort. Conflict conducted in a perfectly rational manner is psychologically no more interesting than a standard Western. For example,Tic-tac-toe, played perfectly by both players, is completely devoid of psychological interest. Chess may be psychologically interesting but only to the extent that it is played not quite rationally. Played completely rationally, chess would not be different from Tic-tac-toe. In short, a pure conflict of interest (what is called a zero-sum game) although it offers a wealth of interesting conceptual problems, is not interesting psychologically, except to the extent that its conduct departs from rational norms.



Ans .

(3)


    Explanation : In para number 2 “Each is torn …” and then further in para 3 “Internal …” These lines in paras 2 and 3 talk about external conflict being psychologically empty, and no psychological problems involved therein. This makes internal conflicts psychologically interesting.





Ans .

(2)


    Explanation : In paragraph 4, refer to line 11, “Chess may be psychologically….. rationally.” According to the author, only when someone acts irrationally will that act be considered psychologically interesting and out of the given choices only option (2) qualifies, wherein adopting a defensive strategy against an aggressive opponent will be irrational. Option (3) is incorrect as the choice that the mountaineer would make would depend on external conditions and there would not be any internal conflicts as such, and the decisions that he would need to make would have to be rational.





Ans .

(2)


    Explanation : In the first paragraph refer to line 4- “Thus the “interests” of the players are generally in conflict.” Choice (3) may also be correct but choice (2) is more appropriate as it is stated directly in the passage whereas choice (3) is an inference. Choice (1) is a consequence of applying game theory to a situation, not one of its pre-requisites, Therefore option 4 is also ruled out.





Ans .

(3)


    Explanation : In paragraph 4 lines 3 onwards- “The effort… genuine” According to this, in case of the detective , if the criminal remains passive, there is no conflict, whereas the scientist has to unravel the secrets of nature (which is “passive”) by deduction .



Directions for questions 35 to 37:The sentences given in each question, when properly sequenced, form a coherent paragraph. Each sentence is labeled with a letter. Choose the most logical order of sentences from among the given choices to construct a coherent paragraph.



Ans .

(2)


    Explanation : DC is the mandatory pair, which makes 3 and 4 incorrect. E is the opening statement. A concludes the argument by substantiating the argument in EBCD. Therefore, the analogy from the previous argument is being extended in ‘A’ (keyword – “similarly”)





Ans .

(4)


    Explanation : From the options, it can be ascertained that ‘B’ is the opening statement. Also, B explains “greater interest… than”, hence ‘C’ is the natural antecedent to ‘B’, wherein “a similar neglect” has been talked, about. Hence (4) is the correct option.





Ans .

(2)


    Explanation : After reading statement B the first question that comes to mind is what does ‘it’ stand for. The question is answered by statement (E) which should be the logical antecedent. This makes EB a mandatory pair and that is present only in option (2).



Directions for questions 38 to 40: In each question, the word at the top of the table is used in four different ways, numbered 1 to 4. Choose the option in which the usage of the word is incorrect or inappropriate.



Ans .

(2)


    Explanation : Option (2) talks about a ‘near’ friend. There is nothing like a near friend. It should have been ‘close’ friend





Ans .

(1)


    Explanation : It should have been “I have my hands full”.





Ans .

(3)


    Explanation : It should have been “I can’t bear her being angry”.




Sub-Section II-B:Number of questions =20


Note: Question 41 to 60 carry two marks each.

Directions for questions 41 to 48: Each of the two passages given below is followed by a set of four questions. Choose the best answer to each question.

Passage-I

Crinoline and croquet are out. As yet, no political activists have thrown themselves in front of the royal horse on Derby Day. Even so, some historians can spot the parallels. It is a time of rapid technological change. It is a period when the dominance of the world’s superpower is coming under threat. It is an epoch when prosperity masks underlying economic strain. And, crucially, it is a time when policy-makers are confident that all is for the best in the best of all possible worlds. Welcome to the Edwardian Summer of the second age of globalisation. Spare a moment to take stock of what’s been happening in the past few months. Let’s start with the oil price, which has rocketed to more than $65 a barrel, more than double its level 18 months ago. Theaccepted wisdom is that we shouldn’t worry our little heads about that, because the incentives are there for business to build new production and refining capacity, which will effortlessly bring demand and supply back into balance and bring crude prices back to $25 a barrel. As Tommy Copper used to say, ‘just like that’. Then there is the result of the French referendum on the European Constitution, seen as thick-headed luddites railing vainly against the modern world. What the French needed to realize, the argument went, was that there was no alternative to the reforms that would make the country more flexible, more competitive, more dynamic. Just the sort of reforms that allowed Gate Gourmet to sack hundreds of its staff at Heathrow after the sort of ultimatum that used to be handed out by Victorian mill owners. An alternative way of looking at the French “non” is that our neighbours translate “flexibility” as “you’re fired”. Finally, take a squint at the United States. Just like Britian a century ago, a period of unquestioned superiority is drawing to a close. China is still a long way from matching America’s wealth, but it is growing at a stupendous rate and economic strength brings geo-political clout. Already, there is evidence of a new scramble for Africa as Washington and Beijing compete for oil stocks. Moreover, beneath the surface of the US economy, all is not well. Growth looks healthy enough, but the competition from China and elsewhere has meant the world’s biggest economy now imports far more than it exports. The US is living beyond its means, but in this time of studied complacency a current account deficit worth 6 perfect of gross domestic product is seen as a sign of strength, not weakness. In this new Edwardian summer, comfort is taken from the fact that dearer oil has not had the savage inflationary consequences of 1973-1974, when a fourfold increase in the cost of crude brought an abrupt end to a postwar boom that had gone on uninterrupted for a quarter of a century. True, the cost of living has been affected by higher transport costs, but we are talking of inflation at 2.3 per cent and not 27 per cent. Yet the idea that higher oil prices are of little consequence is fanciful. If people are paying more to fill up their cars it leaves them with less to spend on everything else, but there is a reluctance to consume less. In the 1970s unions were strong and able to negotiate large, compensatory pay deals that served to intensify inflationary pressure. In 2005, that avenue is pretty much closed off, but the abolition of all the controls on credit that existed in the 1970s means that households are invited to borrow more rather than consume less. The knock-on effects of higher oil prices are thus felt in different ways – through high levels of indebtedness, in inflated asset prices, and in balance of payments deficits. There are those who point out, rightly, that modern industrial capitalism has proved mightily resilient these past 250 years, and that a sign of the enduring strength of the system has been the way it apparently shrugged off everything – a stock market crash, 9/11, rising oil prices – that have been thrown at it in the half decade since the millennium. Even so, there are at least three reasons for concern. First, we have been here before. In terms of political economy, the first era of globalisation mirrored our own. There was a belief in unfettered capital flows, in free migration. Eventually, though, there was a backlash, manifested in a struggle between free traders and protectionists, and in rising labour militancy.Second, the world is traditionally as its most fragile at times when the global balance of power is in flux. By the end of the nineteenth century, Britain’s role as the hegemonic power was being challenged by the rise of the United States, Germany, and Japan while the Ottoman and Hapsburg empires were clearly in rapid decline. Looking ahead from 2005, it is clear that over the next two or three decades, both China and India – which together account for half the world’s population – will flex their muscles. Finally, there is the question of what rising oil prices tell us. The emergence of China and India means global demand for crude is likely to remain high a t a time when experts say production is about to top out. If supply constraints start to bite, any decline in the prices are likely to be short-term cyclical affairs punctuating a long upward trend.



Ans .

(2)


    Explanation : Answer choice (4), says that the danger being talked about is ‘imminent’, which is not necessarily the case as per the author in the passage, whereas the fact that everyone is complacent about it, is being talked about throughout the passage, which makes option (2) correct.





Ans .

(3)


    Explanation : In the sixth paragraph, the author explains why a belief in the "enduring strength of the system" might not be warranted. He also explains the reason behind such a belief-"… a sign of the enduring strength of the system …since the millennium."





Ans .

(1)


    Explanation : This is the correct option as choice (2) is too narrow. Choice (3) is a universal truth which may not be the case. There could be a problem between 1 and 4 but 4 is ruled out because this option is one of the reasons supporting the author’s argument but is not his key argument as such. Moreover, the author does not say that the crisis is imminent.





Ans .

(4)


    Explanation : In the 2nd paragraph, the author is being sarcastic about the fact that the new production and refining capacity will effortlessly bring demand and supply back to balance. (line 2 onwards “the accepted …just like that”) and he quotes Tommy Cooper to emphasize his sarcasm. It must be remembered that we have to consider the author’s point of view, not Tommy Cooper’s. Therefore option (4) is correct



Passage-II

While complex in the extreme, Derrida’s work has proven to be a particularly influential approach to the analysis of the ways in which language structures our understanding of ourselves and the world we inhabit, an approach he termed deconstruction. In its simplest formulation, deconstruction can be taken to refer to a methodological strategy which seeks to uncover layers of hidden meaning in a text that have been denied or suppressed. The term ‘text’, in this respect, does not refer simply to a written form of communication, however. Rather, texts are something we all produce and reproduce constantly in our every day social relations, be they spoken, written or embedded in the construction of material artifacts. At the heart of Derrida’s deconstructive approach is his critique of what he perceives to be the totalitarian impulse of the Enlightenment pursuit to bring all that exists in the world under the domain of representative language, a pursuit he refers to as logocentrism. Logocentrism is the search for a rational language that is able to know and represent the world and all its aspects perfectly and accurately. Its totalitarian dimension, for Derrida at least, lies primarily in its tendency to marginalize or dismiss all that does not neatly comply with its particular linguistic representations, a tendency that, throughout history, has all too frequently been manifested in the form of authoritarian institutions. Thus logocentrism has, in its search for the truth of absolute representation, subsumed difference and oppressed that which it designates as its alien ‘other’. For Derrida, western civilization has been built upon such a systematic assault on alien cultures and ways of life, typically in the name of reason and progress. In response to logocentrism, deconstruction posits the idea that the mechanism by which this process of marginalization and the ordering of truth occurs is through establishing systems of binary opposition. Oppositional linguistic dualisms, such as rational/irrational, culture/nature and good/bad are not, however, construed as equal partners as they are in, say, the semiological structuralism of Saussure. Rather, they exist, for Derrida, in a series of hierarchical relationships with the first term normally occupying a superior position. Derrida defines the relationship between such oppositional terms using the neologism différance. This refers to the realization that in any statement, oppositional terms differ from each other (for instance, the difference between rationality and irrationality is constructed through oppositional usage), and at the same time, a hierarchical relationship is maintained by the deference of one term to the other (in the positing of rationality over irrationality, for instance). It is this latter point which is perhaps the key to understanding Derrida’s approach to deconstruction.For the fact at any given time one term must defer to its oppositional ‘other’, means that the two terms are constantly in a state of interdependence. The presence of one is dependent upon the absence or ‘absentpresence’ of the ‘other’, such as in the case of good and evil, whereby to understand the nature of one, we must constantly relate it to the absent term in order to grasp its meaning. That is, to do good, we must understand that our act is not evil, for without that comparison the term becomes meaningless. Put simply, deconstruction represents an attempt to demonstrate the absent-presence of this oppositional ‘other’, to show that what we say or write is in itself not expressive simply of what is present, but also of what is absent. Thus, deconstruction seeks to reveal the interdependence of apparently dichotomous terms and their meanings relative to their textual context; that is, within the linguistic power relations which structure dichotomous terms hierarchically. In Derrida’s own words, a deconstructive reading “must always aim at a certain relationship, unperceived by the writer, between what he commands and what he does not command of the patterns of a language that he uses. …[It] attempts to make the not-seen accessible to sight.” Meaning, then, is never fixed or stable, whatever the intention of the author of a text. For Derrida, language is a system of relations that are dynamic, in that all meanings we ascribe to the world are dependent not only on what we believe to be present but also on what is absent. Thus, any act of interpretation must refer not only to what the author of a text intends, but also to what is absent from his or her intention. This insight leads, once again, to Derrida’s further rejection of the idea of the definitive authority of the intentional agent or subject. The subject is decentred; it is conceived as the outcome of relations of différance. As author of its own biography, the subject thus becomes the ideological fiction of modernity and its logocentric philosophy, one that depends upon the formation of hierarchical dualisms, which repress and deny the presence of the absent ‘other’. No meaning can, therefore, even be definitive, but is merely an outcome of a particular interpretation.



Ans .

(4)


    Explanation : Option (1) and (3) are contrary to what Derrida says in the passage which makes them incorrect. There can be a confusion between 2 and 4. Option (2) could have been an inference if the statement had been “Language limits our interpretations of reality”. But the word ‘construction’ is incorrect. Therefore only option (4) according to the passage, is correct.





Ans .

(3)


    Explanation : According to the passage, Derrida is against logocentrism and choices (1), (2) and (4) are pro logocentrism which leaves option (3) which is different from logocentrism.





Ans .

(1)


    Explanation : This is a fact based question. In paragraph 2, refers to line 5 “Rather, they exist … position”. Option (1) directly follows from this line.





Ans .

(1)


    Explanation : Answer choice (4) is contrary to what is being said. Answer choice (3) is irrelevant. There can be a confusion between 1 and 2 but it must be noted that it is not the meaning of the text which is based on binary opposites but the interpretation. This leaves us only with answer choice (1).



Directions for questions 49 to 52: Each of the following questions has a paragraph from which the last sentence has been deleted. From the given options, choose that one that completes the paragraph in the most appropriate way.



Ans .

(1)


    Explanation : The passage contrasts crosswords with Sudoku. A crossword touches numerous areas of life and provides a few surprises along the way. So the next sentence needs to talk about Sudoku along these lines. Option (1) which describes Sudoku as "just a logical exercise" (unlike the crossword which touches numerous areas of life) with each one similar to the last (unlike the surprises that a good crossword can provide).





Ans .

(2)


    Explanation : Since expert individuals are left out of such groups, the result is most likely to be mediocrity.





Ans .

(3)


    Explanation : Option (2) talks about humility which is not talked about in the passage, option (3) is an extension of the concept of being a minnow.





Ans .

(1)


    Explanation : The passage has a decidedly negative tone. The author states that just like other generations before it, this generation has also struggled to understand the organizational laws of the frontier, has suffered from unwarranted pride, and has also failed like those before. So only (1) can complete the paragraph by stating the need for humility in front of this failure.



Directions for questions 53 to 56: Each question consists of four sentences on a topic. Some sentences are grammatically incorrect or inappropriate. Select the option that indicates the grammatically correct and appropriate sentence(s).



Ans .

(2)


    Explanation : The second sentence does not use the article. It should be ‘As a/the project progresses” in sentence C there should be the indefinite article ‘a’ before single-minded which leaves us with option (2) as the correct answer.





Ans .

(3)


    Explanation : Sentence B should have “making them break apart”. Sentence C should have “many offending chemicals”.





Ans .

(2)


    Explanation : B should be “rarely has …” C should begin with ‘The’.





Ans .

(1)


    Explanation : Option B should be “since the Enlightenment. Option C should be “in the 1820’s”



Directions for questions 57 to 60: Each of the following questions has a paragraph with one italicized word that does not make sense. Choose the most appropriate replacement for that word from the options given below the paragraph.



Ans .

(3)


    Explanation : Resurrecting i.e. bring back to practice is the best choice. (1), (2) and (4) are negative options.





Ans .

(3)


    Explanation : Sputtering is a light popping sound of a flame which is dying out. The ideas conveyed are dim and grim so ‘shining’, bright and effulgent are out.





Ans .

(4)


    Explanation : Such a scene should be distressing to a sensitive traveler. Irritating and disgusting are negative options. 1 can be clearly ruled out.





Ans .

(4)


    Explanation : The one word reply conveys that it is terse. As it has no element of humour we can easily rule out – “witty”.




Section - III


Sub–Section III-A : Number of questions = 10


Note: Questions 61 to 70 carry one mark each.

Directions for questions 61 to 64: Answer the questions on the basis of the information given below: A management institute was established on January 1, 2000 with 3, 4, 5, and 6 faculty members in the Marketing, Organisational Behaviour (OB), Finance, and Operations Management (OM) areas respectively, to start with. No faculty member retired or joined the institute in the first three months of the year 2000. In the next four years, the institute recruited one faculty member in each of the four areas. All these new faculty members, who joined the institute subsequently over the years, were 25 years old at the time of their joining the institute. All of them joined the institute on April 1. During these four years, one of the faculty members retired at the age of 60. The following diagram gives the area-wise average age (in terms of number of completed years) of faculty members as on April 1 of 2000, 2001, 2002, and 2003.

CAT 2005


Ans .

(3)


    Explanation : In the bar graph, one dip corresponds to the new 25 year old joinee. However, two dips in the trend implies joining of a 25 year old and the retirement of a 60 year old employee. This trait is observed only in Finance department. Hence, the faculty member who retired belonged to Finance.





Ans .

(4)


    Explanation : From the graph of Marketing, it is clear that the new faculty joined in 2001. On April 1, 2000, completed age of Professor Naresh and Devesh were 52 years and 49 years, in no particular order. ∴Age of the third Professor on April 1, 2000 = 49.33 × 3 – (52 + 49) = 47 years Hence, his age on April 1, 2005 was 52 years.





Ans .

(3)


    Explanation : As the dip will be less in case a faculty retired compared to that when a new faculty joined in, so the new faculty member joined the Finance area in 2002.





Ans .

(3)


    Explanation : For the OM area, the only dip comes in the year 2001. So the new 25 year old faculty joined in 2001. Hence, on April 1, 2003, his age will be 27 years old.



Directions for questions 65 to 67: Answer the questions on the basis of the information given below: The table below reports annual statistics related to rice production in select states of India for a particular year.

CAT 2005


Ans .

(1)


    Explanation : State:Haryana , Productivity(tons per hectare): \(\frac{19.2}{3.2} \) = 6
    State:Punjab , Productivity(tons per hectare): \(\frac{24}{4} \) = 6
    State:AndraPradesh , Productivity(tons per hectare): \(\frac{112}{22.4} \) = 5
    State:Haryana , Productivity(tons per hectare): \(\frac{67.2}{16.8} \) = 4 Hence, Haryana and Punjab have the highest productivity.





Ans .

(2)


    Explanation : Gujrat : \(\frac{24}{51}\) = 0.47
    Only per capita production of rice for Haryana, Punjab, Maharashtra and Andhra Pradesh are greater than 0.47.





Ans .

(4)


    Explanation : As seen from the table Haryana, Gujarat, Punjab, MP, Tamil Nadu, Maharashtra, UP and AP are intensive rice producing states.



Directions for questions 68 to 70: Answer the questions on the basis of the information given below: The table below reports the gender, designation and age-group of the employees in an organization. It also provides information on their commitment to projects coming up in the months of January (Jan), February (Feb), March (Mar) and April (Apr), as well as their interest in attending workshops on: Business Opportunities (BO), Communication Skills (CS), and E-Governance (EG).

CAT 2005


Ans .

(1)


    Explanation : Rahul and Yamini.





Ans .

(2)


    Explanation : Gayatri, Urvashi and Zeena cannot attend more than one workshop.





Ans .

(2)


    Explanation : Anshul, Bushkant, Gayatri and Urvashi cannot attend any of the workshops.




Sub–Section III-B : Number of questions = 20


Note: Questions 71 to 90 carry two marks each.

Directions for questions 71 to 74: Answer the questions on the basis of the information given below: In the table below is the listing of players, seeded from highest (#1) to lowest (#32), who are due to play in an Association of Tennis Players (ATP) tournament for women. This tournament has four knockout rounds before the final, i.e., first round, second round, quarterfinals, and semi-finals. In the first round, the highest seeded player plays the lowest seeded player (seed #32) which is designated match No. 1 of first round; the 2nd seeded player plays the 31st seeded player which is designated match No. 2 of the first round, and so on. Thus, for instance, match No. 16 of first round is to be played between 16th seeded player and the 17th seeded player. In the second round, the winner of match No. 1 of first round plays the winner of match No. 16 of first round and is designated match No. 1 of second round. Similarly, the winner of match No. 2 of first round plays the winner of match No. 15 of first round, and is designated match No. 2 of second round. Thus, for instance, match No. 8 of the second round is to be played between the winner of match No. 8 of first round and the winner of match No. 9 of first round. The same pattern is followed for later rounds as well.

CAT 2005


Ans .

(4)


    Explanation :CAT 2005
    Winners after round two would be 1, 2, 3, 4, 5, 11, 10, 9 for 8 rounds respectively. As Lindsay is number two, she will play Venus Williams in quarter finals.





Ans .

(3)


    Explanation : Elena is at number 6 and Serena is at number 8. If they lose, then table would be: CAT 2005
    Maria is at number 1 and she will play the player at number 9. i.e., Nadia Petrova.





Ans .

(1)


    Explanation : CAT 2005
    Matches in bold letters had upsets.
    Then, from the table, winners would be: 1, 31, 3, 29, 5, 27, 7, 25, 9, 23, 11, 21, 13, 19, 15 and 17.
    So for the next round, table would look like: CAT 2005
    Since, there was no upset in the second round, so the table in the next round would look like:
    CAT 2005
    We are given Maria is in the semi-finals. As we are not sure what is the result of other games, table for the next round can be drawn as follows:
    CAT 2005
    Hence, Anastasia will play with Maria Sharapova.





Ans .

(3)


    Explanation :CAT 2005
    In this case, Kim Clijster will either not reach semifinals or she will play Maria in semi-finals. Hence, she cannot play Maria in finals.



Directions for questions 75 to 78: Answer the questions on the basis of the information given below: Venkat, a stockbroker, invested a part of his money in the stock of four companies — A, B, C and D. Each of these companies belonged to different industries, viz., Cement, Information Technology (IT), Auto, and Steel, in no particular order. At the time of investment, the price of each stock was Rs. 100. Venkat purchased only one stock of each of these companies. He was expecting returns of 20%, 10%, 30% and 40% from the stock of companies A, B, C and D, respectively. Returns are defined as the change in the value of the stock after one year, expressed as a percentage of the initial value. During the year, two of these companies announced extraordinarily good results. One of these two companies belonged to the Cement or the IT industry, while the other one belonged to either the Steel or the Auto industry. As a result, the returns on the stocks of these two companies were higher than the initially expected returns. For the company belonging to the Cement or the IT industry with extraordinarily good results, the returns were twice that of the initially expected returns. For the company belonging to the Steel or the Auto industry, the returns on announcement of extraordinarily good results were only one and a half times that of the initially expected returns. For the remaining two companies which did not announce extraordinarily good results, the returns realized during the year were the same as initially expected.



Ans .

(1)


    Explanation : The minimum return will be gained if the extraordinary performing stocks (double & 1.5 growth) are the ones whose expected returns are lowest (i.e. 10% & 20%). Taking the minimum value of the expected returns as 10. We have to see which of the two values of 10 and 20 multiplied by 2 and 1.5 and vice versa yields the minimum value. Hence comparing the minimum value between 20 × 2 + 10 × 1.5 and 20 × 1.5 + 10 × 2, the 2nd one is minimum. Hence the minimum average return is
    \(\frac{20*1.5+10*2+30+40}{4}\) = 30%





Ans .

(2)


    Explanation : If the average return is 35%, then the total return is 35 × 4 = Rs.140.
    The possible arrangements of 140 are
    (i) 40 × 1.5 + 30 + 20 × 2 + 10.
    ∴A = 20 × 2 (Cement or IT)
    B = 10
    C = 30
    D = 40 (1.5) (Steel or Auto)
    or (ii) 40 + 30 × 2 + 20 × 1.5 + 10 ∴A = 20 × 1.5 = 30 (Steel or Auto)
    B = 10
    C = 30 × 2 = 60 (Cement or IT)
    D = 40
    From the data given in the question, we see that statements (II) and (III) are correct.





Ans .

(3)


    Explanation : Total return is 38.75 × 4 = Rs.155
    The possible arrangement is 20 + 10 + 30 × 1.5 + 40 × 2
    Therefore, A = 20, B = 10, C = 30 (Steel or Auto) D = 40 (Cement or IT)
    Hence, statements (I) and (IV) are correct. Hence, (3) is the correct option.





Ans .

(2)


    Explanation : Given Company C is either Cement or IT industry C’s Return is 30 × 2 = 60%
    Among the other values we see that the possible arrangements can be 10 × 1.5 + 20 + 40, 10 + 20 × 1.5 + 40, 40 + 20 + 40 × 1.5
    The average returns will be in each case
    \(\frac{10*1.5+20+40+60}{4}\)= 33.75%
    \(\frac{10+20*1.5+40+60}{4}\) = 35%
    \(\frac{40+20+40*1.5+60}{4}\)=45%
    Considering 33.75% as the valid value, then B belongs to the Auto industry. Hence, (II) and (IV) are correct. Hence, (b) is the correct option.



Directions for questions 79 to 82: Answer the questions on the basis of the information given below: The year is 2089. Beijing, London, New York, and Paris are in contention to host the 2096 Olympics. The eventual winner is determined through several rounds of voting by members of the IOC with each member representing a different city. All the four cities in contention are also represented in IOC.
(a) In any round of voting, the city receiving the lowest number of votes in that round gets eliminated. The survivor after the last round of voting gets to host the event.
(b) A member is allowed to cast votes for at most two different cities in all rounds of voting combined. (Hence, a member becomes ineligible to cast a vote in a given round if both the cities (s)he voted for in earlier rounds are out of contention in that round of voting.)
(c) A member is also ineligible to cast a vote in a round if the city (s)he represents is in contention in that round of voting.
(d) As long as the member is eligible, (s)he must vote and vote for only one candidate city in any round of voting.
The following incomplete table shows the information on cities that received the maximum and minimum votes in different rounds, the number of votes cast in their favour, and the total votes that were cast in those rounds.
CAT 2005
It is also known that:
• All those who voted for London and Paris in round, 1 continued to vote for the same cities in subsequent rounds as long as these cities were in contention. 75% of those who voted for Beijing in round 1, voted for Beijing in round 2 as well.
• Those who voted for New York in round 1, voted either for Beijing or Paris in round 2.
• The difference in votes cast for the two contending cities in the last round was 1.
• 50% of those who voted for Beijing in round 1, voted for Paris in round 3.



    L = London, Paris = P, New York= NY, Beijing = B In round III, one of the two cities, either London or Paris will get 38 votes and the other 37. Further:
    (1) The persons representing London, Paris, Beijing and New York can not vote as long as their own cities are in contention. In round I, New York gets eliminated and hence the representative from NY becomes eligible for voting in the II round hence increasing the total votes by 1. This means the total votes in the first round must be 83 – 1 = 82.
    (2) After round II, the representative from Beijing votes in the III round. This should have increased the number of total votes by 1 and the total votes must have become 83 + 1 = 84. We are given that the total votes in round III are 75 only. We conclude that 84 – 75 = 9 people who voted in round I and II have become ineligible for voting in round III.
    (3) 9 people who have voted in round I and II become ineligible for voting in round III. The reason of their ineligibility is that till round I and II, they have already voted for two different cities which are not available for contention in round III. All of these 9 voters are those who voted for NY in round I and then voted for Beijing in round II.
    (4) Beijing’s vote in round II is 21. This includes 9 votes from people who voted for NY in the first round. So 21 – 9 = 12 people voted for Beijing in both round I and II.
    (5) We are given that 75% of the people who voted for Beijing in round I, voted again for Beijing in round II as well. So, 16 people must have voted for Beijing in round I.
    (6) In round I we have: 82 = L + P + B + NY Or 82 = 30 + P + 16 + 12 Giving P = 24
    (7) In round II, we have: 83 = L + 32 + 21, giving L = 30
    (8) NY had 12 votes in round I. 9 of these votes went to B(see point 2 , again). The rest 3 went to P.
    (9) 16 votes for B in round I. 12 of them still vote for B. The rest 4 voted for either L or P. L has the same number of votes in both the rounds I and II. This means in round II, these 4 votes must have gone to Paris only.
    (10) The representative from NY did not vote in round I but has voted in round II. As L has the same people voting for it (30 votes in both the rounds I and II) and we know the exact break up of B in II. This NY-representative vote must go to Paris only. Further, in order to avoid ineligibility, this NY rep must vote for Paris only in round III also.
    (11) Paris (in round II) break up is: 32 = 24 ( from round I, who voted for Paris ) + 4 ( out of the 16, who voted for Beijing in round I) + 3( out of 12, who voted for NY in round I ) + 1 (NY -Rep)
    (12) Beijing gets eliminated in round II. So the rep of Beijing can vote in round III.
    (13) 12 People (out of 21) who voted for Beijing in round II are still eligible for vote in round III.
    (14) 50% of people who voted for Beijing in I ( i.e. 8 People) voted for Paris in round III. These 8 People include 4 of those who voted for Paris in round II also. Therefore 4(out of 12 who voted for Beijing in round II and are still eligible for vote in round III ) people have voted for Paris in round III.
    (15 ) This implies that the rest 8(out of 12 who voted for Beijing in round II and are still eligible for vote in round III ) can vote for London only. This makes London’s vote = 30+ 8 or 38 in round III. Which implies that Paris got 37 votes.
    (16) The Beijing Rep who is eligible to vote in round III must have voted for Paris only.
    The following table sums up the Vote Pattern: CAT 2005
    (The data shown in Bold was already provided in the problem. The other data is deduced from the solution.)





Ans .

(4)


    Explanation : Required percentage = \(\frac{9}{12}\) *100 = 75%





Ans .

(4)


    Explanation : As seen from the table, Paris got 24 votes.





Ans .

(4)


    Explanation : Required percentage = \(\frac{8}{12}\) * 100 = 66.67%





Ans .

(1)


    Explanation : Based on the table, IOC members from New York must have voted for Paris in Round II.



Directions for questions 83 to 86: Answer the questions on the basis of the information given below:
The table below presents the revenue (in million rupees) of four firms in three states. These firms, Honest Ltd., Aggressive Ltd., Truthful Ltd. and Profitable Ltd. are disguised in the table as A, B, C and D, in no particular order.
CAT 2005
Further, it is known that:
• In the state of MP, Truthful Ltd. has the highest market share.
• Aggressive Ltd.’s aggregate revenue differs from Honest Ltd.’s by Rs. 5 million.


Explanation : The given information can be tabulated as follows:
CAT 2005





Ans .

(2)


    Explanation : As Truthful Ltd. has the highest market share, so Truthful Ltd. can be A or C. From neutral statement, either B and C are Aggressive and Honest or A and D are Aggressive and Honest. According to statement 1, B is Profitable. Then, A and D are Aggressive and Honest.Then, Honest’s total revenue cannot be more than that of Profitable. Hence, statement 2 is false.





Ans .

(3)


    Explanation : According to statement 1, Aggressive is B. Then, Honest has to be C (as given in the neutral statement). Then, statement 2 is also true as Honest Ltd’s. lowest revenue is from Bihar.





Ans .

(3)


    Explanation : B is Honest according to statement 1. Atmost one statement can be true as both give Aggressive and Honest as firm B and firm B cannot have two names.





Ans .

(3)


    Explanation : Profitable can be either A or D. Then, Aggressive and Honest have to be B and C. Hence, Truthful is D or A. For both A and D, lowest revenue is from UP. Hence, (3) is the correct option.



Directions for questions 87 to 90: Answer the questions on the basis of the information given below: Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamil Nadu, Flood Relief (FR in Maharashtra, and Earthquake Relief (ER) in Gujarat. Each volunteer working with Help Distress has to be involved in at least one relief work project.
• A maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.
• The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.
• 17 volunteers are involved in the TR project.
• The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER project alone.
• Ten volunteers involved in the TR project are also involved in at least one more project.


Explanation : CAT 2005
10 in TR also in at least one more ⇒ 7 in TR alone
CAT 2005
TR alone = one less than ER alone ⇒ ER alone = 8
ER alone = double of all 3 ⇒ In all three = \(\frac{8}{2}\)=4
CAT 2005
FR alone = (FR and ER)
CAT 2005
⇒ p = q + 4 ...(1)
Total = 37 [7 + 8 + p + (x + y + q) + 4] = 37
[p + q = 12]
⇒ p – a = 4 ⇒ p = 8 and q = 4
CAT 2005
Now, total number of FR is maximum
⇒ 8 + 4 + 4 + x > 8 + 4 + 4 + y
⇒ x > y and x + y = 6
as n(TR) = 17
= x + y + 4 + 7
⇒ x = {4, 5, 6}
y = {0, 1, 2}





Ans .

(3)


    Explanation : Both FR and TR but not ER = x Minimum x = 4





Ans .

(1)


    Explanation : Option (2) and option (3) are superfluous. They are not required. Option (1), if given, would tell us the value of x = 4 and hence y = 2.





Ans .

(2)


    Explanation : Out of 4 who are in all three projects, 2 move out of FR and one-one move out of ER and TR.
    CAT 2005
    Minimum in FR = 14 + x = 14 + 4 = 18
    Maximum in ER = 15 + y = 15 + 2 = 17
    As x=(4,5,6) & y=(0,1,2)
    Hence, option (2).





Ans .

(4)


    Explanation : FR and ER = 5
    ER and TR = y + 2
    ⇒ 5 = y + 2
    ⇒ y = 3;
    which is not a possible value as y is 0, 1, or 2 only.
    ⇒ option (4)
    Inconsistent data.