Cl Mock Cat 6 Key 2008

Mock CAT – 6

Answers and Explanations 1

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60

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

Section III

20 20 20 60

150

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MCT-0010/08 006

1

1. 1

C introduces women who are the subject of the paragraph. CE forms a mandatory pair with the description of the condition interlinked, B takes it forward, and D forms the conclusion.

2. 4

E forms a pair with A. A provides the background of the start and E traces the rise. B & D describe Barbie’s iconization and C marks the conclusion.

3. 4

AABBA ‘Borough’ means a town, area, or constituency represented by a Member of Parliament, ‘burrow’ means a rabbit hole. ‘Foul’ means something unfavourable, ‘fowl’ is a type of a bird. ‘At’ is the correct preposition used to indicate location. ‘Islet’ means a very small island and ‘eyelet’ means a small hole, usually round and finished along the edge. ‘Meed’ means reward or to recompense and ‘mead’ means alcoholic liquor made by fermenting honey and water.

4. 1

BBAAB ‘Sashay’ means to glide, move, or proceed easily or nonchalantly, ‘sachet’ means a small bag, case, or pad containing perfuming powder. ‘Calender’ means a machine in which cloth, paper, or the like, is smoothed, glazed and ‘calendar’ means a table or register with the days of each month and week in a year. ‘Claque’ means a group of sycophants and ‘clack’ is to make a quick, sharp sound, or a succession of such sounds. ‘Its’ is the possessive form and ‘it’s’ is the contraction of it is. ‘Bridal’ means pertaining to a bride or a wedding and ‘bridle’ means part of the tack or harness of a horse.

5. 3

B is incorrect as it should be ‘structure to that of the traditional…..’ C is incorrect should be ‘drawn between’, E is incorrect as ‘the’ is missing before “climactic”, options A, B & D are correct.

6. 4

B is incorrect, should be ‘changes’ instead of the singular ‘change’, E is incorrect, should be ‘at’ instead of ‘on’. A, C and D are correct.

7. 3

A is incorrect, use the singular form of ‘step’ here, E is incorrect, should be increasing consumption ‘of ’ meat. B, C and D are correct.

8. 1

Since the paragraphs are a part of the same passage, we have to consider the theme of the whole passage while choosing the answer options. In this paragraph the author discusses the changes that happened in the genre of European Literature. The author talks about a ‘new kind of writing was born’. He further illustrates that the old forms of storytelling would have to compete against a ‘young rival’. Therefore, we have to pick up an answer option which introduces the ‘young rival’. Option 1 accomplishes this, hence it is the correct answer. The reader is not a ‘literary force’ as it is mentioned in option 2, this makes option 2 incorrect. Option 3 revolves around the expectations of the reader which are his personal expectations and show no relevance to the passage, this makes option 3 incorrect. In Option 4 the word ‘similar’ tends to compare the reader with someone/something, which is ironical, serious and has a principled reticence. The paragraph does not mention any of these attributes anywhere, which makes option 4 incorrect. Option 5 does not follow the theme of the paragraph, hence it is incorrect.

9. 4

2

This paragraph discusses the reasons for the ‘novelty of the novel’ and why the readers “identify” with the novel. Towards the end of the paragraph the author gives the unique reason which makes the novel different from any other genre of

writing. The author highlights the importance of the narrator’s role in storytelling. In the last two lines of the paragraph, the author tells us what kind of stories invoke our imagination, however he says there is something more which is required to get completely into the novel. Option 1 falls out of the scope of the argument as it talks of the search of certain individuals who may not even be reading the novel, this makes option 1 incorrect. Option 2 cannot be a part of the incomplete sentence as it starts with a present continuous tense whereas the given part of the sentence is in simple present tense; moreover it talks of ‘political aspirations’ which do not match the theme of the passage. Option 3 moves away from the theme and has a admonishing tone that doesn’t match the tone in the first part of the sentence. Option 4 points towards multiple characters who are storytellers and suggests that the reader should ‘identify’ with them, which aptly fulfills the requirements for a reader to ‘engage’ with the novel. This makes Option 4 the correct answer choice. Option 5 is merely a repetition of the last two sentences; hence it cannot be the answer. 10. 3

The paragraph starts by the author describing the ‘art of reading the novel’. Then it draws similarities between the novel and essay. The paragraph progresses further by establishing the superiority of the essay over the novel. In the last line the author talks of the novel aspiring to the ‘condition of the essay’, hence it would require an answer option that continues the discussion further and also relates it to the next paragraph. Option 1 does not fulfill the requirement as it suggests a regression rather than a progression, moreover the last word in the sentence ‘You might even say…..’ is ‘and’ which would not suggest any dramatic turn in the tone of the original sentence. This makes option 1 incorrect. Similarly, options 2, 4 and 5 talk of the novel whereas the next paragraph starts with the names of renowned essay-writing novelists. Option 3 mentions both novelists and essayists; it carries the theme mentioned in this paragraph and even relates to the next paragraph, thus it is the right answer.

11. 5

The paragraph focuses on “prosaic beauty” where there is an amalgamation of opposites- ‘ugliness flirts with beauty, and reason courts the absurd’. Towards the end of the paragraph the author talks of a positive attribute while discussing the technique of Henry Fielding. We have to provide an answer choice that describes the completion of the development by Laurence Sterne. Keeping in view the structure of the paragraph we have to pick an answer choice which talks of a negative development. Option 5 does this aptly. Option 1 is contradictory to the theme of the paragraph. Options 2, 3 and 4 would be applicable in case of a person; they do not pertain to the ‘story’.

12. 2

Refer to paras 1 to 3 where Option 2 is said to be one of the factors which adequately account for functional autonomy. The other options are either incorrect or part of objections to functional autonomy.

13. 4

Option 1 would meet with agreement from the behaviorist. It is not the author’s opinion. Options 2 and 3 are distorted. Option 4 is stated by the author at the end of Para 3. Hence option 4 is correct.

14. 5

austere: “with his celebration of austere and productive virtues, and recommendations for”

15. 5

denying: “worked for the sake of work, denying himself the fruits of his labour.” The pre-modern

16. 1

light: “time is money; the light purse means a heavy heart-were by now ringing hollow”.

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17. 3

century: “The question of what is the contemporary, twenty-first century ethic of capitalism is one”

If “Jersey 6” is not included in the loop, he will not get any chance to pass the ball to any of his team members and hence the minimum number of passes that he can make is 0. When he is included in the loop, he gets to pass the ball back after every fifth pass made by his team.Given that “Jersey 6” is in the loop, the minimum number of passes that he can

18. 3

The passage largely focuses on the fact that sensation and perception are separable.

19. 2

According to the passage the body was initially bound by the sensory phase, which was a very stiff stage, perception had subsequently followed.

1001 make, out of the given 1001 is   = 200 and the maximum  5 

20. 5

According to the passage the process gets enclosed and is now more personal and private, making option 5 correct.

1001 number of passes that he can make is   +1 = 201.  5 

21. 4

If we compare Selection I and Selection III, we can conclude that Tim is definitely not the child that has a sibling amongst the given ten children.

12

24. 5

The alphabets selected by Professor Chaurasia are A(1), B(2), C(3), E(5), G(7), I(9), K(11), M(13), O(15), Q(17), S(19), U(21) and W(23). Therefore, OGVUE is definitely not a word formed by him, as it contains V. Hence, option (5) is the correct choice.

25. 2

As per the information given in the question, the following table can be drawn.

The other two pairs of siblings could be (Sam and Den); (Den and May); (Sam and May); (Ken and Ron); (Ken and Joe) and (Joe and Ron). Option (1): Possible if the other two pairs of siblings are (Sam and May) and (Joe and Ron) Option (2): Possible if the other two pairs of siblings are (Sam and Den) and (Ken and Joe)

Containers Liquids

Option (3): Possible as along with Tim one out of the remaining three children does not have any sibling.

22. 1

As per the information given in the question, the following table can be drawn.

Amar A

×

B

×

Bhuvan

Chirag

×

Elan

×

×

×

×

×

C

×

D E

Dhruv

× ×

×

×

Given that the code assigned to Amar is D. This implies that code assigned to Elan is B and code assigned to Dhruv is E. Also, the code assigned to Chirag is A. Therefore, the code assigned to Bhuvan is C. 23. 3

C1

C2

F

A

C3

C4

C5

E

C6

C7

C

Now, since C7 is neither filled with B nor D, therefore, it has to be filled with G.

Option (5): Possible if the other two pairs of siblings are (Sam and Den) and (Joe and Ron). Option (4): Is not possible because if Ron and Joe are siblings, then both Sam and May do not have any sibling which is not possible, same way if Sam and May are siblings, then both Ron and Joe do not have any sibling, which is not possible. Hence, option (4) is the correct choice.

9

15

Following the same logic if we compare Selection II and Selection IV, we can conclude that Bob is definitely not the child that has a sibling amongst the given ten children. Now, in selection III we can conclude that Ian and Emy are siblings.

6

3

For questions 26 to 29: For questions 26 and 27: For Gaurav to earn minimum possible number of points in each of the weeks, the substitutions made by Gaurav have to be made in such a way that the number of points is minimized in each of the weeks keeping in mind that at most three substitutions could be made each week. The following table provides information about the substitutions made by Gaurav to meet the desired scenario. Note that in weeks 2, 3, 4 and 6 the teams selected by Gaurav are the five players who have earned the five lowest points for that week. For example in week 2, the least five points for that week are earned by H, C, F, D and I. The same holds true for weeks 3, 4 and 6 also. But in week 5 since only three substitutions can be made and also because each of D, F and I have earned more points than B in week 5, the players who were substituted by Gaurav for week 5 were D, F and I.

Week 1 Week 2 Week 3 Week 4 Week 5 Week 6

When “Jersey 1” begins the chains of passes, some of the players repeatedly get the ball, making a loop while others never recieve the ball as illustrated in the figure below:

B

H

J

B

B

C

C

E

D

J

B J

F

F

A

F

C

H

I

I

I

I

E

E

E

D

D

G

G

G

1 13

Hence the number of points earned by Gaurav in week 5 is 100 + 25 + 40 + 20 + 25 = 210.

27. 2

The number of weeks in which B, J and E are together in the same team is 1, i.e. in week 6.

4

10 7

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26. 4

3

For questions 28 and 29: Since Sanjay did not substitute any player at the end of weeks 1 and 2, therefore the team selected by Sanjay in week 2 and week 3 is same as the team selected by him in week 1.

Maximum possible selling price of Shirt-2 will be in January, i.e. 1.05 × 400 = Rs.420. For Trouser-1: Minimum possible percentage decrease in the selling price of Trouser-1in any of the given months

At the end of week 3: Sanjay substituted F with E because E and F earned 140 and 40 points respectively in week 4. He also substituted G with H because G and H earned 60 and 110 points respectively in week 4.

 500 – 425  =  × 100 = 15% 500   Maximum possible percentage decrease in the selling price of

Since the number of points earned by E in weeks 5 and 6 is 20 and 60 respectively and that by F in weeks 5 and 6 is 180 and 150 respectively, Sanjay substituted E with F at the end of week 4.

 575 – 500  Trouser-1 in any of the given months =   = 15% 500   The different possibilities are

For the same reason as stated above Sanjay substituted C with D at the end of week 4.

January

February

March

1

+ 8%

– 12%

+ 10%

2

+ 8%

– 12%

– 10%

Week 6

3

– 8%

+ 12%

+ 10%

4

– 8%

+ 12%

– 10%

The following table provides information about the teams and the number of points earned by Sanjay.

Team

Week 1

Total

Week 2

Week 3

Week 4

Week 5

A

A

A

A

A

A

C

C

C

C

D

D

F

F

F

E

F

F

G

G

G

H

H

H

I

I

I

I

I

I

310

345

660

615

590

625

28. 5

Number of points earned by Sanjay in week 5 is 590.

29. 2

G and H are the two players that were present in the team selected by Sanjay for three consecutive weeks, i.e. G in weeks 1, 2 and 3 and H in weeks 4, 5 and 6.

Maximum possible price of Trouser-1 will be in the month of march when the possibility that is stated in (3) in the table given above holds true. Maximum possible price of Trouser-1

= Rs. (500 × 0.92 × 1.12 × 1.1) = Rs.567 Therefore, the maximum possible amount paid by Rahim for the purchase given in the question = Rs.420 + Rs.567 = Rs.987. 32. 1

For questions 30 to 32: 30. 2

Minimum possible percentage decrease in the selling price of 25  300 − 275  Shirt-1 in any of the given months =   × 100 = 3 % 300   Maximum possible percentage increase in the selling price of

For Shirt-2: Minimum possible percentage decrease in the selling price of 55  400 − 345  % Shirt-2 in any of the given months =   × 100 = 400 4   Maximum possible percentage increase in the selling price of 25  450 − 400  Shirt-2 in any of the given months =   × 100 = 2 % 400   The different possibilities are

4

January

February

March

1

+ 5%

– 10%

+ 8%

2

+ 5%

– 10%

– 8%

3

– 5%

+ 10%

– 8%

February

March

– 15%

+ 10%

+ 10%

2

– 15%

+ 10%

– 10%

1

January

February

March

– 10%

+ 10%

– 10%

Minimum possible difference between the selling price of Shoe-1 and Shoe-2 at any point of time in the given three months will be in March. This will be possible when possibility stated in (2) holds true for Shoe-1. Required difference = Rs.757.35 – Rs.712.8 = Rs.44.55.

= Rs. (300 × 1.1× 0.95 × 0.9 ) = Rs.282.15 Since, we need to calculate the maximum possible amount paid by Rahim for his mentioned purchase, we can easily conclude that he bought Shirt-2 and Trouser-1.

January 1

For Shoe-2: The different possibilities are

35  335 − 300  Shirt-1 in any of the given months =  %  × 100 = 300  3  So the only possible combination of the percentage changes in the selling price of Shirt-1 is +10%(in January), –5%(in February) and –10%(in March). Therefore, the selling price of Shirt-1 in the month of March

31. 4

For Shoe-1: The different possibilities are

For questions 33 to 36: It is given that out of the 15 representatives, three representatives speak two languages each; two representatives speak three languages each; five representatives speak five languages each; one representative speaks one language and four representatives speak four languages each. Let the number of representatives that speak Language 1, Language 2, Language 3, Language 4, Language 5, Language 6 and Language 7 be A, B, C, D, E, F, G and H A + B + C + D + E + F + G + H = 3 × 2 + 2 × 3 + 5 × 5 + 1× 1 + 4 × 4

= 6 + 6 + 25 + 1 + 16 = 54 A + B + C = 9 + 7 + 10 = 26. Therefore, D + E + F + G + H = 54 – 26 = 28. 33. 5

Since we need to maximize the number of representatives that speak a particular language, we will assume that three languages (let it be language 4, language 5 and language 6) are spoken by 5 representatives each. Therefore, at most 28 – (5 + 5 + 5) = 13 representatives speak a particular language.

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34. 3

It is given that the minimum possible number of representatives that speak a particular language is 7.

Therefore, Aamna and Alice bought 2 diaries and 1 diary respectively from shop II.

So, for each of the languages other than the language 1, language 2 and language 3, the number of representatives who speak it cannot be less than 7.

Aamna: 2 Diaries (Shop II); Alice: 1 Diary (Shop II); Anisha: 1 Diary (Shop I or Shop III) Arjun: 1 Diary (Shop I or Shop III) Since the pens sold from shop I and shop II are bought by Alice and Ajay respectively, therefore Arjun and Anisha bought 1 and 2 pens respectively from shop III.

Since, there are four other languages and the aggregate number of representatives remaining is 28, therefore the other

28 = 7 representatives each. 4 So, the number of languages that are spoken by seven four languages are spoken by

representatives each is 5. 35. 4

Option (1) can be true. Let’s assume that Language 1 is German, therefore the aggregate number of representatives that speak French, Spanish and English is 9 + 9 + 9 = 27. So, the remaining one language is spoken by 28 – 27 = 1 representative. Option (2) can be true. Let the number of representatives that speak English and Arabic be X and Y respectively. X + Y 15 + 11 = 26.

Arjun: 1 Pen (Shop III); Anisha: 2 Pens (Shop III) Since, Alice has already bought three articles from shop I (1 Pencil and 2 Pens); therefore Alice bought 2 Erasers from shop II. So, Aamna and Anisha bought 1 eraser each from shop I. Alice: 2 Erasers (Shop II); Aamna 1 Eraser (Shop I); Anisha 1 Eraser (Shop I) Since, Alice has bought 1 pencil from shop I, therefore Aamna bought 2 pencils from shop III. Aamna: 2 Pencils (Shop III) Now, since Anisha has bought 1 eraser and 1 sharpener from shop I, therefore Anisha bought 1 Diary from Shop III and Arjun bought 1 Diary from Shop I. Now, the following table can be made.

So there will be two languages spoken by one representative each. Option (3) can be true. There could be one representative speaking Portuguese. (Logic explained for options (1) and (2)). Option (4) cannot be true. Since Hindi, English and Portuguese are spoken by 8 representatives each, therefore these languages cannot be either of language 1 or language 2 or language 3. Let the number of representatives that speak Hindi, English and Portuguese be M, N and P respectively. M + N + P = 8 + 8 + 8 = 24. Hence, Spanish can only be spoken by 28 – 24 = 4 representatives and not 5. 36. 1

Number of representatives speaking Language 1 is 9. Out of the 28 representatives, 10 representatives each can possibly speak at most two languages. Hence, for at most 3 languages, the number of representative speaking it can be greater than the number of representatives speaking Language 1.

Pencils

Erasers

Pens

Sharpeners

Diaries

0

1 (Shop II)

2 (Shop II)

1 (Shop III)

0

1 (Shop I)

1 (Shop I) 2 (Shop II) 2 (Shop I)

0

1 (Shop II)

Aamna 2 (Shop III) 1 (Shop I) Arjun Alice

0

0

Anisha

0

1 (Shop I) 2 (Shop III)

1 (Shop I)

1 (Shop III)

Ajay

0

1 (Shop III) 1 (Shop II)

0

0

37. 5

For each friend it can be exactly determined which article they bought from which shop.

38. 2

Aamna bought both the pencils from shop III.

39. 2

Total amount spent by Alice = 1 × 5 + 2 × 10 + 2 × 3 + 1 × 45 = Rs. 76

40. 1

Amount spent by Aamna = 2 × 3 + 1 × 1 + 1 × 1 + 2 × 45 = Rs. 98 Amount spent by Arjun = 1 × 40 + 1 × 16 = Rs. 56 Amount spent by Alice = 1 × 5 + 2 × 10 + 2 × 3 + 1 × 45 = Rs. 76 Amount spent by Anisha = 1 × 1 + 2 × 16 + 1 × 2 + 1 × 60 = Rs. 95 Amount spent by Ajay = 1 × 4 + 1 × 20 = Rs. 24

For questions 37 to 40: Given that Ajay did not buy a eraser from Shop II, which means that Ajay bought 1 eraser from Shop III and 1 pen from Shop II. Ajay: 1 Eraser (Shop III) and 1 Pen (Shop II).

For questions 41 and 42: Since, Anisha did not buy any article from shop II, therefore Anisha bought 1 sharpener from shop I. Also, since Anisha bought 1 sharpener from shop I, therefore Aamna bought 1 sharpener from shop II Anisha: 1 Sharpener (Shop I); Aamna 1 Sharpener (Shop II). Since, Alice did not buy any article from shop III and there is no pencil sold from shop II, therefore Alice bought 1 pencil from shop I. For the same reason Alice bought 2 pens from Shop I as the pen sold from shop II is bought by Ajay. Alice: 1 Pencil (Shop I); 2 Pens (Shop I) Anisha and Arjun did not buy any article from shop II. Since, each of Anisha and Arjun bought 1 diary each, therefore they bought it either from shop I or shop III.

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A

C

A

B

B Case I

C Case II

Let Anil, Biswas and Chandra be denoted by A, B and C.

5

Since the time taken by A, B and C to complete one lap of the circular track is 4, 6 and 12 minutes respectively, therefore the ratio of speeds of A, B and C is 3:2:1.

Case III: 153 – 3x = 310 – 10y ⇒ 10y – 3x = 157 The minimum positive integral value of (x + y) for which x and y satisfy the above equation, is obtained when y = 16 and x = 1.

Now, it is given that at time t = T the positions of A, B and C is at the positions of B, C and A respectively at time t = 0.

⇒ (x + y)min = (16 + 1) = 17.

Therefore, case II is not possible as the ratio of the distances covered by A, B and C in the same time is 3:2:1.

Hence, the required minimum number is 17. 45. 3

⇒ y = 11

Length of the circular track = 3x + 2x + x = 6x.

⇒ A total of 11 Skilled workers were hired that day. The number of supervisors = 3

Time taken by A to cover 6x = 4 minutes, therefore time taken by A to cover 3x will be equal to 2 minutes. 41. 4

Minimum possible value of T = 2 minutes = 120 seconds.

42. 1

It is given that 6x = 240 metres. LCM of 4, 6 and 12 is 12. Therefore, after 720 seconds, each of A, B and C will be at their respective initial positions. Initial shortest distance between A and C is ‘x’. So, we can say that after 720 seconds the shortest distance between A and C will be ‘x’. In the remaining 180 seconds, the distance covered by A will

 x + y x + y ⇒ max  1,    = 3 ⇒  10  = 3     10   x+y ⇒3≤ < 4 ⇒ 30 ≤ (x + y) < 40 10

⇒ 19 ≤ x < 29 Let w denote the total amount of daily wages paid to all the workers, including the supervisors. w = x × max(75, 153 – 3x)] + 200 × 11 +3 [2 × max (75,153 − 3x) + 200]

or w = (6 + x)[max (75,153 − 3x)] + 2800

3 × (6x ) = 4.5x. In other words, A is 6x – 4.5x = 1.5x 4 metres behind the initial position of A. In the remaining 180 seconds, the distance covered by C will be

We have to find, at which value of x (such that 19 ≤ x ≤ 28 ) is the value of w, the maximum. If x = 28 w = 34 × 75 + 2800 = 2550 + 2800 = Rs.5350 If x = 27 w = 33 × 75 + 2800 = Rs.5275 If x = 24 w = 30 × 81 + 2800 = Rs.5230 Checking for few more values The maximum value of w is found at x = 28. The maximum value is Rs.5350.

1 × (6x ) = 1.5x. In other words, C is 1.5x – x = 0.5x metres 4 ahead the initial position of A.

be

Therefore, the required answer is 1.5x + 0.5x=2x = 80 metres. For questions 43 to 45: Given, x + y = 30 ⇒ Number of supervisor(s) = max (1, 3) = 3 w(s) = 2w(x) + w(y) = 2 max(75, 153 – 3x) + max(120, 310 –10y) y = 30 – x w(s) = 2 max(75, 153 – 3x) + max(120, 10 + 10x) As y ≠ 0 , therefore the range of values of x is 0 ≤ x ≤ 29. As the value of x increases, the difference between (153 – 3x) and 75, becomes smaller.

Similarly the difference between 120 and (10 + 10x) also becomes smaller, with the increase in the value of x. At x = 10: w(s) = 2 max (75, 123) + max (120, 110) = Rs.366 At x = 11: w(s) = 2 max (75, 120) + max (120, 120) = Rs. 360 At x = 12: w(s) = 2 max (75, 117) + max (120, 130) = Rs.364 Hence for the daily wage of the supervisor to be the minimum, x must be 11, y must be 19 and the daily wage of one supervisor will be Rs.360. Hence, the daily wage of all three supervisors, together, will be Rs.1080. 44. 4

For this, we must have w(x) = w(y). or, max (75, 153 – 3x) = max (120, 310 – 10y) As 75 ≠ 120, only three possibilities are there Case I: 75 = 310 – 10y ⇒ y = 23.5; not a relevant solution as y must be an integer.

For questions 46 and 47: AA Given that

BB CC

DD0 Since A, B and C are distinct digits greater than 1, therefore the base in which these numbers are written cannot be less than 4. For example, let us assume that the numbers are in base 6. Therefore, two cases are possible. Case I: A + B + C = 6 Case II: A + B + C = 12 Similarly, for base 7, base 8 and base 9, the possible values of A, B and C not necessarily in that particular order can be found out. Base 4 Possible values of A, B and C

43. 4

w(y) = Daily wage of a skilled worker = Rs. 200

⇒ max (120, 310 − 10y) = 200 ⇒ 310 − 10y = 200

Let the distances covered by A, B and C at time t = T be 3x, 2x and x respectively.

Base 5

No No values values possible possible

Base 6 Base 7 Base 8 Base 9 1, 2 and 3

1, 2 and 4

1, 2 and 5

1, 2 and 6

3, 4 and 5

3, 5 and 6

1, 3 and 4 4, 5 and 7 3, 6 and 7

1, 3 and 5 2, 3 and 4 5, 6 and 7 4, 6 and 8 3, 7 and 8

Case II: 153 – 3x = 120 ⇒ x = 11

6

⇒ y ≥ 19 ⇒ (x + y)min = 11 + 19 = 30

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46. 3

There are 14 possible values of (A, B and C).

47. 4

Maximum possible value of (A + B + C) = 18.

Alternate method: OE = OF + 2r

OF = FB = 2 2

For questions 48 to 50: r=

(4–2 2 ) = (2 − 2

O

2) units

For questions 51 and 52: K

51. 1

C

A C1

E C2

C3

C4

C5

C6

HC 7

C8

I C9

C 10

D

2

G A

B

3

C 11

AG AB 4 = = GD CD 3

As OC:OA = 3:4 and ∆ AGB ~ ∆DGC and ∆OCD ~ ∆OAB

C

Case II: The numbers filled in the cell C1 and C3 is 4 and 1 respectively. Therefore, the numbers filled in the cell C2, C6, C7, C10, C11 and C12 is 1, 1, 4, 4, 1 and 4 respectively. Therefore, there are only 2 ways in which the numbers can be filled in the cells of the matrix.

 AG AB  ⇒ =   GD CD   AB OA   CD = OC    ⇒ AG:CD = OA:OC = 4:3.

49. 4

2

Case I: The numbers filled in the cell C1 and C3 is 1 and 4 respectively. Therefore, the numbers filled in the cell C2, C6, C7, C10, C11 and C12 is 2, 1, 1, 4, 4 and 1 respectively.

Alternate method:

48. 2

F

Now, as per the information given in the question the number filled in the cell C5 has to be 2. Similarly, the number filled in cell C8, C4 and C9 have to be 2, 3 and 3 respectively.

AB = OA 2 + OB2 = 4 2 units, CD = OC2 + OD2 = 3 2 units

⇒

3

G

D

As CD is parallel to AB, therefore ∆AGB ∼ ∆DGC

C 12

B

Required ratio is 4:3.

52. 4

A C1

C2

E C3

C4

C5

C6

C7

C8

B

From point G, drop a perpendiculer GK, on OA. Now, as KG is parallel to OD, therefore ∆AKG ∼ ∆AOD

⇒

AG KG 12  AG  4 3 = ⇒ KG =   × OD =  7 × 4  × 4 = 7 units AD OD  AD   

1 1 12 6 ⇒ ∆ AGC = × AC × KG = × 1× = square units 2 2 7 7 50. 1

O

F

A

B

r E

Let the radius of the cirlce be r units. In the right triangle OAF, OA 2 = AF2 + OF2 ⇒ 42 = 2 2

(

)

(

)

2

+ ( 4 − 2r ) ⇒ r 2 − 4r + 2 = 0 2

HC 9 C 13 D

C 10 I C 11

C 12

C 15

C 16

C 14

G

F

C

Case I: The numbers filled in the cells C4, C8, C12 and C16 be 1, 2, 3 and 4 respectively. The number of ways in which the numbers can be filled in the cells C3 and C7 is 2, i.e. (C3 = 3, C7 = 4) and (C3 = 4, C7 = 3). For each of these 2 ways, the number of ways in which the numbers can be filled in the cells C11 and C15 is 2, i.e. (C11 = 1, C15 = 2) and (C11 = 2, C15 = 1). Therefore, the number of ways in which the numbers can be filled in the cells C3, C7, C11 and C15 = 2 × 2 = 4. For each of these 4 ways, the number of ways in which the numbers can be filled in the cells C1 and C2 is 2, i.e. (C1 = 4, C2 = 2) and (C1 = 2, C2 = 4). For each of these 2 ways, the number of ways in which the numbers can be filled in the cells C5 and C6 is 2, i.e. (C5 = 1, C6 = 3) and (C5 = 3, C6 = 1). So, the number of ways in which the numbers can be filled in the cells C1, C2, C5 and C6 = 2 × 2 = 4. For each of these 4 ways, there is only 1 way in which the numbers can be filled in the cells C9, C10, C13 and C14. Therefore, the total number of ways is 4 × 4 = 16.

∴ r = 2 − 2 units

006

7

Possible Cases: Now, the numbers can be filled in the cells C4, C8, C12 and C16 in 4! ways. Therefore, the total number of ways in which the numbers can be filled in the cells = 4! × 16 = 384.

56. 4

A D

C2

4!

1

C5

2!

2

HC 9

E C3

C4

C7

C6

2!

3

C 10 I C 11

4

6

5

C 13

2!

C 14

2!

C 15

B

C8 C 12

Q

With point A as one of the three vertices, 8 distinct isosceles triangles can be drawn:

F

∆ABC, ∆ABE, ∆ABD, ∆AED, ∆APR, ∆ABP, ∆ACR and ∆ACE. With point B as one of the three vertices, 10 distinct isosceles triangles including those having A as one of the three vertices. The triangles having point B as the vertex and not having point A as the vertex are 8 in number.

C 16 1

∆BDE, ∆PBC, ∆PBQ, ∆BCR, ∆BQR, ∆BQC, ∆PBD and ∆BCD With point C as one of the three vertices, 3 distinct isosceles triangles(excluding those having one or both of A or B as their

For questions 53 and 54:

54. 3

Let ‘N’ = AB, where A and B are digits. AB = 10A + B. Now, since A is a factor of AB, therefore A should divide 10A + B completely. This also means that A should divide B completely. B = 9: The possible values of A are 1, 3 and 9. B = 8: The possible values of A are 1, 2, 4 and 8. Similarly, when the value of B is 7, 6, 5, 4, 3, 2 and 1 the values of A are the number of factors of each of the possible values of B. B = 0: The possible values of A are 1, 2, 3 ......., 8 and 9. Sum of all such two-digit numbers more than 35 = S S = 40 + 50 + 60 + 70 + 80 + 90 + 44 + 55 + 36 + 66 + 77 + 48 + 88 + 39 + 99 = 942. Let ‘N’ = ABC As per the information given in the question AB should be a factor of 100A + 10B + C. This means that 10A + B should divide 100A + 10B + C completely. Since 100A + 10B is divisible by 10A + B, therefore C should be divisible by 10A + B. This is only possible if C = 0. Therefore, number of such three-digit numbers are 9 × 10 = 90.

vertices) can be drawn: ∆CDE, ∆CRQ and ∆PCQ Therefore, a total of 8 + 8 + 3 = 19 such isosceles triangles can be drawn. For questions 57 and 58: There is more than one way to solve this one of the methods is f(1) = 1; f(2) = 4(f(1) + 6 = 10 f(3) = f(1 + 2) = f(1) + 12(1) + 12 = 25 f(5) = f(3 + 2) = f(3) + 12(3) + 12 = 73 f(10) = 4(f(5) + 6 f(20) = 42f(5) + 24 + 6 = 1198 f(22) = 1198 + 12 × 20 + 12 = 1450 57. 3

f(20) = 1198

58. 5

f(22) – f(5) = 1450 – 73 = 1377

59. 4

Let the cost of 1 desk, 1 chair and 1 table be ‘D’, ‘C’ and ‘T’. 6D = 6C + T ...(i) 12C = 6D + 3T ...(ii) Solving equation (i) and (ii) we get that T =

For questions 55 and 56: 55. 2

3 (C) ⇒ C : T : D = 4 : 6 : 5 2 Let, the value of C, T and D be 4x, 6x and 5x respectively. All the options except option (4) are correct. Hence, option (4) is the correct choice.

A D C

60. 2 R

P Q

By symmetry of the given figure, we can say that the area of the triangle DBC is equal to the area of the triangle AEC. Area of the pentagon CAEDB = 1 + Area of the triangle DEC =

∴ ∆AEC =

8

6 (D) 5

⇒T=

E

B

R

P

C G D Following six steps as shown in the figure above the matix ABCD can be filled in 4! × 2! × 2! × 2! × 2! × 1 = 384

53. 1

C

B

Alternate method: A C1

E

3 square units 4

Just the data that the total number of diamonds is between 500 and 600 is enough to find n. This would mean that there exists a number between 500 and 600 that can be uniquely split into 2 factors. This can happen if the number is a product of two prime numbers. But “n” can be uniquely identified. This must imply that both the prime numbers are equal. Hence look for a square of a prime number between 500 and 600. This happens to be 529. Hence the number of necklaces = n = number of diamonds in each necklace = 23.

1 3 1 +  square units 2  2 

1 ( Area (CAEDB ) − Area (DEC)) = 41 square units 2

006