Cl Mock Cat 3 Key 2008

Mock CAT – 3

Answers and Explanations 1

2

2

1

3

5

4

4

5

1

6

2

7

3

8

3

9

4

10

1

11

5

12

4

13

1

14

3

15

2

16

4

17

2

18

3

19

5

20

4

21

2

22

2

23

1

24

3

25

2

26

4

27

2

28

4

29

3

30

1

31

3

32

1

33

5

34

4

35

3

36

1

37

5

38

2

39

5

40

4

41

4

42

5

43

3

44

4

45

3

46

1

47

3

48

1

49

4

50

2

51

3

52

1

53

2

54

5

55

4

56

1

57

5

58

4

59

3

60

2

61

5

62

4

63

2

64

4

65

3

66

2

67

1

68

5

69

1

70

4

71

2

72

1

73

5

74

2

75

4

MY PERFORMANCE Total Time Taken Total Correct Incorrect Net Questions (Min) Attempts Attempts Attempts Score Language Comprehension Section I and English Usage Quantitative Ability

Section II

Logical Reasoning based Section III Data Interpretation TOTAL

25 25 25 75

150

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MCT-0007/08 003

1

1. 2

‘2’ is the correct option as it goes on to the next step after the para has ended with ‘the single most important thing’. ‘3’ elaborates ‘2’. ‘4’ is one step further whereas ‘5’ is a general statement. ‘1’ is a general statement about children which does not fit in immediately after the paragraph. It could have come earlier.

2. 1

‘1’ is the correct option as it continues with the problem left off in the last line of the paragraph. ‘Health conscious population’ in option 1 connects to ‘salad vegetables......’ in the last line. The paragraph is talking of a future hypothetical situation whereas options 2, 3 and 4 reflect on the current changed scenario. They suddenly make the paragraph move from general to specific. Option 5 is relevant. But option 1 is better than option 5 when it comes to idea continuation.

3. 5

‘5’ is the correct option since it gives the additional relevant details regarding the directive. ‘1’ is ruled out because it is not clear as to what ‘it’ is ‘contrary’ to. It is highly unlikely that ‘2’ will complete the paragraph as it takes a abrupt negative tone. ‘3’ is incorrect since it contradicts the last statement which refers to ‘call for the promotion of only sustainable biofuels’, whereas ‘3’ talks about ‘wind energy’. ‘4’ digresses from the theme of the paragraph.

4. 4

‘4’ is the correct option since it is obviously a question from the same set of sentiments, which deal with the dilemma that America faces. It deals with foreign policy. ‘1’, ‘3’ and ‘5’ are conveying different types of dilemma but away from foreign policy. ‘2’ can come later but it does not immediately continue the idea left off in the paragraph.

5. 1

‘1’ is the correct option since it continues with the comparison in the political arena, though rest of the options also talks about the typical British and American reactions and ‘3’ is a conclusive question which can come later.

6. 2

‘2’ is the correct option. The statement means that people are in doubt as to how the material and the immaterial can impact each other.

7. 3

‘3’ is the correct option because according to the last few sentences of the third paragraph, ‘but it is not justified if causal over determination of behaviour is possible. There could then be a complete physical cause of behaviour, and a mental one. The strongest intuitive objection against over determination..’

8. 3

‘3’ is the correct option according to the last paragraph. The question requires a general answer. Option 2 is not implied by the theory. Options 1 and 4 are too specific, though they are relevant. Option 5 is exaggerated. Option 3 correctly talks about the implication.

9. 4

4 can be inferred from the discussion in para 3, it can be inferred that some basic assumptions of interactionism conflict with the ‘closure of physics’.

10. 1

Statement a cannot be inferred from the passage. Statement b can be inferred from the discussion in the 3rd para. Statement c cannot be inferred as in para 6 it is clearly stated that ‘physical closure leaves no room for this’ and that ‘these matters are still controversial’. Hence, ‘1’ is the correct option.

11. 5

2

The last paragraph mentions that anti-communism was popular and that people had an insane fear of socialism. This resulted in their surrendering to the arms of despotism. This means that they viewed socialism and communism as enemies and very easily dismissed them. Now, when the red party’s weakness was proved, people would regret how easily they dismissed these as enemies.

12. 4

The author’s attitude cannot be called as immature. Contempt is an extreme tone. The author is also not angry. His attitude cannot be called as superficial. Yes, he is derogatory. But yes, the author mentions in the 5th paragraph that he is unsympathetic. He is also critical of the American communists as regards their assumed importance - as seen from words like ‘fateful immaturity’ and ‘playing at being revolutionaries’. Hence, option 4 is correct.

13. 1

Statement a cannot be conclusively inferred from the passage. In fact the author indicates the opposite at the start of para 5. Statement b can be concluded from the passage. Statement c cannot be inferred as the talk is of ‘fateful immaturity’ and not of Intelligence quotient or Brain functions.

14. 3

Germane means Relevant. Qualify in the context meanssomething less than reduce. Moderate is the closest option.

15. 2

Option 2 cannot be conclusively inferred as the passage does not indicate any cause effect relation. The author uses the ‘brutalities’ as evidence.

16. 4

The author definitely offers a view of the field of molecular biology. But the primary purpose is to show that the advances in molecular biology have not brought biologists any closer to answering the question posed in the title of Schrodinger’s book. So, the primary purpose is not just option 3. Option 3 is narrow. Option 4 correctly qualifies as the primary purpose as the passage is exploring the consequences of the book and its title question as regards the scientific world.

17. 2

Refer to the last para where it is stated “developmental biology has no such support... Not all the answers are available at the molecular level...”. Option 2 can be inferred. The other options cannot be inferred.

18. 3

The very simple language and readability of the passage suggests that the author is addressing a very general audience. Even scientific terminologies are explained in a succinct manner.

19. 5

The author has presented certain scientific developments in a non-judgemental manner.

20. 4

Sydney Brenner’s comments in the last para help explain the point mentioned in the second para that we are not closer to answering the question posed in the title of Schrodinger’s book

21. 2

The correct answer is 2 (IFJJ). Sentence 1 is definitely an inference on the government’s intent based on the fact that it has doled out a generous package of sops to farmers in this year’s budget. Sentence 2 is a fact which is verifiable. Sentence 3 is a judgement as it is an individual subjective opinion that the quote is controversial. Sentence 4 is again an individual opinion - Judgement. This makes option 2 the right option to choose.

22. 2

Sentence 2 is clearly a fact about what sections of the media have expressed about the Indo-US nuclear deal. Sentence 1 is an inference drawn about technology based on the event of the recent Arctic oil spillage. This reduces our choices to 2 and 5. Sentence 3 is a judgement on the Indo-US Nuclear deal. That it is a classic example is a judgement. Sentence 4 may appear to be a judgment but it is actually only stating the fact that there are lies, opinions and statistics making option 2 (IFJF) correct.

003

23. 1

24. 3

25. 2

Sentence 2 is clearly a verifiable fact. Sentence 3 is clearly an inference on what would happen to short term interest rates if the government funds its budget deficit through increased borrowing. Sentence 1 which talks about ‘inspired fool hardiness’ is expressing an opinion, a disapproval, and hence is a judgment. Our choices are therefore narrowed to options 1 and 3. Sentence 4 is not a judgment. It is an inference based on the fact that the prices of English manor houses on sale have trebled. This makes option 1 (JFII) the correct option to choose. Sentence 3 is clearly a fact since it is mentioning the definition of wrongful gain according to the Indian Penal Code. Sentence 4 is also a fact, stating what has happened since the project has started. Sentence 2, on close inspection will be seen to be a judgement, an individual opinion. Sentence 1 is an opinion about the reason for the liberalization of the Indian economy and hence qualifies as a judgment. The correct sequence is then given by JJFF (Option 3). Sentence 2 is clearly a fact as it is quoting statistics related to a form of disease in a village. This eliminates option 4. Sentence 1 is also a fact since it is verifiable. Sentence 3 is a judgement, an individual opinion. Sentence 4 is a fact as it can be verified whether Charumitra has said the quote. The correct sequence is therefore given by option 2 (FFJF).

For questions 26 and 27: Let us mark each of the squares from 1 to 17 as shown in the figure given below:

A

Therefore, the length of the side of square number 2 has to be 11 units. So, the length of the side of the square ABCD is 20 units. ⇒ Length of the side of the square number 8 = 9 units. ⇒ Length of the side of the square number 17 = 5 units. ⇒ Aggregate length of the sides of the square number 15 and square number 16 = 5 units.

The only possible value of the length of the side of the square number 15 = 2 units and the length of the side of square number 16 = 3 units ⇒ Length of the side of the square number 13 and square number 14 = 1 unit each. ⇒ Length of the side of the square number 6 and square number 7 = 2 units each ⇒ Length of the side of the square number 9 = 3 units ⇒ Length of the side of the square number 10 = 2 units and length of the side of the square number 11 and square number 12 = 1 unit each.

26. 4

The length of the side of the square ABCD = 20 units.

27. 2

Sum of the areas of the square number 17 and square number 8 = 92 + 52 = 106 square units.

28. 4

Let P = 7Y + 3 and Q = 5X + 1, where Y and X are natural numbers. Given P – Q > 177 and P + Q < 203 So when we take Q = 1 the value of P can lie in the range 178 < P < 202 i.e. P = 185, 192, 199 Q = 6 the value of P can lie in the range 183 < P < 197 i.e. P = 185, 192 Q = 11 the value of P can lie in the range 188 < P < 192 therefore no value of P exist in this range. Given Q – P < 177 and P + Q < 203 So when we take P = 3 the value of Q can lie in the range 180 < Q < 200 i.e. Q = 181, 186, 191, 196 P = 10 the value of Q can lie in the range 187 < Q < 193 i.e. P = 192 The value of P cannot be equal to 17. Therefore, the number of ordered pairs (P, Q) is 10.

29. 3

If a number is divisible by 3, then the sum of the digits of that number is divisible by 3. Similarly, a number is divisible by 9 if the sum of the digits of the that number is divisible by 9. For a number to be divisible by 3 but not by 9, the sum of the digits of the number should be 6 or 12 or 15 or 21 or 24.

B

1 2

4 6

3 5 15

13 14

7 8 9

17 16 D

11 10 12 C

Given that the length of the side of the square number 4 = 3 units ⇒ Length of the side of the square number 5 = 3 units ⇒ Length of the side of the square number 3 = 6 units ⇒ Length of the side of the square number 1 = 9 units Now, the length of the side of the square number 2 has to be greater than 9 units but less than 12 units, so the possible value of the length of the side of square number 2 can be either 10 units or 11 units. Assume that the length of the side of the square number 2 = 10 units. Since, the length of the side of the square number 6 and square number 7 is the same, therefore the length of the side of the square number 6 is 2.5 units. But, this is not possible because the length of the side of each of the squares has to be an integer.

003

We exclude the case of the sum of the digits being 9 and 18 as the number should not be divisible by 9. Also, maximum possible sum of a four-digit number that can be formed = 9 + 7 + 6 + 4 = 26 Sum Possible number Possibile cases 21

9 7 4 1

4!

21

9 6 4 2

4!

24

9 7 6 2

4!

Therefore, the number of such four-digit numbers = 4! × 3 = 72 Note: No such number can be formed where the sum of the digits is 6, 12 or 15.

3

30. 1

We try to select prime numbers such that the difference between any two consecutive points to which he travels is an integer and is minimum. So out of a total of 6 points (including the initial point), the co-ordinates of the 4 points should be (2, 2), (2, 3) (3, 2) and (3, 3). The remaining two points should be chosen in the vicinity of above 4 points to ensure that the total area enclosed is minimum. Possible co-ordinates of these two points are as follows. (a) (2,5),(3,5) (b) (3,5),(5,5) (c) (5,5),(5,3) (d) (5,3),(5,2) Total area enclosed in case (a), (b) and (d) will be equal to 3 square units, whereas in case (c), the total area enclosed will be equal to 4 square units. Below is the description of the path followed by Nangru as in case (a).

Q PS is themedian

Q

PQ QS = =1 PR SR

∴ ∆PQR is an isosceles ∆ with PQ = PR

∴ PS is perpendicular to QR ⇒ ∠PSR = 90°

(

)

∴ ∠QPR = 180o − 40o + 40o = 100o

Hence, statement (B) alone is sufficient to answer the question. Hence, option (3) is the correct choice. 32. 1

Let, y = Number of people using all the three mobile phones and x = Number of people using Samsung and Motorola. Using statement A:

S a m su ng

2x y

N o kia 5

(x–y) x 34

y + 10

3 M otoro la (50 ) In itial P o in t

The number of people using only ‘Nokia and Motorola’. ⇒ (34 – y) = 50 – (x) – (y + 10)

2

⇒x=6 3

2

31. 3

⇒ Number of people using only ‘Nokia and Samsung’ = 2x = 12 Hence, statement A alone is sufficient to answer the question.

5

Using statement B:

Using statement A: P

S a m su ng

N o kia

x

y

3 31

x–3 x

Q

y

X

S

34 R M otoro la (50 )

In ∆PSR : y + y + 2x = 180° ⇒ x + y = 90°

Given that the value of y = 3. There is no information available, therefore we cannot find the number of people who use only ‘Nokia and Samsung’. Hence, statement B alone is not sufficient to answer the question. Hence, option (1) is the correct choice.

o

∴ ∠QPR = x + y = 90 Hence, statement A alone is sufficient to answer the question.

Using statement B: P

33. 5

Given that ⇒

Q

PQ QS = PR SR

4

S

R

1 2 1 + = x y 24

y + 2x 1 = ⇒ 24y = xy – 48x xy 24

⇒ x=

24y y − 48

...(i)

For obtaining integral values of x and y that satisfy the given equation either of the following 2 cases must be true.

003

Case 1: y – 48 is a factor of 24. Case 2: y – 48 is a factor of y.

In this case dividing by 9, remainder is

Using Statement A: x=

24y y − 48

The possible values of y > 52 that satisfy the above equation are 54, 56, 60, 72, ....... Hence, we do not have a unique value of x and y that satisfies the given equation. Hence, statement A alone is not sufficient to answer the question.

=

4(63 + 1)14 9

⇒ Remainder is 4 × 1 = 4. Therefore, the digit sum of the given number is 4.

36. 1

If there are 2 circles, then

1

Using Statement B: x=

24y y − 48

3

Using Statement B: p2 – 9 = 3q + 2r ⇒ p2 = 3q + 2r + 9 ‘2r’ is an even number and ‘3q’ is an odd number if ‘q’ is odd and ‘3q’ is an even number if ‘q’ is an even number. Hence, statement B alone is not sufficient to answer the question Combining Statements A and B: From statement A we know that ‘p’ is and odd number. Statement B can be re-written as p2 – 2r – 9 = 3q. Odd – Even – Odd = Even. Therefore, ‘q’ is an even number. p2 + q2r = Odd + Even × r = Odd + Even = Odd. Hence, using both the statements together, the question can be answered. Option (4) is the correct choice. 35. 3

003

Note that if the digit sum of any number is divisible by 9, then the number is divisible by 9 and if the digit sum is not divisible by 9, then the digit sum is actually the remainder when that number is divided by 9. For example: A number 79 when divided by 9 remainder is 7 which is same as the digits sum of 79 = 7 + 9 = 16 = 1 + 6 = 7 So we can say digit sum of any number is always equal to the remainder when the same number is divided by 9. Digit sum of (ab3cdefghi1) = 45 + 3 + 1 = 49 = 4 + 9 = 13 = 1 + 3=4

3 co lo u rs are re qu ire d

3

1 4

2

2

4 co lo u rs are re qu ire d

3

1

If there are 4 circles then

Combining Statements A and B: We get that 54 < y < 72. The 2 integral values of y within the above range that satisfies the given equation are y = 56 and y = 60. Hence, again we do not get a unique solution for the given equation. Hence, option (5) is the correct choice. Using Statement A: p – 4r = 1 – 2q ⇒ p = 4r + 1 – 2q ‘4r’ and ‘2q’ are even numbers. Also, Even + Odd – Even = Old. Therefore, ‘p’ is an odd number. Since no information is available about ‘q’ and ‘r’, statement A alone is not sufficient to answer the question.

2

If th ere a re 3 circles, the n

The possible values of y < 72 that satisfy the above equation are 49, 50, 51, 52, 54, 56 and 60. Hence, we do not have a unique value of x and y that satisfies the given equation. Hence, statement B alone is not sufficient to answer the question.

34. 4

4 43 4 × 4 42 = 9 9

4

1 3 5

2

2

1 5

3

5 co lo u rs are re qu ire d

4 3

4



2

Therefore we can conclude that for 4 circles, a minimum of 5 colours are required. Hence, option (1) is the correct choice. For questions 37 and 38: Let us assume that the value of N = 3. If we take the address of B5 as (0, 0), there will be 9 cells which can be marked in the matrix. y B1

B2 (– 1, 1)

B4

B3 (0 , 1 )

B5 (– 1, 0)

B7

B6 (0 , 0 )

B8 (– 1, –1 )

(1 , 1 )

(0 , – 1)

(1 , 0 )

x

B9 (1 , – 1)

If we choose the cell B3, the cells that share a common point or common boundary with the cell B3 are B2, B6 and B5. The value of the required sum = |1 – 0| + |1 – 0| + |1 – 1| = 2 If we choose the cell B6, the cells that share a common point or common boundary with the cell B3, B2, B5, B8 and B9. The value of the required sum = |1 – 1| + |1 – 0| + |1 – 0| + |1 – 0| + |1 – 1| = 3. If we choose the cell B5, the cells that share a common point or common boundary with the cell. The value of the required sum = 6. For any value of N, the value of the required sum will always be either 2 or 3 or 6.

5

37. 5

Hence the value of |(i – i1)| + ……. |(i – im)| is either 2 or 3 or 6.

38. 2

If the total number of cells in the matrix is less than 100, then the value of N2 can be one out of 9, 25, 49 and 81.

41. 4

Since angles TPU and UST are equal, the quadrilateral PTUS is a cyclic quadrilateral. Therefore, ∠PTU + ∠USP = 180°. Q

Since, we need to maximise the number of cells for which (i + j) is a perfect square, we need to maximise the value of N.

T

P

U

The following is the matrix if the value of N = 9.

R

S (4 , 4 )

(– 4 , 4 )

Also, since ∠QPT = ∠RSU, therefore ∠PSR + ∠PQR = ∠USP + ∠RSU + ∠PTU – ∠QPT = 180° ⇒ PQRS is a cyclic quadrilateral.

Therefore, ∠QSR = ∠QRR = 70° Hence option (4) is the correct choice. (0 , 0 )

(4 , 0 )

42. 5

10

2

Let, S =

100

10

2

+

log2 3

100

10

2

+…

log3 3

100

log100 3

10  log2 + log3 + …log100 

⇒S=2 (– 4 , 4 )

Maximum possible sum of (i + j) = 8 and minimum possible sum of (i + j) = 0. The only possible value of (i + j) is either 1 or 4. The following is the complete list of addresses of all the cells for which the value of (i + j) is either 1 or 4. (0, 1), (1, 0), (2, –1), (–1, 2), (3, –2), (–2, 3), (–3, 4), (4, –3) (0, 4), (4, 0), (1, 3), (3, 1), (2, 2) Maximum possible number of cells for which the value of (i + j) is a perfect square = 13. 39. 5

 

10

⇒S=

2

log100! 100log3

⇒S=

2 log100!

8

log3

43. 3

We know that EC2 = EA × EB. ⇒ 16 = EA × 2 ⇒ EA = 8 units

Option (1): The total amount at the end of 1 year = 0.97 × 1.1 × 5 + 1.15 × 3 × 0.95 + 2 × 1.09 × 0.98 = Rs. 10.7489 cr.

Option (5): The total amount at the end of 1 year = 10 × 1.15 × 0.95 = Rs. 10.925 cr. Hence, option (5) is the correct choice.

6

After the prices have been reduced let, the price per book be Rs. x. Let, the total number of old books with Ravinder be ‘y’ Therefore, xy = 4914 = 2 × 33 × 7 × 13. Here we need to maximize the value of ‘x’ . Since, the maximum possible value of ‘x’ has to be less than 63, therefore, the maximum value ‘x’ can be 54. For x = 54, the value of y = 91. Hence, option (4) is the correct choice.

E

2

B

A

θ θ

Option (3): The total amount at the end of 1 year = 0.33 × 10 × ( 1.15 × 0.95 + 1.1 × 0.97 + 1.09 × 0.98) = Rs. 10.759 cr. Option (4): The total amount at the end of 1 year = 10 × 1.09 × 0.98 = Rs. 10.682 cr.

25

Hence, option (5) is the correct choice.

Option (2): The total amount at the end of 1 year = 1.15 × 0.95 × 5 + 1.09 × 0.98 × 5 = Rs. 10.8035 cr.

40. 4

 

100log3

(4 , – 4 )

θ

4

C

D

Since, ∆ BEC is an isosceles triangle, therefore the length of BC will be equal to 4. The length of BC cannot 2 because then the rule of triangle inequality will get violated. Also, ∠ ADC = ∠ CBE (since, ABCD is also a cyclic quadrilateral) and ∠CBE = ∠BEC . Therefore, ∠AEC = ∠ADC. Hence, AECD is a parallelogram. Therefore, AE = DC = 8 units. C

4

B

4

1

1

E

003

Case IV: n = 3, i.e. they weigh their weights in groups of three each. 4C = 4, different readings are noted and we must have: 3 R1 + R2 + R3 + R4 = (A + B + C) + (A + B + D) + ( A + C + D) + (B + C + D) = 882 Or 3 × (A + B + C + D) = 882

Perpendicular distance between AB and CD = 42 − 12 (Length of the perpendicular drawn from the point C on the side BE) Area of the trapezium ABCD =

1 (AB + DC) × 15 2

1 × 14 × 15 2 = 7 15 square units Hence option (3) is the correct choice. =

44. 4

882 =73.5 kg. 4×3 As each of them weighs less than 100 kg, this is an invalid case. Hence the friends can weigh their weight in groups of 2 or 3, and their average weight is 73.5 kg. ⇒ The average weight of the four friends =

Let the number of men and women be ‘m’ and ‘w’ in country A. Therefore,

6 7 49 m= w ⇒ m= w. 7 8 48

46. 1

B

49 97 w+w= w. 48 48 Total number of married people is twice the number of The total population of country A = m + w =

married women = 2 ×

P

7 w 8

R N

7  4w  84    =  . Therefore, X =  97   97   48  w  

O

60 . 71

A

Let, OQ = x, then OR = x and RQ = 2 x. Also, ON + NP = 1 (Since radius of the circle is 1 unit)

X 84 71 = × = 1.02(approx) Y 97 60 Hence, option (4) is the correct choice.

⇒

Out of four, a group of n, can be made in 4Cn number of ways.

 2  2 ⇒ RQ = 2   = 1 + 3 1 3 +  

Case I: n = 1, i.e. all of them weigh their weights, one by one. must have:

Hence, option (1) is the correct choice.

⇒

45. 3

Q

It is known that PQ = QR = PR. Now if P is the mid-point of arc AB, then the line OP must divide the sector OAB in two equal parts which implies that line OP bisects RQ at 90°.

Similarly, applying the same concept in case of country B, we get that the value of Y =

From the information given in the question, the figure given below can be drawn.

4C = 4 different readings are noted and we 1 R1 + R2 + R3 + R4 = (A + B + C + D) = 882

882 = 220.5 kg. ⇒ The average weight of the four friends = 4 As each of them weighs less than 100 kg, this is an invalid case. Case II: n = 4, i.e. all of them weigh their weights, simultaneously. 4C = 1, reading is noted and we must have: 4 R1 = (A + B + C + D) = 882

882 4 = 220.5 kg. As each of them weighs less than 100 kg, this is an invalid case. ⇒ The average weight of the four friends =

Case III: n = 2, i.e. they weigh their weights in groups of two each. 4C = 6, different readings are noted and we must have: 2 R1 + R2 + R3 + R4 + R5 + R6 = (A + B) + (B + C) + (C + D) + (A + D) + (A + C) + (B + D) = 882 Or 3 × (A + B + C + D) = 882

47. 3

x 3 2 + x =1 ⇒ x = 2 2 1+ 3

Let the three roots of this cubic equation be α, α and β. We can write: (x – α)(x – α)(x – β) = x3 – A.x2 + Bx – C = 0 or x3 – (2α + β)x2 + (α2 + 2αβ)x – β.α2 = x3 – Ax2 + Bx – C ⇒ A = 2α + β B = α2 + 2αβ C = α2.β Option (1): If at least one of α and β is an even number, then C will be an even number. If only β is even, then B will be an odd number. Hence, (1) is incorrect. Option (2): If α is an even number and β is an odd number, then B will be an even number but A will be an odd number. Hence (2) is incorrect. Option (3): If A is an even number then β must be an even number. Hence, C must be an even number. Hence (3) is correct. Option (4): If C is an odd number, then A is an odd number, as it will be the sum of an even number and an odd number. Hence, option (4) is incorrect. Hence, option (3) is the correct choice.

882 4×3 = 73.5 kg. As each of them weighs less than 100 kg, this is a valid case. ⇒ The average weight of the four friends =

003

7

For questions 48 and 49:

T(40) = F + 1600A + 40B Given that T(40) – T(10) = 4800 and T(30) – T(10) = 2600

A

3 0° 3 0°

This gives two linear equations in variables A and B. 1500A + 30B = 4800 and 800 A + 20B = 2600 Solving, we get A = 3 and B = 10; On Thursday the value of n = 20 units ⇒ T(20) – T(10) = [(20)2 × 3 + 10 × 20] – [(10)2 × 3 + 10 × 10] = Rs. 1000 Hence, option (2) is the correct choice.

E

6 0°

°

9 0°

30

6 0°

F 9 0° 6 0°

B

3 0° D

C

In the right angled triangle ABC, ∠BAC = 90°, ∠ACB = 30° and ∠ABC = 60°.

51. 3

In Section A: Number of students who failed in only one course = (56 + 45 + 60 – 2 × 49) = 63

52. 1

In Section C: Number of students who failed in at least one course = (28 + 61 + 56 – 2 × 43) + 43 = 102

53. 2

In Section A: Number of such students = 115 – (56 + 45 + 60 – 49) = 3 In Section B: Number of such students = 99 – (48 + 23 + 39 – 26) = 15 In Section C: Number of such students = 139 – (28 + 61 + 56 – 43) = 37 In Section D: Number of such students = 65 – (19 + 24 + 37 – 21) = 6 Therefore, total number of students that have passed in all the three courses = 3 + 15 + 37 + 6 = 61

54. 5

In order to find the answer to the question we need to maximise the number of students who failed only in Statistics. In Section A: Out of the 49 students who failed in two courses, let us assume that 45 students failed in Economics and Business Maths. This means that 49 – 45 = 4 students failed in Statistics and Business Maths. So, maximum possible number of students who failed only in Statistics = 56 – 4 = 52. Therefore, at least 115 – 52 = 63 students did not fail only in Statistics. In Section B: Following the same logic as given for Section A, at least 99 – (48 – (26 – 23)) = 99 – 45 = 54 students did not fail only in Statistics. In Section C: Out of the 43 students who failed in two courses, let us assume that all 43 failed in Economics and Business Maths. Therefore a maximum of 28 students failed only in Statistics. So, at least 139 – 28 = 111 students did not fail only in Statistics. In Section D: Out of the 21 students who failed in two courses, let us assume that all 21 failed in Economics and Business Maths. Therefore a maximum of 19 students failed only in Statistics. So, at least 65 – 19 = 46 students did not fail only in Statistics. So across all the four sections, at least 63 + 54 + 111 + 46 = 274 students did not fail only in Statistics.

55. 4

No. of students who passed in Statistics and BM is Equivalent to no of students who failed only in Economics & the no of students who passed in all the three subjects. In Section A: Out of the 49 students who failed in two courses, let us assume that 49 students failed in BM & Statistics. Maximum possible no. of students who failed only on Economics is 45.

⇒ ∠ABF = 60° – 30° = 30° and ∠BAD = 90° – 30° = 60° and ∠AFB = 90° ⇒ ∠AEF = 60°

⇒ ∆ABD is an equilateral triangle and ∠ADB = 60° Now, we write here ∠AEB = ∠ADB and ∠DBE = ∠DAE ⇒ Quadrilateral AEBD is a cyclic quadrilaterals.

⇒ ∠BAE + ∠EDB = 180° ⇒ ∠ADB + ∠ADE = 90° ⇒ ∠ADE = 30° 48.1

In ∆AED, ∠DAE = ∠EDA = 30° and ∠EFD = 90° ⇒ Chord BE is the perpendicular bisector of the chord AD. ⇒ Chord BE must be the diameter of the circum circle of quadrilateral ABDE and hence,

BE 10 = = 5 cm 2 2 Hence, option (1) is the correct choice. circum radius of ∆AED =

49. 4

In ∆BEC, ∠EBC = ∠ECB = 30° and ED is the perpendicular from E on BC. ⇒ Point D is the midpoint of side BC. CD = BD = BE × cos30° = 10 × and ED = BE × sin30° = 10 ×

⇒ area ( ∆EDC ) =

3 = 5 3 cm 2

1 = 5 cm 2

1 1 × CD × ED = × 5 3 × 5 2 2

25 3 cm2 2 Hence option (4) is the correct choice. =

50. 2

8

Let n be the number of units of the machine produced and let T(n) be the total cost of production. T(n) = Fixed cost + Variable cost Fixed cost is independent of the number of units but variable cost is a function of ‘n’. We can write T(n) = F + V(n) As T(n) and n follow a quadratic relation, we can write T(n) = F + (An2 + Bn); for some real numbers, A and B On Monday: n = 10 units T(10) = F + (100 A + 10B) On Tuesday: n = (10 + 20) = 30 units T(30) = F + 900 A + 30B On Wednesday: n = (10 + 30) = 40 units

003

In Section B: Similarly in section B maximum possible no of students who failed only in Economics is 23. In Section C: Out of the 43 students who failed in two courses, let us assume that 28 students failed in BM & Statistics. This means that 43 – 28 = 15 students failed in Economics and one of the other course. Maximum possible no. of students who failed only on Economics is 61–15 = 46 In Section D: Following the same logic as given for section C, at most (24 – (21 – 19)) =22 So across all the four sections the no of students who passed in statistics & BM = (45 + 23 + 46 + 22) + 61 = 197 For questions 56 to 59: Let the persons who wear a blue, red and green shirt be denoted by b, r and g respectively. Restrictions on the seating arrangement: 1. Two b’s must not be together. 2. Three r’s must be together. 3. A ‘b’ and a ‘g’ must not be together. 4. A ‘g’ cannot sit on chair numbered 2 or 9. Case 1: A person wearing a red shirt sits on chair numbered 1. Keeping all the restrictions stated above it is not possible to make a seating arrangement. Case 2: A person wearing a green shirt sits on chair numbered 1. It is only possible if another person wearing a green shirt sits on chair numbered 2, but this violates restriction number 4. Hence, this is also not possible. Case 3: A person wearing a blue shirt sits on chair numbered 1. The six seating arrangements that are possible are as follows.

1

2

3

4

5

6

7

8

9

10

Case 1

b

r

b

r

g

g

r

r

r

b

Case 2

b

r

r

r

b

r

g

g

r

b

Case 3

b

r

g

g

r

r

r

b

r

b

Case 4

b

r

r

r

g

g

r

b

r

b

Case 5

b

r

g

g

r

b

r

r

r

b

Case 6

b

r

b

r

r

r

g

g

r

b

Now, we see that case 4, 5 and 6 are just obtained by reversing case 1, 2 and 3 respectively. It can be concluded that in any possible seating arrangement, the persons who wear a blue shirt will sit on chair numbered 1 and 10. It is given that number of persons wearing a blue shirt is 3. Looking at the table given in case 3, we can conclude that in each of the six arrangements two out of the three different persons i.e. A, B and N always sit on chair numbered 1 and 10. Hence it can be concluded that the persons who wear a blue shirt are A, B and N. From the given table the person wearing a blue shirt can never sit on chair number 2, 4, 7 and 10. So, in arrangement I: A, B and N are sitting at chair numbered 7, 1 and 10 and hence it is inconsistent.

003

Also, the person wearing red shirt sits on chair numbered 2 and 9 of and in all the possible arrangements five different persons namely P, Q, M, Z and L are either sitting on chair numbered 2 or 9. Therefore, P, Q, M, Z and L are wearing a red shirt. Therefore, K and L are wearing green shirt. 56. 1 57. 5 58. 4 59. 3

Option 1: A (Blue), P (Red), R (Red) and L (Green): Permissible Option 2: N (Blue), Q (Red), K (Green) and Z (Red): Permissible. Option 3: K (Green), A (Blue), N (Blue) and Z (Red): Not Permissible. Option 4: B (Blue), L (Green), M (Red) and Q (Red): Permissible. Option 5: A (Blue), L (Green), P (Red) and M (Red): Permissible. Hence option (3) is the correct choice.

For questions 60 to 63: It is given that if a particular friend has given 2 different information in a statement, then either both the information given in that statement will be true or both would be false. This will be an important criteria for solving the question. If we look at the 2nd statement of B, then this statement cannot be true because if A had the costliest item, then the total money spent by A would be = (6 + 5.5) × 2 = 23 or = (6 + 5) × 2 = 22 or = (6 + 4) × 2 = 20 whereas the first part of the same statement says that A spent Rs.18. Therefore both parts of the 2 nd statement of B is false and the 1st statement of B is true. From the first statement B spent the maximum possible amount, which means that B had 2 of each item costing 6 and 5.5 i.e. B spent a total of 23. This means that the 2nd statement of D is true and the first statement is false which indicates that D did not have Samosa and the cost of a Samosa is not equal to 5.5. Also, because the 2 nd statement of B is false, therefore the 1st statement of A is false and therefore the 2nd statement of A is true which means that D has actually spent Rs. 20 without having Jalebi. From statement 1st of D and statement 2nd of A we can conclude that D spent Rs. 20 in having 2 Kachoris and 2 Imartis (As he cannot have Samosa and Jalebi). Therefore, the cost of a Kachori and a Imarti = Rs.10. A combination of 10 can only be possible using 6 + 4. This further concludes that the first statement of C that D did not have either Samosa or Jalebi is true. Also, the 2nd statement is false which means that the cost of a Kachori is not Rs. 6. Therefore the cost of a Kachori is Rs. 4 while the cost of a Imarti is Rs. 6. B has spent Rs. 23 which is only possible using (6 + 5.5) × 2 = 23. Therefore, the first item that B had would have been a Imarti while the 2nd item cost Rs. 5.5. This item is not Samosa because the cost of a Samosa is not Rs. 5.5. This means that the cost of a Samosa is Rs. 5 while the cost of a Jalebi is Rs. 5.5.

9

The situation so far as to who had what items is as follows:

SL. NO NAME 1 A 2

B

3

C

4

D

ITEM 1

ITEM 2

TOTAL VALUE

IMARTI 6

JALEBI 5.5

23

KACHORI IMARTI 4 6

20

A did not have the most costly item which happens to be Imarti. Therefore, A must have had any 2 from Samosa , Kachori and Jalebi . This is the available information so far. Based on this information, let us try to answer the questions. 60. 2

If A spent a total of Rs. 21 which is possible if he had 2 pieces each of the items costing Rs. 5 and Rs. 5.5 which happen to be Samosa and Jalebi.

61. 5

From the table above, it is confirmed that 2 people B and D definitely had Imarti. Also A did not have Imarti but we cannot say anything about C and therefore, either 2 friends or 3 friends would have had Imarti. Hence, option (5), i.e. Cannot be determined is the correct choice.

62. 4

63. 2

The total amount spent by C is more than D (20) but less than B(23). Which means that the total amount spent by C can be either 21 i.e. Samosa and Jalebi or 22 i.e. Imarti and Samosa. Hence, option (4) is the correct choice. The amount spent in Rs. By 2 of the friends is already known, which is given in the table and adds upto Rs. 43. If we want to maximise the total amount spent, then A should have 2 pieces each of Samosa and Jalebi which adds upto Rs. 21. Further C can have Imarti and Samosa which will add upto 22. (Please note that C cannot have the same combination as B and therefore Imarti and Jalebi is ruled out. Maximum money that can be spent = Rs. 20 + 23 + 21 + 22 = Rs. 86.

For questions 64 to 68: As G was given a “Pass” in the group task IX, he must have been selected for the rank, finally. Further, as E was given a “Pass” in the previous group task, i.e. in the group task VIII, he must have participated along with G in the group task IX. Applying this logic for other group tasks, we can list all the subordinates who had definitely participated in each of the nine tasks. At the end of each group task, one subordinate is eliminated. Task

Number of Participants

I

10

II

9

III

8

IV

7

V

6

Pass

Participants

Eliminated

G, E, F, D, J, B, A, C, D, I, J H C, I, A, H G, E, F, D, J, B, C,I A C, I ,A G, E, F, D, J, B, B (I or C) C, I G, E, F, D, J, B, D, F, G (B or C or I) (I or C) G, E, F, D, J, (B J (B or C or I) or C or I)

VI

5

F, D

G, E, F, D, J

J

VII

4

G

G, E, F, D

(F or D)

VIII

3

E

G, E, (D or F)

(D or F)

IX

2

G

G, E

E

10

64. 4

Total number of participants in round II was 9. In this round, A got eliminated and C and I were given a “Pass”. Hence votes were cast against only 7 people. The maximum number of elimination votes against A could be 8. For the case of minimum votes against A, 6 out of these 8 votes must have been against the other 6 subordinates. Out of the remaining 2 votes, if any is given to a subordinate other than A, it will either result in a tie or will result in the elimination of some other subordinate. Hence the remaining two votes also, must go against A only. Hence the minimum number of votes against A could be 3.

65. 3

As can be seen in the above table, all participants in groups tasks I, II, III, VI, VII and IX can be determined. Hence 6 is the required answer.

66. 2

Subordinate C had a “Pass” in group task I and II. He got eliminated in group task III or IV or V. To receive the maximum possible number of votes against him, C must be eliminated in group task V only. Group Task

Number of Participants

Number of “Pass”

Maximum Possible Votes against C 3

III

8

1

IV

7

3

3

V

6

1

5

So the maximum number of votes is (3 + 3 + 5) = 11. Hence (2) is correct. 67. 1

In the group task III, either C or I could have been eliminated. If C got eliminated in group task V and not in group task III, then I must have been eliminated in group task III. Hence (1) is correct.

68. 5

F got eliminated in group task VII. In group task IV, any one of B or C or I could have been eliminated. Hence the correct answer is (5).

For questions 69 to 71: From the two bar graphs, following data can be derived.

Period

Number of Customers

Percentage of Customers Who went Abroad

Number of Customers Who Went Abroad

January –April

120

40%

48

February-May

200

25%

50

March-June

170

40%

68

April-July

180

20%

36

May-August

250

10%

25

The following representation will help in breaking the data month wise. Number of Customers Who Went Abroad: 68 48

January

February

March April May June July August 50 36 25

003

Total number of Customers: January

February

For questions 72 to 75:

March April May June July August 250

120

69. 1

1 1 1 1  Kenya =  + + +  × 400 ; 176 sec onds  9 14 6 11

73 = 19.73% 370

1 1 1  1 Russia =  + + +  × 400 ; 178 sec onds  10 15 9 6 

As March to June = 68 ⇒ (January + February + July + August) = Total Customers – (March to June) = 73 – 68 = 5 April to July = 36 May to August = 25 ⇒ Assuming that no customer went abroad in the month of August, i.e. taking August = 0, we can get the minimum number of customer that could have gone abroad in April = 36 – 25 = 11 For the case of a maximum number of customers going abroad in the month of March, we must have: January + February = 0 ⇒ March + April = 48 – (January + February) = 48 (March)max = 48 – (April)min = 48 – 11 = 37 As February to May = 50, now we get May = 2. As February to June = 68, June = 68 – (37 + 11 + 2) = 18. For each month, we can now determine the number of customers who went abroad:

1 1 1  1 Australia =  + + +  × 400 ; 168 sec onds  10 13 10 7 

73. 5

0

37

11

2

18

5

covered a distance of =

50 × 11 metres 9

550 metres 9 In other words, the Second Runner from Kenya has a start of =

550 metres. 9 To overtake the Second Runner from Kenya, the Second  550   9   Runner from USA will take =  (12 − 11)

0

Hence (4). 71. 2

First Runner from USA will complete one lap in 50 seconds. First Runner from Kenya will complete one lap in

400 = 44.44 sec onds. 9 By the time the Second Runner from USA starts running at the rate of 12 m/s, the Second Runner from Kenya has already

January February March April May June July August 0

The following lists down the time taken by each of the 4 mentioned countries to complete the 4 × 400 metres relay. 1 1 1 1  US A =  + + +  × 400 ; 164 sec onds  8 12 9 11

Total number of customer till August = 120 + 250 = 370. Total Number of customers who went abroad, till August = 48 + 25 = 73 The required percentage =

70. 4

72. 1

550 sec onds 9 This implies that the Second Runner from USA will have to run =

May to August = 25. For a maximum number in June, July + August = 0 ⇒ May + June = 25 and, January + Feb = 5 and, March + April = 48 – 5 = 43 As, February + March + April + May = 50 Or February + 43 + May = 50 ⇒ Feb + May = 7 (May)min = 7 – (February)max =7–5=2 ⇒ (June)max = 25 – (May)min = 23 The number of customers in different months is given below:

550 2200 × 12 = = 733.33 metres 9 3 This is impossible because each runner can cover a maximum of 400 metres. This implies that the Second Runners from USA and Kenya will never meet. Hence, option (5) is the correct choice. =

74. 2

Third Runner from Russia and Kenya do not meet as the speed of each of the corresponding runners from Russia is greater than the speed of the corresponding runner from Kenya.

75. 4

Russia has 3 Intermediate Victories for the first, second and the third runner and USA has one Intermediate Victory for the fourth runner.

January February March April May June July August 0

5

Hence (2) is correct.

003

43

2

23

0

0

11