NC5 Answer

8/14/2019 NC5 Answer

1/14

A888 A 1

NAT CAT Test 1: Answers

1. 1 19. 5 37. 4 55. 4 73. 5

2. 2 20. 3 38. 2 56. 5 74. 1

3. 4 21. 4 39. 4 57. 2 75. 2

4. 1 22. 5 40. 4 58. 4 76. 3

5. 2 23. 2 41. 3 59. 2 77. 2

6. 1 24. 3 42. 2 60. 5 78. 2

7. 3 25. 1 43. 4 61. 2 79. 3

8. 2 26. 1 44. 1 62. 5 80. 4

9. 3 27. 2 45. 3 63. 4 81. 4

10. 3 28. 4 46. 1 64. 2 82. 2

11. 3 29. 2 47. 5 65. 1 83. 4

12. 3 30. 2 48. 4 66. 3 84. 5

13. 2 31. 1 49. 2 67. 2 85. 2

14. 3 32. 2 50. 2 68. 2 86. 4

15. 2 33. 2 51. 1 69. 5 87. 4

16. 3 34. 3 52. 3 70. 2 88. 1

17. 2 35. 4 53. 2 71. 1 89. 4

18. 1 36. 5 54. 4 72. 2 90. 4


2/14

A888 A 2

NAT CAT ANSWERS

(ENGLISH)1. (1)

2. (2)

3. (4)

4. (2)

5. (2)

6. (1)

7. (3)

Answer for Questions (8 10)

In the first blank Alienations is the correct answer as the sentence talks about the Kashmiris association withIndia. Trampled on their rights is the correct phrase.In third blank, one can slide back to the bottom hence Precipice is the correct option.

8. (2)

9. (3)

10. (3)

11. (3) In line 20, it is clearly given that God is all powerful and well meaning, hence the correct answer isthird choice.

12.

(3) First and second option can be eliminated. The third option is correct as from the context, it can beinferred that Camerlengo had logical way to explain the power of God.

13. (2) Second option is the most appropriate and can be inferred from last few lines.

14. (3) The statement given in third choice contradicts the information given in the passage.

15. (2) He is very logical and practical as well as friendly in nature, hence option 2 is correct.

16. (3) 1 st can be ruled out as Ben Johnson never stood for Shakespeare. 2 nd is also eliminated because it isframed wrongly. 4 th does not have any mention in the passage. Hence 3 rd is the most appropriate answer.

17. (2) In last paragraph, Mr. Bate describes that constant theme is Shakespeares resistance to definition. Hence2 is the most appropriate answer.

18. (1) 2 nd and 3 rd can be eliminated. 1st can be inferred from the para 3. Hence 1 is the answer.

19. (5) Statement b andd are given in the passage hence 5 is the answer.

20. (3) From first line where author says that Fund managers .to be doubtful it can be inferred thatattitude of fund managers is biased. Hence prejudiced is the most appropriate answer.


3/14

A888 A 3

21. (4) 4 th option contradicts the information given in the last 3 lines of the passage, but our policy makersrespond well to the crisis. Hence 4 is correct answer.

22. (5) All the three statements are visible; hence 5 is the correct answer.23. (2) The whole passage is focused on deficit and risk associated with it . Hence 2 is the correct answer.

24. (3) Choice 3 can be logically inferred from the passage.

25. (1) CB is linked as one introduces about hate speech and other logically continues it. A is the concludingstatement and C is the opening sentence. Hence choice 1 is correct answer.

26. (1) B is an obvious starter therefore choice 2, 3 and 5 are eliminated. DE is also linked and C must follow B.Hence 1 is correct answer.

27. (2) A logically continues C which is the most appropriate introduction. Hence 2 is correct answer.

28. (4) BAE are linked and F is the most fitting opening sentence hence 4 is correct answer.

29. (2) 2 is the correct answer as A must precede D and E.

30. (2) A can not be opening sentence hence 3 and 4 are eliminated. B is the most appropriate starting sentence.Hence 2 is correct answer.


4/14

A888 A 4

MATHS

31. (1) x = ( )

+++++

bad c

d cba

1111)(

x =bd

bc

ad

ac

d b

d a

cb

ca +++++++

b

d

d

b

a

d

d

a

b

c

c

b

a

c

c

a+++++++

Min value of each group is 2. So min value of n is 8.So option a is correct.

32. (3) Number should be greater than 2986 and less than 5017All digits should be different and last digit should be odd.a. Number below 3000 = 1 (2987)b. Numbers between 3000 and 4000 = 1 x 8 x 7 x 4 = 224Because at first place 3 is fix and for last place only 4 choices 1,5,7 & 9 are possible.

c. Numbers between 4000 and 5000 = 1 x 8 x 7 x 5 = 280Because at first place 4 is fix and for the last place 5 choices 1,3,5,7 and 9 are possible.d. Number greater than 5000 = 1 (5013)So total numbers are = 1 + 224 + 280 + 1 = 506

33. (2)24

1

!8

!448

= xC

34. (3) Routes real sob2 4ac > 064 + 4 log 4 N > 04 log 4 N > -64

log 4 N > -16n > 4 -16

35. (4)1

1:

6

1:

5

1:

4

1:

3

1:

2

1= 60 : 30 : 20 : 15 : 12 : 10

So minimum number of coins is 60 + 30 + 20 + 15 + 12 + 10 = 147Second lowest number of coins is 147 x 2And total value = 147 x 2 x 5

= Rs. 1470

36. (5) Some of digits of this number is = 1 x 50 + 2 x 50 + 3 x 50 + 4 x 50 500 5While the digital some of any perfect square can be 1, 4, 3, 7 or 0 only.So these can not be any value which is perfect square.So option (5) is correct.

37. (4) Min hand 10Hour hand 16 -17 (more towards 17)Angle between 10 -16 = 6 x 20 = 120 0 It is more than 16:45 so angle is more than 130. But less than 140.So 131 0 is correct.

38. (2) 3 2,33 2,333 2. each number is divisible by 9 so seed of each number is 9


5/14

A888 A 5

So the value of seed (seed (3 2) + seed (33 2) + seed (333 2) + ..+ seed (3. 200 times 3

2))= seed (1800)= 9

39. (4) let the initial no of workers are X so now total work is 9X but now work done in following manner asX,X-2,X-1,X-3,X-2,X-4,X-3,X-5,X-49X=18X-2-1-3-2-4-3-5-4-6-5-7-6-8-7-9-8-109X=18X-90X = 10

40. (4) General term in the expansion of (a+b+c) 30 will be K a xbycz where K is a constant. The value of x+y+zmust be 30.Now we need number of solution of x+y+z=30 where x, y, z are natural numbers.Which is = 28

=2

2928 = 406

41. (3)

Area of the field = x 40 2 /2 = 800 Area of the grazed part of the field = 4 x x 10 2 common area = 400 - common areaCommon area = 4 ( 25 -50) = 100 - 200Ungrazed area = 800 400 100 + 200 = 300 + 200 m 2

42. (2)

Area of ABCD = 16cm 2

Area of C2, C3, C4 & C5 in side the square ABCD =2

4224

14 cm =

AD = 24 = AE + EG + GD

24 = 2 + EG + 2

EG = 24 - 4

EH = 2242

= EG


6/14

A888 A 6

Area of square EFGH = ( )2224 = 24 16 2

Required Area = 16 4 24 + 16 2

= 16 2 - 8 4

2cm 2

Answer for Question (43 & 44)

9.09% = 1/11Now we can form a table for the clarity of the question

Volumein

Morning(Vm)

VolumeSale(Vs)

Remaining inEvening

(Vr)

Evaporate(Ve)

Selling price(in Rs.)

(S)

RevenueGenerated (in Rs)

Day 1 11000 1000 10000 1000 10 10000

Day 2 9000 9000/11 90000/11 9000/11 11 9000Day 3 8100/11 8100/121 81000/121 8100/121 12.1 8100GeneralConcept

Vm

11Vm

Vs = VsVmVr =

Ve =10Vr

S venueSVs Re=

43. (4) It is very clear from the table that revenue generated is in Geometric Progression with first term 10000/-

and Common ratio 0.9. So total selling price =9.01

10000

= 100000/-

Cost Price = 88000110008 =

So percentage profit = 10088000

12000 = 13.63%

44. (1) It is clear from the table that volume sale and volume evaporated is same on each day. So answer is50%.

45. (3)

46. (1) After factorization we can writex4 + (a + 1)x 3 + (a + b - 2)x 2 + (b 2a) x 2b =0= (x 2 + bx + ax + b) (x -1) (x + 2) = 0 ..(1)And x 4 + (b - 1) x 3 + (a - b - 2)x 2 + (a + 2b) x 2a =0= (x 2 + bx + a) (x +1) (x - 2) ..(2)In equation (1) and (2) root is common inx2 + ax + b = 0and x 2 + bx + a = 0Now applying the condition of common root

abab

x

ba

x

=

=

122

2

= a + b + 1 = 0

47. (5) For the minimum time first pipe should take 20 min. and second pipe should take 30 min.And let the volume of the tank is 60 litre.Rate of the first pipe = 3 litre/min, Second pipe = 2 litre/min. both at a time = 5 litre/ min.


7/14


8/14

A888 A 8

Answer for Question (53 & 54)

Sum of the apples is 72. So one person gets exactly 24 apples. Apples can be distributed in followingmanner

24 - X 24 24 + XCASE 1 Hari Shyam RamCASE -2 Hari Ram ShyamCASE -3 Ram Hari Shyam

53. (2) 24 + X can not be more then 34 so X has 10 possible values in each case and there are three cases sototal number of ways are 30 .

54. (4) It is sure that one person will get exactly 24 apples. One person can get minimum one apple which cannever be Shyam so Shyam receives at most 72-24-1=47 apples. So first statement is correct.Hari can receive 24 apples also so 2 statements is falls but third statement is correct.

Answer for Question (55 56)

Given that P = 658, C = 372, B = 590, P C =166, P B = 434, B C = 126 Using

P B C = P + B + C - P B - P C - B C + P B C P B C = 106

55. (4) The number of students who failed in Chemistry but not physics = 186 +20= 206

56. (5) The number of students who failed in Physics or biology but not in chemistry =164+328 +136= 628

57. (2) With the help of options it is sure that 1 can not be the part of answer and 4/3 must be the part of answer. After answer elimination only second option is correct.

58. (4) px2

+ qx +r=0 let the roots of this equation are a and b. now 2

7

2 =+ ba

or a+b =7

7=

p

qAnd ab= 10=

p

r now we have 3 unknown and 2 equation so we can not find the value of pqr.

59. (2) Through statement one answer is always zero but at n=4 it is not correct so statement one is notsufficient. while for second statement it is always one. So statement alone is sufficient. Option (2) iscorrect

60. (5) With the help of both the statement we can say that top eight runners are either 1 to 8 or 2 to 9 due totwo cases nothing is sure question can not be answer. Option (5) is correct


9/14

A888 A 9

DATA INTERPRETATION

Answer for Question (61- 65)

From the conditions given in the questions following table can be drawn:

Table 1:Shastri Jones Gavaskar SidhuJaisurya(a) Sehwag(a) Flintoff(a) Sachin(a)Mccullum(w) Johnson(bow) Jaheer(bow) Harbhajan(bow)Yuvraj(bat) Jaheer(bow) Dhoni(w) Mccullum(w)

Dhoni(w) Smith(bat) Yuvraj(bat)Smith(bat)

61. (2) Jaheer and Johnson

62. (5)

1st caseShastri Jones Gavaskar SidhuJaisurya(a) Sehwag(a) Flintoff(a) Sachin(a)Mccullum(w) Johnson(bow) Jaheer(bow) Harbhajan(bow)Yuvraj(bat) Jaheer(bow) Dhoni(w) Mccullum(w)

Dhoni(w) Smith(bat) Yuvraj(bat)Smith(bat) Gambhir(bat)Gambhir(bat) PointingPieterson

2nd caseShastri Jones Gavaskar SidhuJaisurya(a) Sehwag(a) Flintoff(a) Sachin(a)Mccullum(w) Johnson(bow) Jaheer(bow) Harbhajan(bow)Yuvraj(bat) Jaheer(bow) Dhoni(w) Mccullum(w)

Dhoni(w) Smith(bat) Yuvraj(bat)Smith(bat) Pieterson(bat)Pointing(bat) Gambhir(bat)Gambhir(bat)

63. (4) 8 ways, after arranging the different cases the answer comes out to be 8.

64. (2) Only condition 1& 3 satisfy as mentioned in the table 1.

65. (1) From table 1 it was clear that smith is surely in Sehwags team.

Answer for Question (66-70)

66. (3)Let others have x% share in total populationThen (x + (5*x)/100)=63Solving x=60%Therefore population of karnataka is 3% and of others is 60%


10/14

A888 A 10

Let number of others be y per 1000Then applying the concept of weighted averageY*0.6 + 0.08*960 + 0.1*945 + 0.12*910 + 0.07*950 + 0.03*975 = 1*940Solving we get y = 940

67. (2) 4 pairs are possiblePairs are1) Gujrat Rajasthan2) Gujrat Tamilnadu3) Gujrat Karnataka4) Karnataka Tamilnadu

68. (2)The difference in number of females in Gujrat and Tamilnadu

5071950950

81960960

= lakhs

69. (5) cannot be determined.Since the population of Karnataka is not known therefore we cannot determine.

70. (2)

Percentage change = ( ) %1091090

100910

9101000=

=


Pink 3 Cream -3 White- 2Using the condition 1: P>V>S, QV>S>U>W>R>T>QUsing the condition 5: V not whiteUsing the condition 6: P and V not pink from condition 5 V is cream

Using the above condition the colours can be foundQ- cream : T- white : R- pink : W- cream : U- pink: S- pink : V cream : P- white

71. (1)

72. (2)

73. (5)

74. (1)

75. (2)


11/14

A888 A 11


Prelims Mains GD Phy. Fit. InterviewFresh CandidateMonday 30 30 28 25 19 16

+2| +3 | +6 | +3 | 0|Tuesday 31 33 37 29 31 25

+9 | +4 | +1 | +6 | 0|Wednesday 16 25 23 18 18 18

+6 | +6 | +6 | 0 | 0|Thursday 24 30 32 25 25 17

+4 | +13 | 0 | +8 | 0|Friday 23 27 23 27 27 27

The horizontal shows the fresh candidates and the vertical arrows shows the Rollover participants.Using the data from the above table the question can be answered

76. (3) 124The fresh participants = 30 + 31 + 16 + 24 + 23

77. (2) 12The following table provides the data for Wednesday

From round 1 to round 2 Max. possible fresh candidatePrelims -Mains 16(25-6)Mains -GD 19(23-4)GD -Phy. Fit. 17(18-1)Phy. Fit. - Interview 12(18-6)

Hence the maximum fresh candidate reaching to final interview can be twelve.

78. (2) 29Since the candidate on Tuesday only participated on Thursday. Hence in such a new situation the followingtables provide the Rollover candidates on Tuesday and Wednesday.The Rollover candidates for Tuesday and Wednesday are given below:

Prelims Mains G.D. Phy. Fit. InterviewTuesday 9 4 1 6 0Wednesday 2 5 0 0 0Total 11 9 1 6 0

The following table provides data for Thursday.

| +11 |+9 |+1 |+6 |+0Fresh 30 32 25 25 17

| +7 |+8 |+6 |+8 |+0

Hence the total rollover participants on Thursday = 7 + 8 + 6 + 8 + 0 = 29

79. (3) 25From the initial table the maximum Rollover participants on Thursday are 25.

80. (4) 0


81. (4) 798


12/14


13/14

A888 A 13

solving the above equation we get x = 172 and z = 28or 25 < a < = 30 x + 2y + 3z = 275x +3z + 2 (200 x - z) = 275x z = 125For z to be maximum y and x are to be minimumy = 1 x + z = 199x - z = 125solving the above equation we get x = 162 and z = 38For 30< a < = 40 x + 2y + 3z = 218x +3z +2 (200-x - z) = 218x z = 182For z to be maximum y and x are to be minimumy=0 x + z = 200x- z = 182solving the above equation we get x = 191 and z = 9For 40 < a < = 50 x + 2y + 3z = 213x + 3z + 2(200- x - z) = 213x z = 187For z to be maximum y and x are to be minimumy=1 x + z = 199x - z = 187solving the above equation we get x = 193 and z = 7For 50 < a < = 60 x + 2y + 3z = 240x + 3z + 2 (200- x - z) = 240x z = 160For z to be maximum y and x are to be minimumy = 0 x + z = 200x z = 160solving the above equation we get x = 180 and z = 20

85. (2) 25< a < = 30For atleast two to be minimum y and z is to be minimum

it is the similar case as in as in above question so calculating the sum of yand z in all the casesFor 20 < a < = 25 y + z = 29For 25 < a < = 30 y + z = 39For 30 < a < = 40 y + z = 9For 40 < a < = 50 y + z = 8For 50 < a < = 60 y + z = 20

Answer for Question (86 to 90):Since A played all the four matches which means he participated in all the singles.D played two matches which means he played both the doubles.B & C played three match which means they played one double and two singles each.The following tables provides all the possible cases of quarter final and semi final.

CASES FOR QUARTER FINAL.

Case 1 : India Australia Case-2 India AustraliaPhases Match Played Won Phases Match Played WonPhase -1 A H India Phase -1 A H India

B (E/G) India C (E/G) IndiaPhase-2 D & C F&(E/G) Australia Phase-2 D & B F&(E/G) AustraliaPhase-3 A (E/G) India Phase-3 A (E/G) India

B-H Australia C-H Australia


14/14

A888 A 14

CASES FOR SEMIFINAL.

Case 1 : India France Case-2 India FrancePhases Match Played Won Phases Match Played Won

Phase -1 A Q India Phase -1 A Q IndiaB P India C P IndiaPhase-2 D & C R & S India Phase-2 D & B R & S IndiaPhase-3 A P France Phase-3 A P France

B - Q India C-Q India

From the above given tables of the different cases all the questions can be answers.

86. (4) Either (1) or (2)

87. (4) Either (1) or (2)

88. (1) P, Q

89. (4) (1) & (3)

90. (4) France & India

Documents

NC5 Answer