NIMCET Previous Year Questions 2022- Infomaths

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N NIIMMCCEET T A ACCTTUUAAL L P PAAPPEER R 2 200222 2 MATHS

1. If the angle of elevation of the top of a hill from each of the vertices A B and C of a horizontal triangle is α, then the height of the hill is

(a) 1 tansec 2 bB  (b) 1 tancos 2 becA 

(c) 1 tansin 2 cC  (d) 1 tancos 2 aecA 

2. If xmyn = (x + y)m+n, then dy/dx is

(a) xy xy  (b) xy (c) x/y (d) y/x

3. The value of 3 3log53

(a) 5 27 (b) 27 5 (c) 9 5 (d) 5 9

4. The value of   2 342

xdx xxx 

1 221

(a) 24 21 22 C xx  (b) 24 21 22 C xx 

(c) 24 121 2 2 C xx  (d) None of these

5. If the volume of a parallelopiped whose adjacent edges are ^^^ 234, aijk 

11. In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is

(a) 2 : 1 (b) 1 : 2 (c) 3 : 1 (d) 1 : 1

4 xn    (b) 3 xn   

(c)

(a) 3x + 4y = 20 (b) 3x – 4y + 7 = 0 (c) 5x – 4y = 40 (d) 5x + 4y = 40

cos x fx

(d)

    1

 is

None of these

15. If the roots of the quadratic equation x2 + px + q = 0 are tan 30 and tan 15 respectively, then the value of 2 + p – q is

(a) 3 (b) 0 (c) 1 (d) 2

2,bijk 

^^^ 2 cijk  is 15, then  is equal to (a) 1 (b) 5/2 (c) 9/2 (d) 0

6. The area enclosed within the curve |x| + |y| = 2 is

(a) 16 sq. unit (b) 24 sq. unit (c) 32 sq. unit (d) 8 sq. unit

xaxbdx   is equal to

8. If 2,aijk   2,bijk   ,cijk  and 7, abc    then the value of  are

(a) 2, - 6 (b) 6, - 2 (c) 4, - 2 (d) – 4, 2

9. There are two sets A and B with |A| = m and |B| = n. If |P(A)| - |P(B)| = 112 then choose the wrong option (where |A| denotes the cardinality of A, and P(A) denotes the power set of A)

(a) m + n = 11 (b) 2n – m = 1

(c) 2m – n = 1

(d) 3n – m = 5

10. A particle is at rest at the origin. It moves along the x-axis with an acceleration x – x2, where x is the distance of the particle at time t. The particle next comes to rest after it has covered a distance

(a) 1 (b) 1/2 (c) 3/2 (d) 2

(a)

lim0 x x x e   (b) 1/ 0 1 lim0 x x xe    (c) 0 sin lim0 12 x x x     (d) 0 cos lim0 12 x x x    

(a)

18. The first three moments of a distribution about 2 are 1, 16, -40 respectively. Then mean and variance of the distribution are

(a) (2, 16) (b) (2, 15) (c) (3, 15) (d) (1, 16)

19. If a1, a2 ….. an, are in Arithmetic Progression with common difference, d, then the sum (sin d) (cosec a1 . cosec a2 + cosec a2 . cosec a3 + …… + cosec an-1 . cosec an) is equal to

(a) cot a1 – cot an (b) sin a1 – sin an

(c) cosec a1 – cosec an (d) a1 – an

20. Solution of the equation

1212

tansin1 2 xxxx   are

(a) 0, 1 (b) 1, - 1 (c) 0, - 1 (d) 0, - 2

21. If cosec  - cot  = 2, then the value of cosec  is

(a) 5 3 (b) 3 5 (c) 4 5 (d) 5 4

1 P PRREEVVIIOOUUS S Y YEEAAR R S SOOLLVVEED D P PAAPPEERRS S
^^^

(a)
2
(b)
7. If a < b, then   b a (d)
2 ba
  22 2 ba (c)   33 2 ab
(b – a)2
2 xn    (d) 2 3 xn   
12. The solutions of the equation 4 cos2x + 6 sin2x = 5 are (a)
13. A straight line through the point A(4, 5) is such that its intercept between the axes is bisected at A, then its equation is
x
14. The domain of the function
(a) [-1, 0)  {1} (b) [-1, 1] (c) [-1, 1)
16. Which of the following is NOT TRUE?
  1010 1010 xx xx fx   is
log10 (2 – x) (b) 10 11 log 21 x x    
17. Inverse of the function
(c)   10 1 2log21 x (d) 10 12 log 42 x x   

22. The function

(a) differentiable at x = 0 (b) continuous at x = 0

(c) discontinuous at x = 0

(d) not differentiable at x = 0

23. If the foci of the ellipse 22 2 1 25 xy b  and the hyperbola 22 1 1448125 xy are coincide, then the value of b2 is

(a) 25 (b) 16 (c) 64 (d) 49

24. If    , abcabc  then

(a) a and b are collinear

(b) a and b are perpendicular

(c) a and c are collinear

(d) a and c are perpendicular

25. Suppose that the temperature at a point (x, y) on a metal plate is T(x, y) = 4x2 – 4xy + y2. An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What is the highest temperature encountered by the ant?

(a) 125 (b) 120 (c) 0 (d) 25

26. The function     2 log1fxxx is

(a) an even function (b) an odd function

(c) a periodic function

(d) neither an even nor an odd function

27. Angles of elevation of the top of a tower from three points (collinear) A, B and C on a road leading to the foot of the tower are 30, 45 and 60 respectively. The ratio of AB and BC is

(a) 3:1 (b) 3:2 (c) 1 : 2 (d) 2:3

28. The mean of 25 observations are found to be 38. It was later discovered that 23 and 38 were misread at 25 and 36, then the mean is

(a) 32 (b) 36 (c) 38 (d) 42

29. In a Harmonic Progression, pth term is q and the qth term is p. Then pqth term is

(a) 0 (b) 1 (c) pq (d) pq (p + q)

30. A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition. The probability that it is divisible by 3 is

(a) 1/3 (b) 1/4 (c) 1/5 (d) 1/6

31. Let 22 aijk  and b be another vector such that .14ab  and 38 abijk  the vector b =

(a) 5i + j + 2k (b) 5i – j – 2k

(c) 5i + j – 2k (d) 3i + j + 4k

32. A survey is done among a population of 200 people who like either tea or coffee. It is found that 60% of the population like tea and 72% of the population like coffee. Let x be the number of people who like both tea and coffee. Let m ≤ x ≤ n, then choose the correct option

(a) n – m = 56 (b) n – m = 28

(c) n – m = 32 (d) n + m = 92

33. The eccentricity of an ellipse, with its centre at the origin is 1/3. If one of the directrices is x = 9, then the equation of ellipse is :

(a) 9x2 + 8y2 = 72 (b) 8x2 + 9y2 = 72

(c) 8x2 + 7y2 = 56 (d) 7x2 + 8y2 = 56

34. For a  R (the set of all real numbers), a  - 1,

Then one of the values of a is

(a) 5 (b) 8 (c) -15/2 (d) -17/2

35. Let a be the distance between the lines – 2x + y = 2 and 2x – y = 2, and b be the distance between the lines 4x – 3y = 5 and 6y – 8x = 1, then

(a) 40115ba  (b) 40211ab 

(c) 11240ba  (d) 11240ab 

36. If 111 121 112 Dx y 

(a) Divisible by x and y

for x  0, y  0, then D is

(b) Divisible by x but not by y

(c) Divisible by (1 + x) and (1 + y)

(d) Divisible by (1 + x) but not (1 + y)

37. If 0 < P (A) < 1 and 0 < P(B) < 1, and P(A  B) = P(A) P(B), then

(a) P(B|A) = P(B) – P(A)

(b) P(AC – BC) = P(AC) – P(BC)

(c) P (A  B)C = P(AC) P(BC)

(d) P(A|B) = P(A) – P(B)

38. Let a, b, c be distinct non-negative numbers. If the vectors , aiajckik  and cicjbk  lie in a plane, then c is

(a) The Arithmetic Mean of a and b

(b) The Geometric Mean of a and b

(c) The Harmonic Mean of a and b

(d) Equal to zero

39. The value of 1152 cotcostan 33 ec  

is

(a) 6/17 (b) 3/17 (c) 4/17 (d) 5/17

40. Which term of the series 551 34,,,........ 5 is 5 ? 13

(a) 12 (b) 11 (c) 10 (d) 9

41. Coordinate of focus of the parabola 4y2 + 12x – 20y + 67 = 0 is

(a) 517 , 42 

42. The 10th and 50th percentiles of the observations 32, 49, 23, 29, 118 respectively are

(a) 21, 32 (b) 23, 32 (c) 23, 33 (d) 22, 31

43. Area of the parallelogram formed by the lines y = 4x, y = 4x + 1, x + y = 0 and x + y = 1 is

(a) 1/5 (b) 2/5 (c) 5 (d) 10

44. If a1, a2, …… an are any real numbers and n is any positive integer, then (a)

(c)

(d) None of these

45. There are two circles in xy-plane whose equations are x

and

2
   1/ 2 12,0 , ,0 x xx fx ex         is
          1 12.... 1 lim 60 112.... aaa a n n nnananan      .
 
 
  
   (d) 517
  
(b) 175 , 24
(c) 175 , 42
, 24
 2 2 11 nn iiii naa   (b)  2 2 11 nn iiii naa  
 2 2 11 nn iiii aa   
2 + y2 – 2y
0
x2 + y2 – 2y – 3
=
= 0. A point

(x, y) is chosen at random inside the larger circle. Then the probability that the point has been taken from smaller circle is

(a) 1/3 (b) 2/3 (c) 1/2 (d) 1/4

46. f(x) = x + |x| is continuous for

(a) x  (-,) (b) x  (-,) – {0}

(c) only x > 0 (d) No value of x

47. The correct expression for cos-1 (-x) is

(a) /2 – cos-1 x (b)  - cos-1x

(c)  + cos-1 x (d) /2 + cos-1 x

48. If 11 coscos, 23 xy then 9x2 – 12xycos +4y

49. If   22 1, xy ab ab 

right angles, then

(a) a2 + b2 = 2c2 (b) b

(c) a2 – b2 = 2c2 (d) a2 – b

50. If ,  are the roots of x2 – x – 1 = 0, and An = n + n, then Arithmetic Mean of An-1 and An is

(a) 2An-1 (b) 1 1 2 n A  (c) 2An-2 (d) None of these

ANALYTICAL ABILITY AND LOGICAL REASONING

51. U, V, W, X and T are sitting on a bench. T is sitting next to U, V is sitting next to W, W is not sitting with X who is on the left end of the bench. V is in the second position from the right. T is to the right of U and X. T and V are sitting together. In which position T is sitting?

(a) Between U and W (b) Between V and X

(c) Between X and W (d) Between U and V

52 DNN, FQQ, HTT, ……., LZZ

(a) JWW (b) JVV (c) JXX (d) IWW

53 Find the synonym that is most nearly similar in meaning to the word : DEBACLE

(a) Catastrophe (b) Corker

(c) Opulence (d) Dandy

54 Fill in the blank : HEC, JGE, LIG, NKI,

(a) ONM (b) PMK (c) KMP (d) HGF

55 Replace the question mark with an appropriate image to complete the analogous pair.

56. Which of the following is the odd one from the given alternatives?

(a) Driving (b) Sailing

(c) Diving (d) Swimming

57 Select the pair of words, which are related in the same way as the capitalized words are related to each other.

Frugal : Extravagant

(a) Predecessor : Precursor (b) Hermit : Philosopher

(c) Teacher : Philanthropist (d) Criticise : Advocate

58. Identify the sixth number in the series : 6, 11, 21, 36, 56, ?

(a) 52 (b) 81 (c) 82 (d) 21

59 Looking at the portrait of a woman, Manu said, “Her mother is the wife of my father’s son and I have no brother and sister.” Whose portrait was Manu looking at?

(a) His son (b) His daughter

(c) His father (d) His nephew

60. Select the pair of words, which are related in the same way as the capitalized words are related to each other.

BUTTERFLY : FREEDOM

(a) chicken : Rooster (b) Self-reliant : Buoyant

(c) Frog : Water (d) Horse : Speed

61 If in a certain language, KOLKOTA is coded as LPMLPUB, how is MUMBAI coded in that code?

(a) NVNBCJ (b) OVNBBH

(c) NVNCBJ (d) NUNBCH

62 Select the related word from the given alternatives. MELT : LIQUID : : FREEZE : ?

(a) PUSH (b) CONDENSE

(c) SOLID (d) ICE

63 In the following figure, find the total number of squares.

(a) 20 (b) 36 (c) 18 (d) 24

64 Which of the following diagrams correctly represents lions, elephants, and animals? (a) (b) (c) (d)

65. Identity the fifth number in the series: 122, 144, 166, 188, ?

(a) 210 (b) 234 (c) 345 (d) 310

66. Fill in the blank : JAK, KBL, LCM, MDN, …….

(a) MEN (b) NEO (c) OEP (d) PEQ

67 Deepak, Rahul, Manoj, and Vinod are brothers. Who is the heaviest?

I. Rahul is heavier than Deepak and Vinod, but lighter than Manoj

II. Deepak is lighter than Rahul and Manoj, but heavier than Vinod.

(a) Either I or II is sufficient.

3
2
2
(b)
2
(d)
= (a) – 36 sin
36 sin
(c) 36 cos2
36
 and x2 – y2 = c2 cut at

2
2
2 – a
= 2c
2
2 = c
(a) (b) (c) (d)

(b) Data is both the statements together are not sufficient

(c) Statement I alone is sufficient, but statement II alone is not sufficient

(d) Statement II alone is sufficient, but statement I alone is not sufficient

68 Today is Wednesday. What would be the dady after 61 days?

(a) Monday (b) Saturday

(c) Sunday (d) Tuesday

69. Choose the word opposite in meaning to the given word : MITIGATE

(a) Tranquilize (b) Intensify

(c) Alleviate (d) Abate Directions (Qs. No. 70-72): In each question below are given two statements followed by two conclusion numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts Read the conclusion and then decide which of the following given conclusions logically follows from the two given statements, disregarding commonly known facts.

70 Statements : No women teacher play. Some women teachers are athletes.

Conclusions:

I. Male athletes can play.

II. Some athletes can play.

(a) Only conclusion II follows

(b) Only conclusion I follows

(c) either I or II follows

(d) Neither I nor II follows

71. Statements : All mangoes are golden in colour. No golden-coloured things are cheap.

Conclusions: I. All mangoes are cheap.

II. Golden-coloured mangoes are not cheap.

(a) Only conclusion II follows

(b) Only conclusion I follows

(c) either I or II follows

(d) Neither I nor II follows

72 Statements : All young scientists are open-minded. No open-minded men are superstitious.

Conclusion :

I. No scientist is superstitious.

II. No young people are superstitious.

(a) Neither I nor II follows

(b) Only conclusions I follows

(c) Either I or II follows

(d) Only conclusions II follows

73. Find out the wrong number in the FOLLOWING series.

2, 5, 10, 17, 26, 38, 50, 65

(a) 26 (b) 38 (c) 65 (d) 50

74 Find out the wrong number in the FOLLOWING series.

30, –5, –45, –90, –145, –195, –255.

(a) –145 (b) –255 (c) –5 (d) –195.

75. Six books are labelled A, B, C, D, E and F are placed side by side. Books B, C, E and F have green covers while others have yellow covers. Book A, B and D are new while the rest are old volumes. Book A, B and C are law reports while the rest are medical extracts. Which two books are old medical extracts and have green covers?

(a) C and D (b) C and E (c) B and C (d) E and F Direction (Qs. 76-78):

The following questions are based on the pie-charts given below: Percentage-wise distribution of students studying in Arts and Commerce in seven different institutions Different institutions – A, B, C, D, E, F and G.

Total number of students studying Arts = 3800.

Total number of students studying commerce = 4200

76. What is the total number of students studying Arts in Institutes A and G together?

(a) 1206 (b) 1126 (c) 1026 (d) 1226

77 How many students from Institute B study Arts and Commerce?

(a) 1180 (b) 1208 (c) 1018 (d) 1108

78 The ratio of the number of students studying Arts to that studying Commerce in institute E is:

(a) 19 : 27 (b) 19 : 16 (c) 19 : 28 (d) 27 : 14

79. Statement: Many shops in the local market have extended their shops and occupied most part of the footpath in front of their shops.

Cause of Action:

I. The civic authority should immediately activate a task force to clear all the footpaths encroached by the shop owners.

II. The civic authority should charge hefty penalty to the shop owners for occupying the footpath.

III. The civic authority should setup a monitoring system so that encroachments do not recur in future.

(a) I and II follow (b) All I, II and III follows

(c) II and III follows (d) None follows

80. Statement : There is a significant increase in the number of patients affected by some disease in a city Course of action :

I. Municipal Corporation of the city should take immediate action

II. This problem should be raised in the UNESCO.

III. Hospitals in the city should be equipped properly for the treatment of the patients.

(a) I and III follow (b) I and II follow

(c) All follow (d) Only III follow

Direction (Qs. 81-85): Comprehension:

Direction: Read the following information carefully and answer the questions.

Five Dramas A, B, C, D and E have to be staged in 6 hour where 1 hour needs to be given per drama.

4

1. A break of 1 hour has to be taken in third or four hour.

2. Drama show cannot be started with A and cannot end in C

3. D has to follow B immediately with no break in between.

4. A cannot be done immediately after D.

5. A has to precede E immediately with no break in between.

81 Which hour is a break hour?

(a) 5th (b) 2nd (c) 3rd (d) 4th

82 Which is the drama to be staged first?

(a) B (b) None (c) D (d) A

83 Which is the drama staged immediately after the break?

(a) A (b) B (c) D (d) None of these

84 Which drama is staged immediately after D?

(a) E (b) None (c) C (d) B

85 Which drama is staged immediately after E?

(a) A (b) E (c) None (d) C

86 The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is

(a) 127 (b) 235 (c) 305 (d) 123

87 At what time, in minutes, between 3 o’clock and 4 o’clock both the needles will coincide each other?

(a) 1 5 11 (b) 4 12 11 (c) 4 13 11 (d) 4 16 11

88. 1. A is the brother of B.

2. C is the father of A.

3. D is brother of E.

4. E is the daughter of B. Then, the uncle of D is?

(a) E (b) A (c) B (d) C

89. Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

(a) 648 (b) 2700 (c) 10800 (d) 1800

90 A person’s present age two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. What is the present age of the mother?

(a) 40 (b) 30 (c) 50 (d) 60

COMPUTER

91. Suppose the largest n bit number requires, ‘d’ digits in decimal representation. Which of the following relations between ‘n’ and ‘d’ approximately correct.

(a) d = 2n (b) n = 2d

(c) d < n log10 2 (d) d > n log10 2

92. ‘Floating point representation’ is used to represent

(a) Boolean Values (b) Integers

(c) Whole numbers (d) Real numbers

93. Which of the following is equivalent to the Boolean expression :

(a) XY (b) XY (c) XY (d) XY

94. The Boolean expression AB + AB' + A'C + AC is unaffected by the value of the Boolean variable

(a) B (b) None of these

(c) A (d) C

95. Write the simplified form of the Boolean expression

(A + C) (AD +AD') + AC + C:

(a) A + C

(c) A + C'

(b) A' + C

(d) A + D

96. If a processor clock is rated as 2500 million cycles per second, then its clock period is

(a) 2.50 × 10–10 sec (b) 4.00 × 10–10 sec

(c) 1.00 × 10–10 sec (d) None of these

97. The minimum number of NAND gates required for implementing of Boolean expression, ABABCABC  is

(a) 1 (b) 0 (c) 2 (d) 3

98. FFFF will be the last memory location in a memory is size

(a) 1 k (b) 16 k (c) 32 k (d) 64 k

99. The maximum and minimum value represented in signed 16 bit 2’s complement representations are

(a) 0 and 65535 (b) –32768 and 32767

(c) –16384 and 16383 (d) 0 and 32767

100. If a signal passing through a gate is inhibited by sending a low into one of the inputs, and the output is high, the gate is a(n):

(a) NOR (b) AND

(c) NAND (d) OR

ENGLISH

101. Fill in the blanks with the correct option. Technical writing demands _______ use of language.

(a) Dramatic (b) Factual

(c) Figurative (d) Poetic

102 Select the correct form of verb/subject verb agreement.

The principal, along with his assistants, _________ the meeting.

(a) attend

(b) are attending

(c) attending (d) is attending

103. Select correct articles.

He is ........ M.A. with PhD and teaches in ...... university.

(a) a, an (b) a, the

(c) the, the (d) an, the

104. Choose the correct alternative

The county cleared this path and paved it with packed gravel, so they have a peaceful place to hike and bike.

Which of the following alternatives to the underlined portion would NOT be acceptable?

(a) path, paving (b) path and then paved

(c) path before paving (d) path paved

105. “Bite and bullet” means

(a) to analyse your faults

(b) to accept something that is difficult or unpleasant

(c) to stop a conflict

(d) to become mad

106. What can you call a person who leads an unconventional style or living?

(a) Cynic (b) Agnostic

(c) Bohemian (d) Altruist

107. Fill in the blank with the most appropriate option. Kedar ________ this project for a month and now he is about to join a new project.

(a) guided (b) has been guiding

(c) guiding (d) guides

Direction (Qs. 108-110):

Read the following passage carefully and answer the questions:

5
      XYXYXY 

You might think you’ve experienced VR, and you might have been pretty impressed. Particularly, if you’re a gamer, there are some great experiences to be had out there (or rather, in there) today. But over the next few year, in VR as in all fields of technology, we’re going to see things that make what is cutting-edge today look like space lnvaders. And although the games will be amazing, the effects of this transformation will be far broader, touching on our work, education, and social lives.

Today’s most popular VR applications involve taking total control of user’s sense (sight and hearing, particularly) to create a totally immersive experience that places the user in a fully virtual environment that feels pretty realistic. Climb up something high and look down, and you’re likely to get a sense of vertigo. If you see an object moving quickly towards your head, you’ll feel an urge to duck out of the way.

Very soon, VR creators will extend this sensory hijacking to our other facilities – for example, touch and smell – to deepen that sense of immersion. At the same time, the devices we use to visit these virtual worlds will become cheaper and lighter, removing the friction that can currently be a barrier. I believe extended reality (XR) – a term that covers virtual reality (VR), augmented reality (AR), and mixed reality (MR) – will be one of the most transformative tech trends of the next five years. It will be enabled and augmented by other tech trends, including super-fast networking, that will let us experience VR as a cloud service just like we currently consume music and movies. And artificial intelligence (Al) will provide us with more personalized virtual worlds to explore, even giving us realistic virtual characters to share our experiences with.

108 The passage states all the following about VR applications except.

(a) Vertigo is a major feature of all Al applications

(b) VR applications takes control of the user’s senses

(c) VR applications creates a virtual environment that feels pretty realistic

(d) Future Al will allow us to share our experiences with realistic virtual characters

109 ‘Duck out of something’ means

(a) To fall down

(b) To meet with an accident

(c) To avoid doing something

(d) To hit something hard

110 Select an antonym for the word ‘argument’ from the options given below:

(a) Aggrandize (b) Reinforce

(c) Inflate (d) Curtail

111. Which of the phrases given below should replace the phrase printed in bold to make the sentence grammatically correct.

He is addicted to smoke.

(a) used to smoke (b) addicted to smoking

(c) addicted with smoking (d) addict of smoke

112. Select the one which best expresses the same sentence in indirect/direct speech:

He said, “I am glad to be here this evening.”

(a) He said that he was glad to be there that evening

(b) He says he was glad to be here his evening

(c) He asked he is glad to be here this evening

(d) He said he was glad to be here this evening.

113. Choose the synonym.

LIBERAL

(a) Reactionary (b) Generous

(c) Sober (d) Affectionate

114. Identify the word that is similar in meaning to the underlined word.

Raghu make adulatory remarks about the waiter who served the food.

(a) Derogatory (b) Ironic

(c) Slanderous (d) Complimentary

115 Select the word which means the same as the following:

To read something carefully

(a) Presume (b) Perverse

(c) Persiflage (d) Peruse

116. Choose the antonym: BOLD

(a) Nervous (b) Coy (c) Timid (d) Fearful

117. Fill in the blanks with the correct preposition

Jagdish is waiting for me _______ the campus.

(a) On (b) At (c) In (d) Out

118 Choose the correct alternative with the correct choice given below each statement. You are to conform __________ the rules of the institute.

(a) to (b) with (c) of (d) on

119. Read the following sentence to find if there is any error in any part:

If I were him / I would teach / him a lesson / No error

(a) No error

(b) If I were him

(c) I would teach (d) him a lesson

120. Fill in the blank. Neither Peter nor I _________ responsible for this blunder.

(a) am (b) are (c) is (d) were

ANSWERS

6
1 2 3 4 5 6 7 8 9 10 d d b c c d d a c c 11 12 13 14 15 16 17 18 19 20 a a d a c d b c a c 21 22 23 24 25 26 27 28 29 30 d a b d c b a c b c 31 32 33 34 35 36 37 38 39 40 a a b d a c c b a b 41 42 43 44 45 46 47 48 49 50 c b a b d a b b c b 51 52 53 54 55 56 57 58 59 60 d a a b d a d b b b 61 62 63 64 65 66 67 68 69 70 c c d a a b c a b d 71 72 73 74 75 76 77 78 79 80 a a b a d c c a b a 81 82 83 84 85 86 87 88 89 90 c a d c c a d b d a 91 92 93 94 95 96 97 98 99 100 d d a a a b b d b b 101 102 103 104 105 106 107 108 109 110 b d d d b c b a c d 111 112 113 114 115 116 117 118 119 120 b a b d d c a a a a
NIMCET-2022

5. Ans. (c) As volume of parallelepiped with edges

Let h be height of hill and  be a angle of elevation at A, B, C. In this way foot of hill M is circumcentre of ABC  AM = CM = BM = R

Now by sine law

2sin2sin2sin abc R ABC  …(1)

Also tan h R   …(2)

 h = R tan 

Now put 2sin a R A  from (1)

tan 2sin a h A   tancos 2 a ecA   2. Ans. (d) Here xmyn = (x + y)m+n Taking log on both sides m logx + n logy = (m + n) log

 2(2 – 4) – (3 - 8) + (6 - 4

(2- 9) + 1 (2 - 9) = 0

(2 - 9) ( + 1) = 0 9

) = 15

6. Ans. (d) |x| + |y| = 2   x  y = 2 So lines are

 x + y = 2, x + y = - 2, x – y = 2, x – y = - 2 It forms a rectangle

As x + y = 2, x + y = - 2 and x – y = - 2, x – y = 2 are two set of parallel lines and perpendicular to one another Length

7. Ans. (d) Here a < b

So if a < x < b  x – a > 0, x – b < 0

|x – a| = x – a, |x – b| = - (x

b)

8. Ans. (a) As 7 abc

7 NIMCET-2022 (SOLUTIONS)
1. Ans. (d)
differentiate w.r.t. x   1 .1 mndydy mn xydxxydx     dymnnmmn dxxyyxxy            mxymnx dymynynxny dxyxyxxy            dymynxmynx dxyxyxxy      dyy dxx  3. Ans. (b) As 33 3log5log5 3 33.3  1 3 log5 1 27 27.327.5 5  as loga x ax  4. Ans. (c)   2 342 1 221 xdx xxx  = 2 5 24 1 21 2 x dx x xx   = 35 24 11 21 2 xx dx xx   Put 24 21 2 t xx  35 44 dxdt xx     35 11 4 dt dx xx    11 2 442 dtt tc t   24 21 2 2 xx c  
(x + y)
234 aijk  , 2,2 bijkcijk  is abc    =15 234 1215 12   
2  
   
2002822 BD     22 200222
area of rectangle 22.228 lb  sq. unit.
of CD
22
So
 
a xaxbdx          bb aa xaxbdxbadx        2 b a baxbababa 
b
  Here    ,1,2,1,,2ab   1,1,1 c  abc    12 127 111     ( + 2) – (3) + 1 (-2 + 2) = 7  2 + 4 - 12 = 0  2 + 6 - 2- 12 = 0 ( + 6) – 2 ( + 6) = 0 A (( B C D x – y = 2 x – y = - 2 x + y = 2 x + y = - 2 A B C N M   R R h


+ 6) ( - 2) = 0   = - 6, 2

9. Ans. (c) As if |A| = m  |P(A)| = 2m |B| = n  |P(B)| = 2n Also |P(A)| - |P(B)| = 112

 2m – 2n = 112  m = 7, n = 4 Satisfies (a), (b), (d) choices So only (c) choice 2m – n = 1 is wrong.

10. Ans. (c) As . dvdvdx a dtdxdt 

14. Ans. (a) As  

 Here   1 cos 1,1 x D 

D[x] = R Also [x]  0  x  [0, 1)

 Required domain = [-1, 1]  R – [0, 1) = [-1, 0)  {1}

15. Ans. (c) Here quadratic equation is x2 + px + q = 0 and roots are tan30, tan 15

 S= Sum = tan30

11.

(a) According to given condition

13. Ans. (d)



As intercept form of straight line is 1 xy ab 

Also as A is mid. point of line BC 4,5 22 ab   a = 8, b = 10

 Required line 1 810 xy

 5x + 4y = 40

8
(
dv av dx     Here 2 axx 
   2 vdvxxdx  
223 vxx  As 23 00 23 xx v  3 2 x 
2 dv vxx dx
223
      11111 tantantan
ABABC     11 tancot 32232 CC      …(1) As we
ANALOGY tancot 22 ABabC ab      …(2) From (1)  
tancot
C AB  from (1), (2) 1 3 ab ab   By componendo / dividendo 213 213 a b   2 1 a b 
(a)
4 xn   
Ans.
23232
know by NAPIER’s
11
232
12. Ans.
4cos2x + 6sin2x = 5
4cos2x + 6(1 – cos2x) = 5
1 = 2cos2x 222 1 coscoscos 24 xx 

 
/ :0 fgfg DDDxgx
P=
= tan30.tan15 = q   1 23 3 Sp  …(1)   1 .23 3 Pq  …(2) as tan1523  From (1), (2) p – q + 2   123 232 33      123 3131 333  3311 16. Ans. (d) Only (d) is 0 0 limlimcoscosh 1212 h x x xh        cos01 10 1201  
Ans. (b) 1010 1010 xx xx y   Apply componendo / dividendo 2 12.10 10 1 2.10 x x x y y   2 10 11 102log11 x yy x yy   10 11 log 21 y x y   So inverse is 10 11 log 21 x y x      18. Ans. (c) Here rth moment of distribution is   1 r rixx n    Have   1 1 21 ix n    21 ix n   213 ix Meanx n   Also  2 2 1 216 ix n     We have to find     2211 321ii xx nn      2 1 2122ii xx n    (0, b) C A B (a, 0) (4, 5)
 + tan15
= - p
Product
17.

= 2 3 23 0

(cota1 – cota2) + (cota2 – cot a3) + …….. + (cotan-1 11 1 1 122 2 111 12 2...... 3 222 1...... 123 lim x

aaaa aa      

sin sinsin sinsinsinsinsinsin nn 

d dd aaaaaa 1 coscot 2 ec  …(2) Adding

xx xx x xxx x x 

21321 12231 (2) 155 2cos2cos 224 ecec   22. Ans. (a) At x

sinsinsin sinsinsinsinsinsin nn nn 0       2 11 2 000 limlim12lim12xx xxx fxxx 

 11 xx 

aaaaaa aaaaaa    = e2 Also f(0) = e2     2 0 lim0 x fxfe   cont at x = 0. Now to check differentiability       0

2121 0 '0lim 0 x

12 fxf f x     1 2 0

                            (By binomial expansion     2 1 11..... 2 n nn xnxx  )       23 23 0

sincoscossin sinsin aaaa aa 12 lim x x

11112 lim12.48.... .23 x

xxx xx xx 

xx

 1

sincoscossin sinsin nnnn nn  

11 1 xe x 

= 1     222 1...... 12         2 323 0

(1), x xx x 

23. Ans. (b) Foci of Ellipse and Hyperbola are coinciding.

For 22

2 1 25 xy b   b2 = a2 (1 – e2) b2 = a2 – a2e2  b2 = 25 – a2e2 …(1) For Hyperbola 22 1 1448125 xy   b2 = a2 (e2 – 1)  b2 = a2e2 – a2 2222 81144225 9 252525 aeae  … (2)

Put

2(cosec  + cot) = 1 21 222 112... 2323 lim x

(2) in (1) b2 = 25 – 9  b2= 16

24. Ans. (d) As   11 sin abcabcn   

1 is angle between ab  and c but ab  is  to both , ab and if c is also  to a from choice (d) ab and c are parallel   0 abc 

=        = finite value

IIly   22 sin abcabcn   

Again 2 is angle between a and bc  but bc   to both , bc

Also if a is  to c a  and bc  are parallel so are parallel   0 abc 

25. Ans. (c) T(x, y) = 4x2 – 4xy + y2

Also circle of radius 5 centred at origin (x – 0)2 + (y – 0)2 = 52  x2 + y2 = 25

9 
 

        


2 112 212ii xx nnn  
= 16 + 1 – 2(1) = 15 variance = 15 19. Ans. (a) Here a1, a2, ……., an are in A.P., – cotan) =cota1-cotan 20 Ans. (c)   112 tan1sin1 2 xxxx   Now   11 2
d = a2 1 tan1cos 1 xx xx   112 2
a1 = a3 – a2 = …. an 1 cossin1 2 1 xx xx
an-1 Now sin d(cosec a1 cosec a2 + cosec a2 cosec a3 + ….. + cosec an-1 cosec an)    22 2
 12231 1 111 1 xxxx xx    x2 + x = 0  x(x + 1) = 0 x = 0, - 1 21. Ans. (d) If cosec  – cot  = 2 …(1) As (cosec  - cot ) (cosec  + cot ) = cosec2- cot2

  222,442525 Txyxxxx 

  22,253425 Txyxxx    2 2

42 6425 225 xx dT xx dx x  2 2 2

4 6425 25 x xx x

25 64 25

xx x x

  22 2

   =0

  22 62542520 xxx 

 226254252 xxx 

2 222 362516252 xxx 

2 222 9254252 xxx 

225x2 – 9x4 = 4(625 + 4x4 – 100x2)

 25x4 – 625x2 + 2500 = 0

 x4 – 25x2 + 100 = 0 (x2 – 20) (x2 – 5) = 0  x2 = 20, 5

At x2 = 5 dT dx  changes sign from +ve to –ve

 x2 = 5 is a point of maxima

Also x2 + y2 = 25  y2 = 20

Now T(x, y) = 4x2 – 4xy + y2 4.5452020 = 40 – 40 = 0

26. Ans. (b) It’s an odd function

As f(x) + f(-x)

22 log1log1 xxxx

 f(x) + f(-x) = 0  f(-x) = - f(x)

 f(x) is an odd function.

29. Ans. (b) In H.P.,

30. Ans. (c)

Incorrect sum = 950

Now is divisible by 3 if Sum of digits of that no is divisible by 3.

As we have to make 4 digit no. out of 1, 2, 3, 4, 5 If we leave 3, we have 1, 2, 4, 5

Here sum = 12 divisible by 3

So nos. with help of 1, 2, 4, 5 are 4 out of total 5

31. Ans. (a) From choices .14,34 ababijk 

Only (a) choice 52ijk  satisfies both

32. Ans. (a) Here   60 200120 100 nT 

  72 200144 100 nC 

n (T  C) = 200, n (T  C) = x

As n(T) + n(C) – n(T  C)  n(T  C)  min (n(T), n(C))

120 + 144 – 200  n(T  C)  120

64  n(T  C)  120

Given m  x  n  n – m = 120 – 64 = 56

33. Ans. (b)

Ellipse with centre at origin (0, 0) is 22 22 1 xy ab

Here 1 3 e  and directrices a x e 

Also b2 = a2(1 – e

10

 
 
     2
log1log1log10 xxxx    
 
2222
27. Ans. (a) In OCD, tan603 h z  3 h z  …(1) In OBD , tan451 h yz    y + z = h …(2) In OAD, 1 tan30 3 h xyz   3 xyzh  …(3) Put (2)
 31 xh
(1) in (2) 31 33 h yhh          31 3:1 31 3 h ABx BCy h 
in (3)
Put
25 1 2538 i i x   
Incorrect sum-Incorrect values +correct values
28. Ans. (c)
n
25   =38
New Mean =
95025362338
Tp = q, Tq = p  Tpq =
If
1 (results)
Required probability 41 55 

993 1 3 aa a e  2 9 a  …(1)
2 1 91 3         b2 = 8 …(2) A B C 30 45 60 D h
2)

………(2)

From (1) and (2) 11540ab

Divisible by (1 + x), (1 + y)

37. Ans. (c) Here P(A  B) = P(A) . P(B)

 A, B are independent events  Ac, Bc are also independent.

Now (c) choice implies P(A  B)c = P(Ac  Bc) = P(Ac ). P(Bc) as Ac, Bc are independent.

38. Ans. (b) If vectors ,, aiajckikcicjbk  are in same plane. 1010

 a(-c) – 1 (ab – c2) + c(a) = 0

 ab = c2  a, c, b in G.P.  c is G.M. of a, b.

39. Ans. (a)

1152 cotcostan 33 ec

Now 1153 costan 34 ec

1152 cotcostan 33

1132 cottantan 43

32 43                    1 6 cotcot 17        6 17 

 It’s

4y

11

20y = - 12x

11  Ellipse is 22 22 1 xy ab  22 1 98 xy   8x2 + 9y2 = 72 34. Ans. (d)           1 12.... 1 lim 60 112.... aaa a n n nnananan      1 1 12 lim 112 .1..... aaa a a n a n n nnn n nnaaa nnnn            
   1 12 lim 112 1..... aaa a n n nnn n aaa nnnn   
        1 112 ..... lim 1112 1..... aaa a n n nnnn n aaa nnnnn                  1 1 1 1 lim 11 1 a n r a n n r r nn r a nnn               1 1 1 0 12 0 0 1 11 1 2 2 a a x xdx aa x a ax axdx            21 12160aa    (a + 1) (2a + 1) = 120 2a2 + 3a – 119 = 0  2a2 + 17a – 14a – 119 = 0 2a(a – 7) +17 (a – 7) = 0 (a – 7) (2a + 17) = 0  a = 7, - 17/2 35. Ans. (a) Here lines are – 2x + y = 2  2x – y = - 2 and 2x – y = 2 and distance between these two is a     22 22 4 21 5 a   …….(1) other two lines are 4x – 3y = 5 and 6y – 8x = 1 1 43 2 xy  22 5 11 2 10 43 b  


 
36. Ans. (c) D= 111 121 112 x y     
 
100 11011 101 xxy y
ccb 
aac
   
  
ec
  
cottan 1.32 43              
1 98 17 12 cottancottan 6 6 12
term.
40. Ans. (b) As sequence is 5555 ,,,......, 34513
11th
0 4y2 –
– 67. 4(y2 – 5y) = - 12x – 67 2 25 45126725 4 yyx    = - 12x - 42 2 5 4 2 y    42 12 12 x    2 57 3 22yx    It’s of type Y2 = 4a X  4a = - 3  a = - 3/4 4 3 5
41. Ans. (c) Here PARABOLA is
2 + 12x – 20y + 67 =

 focus is (a, 0)  X = a, Y = 0

42. Ans. (b) Here data is 32, 49, 23, 29, 118 In ascending order 23, 29, 32, 49, 118

46. Ans. (a) y = x + |x| is continuous for all x as sum of two continuous functions is always continuous.

43. Ans. (a) Here parallel lines of parallelogram are y = 4x and y = 4x + 1

 4x – y = 0, 4x – y + 1 = 0 and x + y = 0, x + y = 1

As area of parallelogram with parallel sides

a1x + b1y + c1 = 0, a1x + b1y + d1 = 0

a2x + b2y + c2 = 0, a2x + b2y + d2 = 0

44. Ans. (b) By result ; m  [0, 1] 1212 m mmm nn aaaaaa

2

nn ii ii naa 

45. Ans. (d) Here circles be x2+y2 –2y = 0 with centre (0, 1), r = 1 as (x – 0)2 + (y – 1)2 = 1 x2 + y2 – 2y – 3 = 0 with centre (0, 1), r = 2 as (x – 0)2 + (y – 1)2 = 4 = 22

Now area of larger circle is .22 = 4 and area of smaller circle is .12 =  If point (x, y) is taken from smaller circle.  favourable cases / region = 

44   

47. Ans. (b)   11 coscosxx 





222 1, xy

2.0

= c

12
 X =
0  735 ,0 242xy  175 , 42xy 175 , 42   
-3/4, Y =
1 100
k n Pk       term   10 51 10.6 100 P   th
50 16 50503 100100 n P      rd item. = 32.
th
item which is 23.
is   
1221
in our case = Area   1010 1 415  
1122
cdcd abab So
nn    
  
2
Take m = 2 2 222 1212............nn aaaaaa nn     
11
Out of total = 4  required prob. 1
22 1 cos11 2323 xyxy             22 1 49 cos 66 xy xy           2249 cos 6 xyxy   226cos49xyxy  226cos49 xyxy      22222 12cos36cos49 xyxyxy   2222 9436 xyxy  2222 9412cos3636cos36sin xyxy 
xyc ab   
48. Ans. (b) As 11 coscos 23 xy
49. Ans. (c) Here curves are 22
21
xydy aabbdx     22 0 xydy dx ab  2 1 2 dyxb m dxy a  …(1)
Now m1 = slope of first curve
2 – y2 = c2 220 dy xy dx  2 dyxx m dxyy  …(2)
  m1m
1 2 2 .1 xbx yy a 
yb  …(3)
IIly for second curve x
If two curves cut at 90
2 = -
22 22 xa
2
2  b2x2 – b2y2 = b2c2 …(5) Multiply (4) by c2, (5) by (a2)  c2b2x2 + a2c2y2 = a2b2c2 and a2b2x2 – a2b2y2 = a2b2c2  Subtract (c2b2 – a2b2)x2 + (a2c2 + a2b2)y2 = 0 Put 2 22 2 a xy b  in above     2 2222222222 2 0 a cbabyacaby b     (a2c2 – a4) + (a2c2 + a2b2) = 0 (0, 3) (0, 1) (0, 0) (0, -1)
Now first equation is b2x2 + a2y2 = a2b2 …(4) and second equation is x2 – y

(b) As

51. Ans. (d) VTU OR UTV, VW OR WV, (XWX)

52.

+3

53. Ans. (a) The exact synonym of the given word 'Debacle' is 'Catastrophe'. Debacle: a complete failure, especially because of bad planning and organization.

Example: The collapse of the company was described as the greatest financial debacle in US history. Catastrophe: a sudden event that causes very great trouble or destruction. Example: They were warned of the ecological catastrophe to come. By reading the above definitions we can infer that the correct answer is Option (a).

54 Ans. (b)

55. Ans. (d)

56. Ans. (a) All words except ‘driving’ are related to movement activities that are related to water bodies, while driving cannot be done on water bodies. Hence, ‘driving’ is the odd one out.

Word Meaning

Driving Operate and control the direction and speed of a motor vehicle.

Diving The sport or activity of swimming or exploring under water.

Sailing The action of sailing in a ship or boat.

Swimming The sport or activity of propelling oneself through water using the limbs. 57

60. Ans. (b) We know that Round the world, butterflies are seen as the departed souls of our ancestors. Indigenous people recognize the chrysalis as the soul trapped inside the body. The emergence of the adult butterfly symbolizes the freedom of the soul upon death. Hence, Both Butterfly and Freedom are inter-related. The same way 'Self-reliant' is related to the word 'Buoyant'.

Both are of similar meaning.

Therefore, the correct answer is Option (b).

62. Ans. (c) First is the process of formation of the second.

13 2a2c2 = a4 – a2b2  2c2 = a2 – b2
Ans.
of x2 – x – 1 = 0   +  =
 = - 1 As An+1 = n+1 + n+1 = (n + n) ( + ) -  (n-1 + n-1) = An( + ) - (An-1) as An = n + n, An-1 = n-1 + n-1 = An(1) + 1(An-1) = An + An-1  An+1 = An + An-1  A.M. of An, An-1 11 22 nnn AAA  
50.
,  are roots
1,
 X U T V W
U and V.
Hence, T is sitting between
Ans. (a) +3 +3 +3
D N N, F Q Q, H T T,…J W W, L Z Z +2 +2 +2 +2
H E C, T G E, L I G, N K I, P M K
(d)
Ans. (b) 6, 11, 21, 36, 56, 81
Ans. (b)
Ans.
58.
59.
Ans. (c) K O L K O T A +1 +1 +1 +1 +1 +1 +1 L P M L P U B Similarly M U M B A I +1 +1 +1 +1 +1 +1 N V N C B J
61
63. Ans. (d) A B C D E J I H G F No. of squares No. of squares = 4  2 + 3  1 = 11 = 5 + 4 + 4=13 Total number of squares = 11 + 13 = 24 64 Ans. (a) 65 Ans. (a) 122, 144, 166, 188, 210 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 5 +10 +15 +20 +25 +2 2 +22 +22 +22+ PortraitManu + + Lions Elephants Animals

66. Ans. (b)

Hence 38 is wrong number.

74 Ans. (a) 30

30 - 35 = -5

-5 - 40 = -45

-45 - 45 = -90

-90 - 50 = -140

67. Ans. (c) I. Manoj > Rahul > Deepak, Vinod

II. Manoj, Rahul > Deepak > Vinod

So, clearly can see that statement I alone is sufficient to answer the question.

68. Ans. (a)

Wednesday  3 + 61 = 64

7  R = 1  Monday

69 Ans. (b) 'Mitigate' is a verb which means 'to reduce or to diminish'. Option B - to become greater, more serious, or more extreme, or to make something do this:

Therefore, the most appropriate antonym for 'Mitigate' is Option B - Intensify

70. Ans. (d)

Woman Teacher

Athletes

From Venn Diagram we can clearly see that neither I nor II follows.

71. Ans. (a)

Mangoes

From Venn Diagram, we can see that only II follows.

72 Ans. (a)

Young scientists Open minded Superstitious

From Venn Diagram, it is clear that neither I nor II follows.

-140 - 55 = -195

-195 - 60 = -255

Hence, -145 is wrong. -140 should have come in place of -145

75. Ans. (d) Green cover = B, C, E, F

Yellow cover = A and D

New volumes = A, B, D

Old volumes = C, E, F

Law reports = A, B, C

Medical extracts = D, E, F

Clearly, E and F are the two old medical extracts with green covers.

Directions (Solutions 76-78) :

100%  3800  1%  38

Institutes Students studying Arts

A 15%  15  38

B 8%  8  38

C 17%  17  38

D 21%  21  38

E 14%  14  38

F 13%  13  38

G 12%  12  38

100%  4200 1%  42 Institutes Students studying Arts

A 12%  12  42

B 17%  17  42

C 15%  15  42

D 14%  14  42

E 18%  18  42

F 13%  13  42

G 11%  11  42

76 Ans. (c) No. of students studying arts in institutes A and G together = 15% + 12% = 27%  38 = 1026

77. Ans. (c) No. of students from institute B studying Arts and commerce = 8  38 + 17  42 = 304 + 714 = 1018

78 Ans. (a) Arts : Commerce

E  14  38 : 18  42

19 : 27

79 Ans. (b) All of the given options stand true and therefore option (b) is the right answer

80 Ans. (a) I and III follows.

14
A K, K B L, L C M, M D N, N E O
J
73 Ans. (b) 37 2, 5, 10, 17, 26, 38, 50, 65 +3 +5 +7 +9 +12 +12 +15 +11 +13
+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
Golden Cheap 
 Play

After 8 years 3 : 6 Present

(A) + C = AA + AC + C =A + C(A + 1)

15 Solutions (81 to 85): Days Dramas 1 B 2 D 3 Break 4 C 5 A 6 E 81 Ans. (c) 82 Ans. (a) 83 Ans. (d) 84. Ans. (c) 85.Ans. (c) 86 Ans. (a) Required number = H.C.F. of (1657−6) and (2037−5) = H.C.F. of 1651 and 2032 1651=13×127 and 2032=2×2×2×2×127 Common factor 127 Required number = 127 87. Ans. (d)   2 30 11 MH     2 303 11 M  180 11  4 =16 11 min part 3. 88 Ans. (b) C+ | A B | E- D+ From diagram , we can see A is the uncle of D. 89 Ans. (d) 63104 270 x    x = 1800 90. Ans. (a) Person : his mother 3 Present  2 : 5 + 8 +8 (1 : 2) 3 1 3  Person : His mother Present (2 : 5)  8 = 40 + 1 +1 1  8 yrs
age of mother = 40 yrs
Ans.
Take log both sides log1010d > log102n  d > nlog102 92. Ans. (d)
Ans.
        xyxyxy xyxyxy   = xyxyxy    xyyxy  1 yy  xxy  = xyxy  94. Ans. (a) AB + AB' + A'C + AC A(B + B') + C(A' + A) A + C independent of B 95. Ans. (a) (A + C)
+
+ AC + C (A + C)
+
+ 1) (A + C)
=A + C 96. Ans. (b) Clock Period 1 cycle per second  10 10 6 110010 410 25 2510    97. Ans. (b) 98. Ans. (d) 99. Ans. (b) 100. Ans. (b) 101. Ans. (b) 102. Ans. (d) 103. Ans. (d) 104. Ans. (d) 105. Ans. (b) 106. Ans. (c) 107. Ans. (b) 108. Ans. (a) 109. Ans. (c) 110. Ans. (d) 111. Ans. (b) 112. Ans. (a) 113. Ans. (b) 114. Ans. (d) 115. Ans. (d) 116. Ans. (c) 117.Ans. (a) 118. Ans. (a) 119. Ans. (a) 120. Ans. (a)
91.
(d) 10d > 27
93.
(a)
(AD
AD')
(A(D + D')
C(A

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