NIMCET Previous Year Questions 2022- Infomaths

PREVIOUS YEAR SOLVED PAPERS NIMCET ACTUAL PAPER 2022 11. In a triangle ABC, if the tangent of half the MATHS If the angle of elevation of the top of a hill from each difference of two angles is equal to one third of the of the vertices A B and C of a horizontal triangle is tangent of half the sum of the angles, then the ratio α, then the height of the hill is of the sides opposite to the angles is (a) 2 : 1 (b) 1 : 2 (c) 3 : 1 (d) 1 : 1 1 1 (a) b tan  sec B (b) b tan  cos ecA 12.

The solutions of the equation 4 cos2x + 6 sin2x = 5 2 2 are 1 1 (c) c tan  sin C (d) a tan  cos ecA   (a) x  n  (b) x  n  2 2 3 4 2. If xmyn = (x + y)m+n, then dy/dx is 2   x y (c) x  n  (d) x  n  (a) (b) xy (c) x/y (d) y/x 3 2 xy 13. A straight line through the point A(4, 5) is such that 3  log3 5 3. The value of 3 its intercept between the axes is bisected at A, then 27 9 5 5 its equation is (a) (b) (c) (d) (a) 3x + 4y = 20 (b) 3x – 4y + 7 = 0 27 9 5 5 (c) 5x – 4y = 40 (d) 5x + 4y = 40 2 x  1 dx  cos 1 x 4. The value of  3 4 2 14.

The domain of the function is f x    x 2x  2x  1  x 2 1 2 1 (a) [-1, 0)  {1} (b) [-1, 1] (a) 2 2  2  4  C (b) 2 2  2  4  C x x x x (c) [-1, 1) (d) None of these 15.

If the roots of the quadratic equation x2 + px + q = 0 1 2 1 (c) (d) None of these 2 2  4 C are tan 30 and tan 15 respectively, then the value 2 x x of 2 + p – q is 5. If the volume of a parallelopiped whose (a) 3 (b) 0 (c) 1 (d) 2 ^ ^ ^ 16. Which of the following is NOT TRUE? adjacent edges are a  2 i  3 j  4 k , x 1 ^ ^ ^ ^ ^ ^ (a) lim x  0 (b) lim 1/ x  0 b  i   j  2 k , c  i  2 j   k is 15, then  is x  e x  0 xe equal to sin x cos x 0 0 (c) lim (d) lim (a) 1 (b) 5/2 (c) 9/2 (d) 0 x 0 1  2 x x 0 1  2 x 6. The area enclosed within the curve |x| + |y| = 2 is 10x  10 x (a) 16 sq. unit (b) 24 sq. unit 17. Inverse of the function f  x   x is (c) 32 sq. unit (d) 8 sq. unit 10  10 x b 1  1 x  7. If a < b, then   x  a  x  b  dx is equal to (a) log10 (2 – x) (b) log10   2 a  1 x  2 1 1  2x  b2  a2  b  a  (c) log10  2 x  1 (d) log10   (a) (b) 2 4  2 x 2 2 18. The first three moments of a distribution about 2 are a 3  b3   2 1, 16, -40 respectively. Then mean and variance of (c) (d) (b – a) 2 the distribution are (a) (2, 16) (b) (2, 15) (c) (3, 15) (d) (1, 16) 8. If a   i  j  2k , b  i   j  2k , c  i  j  k , and 19. If a1, a2 ….. an, are in Arithmetic Progression with  a b c   7, then the value of  are common difference, d, then the sum (sin d) (cosec a1   . cosec a2 + cosec a2 . cosec a3 + …… + cosec an-1 . (a) 2, - 6 (b) 6, - 2 (c) 4, - 2 (d) – 4, 2 cosec an) is equal to 9. There are two sets A and B with |A| = m and |B| = n. (a) cot a1 – cot an (b) sin a1 – sin an If |P(A)| - |P(B)| = 112 then choose the wrong option (c) cosec a – cosec a (d) a1 – an 1 n (where |A| denotes the cardinality of A, and P(A) 20.

Solution of the equation denotes the power set of A)  (a) m + n = 11 (b) 2n – m = 1 tan 1 x 2  x  sin 1 x 2  x  1  are (c) 2m – n = 1 (d) 3n – m = 5 2 10. A particle is at rest at the origin. It moves along the (a) 0, 1 (b) 1, - 1 (c) 0, - 1 (d) 0, - 2 x-axis with an acceleration x – x2, where x is the 21. If cosec  - cot  = 2, then the value of cosec  is distance of the particle at time t. The particle next 5 3 4 5 (a) (b) (c) (d) comes to rest after it has covered a distance 3 5 5 4 (a) 1 (b) 1/2 (c) 3/2 (d) 2 1.

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1/ x (a) 9x2 + 8y2 = 72 (b) 8x2 + 9y2 = 72  1  2 x  , x  0 2 2 22. The function f  x    , is (c) 8x + 7y = 56 (d) 7x2 + 8y2 = 56 2 e , x  0   34. For a  R (the set of all real numbers), a  - 1, (a) differentiable at x = 0 (b) continuous at x = 0 1a  2a  ....  na  1 (c) discontinuous at x = 0 . lim  a 1 n   n  1  na  1   na  2   ....   na  n  60 (d) not differentiable at x = 0 Then one of the values of a is x2 y 2 23. If the foci of the ellipse  2  1 and the (a) 5 (b) 8 (c) -15/2 (d) -17/2 25 b 35. Let a be the distance between the lines – 2x + y = 2 2 2 x y 1 and 2x – y = 2, and b be the distance between the hyperbola are coincide, then the   144 81 25 lines 4x – 3y = 5 and 6y – 8x = 1, then value of b2 is (a) 40b  11 5a (b) 40 2a  11b (a) 25 (b) 16 (c) 64 (d) 49 (c) 11 2b  40a (d) 11 2a  40b 24. If a  b  c  a  b  c , then 1 1 1 (a) a and b are collinear 36. If D  1 2  x 1 for x  0, y  0, then D is

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1 1 2 y (a) Divisible by x and y (c) a and c are collinear (b) Divisible by x but not by y (d) a and c are perpendicular (c) Divisible by (1 + x) and (1 + y) Suppose that the temperature at a point (x, y) on a (d) Divisible by (1 + x) but not (1 + y) 2 2 metal plate is T(x, y) = 4x – 4xy + y . An ant, 37. If 0 < P (A) < 1 and 0 < P(B) < 1, and P(A  B) = walking on the plate, traverses a circle of radius 5 P(A) P(B), then centered at the origin. What is the highest (a) P(B|A) = P(B) – P(A) temperature encountered by the ant? (b) P(AC – BC) = P(AC) – P(BC) (a) 125 (b) 120 (c) 0 (d) 25 (c) P (A  B)C = P(AC) P(BC) 2 The function f  x   log x  x  1 is (d) P(A|B) = P(A) – P(B) 38. Let a, b, c be distinct non-negative numbers. If the (a) an even function (b) an odd function vectors ai  a j  ck , i  k and ci  c j  bk lie in a (c) a periodic function (d) neither an even nor an odd function plane, then c is Angles of elevation of the top of a tower from three (a) The Arithmetic Mean of a and b points (collinear) A, B and C on a road leading to the (b) The Geometric Mean of a and b foot of the tower are 30, 45 and 60 respectively. (c) The Harmonic Mean of a and b The ratio of AB and BC is (d) Equal to zero 5 2 (a) 3 :1 (b) 3 : 2 (c) 1 : 2 (d) 2 : 3  39. The value of cot  cos ec 1  tan 1  is The mean of 25 observations are found to be 38. It 3 3  was later discovered that 23 and 38 were misread at (a) 6/17 (b) 3/17 (c) 4/17 (d) 5/17 25 and 36, then the mean is 5 5 1 5 (a) 32 (b) 36 (c) 38 (d) 42 , , ,........ is 40. Which term of the series ? th th 3 4 13 5 In a Harmonic Progression, p term is q and the q (a) 12 (b) 11 (c) 10 (d) 9 term is p. Then pqth term is 41. Coordinate of focus of the parabola 4y2 + 12x – 20y (a) 0 (b) 1 (c) pq (d) pq (p + q) + 67 = 0 is A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition. The probability that it is  5 17   17 5  (a)   ,  (b)   ,  divisible by 3 is 4 2   2 4  (a) 1/3 (b) 1/4 (c) 1/5 (d) 1/6  17 5   5 17  (c)   ,  (d)   ,  Let a  2i  2 j  k and b be another vector such  4 2  2 4 that a.b  14 and a  b  3i  j  8k the vector b = 42. The 10th and 50th percentiles of the observations 32, 49, 23, 29, 118 respectively are (a) 5i + j + 2k (b) 5i – j – 2k (a) 21, 32 (b) 23, 32 (c) 23, 33 (d) 22, 31 (c) 5i + j – 2k (d) 3i + j + 4k A survey is done among a population of 200 people 43. Area of the parallelogram formed by the lines y = 4x, y = 4x + 1, x + y = 0 and x + y = 1 is who like either tea or coffee. It is found that 60% of (a) 1/5 (b) 2/5 (c) 5 (d) 10 the population like tea and 72% of the population like coffee. Let x be the number of people who like 44. If a1, a2, …… an are any real numbers and n is any positive integer, then both tea and coffee. Let m ≤ x ≤ n, then choose the 2 2 n n n n correct option (a) n i 1 ai2   i 1 ai (b) n i 1 ai2   i 1 ai (a) n – m = 56 (b) n – m = 28 2 (c) n – m = 32 (d) n + m = 92 n n 2 a  a (c) (d) None of these   i i i 1 i 1 The eccentricity of an ellipse, with its centre at the origin is 1/3. If one of the directrices is x = 9, then 45. There are two circles in xy-plane whose equations the equation of ellipse is : are x2 + y2 – 2y = 0 and x2 + y2 – 2y – 3 = 0. A point

(b) a and b are perpendicular

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(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 f(x) = x + |x| is continuous for (a) x  (-,) (b) x  (-,) – {0} (c) only x > 0 (d) No value of x The correct expression for cos-1 (-x) is (a) /2 – cos-1 x (b)  - cos-1x -1 (c)  + cos x (d) /2 + cos-1 x x y If cos 1  cos 1   , then 9x2 – 12xycos +4y2= 2 3 (a) – 36 sin2 (b) 36 sin2 2 (c) 36 cos  (d) 36 2 2  x  y If       1,  a  b  and x2 – y2 = c2 cut at a b right angles, then (a) a2 + b2 = 2c2 (b) b2 – a2 = 2c2 2 2 2 (c) a – b = 2c (d) a2 – b2 = c2 2 If ,  are the roots of x – x – 1 = 0, and An = n + n, then Arithmetic Mean of An-1 and An is 1 (a) 2An-1 (b) An 1 (c) 2An-2 (d) None of these 2

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.

ANALYTICAL ABILITY AND LOGICAL REASONING 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 DNN, FQQ, HTT, ……., LZZ (a) JWW (b) JVV (c) JXX (d) IWW Find the synonym that is most nearly similar in meaning to the word : DEBACLE (a) Catastrophe (b) Corker (a) 20 (b) 36 (c) 18 (d) 24 (c) Opulence (d) Dandy 64. Which of the following diagrams correctly represents Fill in the blank : HEC, JGE, LIG, NKI, ------lions, elephants, and animals? (a) ONM (b) PMK (c) KMP (d) HGF Replace the question mark with an appropriate image to complete the analogous pair. (a)

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(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.

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(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.

(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

75. Six books are labelled A, B, C, D, E and F are placed Direction (Qs. 81-85): side by side. Books B, C, E and F have green covers Comprehension: while others have yellow covers. Book A, B and D Direction: Read the following information carefully are new while the rest are old volumes. Book A, B and answer the questions. and C are law reports while the rest are medical Five Dramas A, B, C, D and E have to be staged in 6 extracts. Which two books are old medical extracts hour where 1 hour needs to be given per drama. and have green covers?

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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.

Which hour is a break hour? (a) 5th (b) 2nd (c) 3rd (d) 4th Which is the drama to be staged first? (a) B (b) None (c) D (d) A Which is the drama staged immediately after the break? (a) A (b) B (c) D (d) None of these Which drama is staged immediately after D? (a) E (b) None (c) C (d) B Which drama is staged immediately after E? (a) A (b) E (c) None (d) C The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is (a) 127 (b) 235 (c) 305 (d) 123 At what time, in minutes, between 3 o’clock and 4 o’clock both the needles will coincide each other?

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

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 –10 (c) 1.00 × 10 sec (d) None of these 97. The minimum number of NAND gates required for implementing of Boolean expression, AB  ABC  ABC 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

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

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expression :  X  Y  . X  Y . X  Y

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(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 Direction (Qs. 108-110): (A + C) (AD +AD') + AC + C: Read the following passage carefully and answer the (a) A + C (b) A' + C questions: (c) A + C' (d) A + D 5

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

NIMCET-2022 ANSWERS 1 d 11 a 21 d 31 a 41 c 51 d 61 c 71 a 81 c 91 d 101 b 111 b

6

2 d 12 a 22 a 32 a 42 b 52 a 62 c 72 a 82 a 92 d 102 d 112 a

3 b 13 d 23 b 33 b 43 a 53 a 63 d 73 b 83 d 93 a 103 d 113 b

4 c 14 a 24 d 34 d 44 b 54 b 64 a 74 a 84 c 94 a 104 d 114 d

5 c 15 c 25 c 35 a 45 d 55 d 65 a 75 d 85 c 95 a 105 b 115 d

6 d 16 d 26 b 36 c 46 a 56 a 66 b 76 c 86 a 96 b 106 c 116 c

7 d 17 b 27 a 37 c 47 b 57 d 67 c 77 c 87 d 97 b 107 b 117 a

8 a 18 c 28 c 38 b 48 b 58 b 68 a 78 a 88 b 98 d 108 a 118 a

9 c 19 a 29 b 39 a 49 c 59 b 69 b 79 b 89 d 99 b 109 c 119 a

10 c 20 c 30 c 40 b 50 b 60 b 70 d 80 a 90 a 100 b 110 d 120 a

NIMCET-2022 (SOLUTIONS) 1.

Ans. (d)



1 dt 1 t  2 t c  4 t 4 2

B

A

2

N  h R M R

 5.

2 3 4  a b c  =15  1  2  15 1 2 

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 a b c R   …(1) 2sin A 2sin B 2sin C h Also  tan  …(2) R  h = R tan  a Now put R  from (1) 2sin A a a h tan   tan  cos ecA 2sin A 2

 2(2 – 4) – (3 - 8) + (6 - 4) = 15 (2- 9) + 1 (2 - 9) = 0 9 (2 - 9) ( + 1) = 0    2 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 A B (( x + y = - 2 x–y=-2 x–y=2

Ans. (d) Here xmyn = (x + y)m+n Taking log on both sides m logx + n logy = (m + n) log (x + y) differentiate w.r.t. x m n dy 1  dy      m  n. 1   x y dx x  y  dx 

C x+y=2 D 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 of CD 

dy  m  n n  m m  n      dx  x  y y  x x  y



Ans. (c) As volume of parallelepiped with edges a  2i  3 j  4k , b  i   j  2k , c  i  2 j   k is

C

2.

2 1  x2 x4  c 2

BD 

dy  my  ny  nx  ny  m  x  y    m  n  x    dx  y  x  y x x  y 

 2  0   0  2  8  2 2 2

2

 2  0  0  2  2 2 2

2

So area of rectangle  lb  2 2.2 2  8 sq. unit.

7.

dy y dy  my  nx  my  nx     dx x dx  y  x  y   x  x  y 

Ans. (d) Here a < b So if a < x < b  x – a > 0, x – b < 0  |x – a| = x – a, |x – b| = - (x – b) b

3.

   x  a  x  b dx

Ans. (b) As 33log3 5  33.3 log3 5 1 27  27.3log3 5  27.51  as aloga x  x 5

a

b

b

    x  a    x  b  dx    b  a  dx a

 x  1 dx

  b  a  x a   b  a  b  a   b  a 

2

4.

Ans. (c) 

x3 2 x 4  2 x 2  1

a

b

2

= 8.

1 1  5 3 x2  1 x x dx = dx   5 2 1 2 1 2 2  4 x 2 2  4 x x x x 2 1  4 4 Put 2  2  4  t   3  5  dx  dt x x x x  dt 1 1  3  5  dx  4 x x  

Ans. (a) As a b c   7 Here a    ,1, 2  , b  1,  , 2  c  1,1,1

 1 2     a b c   1  2  7 1 1 1  ( + 2) – (3) + 1 (-2 + 2) = 7  2 + 4 - 12 = 0  2 + 6 - 2- 12 = 0 ( + 6) – 2 ( + 6) = 0 7

14. Ans. (a) As D f / g  D f  Dg   x : g  x   0

( + 6) ( - 2) = 0   = - 6, 2 9.

 Here Dcos1 x   1,1

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.

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 + tan15 = - p P= Product = tan30.tan15 = q 1 …(1) S   2 3  p 3 1 …(2) P . 2 3  q 3 as tan15  2  3 From (1), (2)  2 3  1  2  3   p–q+2  2 3 3   1 2 3   3 1    3 1 3 3 3   3  3 1  1

dv dv dx  .

10. Ans. (c) As a  dt dx dt  dv  a   v  dx  dv Here a  x  x 2  v  x  x 2 dx





v 2 x 2 x3 vdv   x  x  dx    2 2 3 3 x 2 x3 As v  0    0  x  2 2 3 2





11. Ans. (a) According to given condition 1 1 1  1 1  tan  A  B   tan   A  B    tan    C   2 3 2 3 2     1 C  C  1 …(1)  tan     cot 3 2 2 2 3 As we know by NAPIER’s ANALOGY C  A B  a b …(2) tan  cot  2  2  ab 1 1 C From (1) tan  A  B   cot 2 3 2 ab 1   from (1), (2) ab 3 2a 1  3 a 2    By componendo / dividendo 2b 1  3 b 1



16. Ans. (d) Only (d) is cos 0 1 cos x cosh   1 0 lim  lim h  0 1  2h x  0 1  2 x 1  2 0 1

10x  10 x 10x  10 x Apply componendo / dividendo y 1 2.10 x   102 x y  1 2.10 x

17. Ans. (b) y 

 102 x 

12. Ans. (a) 4cos2x + 6sin2x = 5  4cos2x + 6(1 – cos2x) = 5 1   1 = 2cos2x  cos 2 x   cos2 x  cos2 2 4  x  n 



x

1 y 1 y  2 x  log10 1 y 1 y

1 1 y log10 2 1 y

1  1 x  So inverse is y  log10   2  1 x 



4

18. Ans. (c) Here rth moment of distribution is r 1 r   xi  x n 1  xi  2  1 Have 1    xi  2   1  n n x  i  2 1  3  Mean  x  n 1 2 Also 2    xi  2   16 n 1 1 2 2  We have to find   xi  3    xi  2  1 n n 1 2    xi  2   1  2  xi  2    n 



13. Ans. (d) C

(0, b) A

(4, 5)

B (a, 0) x y As intercept form of straight line is   1 a b Also as A is mid. point of line BC a b   4,  5  a = 8, b = 10 2 2 x y  Required line   1  5x + 4y = 40 8 10

8



 11     1  1  2 x x  1   2 x   2x  2  x   1 1    1     1  2    3  x  x  x   2 x   ......  3     2 2 2 23     1     ......   1 2 3           lim  x 0 x (By binomial expansion n  n  1 2 n x  ..... ) 1  x   1  nx  2

1 1 2 2   xi  2   n 1  n   xi  2  n = 16 + 1 – 2(1) = 15 variance = 15 

19. Ans. (a) Here a1, a2, ……., an are in A.P.,  d = a2 – a1 = a3 – a2 = …. an – an-1 Now sin d(cosec a1 cosec a2 + cosec a2 cosec a3 + ….. + cosec an-1 cosec an) sin  d  sin d sin d    .......  sin a1 sin a2 sin a2 sin a3 sin an 1 sin an  

sin  a2  a1  sin a1 sin a2



sin  a3  a2 

sin  a2  sin a3

........ 

sin  an  an 1  sin an 1 sin an

sin a2 cos a1  cos a2 sin a1  ...... sin a1 sin a2

sin an cos an 1  cos an sin an 1 sin an 1 sin an = (cota1 – cota2) + (cota2 – cot a3) + …….. + (cotan-1 – cotan) =cota1-cotan 

20. Ans. (c) tan 1 x  x  1  sin 1 x 2  x  1  Now tan

x  x  1  cos

1

x  x  1  1

1 x  x 1



 2 22   1    ...... 1 2  

 2

 22 1  x  23   2 2 23   1  x 1  2 x       ...   2 3    2 3  lim  x 0 x = finite value

1

1

1  x2  x

x  x  1

1

 cos 1

 1  x  11  x 1  2 x  3   lim 1  2  2 .4 x 2  8 x  .... x 0 x .2 x3 3  

2

 sin 1 1  x  x 2 

23. Ans. (b) Foci of Ellipse and Hyperbola are coinciding. x2 y 2 For   1  b2 = a2 (1 – e2) 25 b2 b2 = a2 – a2e2  b2 = 25 – a2e2 …(1) x2 y 2 1   For Hyperbola 144 81 25  b2 = a2 (e2 – 1)  b2 = a2e2 – a2 81 144 225 … (2)  a 2 e2   a 2e2  9 25 25 25 Put (2) in (1) b2 = 25 – 9  b2= 16

 2

 1  x  x2  1  1  x  x2

1 x  x  x + x = 0  x(x + 1) = 0 x = 0, - 1 2

2

21. Ans. (d) If cosec  – cot  = 2 …(1) As (cosec  - cot ) (cosec  + cot ) = cosec2- cot2 = 1  2(cosec  + cot) = 1 1  cos ec  cot   …(2) 2 Adding (1), (2) 1 5 5 2cos ec  2    cos ec  2 2 4



x 0

x 0



1 is angle between a  b and c but a  b is  to both a , b and if c is also  to a from choice (d)





 a  b and c are parallel  a  b  c  0







IIly a  b  c  a b  c sin  2 n2

22. Ans. (a) At x = 0 lim f  x   lim 1  2 x 



24. Ans. (d) As a  b  c  a  b c sin 1 n1

1 x

1    lim  1  2 x  2 x  x 0  

2

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

= e2 Also f(0) = e2  lim f  x   f  0   e 2  cont at x = 0.

Also if a is  to c  a and b  c are parallel so





are parallel  a  b  c  0

x 0

Now to check differentiability

f '  0   lim x 0

f  x   f  0 x0

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

1

 lim x 0

1  2 x  x  e2 x 9

 T  x, y   4x 2  4x 25  x2  25  x2

Put (2) in (3)

T  x, y   25  3x2  4 x 25  x 2

Put (1) in (2)  y  h 

4 x  2 x  dT  6 x  4 25  x 2  dx 2 25  x 2 4x

 6 x  4 25  x 2 



2

25  x 2  25  x 2   x 2   =0  6x  4  2 25  x

Incorrect sum-Incorrect values +correct values New Mean = n 950  25  36  23  38  =38 25

2

2



 log x  x2  1  log  x  x2  1





= 950

i 1

26. Ans. (b) It’s an odd function As f(x) + f(-x)



 3  1 h  3 :1  3  1 h

28. Ans. (c)  xi  25  38  Incorrect sum

29. Ans. (b) In H.P., If Tp = q, Tq = p  Tpq = 1 (results)

225x2 – 9x4 = 4(625 + 4x4 – 100x2)  25x4 – 625x2 + 2500 = 0  x4 – 25x2 + 100 = 0 (x2 – 20) (x2 – 5) = 0  x2 = 20, 5 dT At x2 = 5  changes sign from +ve to –ve dx  x2 = 5 is a point of maxima Also x2 + y2 = 25  y2 = 20 Now T(x, y) = 4x2 – 4xy + y2  4.5  4 5 20  20 = 40 – 40 = 0



 3 1   h   3  3 

h

3

6 x 25  x 2  4  25  2 x 2  9 x 2  25  x 2   4  25  2 x 2 

 3  1 h

25

 6 x 25  x 2  4  25  2 x 2   0 36 x 2  25  x 2   16  25  2 x 2 

AB x   BC y

x

30. Ans. (c) 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  Required probability 



4 1  5 5

31. Ans. (a) From choices a.b  14, a  b  3i  j  4k

   log  x 2  1  x 2   log  x 2  1  x 2   log1  0    f(x) + f(-x) = 0  f(-x) = - f(x)  f(x) is an odd function.

2

Only (a) choice 5i  j  2k satisfies both 32. Ans. (a) Here n T  

60  200  120 100

72  200  144 100 n (T  C) = 200, n (T  C) = x As n(T) + n(C) – n(T  C)  n(T  C)  min (n(T), n(C)) n C  

27. Ans. (a)

h 30

45

A

B

120 + 144 – 200  n(T  C)  120 64  n(T  C)  120 Given m  x  n  n – m = 120 – 64 = 56

60

C

D 33. Ans. (b)

h  tan 60  3 z

In OCD,

z

h

Here e 

h  tan 45  1 yz

y+z=h

…(2)

h 1  tan 30  In OAD, x yz 3

x  y  z  3h

x2 y 2  1 a 2 b2

a 1 and directrices x   e 3 a a   9   9  a  3  a2  9 …(1) 1 e 3   1 2  Also b2 = a2(1 – e2)  9 1      b2 = 8 …(2)   3   

…(1)

3

In OBD ,

Ellipse with centre at origin (0, 0) is

…(3) 10

x2 y 2 x2 y 2  2 1   1 2 9 8 a b  8x2 + 9y2 = 72

1 0 0  1 1 x 0  1  x 1  y  1 0 1 y Divisible by (1 + x), (1 + y)

 Ellipse is

34. Ans. (d)

1  2  ....  n  a

lim

n 

 n  1

a 1

a

a

 na  1   na  2   ....   na  n  



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.

1 60

a   1 a  2 a n  n a        .....      n   n   n    lim a 1 n  1  2 n   1    n a 1 .n 1    a     a    .....   a    n  n n   n   

a

a

38. Ans. (b) If vectors ai  a j  ck , i  k , ci  c j  bk are in same plane. a a c  1 0 1 0 c c b

a

1 2 n       .....    n n  n lim a 1 n  1  2 n   1    1     a     a    .....   a    n n n n        

 lim

n 

a a a 1  1   2  n    .....         n   n   n  n 

 1 1    n

 lim

n 

a 1

1  1  2 n    a     a    .....   a    n   n  n n  

1 n r   n r 1  n   1 1    n

 a(-c) – 1 (ab – c2) + c(a) = 0  ab = c2  a, c, b in G.P.  c is G.M. of a, b.

a 1

39. Ans. (a) 5 2  cot  cos ec1  tan 1  3 3  5 3 Now cos ec1  tan 1 3 4 5 2   cot  cos ec 1  tan 1  3 3    1 3 1 2   cot  tan  tan  4 3 

a

1  r  a   n r 1  n n

1

1

x a 1 1 a  1 a 0  1   1 1 x2 0  a  x dx ax  2 0 a  2 a  x dx



2

 a  1 2a  1



5

3

4

  3 2   1  4  3    cot  tan   3 2    1  .     4 3  

1  (a + 1) (2a + 1) = 120 60

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

  9  8   1  12    1  17    cot  tan     cot  tan    6  6        12   

35. Ans. (a) Here lines are – 2x + y = 2  2x – y = - 2 and 2x – y = 2 and distance between these two is a  2   2    4 …….(1) a 22  12 5

 6  6   cot  cot 1     17 17   

40. Ans. (b) As sequence is

4x – 3y = 5 1 and 6y – 8x = 1  4 x  3 y   2  5 2  11 ………(2) b 2 4  32 10 other two lines are

5 5 5 5 , , ,...... , 3 4 5 13

 It’s 11th term.

41. Ans. (c) Here PARABOLA is 4y2 + 12x – 20y + 67 = 0 4y2 – 20y = - 12x – 67.

4(y2 – 5y) = - 12x – 67 25   4  y 2  5 y    12 x  67  25 = - 12x - 42 4 

From (1) and (2)  11 5a  40b

2

36. Ans. (c)

5 42     4  y    12  x   2 12    

1 1 1 D= 1 2  x 1 1 1 2 y

2

5 7     y    3  x   2 2   2 It’s of type Y = 4a X  4a = - 3  a = - 3/4 11

 focus is (a, 0)  X = a, Y = 0  X = -3/4, Y = 0 7 17 3 5 5  x   ,y 0 x   ,y  2 4 4 2 2  17 5    ,   4 2

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

47. Ans. (b)

48. Ans. (b) As cos 1

th

  n 1 Pk   k    term   100    5  1 P10  10  .6 th item which is 23.

100  n 1  6  P50  50    50    3 rd item. = 32.  100   100 

4 1

 n   n a    ai  i 1  i 1  n

xy 

 4  x 9  y 

2

2

 

6 2



2

6

6 cos   xy  4  x

2

9  y2

 xy  6 cos   4  x 2 9  y 2  x 2 y 2  12 cos  xy  36 cos 2    4  x 2  9  y 2 

 9 x 2  4 y 2  x 2 y 2  36  9 x 2  4 y 2  12 xy cos   36  36 cos 2   36sin 2 

49. Ans. (c) Here curves are 2 2  x  y 2 2 2       1, x  y  c a b Now m1 = slope of first curve x y dy 2  x   y  1 dy 0  0 2  2    2 . a b dx a  a   b  b dx

44. Ans. (b) By result ; m  [0, 1]

a12  a22  ........  an2  a1  a2  ....  an    n n  

 4  x  9  y  

 cos  

1  5

a1m  a2m  ........  anm  a1  a2  ....  an    n n   Take m = 2

 xy  cos    6  1

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  c  d  c  d2  is 1 1 2 a1b2  a2b1 So in our case = Area 

x y  cos 1   2 3

2 2 x y  x  y   cos 1   1   1      2 3 2  3   

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

1  0 1  0 

cos1   x     cos1 x

m

dy x b2  . 2 …(1) dx y a IIly for second curve x2 – y2 = c2 dy  2x  2 y 0 dx dy x x   m2  …(2) dx y y m1 

2

2

2 i

If two curves cut at 90 x b2 x .  1  m1m2 = - 1   y a2 y

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

x2 a2  …(3) y 2 b2 Now first equation is b2x2 + a2y2 = a2b2 …(4) and second equation is x2 – y2 = c2  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 a2 Put x 2  2 y 2 in above b  a2   c 2b2  a 2b2  2 y 2   a 2 c 2  a 2b2 y 2  0 b  

(0, 3)

(0, 1) (0, 0) (0, -1) 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 =   1  Out of total = 4  required prob.  4 4







(a2c2 – a4) + (a2c2 + a2b2) = 0 12



2a2c2 = a4 – a2b2  2c2 = a2 – b2

58. Ans. (b) 6, 11,

50. Ans. (b) As ,  are roots of x2 – x – 1 = 0   +  = 1,  = - 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  An 1 An 1  A.M. of An, An-1  n  2 2 51. Ans. (d) VTU OR UTV, VW OR WV, (XWX)

+ 5

21, 36,

56,

81

+10 +15 +20 +25

59. Ans. (b)

+

-

+

Manu

-

Portrait XUTVW Hence, T is sitting between U and V. 52. Ans. (a) +3

+3

+3

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).

+3

D N N, F Q Q, H T T,…J W W, L Z Z +2

+ +2 + +2 + +2 + 3 3 3 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).

61. 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

54. Ans. (b)

+ + + + 2 2 2 H E C, T G E, L I G, N K I, 2P M K + 2+ 2

+ 2+ 2

+ 2+ 2

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

63. Ans. (d) A B C D E

+ 2+ 2

+ J

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

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

Sailing The action of sailing in a ship or boat.

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

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) Lions

Elephants

Animals 65. Ans. (a) 122, 144, 166, 188, 210

+2 2

57. Ans. (d) 13

+22 +22 +22

66. Ans. (b)

Hence 38 is wrong number.

+ 1

+ 1

+ 1

+ 1

+ 1+ 1

+ 1+ 1

+ 1+ 1

+ 1+ 1

J A K, K B L, L C M, M D N, N E O

74. Ans. (a) 30 30 - 35 = -5 -5 - 40 = -45 -45 - 45 = -90 -90 - 50 = -140 -140 - 55 = -195 -195 - 60 = -255 Hence, -145 is wrong. -140 should have come in place of -145

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.

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

70. Ans. (d) Athletes Woman Teacher

 Play

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

71. Ans. (a) Mangoes Golden



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

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

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

72. Ans. (a) Open Young minded scientists



78. Ans. (a) Arts : Commerce E  14  38 : 18  42 19 : 27

Superstitious

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

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

80. Ans. (a) I and III follows.

73. Ans. (b) 37 2, 5, 10, 17, 26, 38, 50, 65 +3

+5

+7 +9 +12 +12 +15 +11 +13 14

Solutions (81 to 85): Days Dramas 1 B 2 D 3 Break 4 C 5 A 6 E

After 8 years 3 : 6 Present age of mother = 40 yrs 91. Ans. (d) 10d > 27 Take log both sides log1010d > log102n  d > nlog102 92. Ans. (d)

93. Ans. (a) 81. Ans. (c)

82. Ans. (a)

84. Ans. (c)

85. Ans. (c)

83. Ans. (d)





 x  y   x  y  x  y 

 x  y   x  y  x  y 

x y  y  xy

 y  y 1

x  x y = x  y  xy

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

94. Ans. (a) AB + AB' + A'C + AC A(B + B') + C(A' + A) A + C independent of B 95. Ans. (a) (A + C) (AD + AD') + AC + C (A + C) (A(D + D') + C(A + 1) (A + C) (A) + C = AA + AC + C =A + C(A + 1) =A + C

2 2 87. Ans. (d) M   30 H   M   30  3 11 11 180 4 min part 3.  = 16 11 11

96. Ans. (b) Clock Period  88. Ans. (b) C+ | A B | E- D+ From diagram , we can see A is the uncle of D. 89. Ans. (d)

= x y  xy  x y



1 cycle per second

1 100 1010   4 1010 6 25 25 10

97. Ans. (b) 98. Ans. (d) 99. Ans. (b)

6  3 10  4   x = 1800 270 x

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

100. Ans. (b)

101. Ans. (b) 102. Ans. (d)

103. Ans. (d)

104. Ans. (d)

106. Ans. (c)

107. Ans. (b) 108. Ans. (a)

109. Ans. (c)

110. Ans. (d) 111. Ans. (b)

105. Ans. (b)

90. Ans. (a)

112. Ans. (a) 113. Ans. (b)

114. Ans. (d)

115. Ans. (d)

117. Ans. (a)

116. Ans. (c)

118. Ans. (a) 119. Ans. (a) 120. Ans. (a)

15