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Quant Without Maths

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A revolutionary way of attacking CAT Quant without actually knowing it... brought to you by Career Launcher

A revolutionary way of attacking CAT Quant without actually knowing it... brought to you by Career Launcher

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Quant Without Maths Presentation Transcript

  • 1. Quant For Dummies Nitin Jindal
  • 2. You should be attending this session if in a Quant class your state resembles any of the following:
  • 3. Statutory Warning This session is not meant for those who are good at Quant!
  • 4. Session Objectives • To work together in scoring marks in the Quant section • To demonstrate that a relaxed state of (fearless) mind can help us significantly while attempting quant • That all of the above is possible without really knowing Maths
  • 5. Please stand, hold your right hand to your heart and repeat after me I pledge to never be scared while attempting the quant section. Repeat! Where the mind is without fear… …Rabindranath Tagore
  • 6. CAT 2005 Q – 26 Let x = 4+ 4+ 4– 4 – ...to inf inity . Then x equals ( 2 marks ) ₩ 13 – 1 g. 3 b. 2 │ ₩ 13 + 1 13 c. d. 2 │
  • 7. CAT 2005 Q – 26 Let x = 4+ 4+ 4– 4 – ...to inf inity . Then x equals ( 2 marks ) ₩ 13 – 1 g. 3 b. 2 │ ₩ 13 + 1 13 c. d. 2 │ 9 =3 16 = 4 The first term is Rt4 which means > 2 13 = 3.5 ish The second term becomes Rt(4+2) which means it must be less than 3
  • 8. First Commandment When I read a question, I will also always read the options too!
  • 9. Application Question 1 Q – 67 If ab × cd = 1073 and ba × cd = 2117, find the value of (ab + cd) given that ab, ba and cd are all two digit positive integers. 1. 66 2. 65 3. 63 4. 95 Hence , answer is option 1
  • 10. Application Question 2 1  x  4 Q. 64. Find the maximum value of |30 + 9x – 3x2|, where 93 147 1. 2. 4 4 4. 18 3. 30 Let x = 0 Let x = 1 x = 0  value =30 x = 1  value =36 Hence , answer is 147/4
  • 11. First Commandment When I read a question, I will also always read the options too! Always remember one of the options has to be right!
  • 12. Second Commandment Geometry is easy! 3. Geometry questions are always based on b) Angles c) Length of sides d) Area / Volume e) Ratios of the above
  • 13. Second Commandment Steps to solving Geom Qs c) Check how close are the options? b) If no figure is given I will draw it roughly to Scale c) If a figure is given, I will check if it is drawn to scale, if not I’ll redraw
  • 14. Second Commandment My CAT admit card! • I have a ruler. = 5 angles • Protractor 45 60 30 75 15
  • 15. 75° 60° 45° 33.7° 30° 15°
  • 16. Application Question 3 In the given figure, ABCD is a rectangle which is divided into four equal rectangles by PS, QT and RU. If BC = 3 cm and AB = 8 cm then MN is P Q R A B M O N D C U S T 1. 0.33 cm 2. 0.66 cm 3. 1 cm 4. 1.25 cm
  • 17. Application Question 3 P Q R A B M 3 cm O N D C U S T 1. 0.33 cm 2. 0.66 cm 3. 1 cm 4. 1.25 cm RU = BC = 3 and MN is the largest part of RU Hence, answer is 1.25 cm
  • 18. CAT 2005 Q – 9 What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm? ( 1 mark ) a. 1 or 7 b. 2 or 14 6 cm 8 cm c. 3 or 21 d. 4 or 28 10 cm Draw and measure!
  • 19. Mock 2, Question 16 In the given figure, P is a point on the circumcircle of DABC. From P, perpendiculars PD, PE & PF are drawn BPF = 30° on the sides AC, BC and AB respectively. If , then find the measure of <DPC F P B 3.30° 2. 60° 3. 40° 4. 50° E 5. 20° C A D
  • 20. Our Method F x P B E C A D Hence , answer is option 1
  • 21. Mock 3, Question 48 In the figure given below, ABCD is a square. It is given that BP : PM = 4 : 1 and DP : PN = 3 : 2. Find the ratio of lengths of CN to CM. A B N P C D M 1. 3 : 4 2. 3 : 5 3. 4 : 5 4. 2 : 3 5. 1 : 2 Hence , answer is option 4
  • 22. CAT 2005 Q – 22 In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is ( 2 marks ) M ( ) C a. 2 2 – 1 b. 2 2 – 1 2 (3 ) ( ) 2 –1 d. 2 2 – 1 c. O E A B 2 3 L H D F G N
  • 23. CAT 2005 M C b. ( 2 ) a. 2 2 – 1 2 –1 2 (3 ) ( ) 2 –1 E O c. d. 2 2 – 1 A B 2 L 3 H D F G N Hence , answer is option b
  • 24. CAT 2005 Q – 24 P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR? ( 2 marks ) ( ) ( ) b. 2r 2 + 3 a. 2r 1 + 3 ( ) d. 2r + 3 c. r 1 + 5
  • 25. CAT 2005 P Q S R
  • 26. Third Commandment Wherever possible I will assume a particular case for a general case i.e. Hexagon implies a regular hexagon Quadrilateral / Parallelogram=Square Triangle=Equilateral triangle or Right Angled Triangle With sides? Usually with sides 3,4,5 or its multiples.
  • 27. CAT 2005 Q – 24 P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR? ( 2 marks ) ( ) ( ) b. 2r 2 + 3 a. 2r 1 + 3 ( ) d. 2r + 3 c. r 1 + 5
  • 28. CAT 2005 P Q P Q S R R S Hence , answer is option a
  • 29. Application Question 4 An ant starts from point A and crawls along the surface of the cylinder to reach the point B, vertically above A. The path followed by the ant is equal to that of 4 identical spirals as shown in the figure. Find the radius of the circle that circumscribes a square such that the perimeter of the square is equal to the distance traversed by the ant. The diameter of the cylinder is 3/π units and the height is 16 units. 5 5 B 1. 2. 2 2 16 5 4. 5 3. 3 2 A 3 /
  • 30. Application Question 4 contd. In the worst case the ant travels 16 units (the height of the cylinder).  Perimeter of square is atleast 16 and side atleast 4 Now move to the options and eliminate Hence , answer is option 2
  • 31. Fourth Commandment Algebra is my friend! How? 3. In all “series” questions, I will add first 2-3 terms, & eliminate options. 5. In quadratic equations, I will try to assume a simple equation satisfying the given options & eliminate options.
  • 32. Fourth Commandment 1. In log questions, I will assume base 10 and remember that: c) Log is not defined for negative numbers e) Log 10=1, Log 100=2 & so on. g) Base can never be 1 or negative 4. In all a,b,c or x,y,z questions or where n terms are given I will always assume them to be easy numbers like 0,1,-1,2, etc.
  • 33. Question 58 If x = a(b – c), y = b(c – a) and z = c(a – b), then find the value of 1 ←x 2 y 2 z 2 that xyz ᄍ 0) . (Given ↑+ + . abc ↑ yz xz xy → 27 1 1. 2. abc abc 3 9 3. 4. abc abc 1 5. 3abc
  • 34. Our method Expression will hold for all values of x, y & z. So put x = y = z = 1 3 Hence , answer is abc
  • 35. Application Question 4 Q – 77 The natural number 555...5 consisting of 65 fives, is equal to ₩ 65  1 10 I. 5ᄡ │9 ( )( ) II. 5 ᄡ 1 + 1013 + 1026 + ... + 1065 1 + 10 + 102 + .... + 1012 ( )( ) III. 5 ᄡ 1 + 105 + 1010 + 1015 + ... + 1060 ᄡ 1 + 10 + ... + 104 1. Only I 2. Only I and II 3. Only I and III 4. All three There is no rush, take your time answer is option 3
  • 36. Mock 1, Question 64 Find all the values of p, such that 6 lies somewhere between the roots of the equation x 2 + 2 ( p – 3 ) x + 9 = 0 . (‘x’ is a real number) 3 1. p <  2. p > 6 4 3 3 5. p >  4 0<p< 3. 0 < p < 6 4. 4 Let p = 3 Let p = -1 p = 3  x is imaginary and solve! There is no rush, take your time answer is option 1
  • 37. Application Question 5 x2 + (x + 1)(x + 2)(x + 3)(x + 6) = 0, where x is a real number, If then one value of x that satisfies this equation is: 3 3 2. ( 3 + 3) 1. 3+ 3 3. 4. 0 Hence , answer is option 2
  • 38. Mock 2, Question 1 1 1 1 1 + + + ....... + Find the sum of the series . log3 9 log9 9 log27 9 log n 9 3 n(n + 1) 2 1. 2. n(n + 1) 2 n(n + 1) n(n + 1)(2n + 1) 4 3. 4. 5. n(n + 1) 4 12 Let n = 1 Now put n= 2 In both Q & As and solve! Options 1,2 & 5 are eliminated
  • 39. Application Question 6 Q – 50 Sum of the first n terms of a geometric progression n ₩1 Sn = a. +b is given as │3 If the sum of infinite terms of this series is unity, then Let n = 0 1. a+ Alpha + beta = 0 2.3a+b = – 1 b= 1 a+ = –1 3. 4.2a+3b 1 = 2b Let n = infinity and solve! Apply the 4th commandment
  • 40. CAT 2004 Q – 44 If f(x) = x3 – 4x + p , and f(0) and f(1) are of opposite signs, then which of the following is necessarily true ( 1 mark ) 1. –1 < p < 2 2. 0 < p < 3 3. –2 < p < 1 4. –3 < p < 0 Put x = 0 Can p be negative ? Put x = 1
  • 41. CAT 2004 a b c = = = r then r cannot take Q – 47 If b+c c +a a+b any value except: (1mark) 1 1. 2. –1 2 1 1 3. 2 or – 1 4. – or – 1 2 Put a,b,c = 1 Now test for r = -1 Let a=-1, b & c = ½
  • 42. CAT 2004 Q – 58 The total number of integers pairs (x, y) satisfying the equation x + y = xy is ( 1 mark ) 1. 0 2. 1 3. 2 4. None of the above 0,0 & 2,2
  • 43. CAT 2002 Q- 58 The area of the triangle whose vertices are (a, a), (a + 1, a + 1) and (a + 2, a) is a. a3 b. 1 c. 2a d. 21/2 Let a = 0 & plot the As there are two values values and calculate without a, be ultra careful
  • 44. CAT 2005 3065 – 2965 R= Q – 8 If , then ( 1 mark ) 64 64 + 29 30 a. 0 < R ᆪ 0.1 b. 0.5 < R ᆪ 1.0 c. 0.1 < R ᆪ 0.5 d. R > 1.0 Check for lesser powers and generalize
  • 45. Fifth Commandment 1. In questions of probability involving ‘n’ trials I’ll assume n=1 Probability of an event that has not happened is 0 2. I will check all those options where both p and 1-p are given
  • 46. Fifth Commandment Probability of anything lies between the best and the worst case scenarios therefore, I will simplify the question by taking best and worst case scenarios for anything to happen or not to happen
  • 47. Application Question 11 Two dice are thrown together n times in succession. Find the probability of obtaining ‘1’ on the top face of both the dice in at least one of the throws. n 1 ₩ 35 1. 2. 1  36 │ 36 n n ₩1 ₩ 35 3. 4. │ 36 │ 36
  • 48. MOCK 1,Question 54 Bag A contains 6 white balls and 4 black balls and bag B contains 3 white balls and 2 black balls. A white ball is picked from bag A and put into bag B. Then, three balls are picked from bag B and put into bag A. Find the probability that a ball picked now from bag A is black. 1. 1/4 2. 1/3 3. 7/12 4. 5/12 5. 11/24
  • 49. Our method Best & the worst case scenario. The probability will lie between them. Best case for the ball to be black:- The 2 balls picked from bag B and put into bag A are both black, with the 3rd being white. In that case bag A will contain 6 black and 6 white balls and hence the probability of picking a black ball would be 6/12. Worst case for the ball to be black:- All the 3 balls picked from bag B and put into bag A are white. In that case bag A will contain 8 white and 4 black balls and hence the probability of picking a black ball would be 4/12. This eliminates many options like 1,2 & 3.
  • 50. Sixth Commandment Graph questions are not as ugly as they look, in fact in CAT, like in life, looks are deceiving. The more intimidating they look, the easier they are!
  • 51. Question 24 The graph of |y| against |x| is shown below: |y | (0 , 2 ) |x | O (3 , 0 ) (0 , 0 ) Which of the following shows the graph of (y – 1) against (x – 1)? [Note: The graphs are not drawn to the scale]
  • 52. Question 24 Cont. (y – 1 ) (y – 1 ) 2 1 C (1 , 3 ) (0 , 7 /3 ) D B (0 , 1 /3 ) B (4 , 1 ) (– 1 /2 , 0 ) C A ( 5 /2 , 0 ) (x – 1 ) (– 1 /2 , 0 ) E A ( 5 /2 , 0 ) ( x – 1 ) F (4 , – 1 ) F (0 , – 1 /3 ) (0 , – 7 /3 ) D E (1 , – 3 ) (y – 1 ) (y – 1 ) 4 3 (– 1 , 3 ) B A (0 , 7 /3 ) B (0 , 1 /3 ) (x – 1 ) (– 5 /2 , 0 ) C ( – 4 ,1 ) C (x – 1 ) A (1 /2 , 0 ) (– 4 , – 1 )D (– 5 /2 , 0 ) D F (1 /2 , 0 ) F ( 0 , – 7 /3 ) (– 1 , – 3 ) E (0 , – 1 /3 ) E 5. None of these Hence , answer is option 3
  • 53. Question 41 Which of the following graphs, represents the curve y = 2|x| + 3[x]? Here, [x] represents the greatest integer function. [Figures not drawn to the scale] y 1. 2 3 4 y y y 8 8 8 6 7 7 7 5 6 6 6 5 4 5 5 4 4 4 3 3 3 3 2 2 2 2 1 1 1 1 x x x O1 2 3 4 –4 –3 –2 –1 x O1 2 34 –4 –3 –2 –1 –4 –3 –2–1 O1 2 3 4 –1 –4 –3 –2 –1 O 1 2 3 4 –1 –1 –1 –2 –2 –2 –3 –2 –3 –3 –4 –3 –4 –5 –5 –4 –4 –6 –6 –5 –7 –7 –6 –8 –7 5. None of these Hence , answer is option 1
  • 54. Seventh Commandment Bring your relaxed brain with you for the exam, there are some really easy questions we just need to be patient with.
  • 55. Application Question 9 Q – 59 A function f is defined for all whole numbers n by the following relation f(n + 2) + f(n) – 2f(n + 1) = 0 If f(16) = 4 and f(24) = 7, what is the value of f(16 + 24)? 5. 10 2. 13 3. 170 4. 3340
  • 56. MOCK 1 Question 51 Two identical Re. 1 coins are kept on a table touching each other as shown in the figure below. One of the coins is fixed on the table whereas the other coin rolls (without sliding) along the periphery of the fixed coin, touching it at all times. How many complete rotations has the rolling coin made, when it reaches its initial position again for the first time? 1. 1 2. Between 1 and 2 3. 2 4. Between 2 and 3 5. 3
  • 57. Our method Have you got 2, Re 1 coins in your pocket? To nikaalo na, question solve karna hai. . .
  • 58. Mock 3, Question 47 On giving 3 pencils free with every 5 pens bought, a shopkeeper makes a profit of 20% and on giving 6 pencils free with every 2 pens bought, he suffers a loss of 25%. Find the approximate profit percent made by the shopkeeper when he gives 4 pencils free with every 6 pens bought, if the selling price of 1 pen remains the same. (Assume that the pencils are identical and the same applies to the pens) 1. 18% 2. 20% 3. 24% 4. 16% 5. 22% Hence , answer is option 1
  • 59. Mock 1, Question 55 All natural numbers that give remainders 1 and 2 when divided by 6 and 5, respectively, are written in ascending order, side by side from left to right. What is the 99th digit from the left? 1. 2 2. 7 3. 1 4. 9 5. 0 7,37,67,97,127,157,187,217,247,277,307,337,367,397,427 ,457,487,517,547,577,607,637,667,697,727,757,787,817, 847,877,907,937,967,997,1027 Hence , answer is 0
  • 60. Mock 2, Question 5 | 3x + 7 | ᄈ | 5x + 6 | . Find the range of values of x, where (‘x’ is a real number) 1 6 1 13 1 3. x  2.  < x  1.  x 2 5 2 8 2 13 6 13  x< 4. x   5. 8 5 8 Put x=0, eliminates 3,4,5
  • 61. CAT 2005 Answer the questions on the basis of the information given below. Ram and Shyam run a race between points A and B, 5 km apart, Ram starts at 9 a.m from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed, Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed. (1 mark each) Q – 6 At what time do Ram and Shyam first meet each other? a. 10 a.m b. 10:10 a.m c. 10:20 a.m d. 10:30 a.m. Q – 7 At what time does Shyam over take Ram? a. 10:20 a.m b. 10:30 a.m c. 10:40 a.m d. 10:50 a.m
  • 62. Mock 3, Question 39 Two distinct integers m and n satisfy the equation m + n – mn = – 36. Which of the following cannot be the  2m  n  value of  ?  m+n  17 37 1.  2. 2 3. 20 20 4. – 1 5. None of these Hence , answer is option 5
  • 63. CAT 2004 1 Q – 48 Let y = 1 2+ 1 3+ 1 2+ 3 + ... What is the value of y? (1 mark) 11  3 11 + 3 1. 2. 2 2 15 + 3 15  3 3. 4. 2 2 Can’t we do it ?
  • 64. CAT 2003 Leak Q – 133 In the figure below, ABCDEF is a regular hexagon and . FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF? 1 1 A B a. b. 6 12 1 1 F C c. d. O 24 18 E D
  • 65. CAT 2003 Leak A B F C Y E D
  • 66. CAT 2003 Retest Q – 81 Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF? 1 1 a. b. 2 3 5 2 c. d. 6 3
  • 67. CAT 2003 Retest A B F C E D
  • 68. CAT 2002 Q - 57 In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE =7cm2 ; EC = 3(BE). The area of ABCD (in cm2) is A D B C E a. 21 cm2 b. 28 cm2 c. 42 cm2 d. 56 cm2
  • 69. CAT 2001 D C Q–8 In the above diagram, ABCD is a rectangle with AE = EF = FB. What is the ratio of the areas of DCEF and that of the A E F B rectangle? 1 1 a. b. 8 6 1 c. d. None of these 9
  • 70. Finally I will not be afraid of any question in quant, however scary it looks M B C F 16 O E P A B B L H D F E G C A A N D 3 /
  • 71. Application Question 2 ABCD is a square. Arc AC and BD are drawn on the square ABCD with centres at D and C respectively. Find the ratio of area of the shaded region to the area of the square ABCD. 1. 2. 2 3 3   3 4 3 4  3. 3  3 3 4.  2 2 32 A B 3 = 1.732 or 3< 4 =2 Pi = 22/7 or 3.14 Hence, answer is option 1 C D
  • 72. Desperate Quant • This is pure magic • You don’t need anything but power of sight • This applies to all those question where options either have fraction or involve square roots or complex functions • This is very useful to shortlist options that appear right, which we can then put back into the question to double-check It can give wrong results when used in isolation
  • 73. Quick Quant (a + b + c) (b + c – a) = (c + a – b) (a + b – c) where a, b and c are the sides of the triangle ABC. Find the area of the triangle ABC. 1 (a + b – c) (a + b + c) 1 (a – b + c) (a + b + c) 1. 2. 4 4 1 1 3. (b + c – a) (a + b + c) 4. (b + c – a) (a + b – c) 4 4
  • 74. CAT 2005 Q – 22 In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is ( 2 marks ) M ( ) C a. 2 2 – 1 b. 2 2 – 1 2 (3 ) ( ) 2 –1 d. 2 2 – 1 c. O E A B 2 3 L H D F G N
  • 75. CAT 2005 Q – 24 P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR? ( 2 marks ) ( ) ( ) 2r 2 + 3 a. 2r 1 + 3 b. ( ) 2r + 3 r 1+ 5 c. d.
  • 76. CAT 2003 Retest Q – 53 Which of the following represents the numeral for 1995? I. MCMLXXV II. MCMXCV III. MVD IV. MVM a. Only I and II b. Only III and IV c. Only II and IV d. Only IV
  • 77. CAT 2003 Retest Q – 55 Sprinter A traverses distances A1A2, A2A3, and A3A1 at an average speeds of 20, 30 and 15 respectively. B traverses her ( ) entire path at a uniform speed of 10 3 + 20 . C traverses distances ( ) ( ) 40 40 C1C2, C2C3 and C3C1 at an average speeds of 3 +1 , 3 +1 3 3 and 120 respectively. All speeds are in the same unit. Where would B and C be respectively when A finishes her sprint? a. B1, C1 b. B3, C3 c. B1, C3 d. B1, Somewhere between C3 and C1
  • 78. CAT 2003 Retest Q – 56 Sprinters A, B and C traverse their respective paths at uniform speeds of u, v and w respectively. It is known that u2:v2:w2 is equal to Area A: Area B: Area C, where Area A, Area B and Area C are the areas of triangles A1A2A3, B1B2B3, and C1C2C3 respectively. Where would A and C be when B reaches point B3? a. A2, C3 b. A3, C3 c. A3, C2 d. Somewhere between A2 and A3, Somewhere between C3 and C1
  • 79. CAT 2003 Retest Q – 76 Let S1 be a square of side a. Another square S2 is formed by joining the mid-points of the sides of S1. The same process is applied to S2 to form yet another square S3, and so on. If A1, A2, A3, … be the areas and P1, P2, P3, … be the perimeters of S1, S2, P1 + P2 + P3 + L S3, …, respectively, then the ratio equals A1 + A 2 + A 3 + L ( ) ( ) 2 1+ 2 2 2 2 a. b. a a ( ) ( ) 2 1+ 2 2 2 2+ 2 a c. d. a
  • 80. CAT 2003 Leak Q – 130 There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12 square centimeters then the area (in square centimeters) of the triangle ABC would be 9  12 a. b.  93 63 c. d.  
  • 81. CAT 2004 a b c = = = r then r cannot take Q – 47 If b+c c +a a+b any value except. ( 1 mark ) 1 1. 2. –1 2 1 1 3. 2 or – 1 4. – or – 1 2
  • 82. CAT 2004 Q – 71 If group C contains 8 questions and group B carries at least 20% of the total marks, which of the following best describes the number of questions in group B? ( 2 marks ) 1. 11 or 12 2. 12 or 13 3. 13 or 14 4. 14 or 15
  • 83. CAT 2002 Q – 52 In DABC, the internal bisector of A meets BC at D. If AB = 4, AC = 3 and A = 60 , then the length of AD is 12 3 a. 2 3 b. 7 63 15 3 c. d. 7 8
  • 84. Please stand, hold your right hand to your heart and repeat after me I pledge that I will stay relaxed and focused while attempting the CAT paper
  • 85. Please stand, hold your right hand to your heart and repeat after me I pledge that I will stay relaxed and focused while attempting the CAT paper
  • 86. nitin.jindal@careerlauncher.com 9881858764