Maths by Amiya XAT 1-130

March 27, 2018 | Author: ramsan1991 | Category: Circle, Triangle, Elementary Mathematics, Elementary Geometry, Geometry
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3E Learning, 3rd Floor, Anand Complex, Near Lalpur PS, H.B. Road Ranchi, 095 34 002244 Maths By Amiya XAT 1-130 1. 3456 7ℎ9 :;457 ;< 4579=>9?74;5 ;< :@4= ;< >7=@4Aℎ7 B459> 2C D − 5CF + 3F D − 9C + 12F + 9 = 0 a. (0,-1) b. (3,1) c.(9,5) d. (6,3) 2. Total number of integral solution of family of equation C + F = 1 − O, C P + F P = 1 − O D , is a. 3 b. 4 c. 6 d. None of these 3. Total number of real solution pair of (x , y) if C = C D + F D & F = 2CF. a. 3 b. 4 c. 1 d. None of these(???) 4. If (1 + 2C)DS = @S + @T C + @D C D + @P C P + ⋯ + @TV CTV + @DS C DS then 3@S + 2@T + 3@D + 2@P + ⋯ + 2@TV + 3@DS = ? a. W∗PYZ [ P b. D e. None of these W∗PYZ \ P D c. W∗PYZ \T d. D W∗PYZ [T D ] 5. If 0. @25@25@25. . . . . . = ^ ; where b and c are co-prime then what is the difference of maximum and minimum value of (a+b+c). a. 1832 b. 1761 c. 71 d. None of these 6. If c = a. 9 d TSTY e! , What is the value of integral "k" for maximum value of N b. 10 c. 11 d. 101 e. None of these 7. Mr. Pandey has 10 different keys of which only one can open the lock. He tries to open the lock by using the keys one after another (keeping aside the failed ones). What is the chance that 7th key will work. a. 7/10 b. 1/10 c. 1/2^7 d. None of these 8. If three players play a tournament of 9 matches, such that there is only one winner of any match, and winner is awarded with +2 points and others two with -1 (minus 1). In how many ways they can play the tournament (of 9 matches) such that after the tournament they finish each with zero point. a. 84 b. 1680 c. 0 d. P(9,3)*P(6,3) e. None of these Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya ©AMIYA KUMAR 3E Learning, 3rd Floor, Anand Complex, Near Lalpur PS, H.B. Road Ranchi, 095 34 002244 9. In a triangle. two vertices are (2,3) & (4,0) and its circumcentre is (2,y) then what would be circum radius of the triangle is a. l b. √5 D[√TP d. c. 2 TP e. None of these l axis at B. ((m,n) is coordinates of point C( …) on the given line such 10. If 2x+3y=18 intersect y-axis that AB=AC , where A=(10,10). Then 13*(m+ 13*(m+n)=? a. 130 b. 132 c. 122 d. 78 e. None of these 11. How many positive integral value of m are possible , if than 1. a. 8 d. 24 c. 22 b. 20 TTT € is a terminating decimal greater e. None of These 12. Mr. Sood has three 1 rupee coins and three other coins of any integral (>1) value but of same denomination. Then what is the least amount he cannot give to any one with the help of any three coins c. 9 d. 10 e. None of These T a. 7 b. 8 13. If we have to fence a 40m* m*30m ground to keep people 7 meter away from the ground, then find the minimum length of the required fence. fence.________________? 14. Ten basket, numbered 1-10 10 are carrying balls. Each balls weighs either 999gm or 1000gm and each basket sket carries only balls of equal weights. The combined weight of 1 ball selected from first basket, 2 balls from second basket, 4 balls from third basket and so on, and 2V balls from tenth basket is 1022870. Then the baskets that have lighter balls are a. 1,3,5 b. 2,4,5 c. 1,9 d. 2,8 e. None of these 15. 16. If If n P T a. o = n[] o = n[][ [^ T b. P W pp [qp [rp [⋯[(ps))p tp [up [vp [⋯[(ps\t))p a. 99 b. 100 = n[][^[ƒ T c. D l then n ][D^[Pƒ d. 1 =? e. None of these > 1.01 then maximum (n) = ? c. 125 d. 150 e. None of these 17. If BCKH, HKFA & KDEF are unit area square then xy(∆{|}) =? xy(∆{~}) T a. o T b.W T c.l T d.„ Maths By Amiya, 3E Learning, e. None of these www.facebook.com/MathsByAmiya ©AMIYA KUMAR a. -3 e. nY \] Y [^ Y n\][^ e.B. 7ℎ95  =? (45 69A=99). which 20. then among following. 0 c. If S ={1. Near Lalpur PS.50} . None of these 25. a. 4 c. 3 c. d. a. 10 b. 4 d.. 3rd Floor. 30 b.. Cannot be determined e.com/MathsByAmiya ©AMIYA KUMAR . None of these 24.2. 2 cm^2 d.11 c.1). None of These c. If the two trains are travelling with no acceleration then at what time do they cross each other.3. None f these 26.facebook. None of these 23. 29 †∗‡ TD TD = ‡\D e. None of these n√D[] is a rational number. Train-B leaves Jamshedpur at 9 : 30 am and reaches Ranchi at 1 :00 pm. 10 : 26 am d. If product of two roots of C o − 10C P + ŒC D + 838C − 2340 = 0 is -45 the find the value of "m" a. a. 150 d. What would be summation of all possible values of "a" for which quadratic equation C D + @C + @ + 1 = 0 has integral roots. 10 : 16 am c. (1. 14 e.> = 1. then N is the number of subsets of S. b & c . 52 d. 5 b. 7 c. 7 cm^2 c. 49 b. www. a. ŠnY [] Y \^ Y ‹ (n[]\^) nY []Y [^ Y n[][^ b. then ar(∆ABC)=? ar(∆ABC)=? e. D is point on AC such that BD=BC=2cm. Then k=? a. None of these 27. In an isosceles ∆ABC . 4 cm^2 b. 2 b.. -49 c. where A.. 3 b. If ax+12y = 17 & 3x+by= 61 is a pair of parallel lines and a & b are positive real numbers then what would be minimum value of sum of a and b. None of these a.y) exist for a. \nY [] Y [^ Y \n[][^ 21. Data inadequate e. None of these 19. such that it has at least one element divisible by 5. 5 e. For all positive integers a.. There are how many integral ordered pair of (x. If 2>45 + 3?. √7 ?ŒD 22. 2 b. Anand Complex. If line x=k equally divides a triangle whose vertices are (9.7 e. a. (a) or (c) e. AB=AC. then what would be digital sum of N. 6 d. 60 c. 095 34 002244 18. 3E Learning.. Road Ranchi. 4 d.1) and origin.. 9 d. None of these Maths By Amiya.B & C are angles of a triangle. 9: 36 am b. H.3E Learning.>… = 3√2 & 3>45… + 2?. if ]√D[^ should be always an integer. 12 d. Train-A leaves Ranchi at 7 : 30 am and reaches Jamshedpur at 11 : 30 am. Then Abhishek bhishek multiplies the two of his paper numbers number .facebook. 1067 ˜ + ]˜ ^ ˜ + ^˜ n˜ − 3 =? . NoT 33. where a. None of These 35. where a.e)=? a. 1/2 b. assuming all men are equi--capable of doing work? a. A monkey is at origin (0. 5. but by evening there still remained a portion to harvest. 095 34 002244 28. How many men were there in the team. If <(C) = (@ − C ” )– . 0). Then it split into two equal groups: the first remained in the larger field and harvested it by evening. 67 ^ n˜ ] c. a. Road Ranchi. 2 b. he has drawn. None of these 29. 1/3 b. 5/6 e. 3E Learning. 6/5 e. If ABCDEF is a regular hexagon with side 1 unit and ARQB and ASPF are two squares insidee the hexagon then ar( ar(∆PAQ)/ ar(∆SRP).B. @ − C ” e.d. None of these 32.3E Learning.e at 17 past 1 it shows 7:77 7:77. 3 c. 1967 b. and Mahajan adds his two. 1 e. www. A team of harvesters was assigned the task to harvest two fields. One member of team t in a single day’s harvest this portion the next day. 9. 16/5 d. 32 e. 5/8 31. 3 b. None of these a. None of these • 34. 3rd Floor. the second group harvested the smaller field. the on one who gets larger number will win. Then they pick two cards from their papers randomly. 5 e. 2 c. then what fraction of the day will that watch show the correct time ? c. H. then minimum number of jumps monkey required to go from (0. It always jumps from a point having integer coordinates to a point with integer coordinates and always covers a distance of 5 units in each jump. 4 d. 2 d. 4 b. Near Lalpur PS.c.16 d. 6 & Mahajan writes 8. What is the probability that Abhishek wins ? a. Due to low battery the watch shows 7 instead of 1 i. A 12-hour digital watch displays isplays only hour and minute. None of these 30. 1) ? a. 0)origin to (0. 4 c. then f[f(x)] is equal toto a. 10 on three different papers.d& e are real numbers then maximum (a.com/MathsByAmiya ©AMIYA KUMAR .c. C ” d. If n ] ^ ] ^ n + + = 11 & + + = 8 then ] ^ n n ] non zero real numbers. Half a day the team eam worked on the larger field.. Cannot be determined e. Anand Complex.b.b. 5/9 c. C – c. If a+b+c+d+e = 8 & a^2+b^2+c^2+d^2+e^2 = 16. 3/4 d. 4/9 (D)1/9 e. one twice the size of the other. a. 4 b. x • b. where a > 0 and n is a positive integer. None of these Maths By Amiya. Abhishek & Mahajan two players play a game : on three different papers Abhishek writes 3. b & c are d. 8 c. 2 a. for which √5 + 1 − √5 − 1 < 0. None of These + ⋯ … … … . If sides of a triangle is in G. 21 T T + + + T o a.75 d.444 d. 2^23 e. What is the maximum value of sin D ∗ sin D ∗ sin D .   . 29C14 c. 1.. 2.. T a. None of these 37. 5/2 d.. 1.5 c.. 2^22 c. -5/2 c. 3 T b.. 8 d. None of these 40..+ (24C23)*(25C24)+ (24C24)*(25C25) = ? a.75 e. (1 + cot 45) (1 + cot 46) (1 + cot 47) ….. 9216 d. 9 e.P with common ratio "r" then which defines "r" the best a.. 29C10 d. d. -35 T Tl DW b. None of these 43. Nn=9999.. N2 = 9999. 3. What is the coefficient of C o in the expansion of  C Y − C ˜ ¡ a. ¡ D D √W\T √W[T .. None of these £ ¤ d. =? c.B & C are angles of triangle. H.*(N10) a.4608 c..B.. mid point is not possible since domain is an open interval e. 3.555 www..facebook. Ÿ b. None of these • Y „ 44.5 b.999 (2^n digits number) then sum of all digits of (N1)*(N2)*(N3)*. 6 b.. None of These 39. such that PQ=5cm and QR=6 cm then what would be radius of the circle. (1 + cot 89)=? a..3E Learning. None of these 38. o d. Anand Complex. 26C13 b.. What is midpoint of domain of function <(C) = œ5 − √3C + 5 for real value of x a. 2. (24C0)*(25C1)+ (24C1)*(25C2)+ (24C2)*(25C3)+ . . 3E Learning.644 Maths By Amiya.com/MathsByAmiya ©AMIYA KUMAR . ž c.. If N0 = 9. 2^24 ¥ 42. a. 10872 e. D ¡ D e. where A. 2. 7 c. 3rd Floor. What is the digital sum of minimum value of positive integer n. 35 45. D T c... None of these 41. D Ÿ T\√W T[√W d. 2304 b. 0 b. T b. 1. Road Ranchi. 2 b. 095 34 002244 36.44 V T + c. N1 = 99. From a point P outside a circle two tangents PQ and PR are drawn on a circle where Q and R points on the circle. ¦ e. -21 e. T\√W T[√W .   D D √W\T √W[T ž D .. Near Lalpur PS... 095 34 002244 46. c(n... c (n+2. None of these 52. Total number of non negative integral ordered solution of (a. 7 e. (-2. C.10 c.3) e. None of These 49.3) . 3rd Floor. None of these D o ¦ 54. None of these 50. By this routine if his total Friday party is 84 more than his total Friday party with a particular friend. If 3x . Near Lalpur PS. negative e. Non-positive T 48. c (n+2 .c) of m< a+b+c < n . 4x. F ∈ c }.3. 16/9 e. where 0<m<n belongs to integer a.c(m. None of these c.. d.5.3E Learning.B. If an unbiased coin is thrown 50 times then what is the probability of getting atleast 1 head T T DªZ \T b. None of these 51. Non-negative T d. Anand Complex. None of These 47. F) ∶   ¡ +   ¡ = 125. D D D Maths By Amiya.7.3) c.3) b. 16 e. Y© c.facebook. ªZ d. www. 8 b.3) d.4) . 3 ∗ ?.> D C ∗ sinD C − sino C − cosD C = 0 then tanD C =? a. sec o C − 4 tanP C + 4 7@5 C is a. None of these 53. 3E Learning. (2. H.c (m+2 .com/MathsByAmiya ©AMIYA KUMAR .1) b.b.21. c(n. origin e.1) c. 0 b. Road Ranchi.3) .7y=13 . 12 d. 3 ∗ (5D ) ∗ (25 − 1) c. positive b.3y=35 and 14x + 6y = 22 are equation of sides of a triangle the coordinates of orthocentre is a. 3) .23 times. If there three parallel lines in a same plan and there are "n" points on each of the lines. 1 d. If an unbiased coin is thrown 50 times then what is the probability of getting heads 1. b. (5D ) ∗ (45 + 3) e.. None of these a. isa. 0 b. then how many triangles are possible considering these points as vertices a.3) b. (5D ) ∗ (45 − 3) d. † ‡ where N is the set of all natural numbers. The number of elements in the set {(C. 4 d. On Friday party night Mr.c (m+3. T T T a. 3 c.-1) d. (2. c. Then how many friends does he have a. 9/16 c. c(n.c(m. Chetan drink along with his three friends but he always wants minimum one new face in his party so he never drink with the same three friends more than once. B. 188 sq. 2 b.. Anand Complex. No such equation be possible whose roots are b and 1/b since irrational roots always occur in conjugate pair e.. 095 34 002244 55. 11872 & 13152. 9 e. 440 d. 430 c. 148 sq. 7 b. 450 e. If @ + ¬ + ? = 5. (1 − «)(1 − «D ) + (2 − «)(2 − «D ) + (3 − «)(3 − «D ) + ⋯ + (10 − «)(10 − «D ) =? . What is the minimum value of "b" . 864110 d. 3 c. None of these 60. None of these 61. None of These 57. 3rd Floor. None of these 56. 280 d. H. 47 sq. Road Ranchi. 2 d.3E Learning. then what would be digital sum of their product. 320 b. 3E Learning. for which f(x) = 3x^2 . 4C D + 17C + 1 = 0 b. @¬ + ¬? + ?@ = 6 & @¬? = 7. 592 sq. None of these 59. If 4 + √3 and "b" are roots of equation C D − √3 C + ? = 0 & then equation whose roots are b and 1/b is − a. 288 c. 5 d. 160 e. www. 420 b. cm b. 3 b. None of these Maths By Amiya. 43/7 d. 3 e. 1 c. Near Lalpur PS. a. cm e. a. 0 b. 7 nY ^ + ]Y ^ + ^Y n + ]Y n + nY ] + ^Y ] e. Then M-m =? a. None of These =? 58. 998700) and written in ascending order then which number would come at 2005th position from start. None of these 62. if number set a. a. a. a. There are how many integral value(s) of K possible. then a. 864100 c.4k*x + k+3 is always positive. 23/7 c.com/MathsByAmiya ©AMIYA KUMAR . Cannot be determined d. 27 for a and b being elements of Natural e. 864111 e.facebook. cm d. where « is cube root of unity. 12 c. If M and m are maximum and minimum values of divisor set D respectively such that all elements of D always gives remainder 32 when divide any one among 7712.g. If all 6 digits numbers abcdef are taken in where @ ≥ ¬ ≥ ? ≥ 6 ≥ 9 ≥ < (e. 4C D + 17C + 4 = 0 c. 864000 b. More than 3 63. What would be total surface area (approx) of uniform circular ring whose internal diameter and external diameters are 14cm and 16 cm respec. cm c. 17 TV o„ n <] < oP TSP d. If sum of two natural numbers is 384 and their LCM is 4216 . None of these 69. e. None of These 68. If there are "n" black balls . where a. 47/70 e. 2268 e. 23/70 c. 144 c. Then how many pairs {A. If S(M)= 46 then what would be summation of least value of M and fifth least value of M... Cannot get such values e. None of these 65. Out of 5 men and 4 women . C ∈ ®°95 c@7±=@B c±Œ¬9= & C ≤ 20} and set A and B are subset of X but   …. 67. 515 c. a.g.b.c. "n+1" green balls and "n-1" red balls and it is known that one should pick minimum 142 balls (without replacement. There are how many four digits numbers abcd are possible such that @ ≥ ¬ ≥ ? ≥ 6 a.com/MathsByAmiya ©AMIYA KUMAR . 0 d. 3 lies in between roots of C D − C + ? = 0 and satisfy a*b*c = 240 . a. a.d are positive integers then minimum value of (a+b+c+d) a. 495 c.18} a. www.. 242 c. 2540 c. 729 d. 6 b. 494 b. a. 15 e. 70/273 b. 2360 d. in how many ways we can select minimum 2 men and any number of women.728 e. If ® = {C|C. 715 e. Near Lalpur PS. Find the sum of all possible values of "b" for which C D − ¬C + 30 = 0 has integral roots. NoT 71. NoT 70. 36 e. 215 b. 2^8 d. NoT 72. Road Ranchi. None of these 73.facebook... (5c3)*(2^4) e. If 1+2+3+4.+ (2a) = b^2 and 1+2+3+4+. 3E Learning. None of these b. 14 d. 69 c. a. 7 c. 3rd Floor.. 5” is a perfect square. For how many integral value of "c". Anand Complex. 095 34 002244 64. If S(N) is denoted as sum of digits (where N is in base 10) . 73 b.4. NoT Maths By Amiya. 539 e. 500 b. H. 72 b. 736 b. a.B. For how many values of Natural number 5 ≤ 1000. (5c3)*[(2^6) .. 203/273 d.16. 98 d. S(198) = 1+9+8 = 18.+(2c+1)=d^2 . If <(C) = (C − 1)TS & A(C) = C D − 1 then what would be remainder when f(x) divided by g(x). NoT 74.. 799998 b. 516 d. 597998 c.. What is the sum of all natural numbers less than equal to 128 in base 10 which when converted in base 2 has exactly four 1s. (5c3)*(2^6) c.B} are possible such that  ∩ … = {2. where a and b are integers too. 714 d.1] 66. a. 588998 e. in any trial) to get all three color balls then what is the probability of picking one red ball. 499988 d.3E Learning. NoT Ÿ . NoT (???) 79. ((-4. ……. If fig is a square of side 6 cm. 1/3 d. 6 78. (√3-1)/4 e. All Three possible e. NoT ≤0 www. ¸4A47@B ¹±Œ(197 17 = 1 + 7 ≡ 8 c.B. 1 b. 8 d. where [c] is greatest integer less than equal to N.>9? D ¶ a. If T . †∗(†\W)∗Š† Y \TW†[WS‹‹ d.3E Learning. (0. 8 e. then find the radius of smaller circle. Anand Complex. 3rd Floor. 8 b. infinite † Y \V e. D . 2 c. and all four circles are tan tangential gential to each other and to respective sides of square. 4 b.. None of these 76. What would be minimum value of 4>9? D ¶ + 9 ?. Both smaller circles are congruent and both big circles are congruent too. √2 cm c.. a. (√3+1)/4 b. 095 34 002244 75.. 9 c. then what would be digital sum of maximum possible value of T º D º P º …………. √3 cm b.facebook. 25 77. Not a.32) b. Road Ranchi. 9 e. digital (197) ≡ 1 + 9 + 7 ≡ sum is continuous summation of digits till single digit. (10.com/MathsByAmiya ©AMIYA KUMAR . 1.11) . a. NoT 80. 3E Learning. -10) c. What is the digital sum of ž TS·Z „ e. 4 d. If coordinates of three points are (3. There are how many integ integral values of x for which a. a. NoT 81. TSS be 100 0 arithmetic means between 20 and 50.5 cm e. H. Near Lalpur PS. 1/4 c. 21 d. 4 b.2) d. 1 cm d. 10 Maths By Amiya. 9 c. If fig is a square of side 1 unit and all five circles are equal then what is the radius of the circles.17) then what would be coordinates of fourth point such that all four become vertices of a parallelogram.-4) 4) & (5. ((-2. º TSS ? a. On a plane there are how many points . "B" starts from the same place and he covers 12 m in his first minute and on subsequent minutes he travels 1 m more than the previous. 2 c. 3rd Floor. -2048 d. he covers 2 m more than the previous minute minute. 180 d. H. 20 green balls and 20 black balls are divided in to two boys such that both will get exactly 30 balls a. www. from where the shortest distance to three lines (one same plane) in which no two are parallel or collinear are equal. -1024 e. 0 b. 60 – 15√3 d. cm b. 4 e. cm c.facebook. NoT 86. then what is the distance between the towers in meters is a. then On how many minutes will the "B" be ahead of "A"? a.30) e. 4 e. -6360 b. 120 b. There are two towers on a plane. 3 d. NoT No 83. 095 34 002244 82. 330 c. Near Lalpur PS. 150 c. Anand Complex.3E Learning. -1540 c. a. 3E Learning. cm d. 300 b.com/MathsByAmiya ©AMIYA KUMAR . In how many ways 20 red balls.B. 30 + 15√3 b. If line ne L1 || L2 and all angles "a" are equal then x+y+z = ? a. 9 e. 7 c. 60 + 15√3 e. 1 c. There are two points P and Q are on the line joining their feet and between them. 45 + 15√3 c. After 3 minutes. a. "A" starts from point A on a certain time and covers 1 m in the first minute and den on subsequent minutes. 15840 e. 16438 sq. If PQ is 30m. NoT 89. 3 d. NoT 87. 1 b. 210 e. 331 d. NoT Maths By Amiya. What would be sum of last 4 digits when we convert 555^555 in to base 2 a. 15400 sq. 6 b. What would d be the area of a 40 sided regular figure whose one side is 11 cm. 1432√440 sq. NoT 88. c(60.. NoT 85. 8 d. Road Ranchi. What would be coefficient of CTV in the expansion of 20?1 20?2 2 20?3 20?20 20 »C − ¼ »C − 2D ∗ ¼ »C − 3D ∗ ¼ ∗ … … … … ∗ »C − 20D ∗ ¼ 20?0 20?1 1 20?2 20?19 20 a. the angles of elevation of the tops of the tower from P are 30° 30 and 60° and from Q are 60° and 45 45°. NoT 84. 11 d. What would be sum of all different remainders when product of a pair or twins prime divided by 9. 2 23@ + 21) = 4 – B. DATERAM wishes to date few girls of JLRI and XXM-C. NoT 95. -1 d. + D” ∗ C D” then (S )^2 − (T )^2 + (D )^2 − (P )^2+ . If ?. For which value of "a" log Dn[P (6@D + a. 0 b.>¶ + >45¶ = √2 ∗ ?.APn[„ (4@2 + 12@ + 9) c. −(D”\T )^2 + (D” )^2 = ? a. 7 c.240 only. 9 e. NoT 96. Anand Complex. If (1 + C + C D )D” = S + T ∗ C + D ∗ C D + .com/MathsByAmiya ©AMIYA KUMAR . d. If both (A) and (R) are true. √2 ∗ >45¶ b. . 0 b. a. If two sides of an integral sided triangle are 5 & 7 unit then how many integral values are possible such that circum centre is outside of the triangle a. 4 b. 5 c. Road Ranchi. . He is expected that he could get 22 kisses by a JLRI girl and 18 by XXM-C girls in return in a date. If (A) is false but (R) is true.−4 e. More than 6 but less than 14 c.>¶ − >45¶ = ? a. More than 21 e. More than 13 but less equal to 21 d.− T o d. . NoT 92. for which P^N would give reminder 1 when it is divided by 32 . NoT 91.¿À”Á √D d. b. 392 d. 3rd Floor. How many two digits numbers "N" . 412 e. a.5760 to spend on date and he can date at most 20 girls. www.>¶ then ?. 7@5¶ e. Assertion (A) : If m and n are integers and roots of C D + ŒC + 5 = 0 are rational then these roots must be integers. H. e. Then find the maximum number of kisses he get in return. 360 c. 6 b. 095 34 002244 90.facebook. NoT 93. Near Lalpur PS. A Date with JLRI would costs him Rs. where P can be any prime greater than 32 a. 1 c. Reason (R) : If m and n are integers then roots of C D + ŒC + 5 = 0 are necessarily integers. If (A) is true but (R) is false. c. 3E Learning.360 only and that with XXM-C Rs. He has only Rs. .B. 6 d. 12 e. ” e. 352 b.NoT 94.3E Learning. If both (A) and (R) are true but (R) is not the correct explanation of (A). (twins prime pair is pair of two primes differ by 2) a. ¿À”Á √D c. NoT 97. . Less than 7 b. and (R) is the correct explanation of (A). . Not Maths By Amiya. . NoT Maths By Amiya. Anand Complex. H. 8 e. 120° d.com/MathsByAmiya ©AMIYA KUMAR . 3rd Floor.13 & 14 remainders are 3. 30^3 d 30^(1/30) e. www.. Then what is the probability ty that f(x) is an increasing function.. If @] + ¬ n = ? . If f(x) = x^3 + a*x^2 2 + bx + c. 99/10000 c. 32 d.19y+ 9z = 0 then for integral x.b and c are different primes then a+b+c= ? a. 1/3 c. 115° c. 0 b.facebook.4. 99/200 c. c and n. 130° 101. 22 c. A bag contains ‘100’’ cards marked 1. 35 c. 99 d. a. 3. Near Lalpur PS. b. then what is the maximum value of • • • (@n ∗ ¬ ] ∗ ? ^ )˜Z + (@ ( ] ∗ ¬ ^ ∗ ? n )˜Z + (@^ ∗ ¬ n ∗ ? ] )˜Z a. 0 e. . a + b + c = 30. 095 34 002244 98. TS∗† •Y If <(C) = † YÄ [D†•Y [P†•Å [ then maximum value of f(x) is [P† Æ [T a.5&6 respectively then what would be sum of remainders when number is successively divided by 14. 72 e. Not Possible e. NoT 99. NoT For positive integer a. where a... 30 b. NoT 103..12. . NoT 102... NoT 100. 3E Learning. If m(ABD)=90° . More than one value d. 99/100 b. NoT b.100. AABAA is a perfect square then What is the value of A+A+B+A+A=? a.. 105. 1 c. c are values taken respectively by throwing a die three times.. NoT 104. ‘Ram’ draws a card from the bag and put back into the bag. NoT 106.21 z = 0 & 4x .12&11. Road Ranchi. If a number is successively divided by 11. Then ‘‘Rahim’ draws a card. 2. AC=BD C=BD and BC=CD then m(AEC)=? a.3E Learning.. b.. 20 e. 23 c. 93 c. The probability that ‘Ram’ ‘ draws a card higher than 'Rahim' is ____ a. 18 b. 1/4 b. 2 d. 30 b. 150° e. 5/12 d.13.. 20 b.B. where a. 1/2 e. a. y & z which option can be possible value of value of x+y++z. If 15x + 20y . 99/20000 d. 3^30 c. 71 b. a. 25 d. 12 + 6√3 c.B. 3rd Floor.. @  0 then which one is correct. Data is inadequate e. 7 c. What is the coefficient of C „ in the expansion of Š1 – C – C D + C P ‹ a. then a+b+c =? a. |a| = 2c e. Maximum number of trail needed to find correct combinations of key-lock out of 10 keys and their 10 locks (one lock has only one key which is in the bunch of given key. 10 c.. If Ç & Ç ” are two roots of @C D + ¬C + ? = 0. H. a. 24 d. 132 e. BC=5cm & CA=4cm. one trail should have unique combination of i and j for (Key-i & Lock-j) ) a. @ ∗ ? ” + @” ∗ ? + ¬ = 0 c. NoT Maths By Amiya.. then find the area of quard(ACDE) a. -144 c. Cant expend Rs 330 e. NoT 113. Money expensed by Bewkoof Ram on his date is directly proportional to smile of his GF in quadratic form. In a ∆ABC . Then tell us if Bewkoof Ram is planning to purchase a bouquet cost Rs. 240/79 d.) in base 4. | a | = c c. www. a+b+c=0 b. 40/49 e.. If the height of shahjahan is 6 feet then find the height of the tower in feet. –132 d..com/MathsByAmiya ©AMIYA KUMAR . 6 b. (@ ∗ ? ” )” + (@” ∗ ?)” + ¬ = 0 e. a. 144 b. and 6feet is the height of water level to his eye) a. NoT 114. 330 then how many time does his GF smile. 3 e. Road Ranchi. NoT 110. 6 b. (? > 0) are tangential. a = 2|c| d.. 1/7 in base 10 = (0. 9/49 b. Anand Complex. If his GF smiles 2 times he spend Rs 35/.abcabcabc. 55 d. Angle of elevation of one tower of Taj-Mahal is 45 degree in the eye of Shahjahan and at the same time angle of depression of image of the same tower in Yamuna is 60 degree for Shahjahan. 45 e.. 10! b.??? a. Among options which is the correct condition if circles C D + F D = @C and C D + F D = ? D . NoT • • d. 4 d. 3E Learning. If D and E are points on side AC and AB respectively such that AD is angle bisector of angle(BAC) and DE||AC. Near Lalpur PS. NoT l 109. NoT 111. NoT 112. 54/49 c. (@ ∗ ? ” )–È• + (@” ∗ ?)–È• + ¬ = 0 108.One day his GF was not in mood to smile then he purchases an Alpenliebe cost Rs 1only.facebook.(consider tower is just on the bank of River Yamuna. 8 d. if 5 smiles then Rs 176/. 12 − 6√3 b. 2|a| = |c| b.3E Learning. 095 34 002244 107. AB = 3cm . 5 c. a. Then it is found that sum of page numbers on remaining pages is 24000. 48 e. digital sum is defined as summation till single digit. statement-2 is true e.3E Learning. If a is one root of x^2 + x+1=0 and (1+ a)^97 = A + B*a. 4 c. 255 c. What is the digital sum of C(100. Statement-1 is true. Then A + B =? a.+ 100* C(100.. positive c. a. Statement-1 is true..B. 60 d.. statement-2 is false d.2) +3* C(100. NoT Maths By Amiya. A page is torn from Novel which has 2n-1 and 2n pages are printed on same paper but opposite side for n being natural number. cm) whose in-radius and circum-radius are 4cm and 6 cm respectively a. 3 c. less than x e. Road Ranchi. NoT 120. 265 d. Anand Complex.6 d. 1 e.facebook. 84 b.. NoT 119. -2 b. a. NoT 117. StatementStatement-1 : For each prime number P greater than 3.. There are how many seven digits numbers are possible whose sum of digits is 59 and are divisible by 11.1)+2* C(100.1 c. 095 34 002244 115. 32 b. a. 2 b. If 11^11 = 2853bba70abb then a=? a. P^(2P+1) – P is divisible by 12. 54 d. Statement-1 is false.. statement-2 is true. NoT 122. 6 d. statement-2 is a correct explanation for statement-1 b. 3E Learning.100) . www. : (P + 1)^(2P+1) – P – 1 is divisible by 12. 2 e.. 2 d. If set M is containing all values of C − |C| for all real values x. NoT 121. negative d. H. then element(s) of set M is (are) a. Then page no on torn page can be .3)+. StatementStatement-2 : For each prime number P greater than 3. 0 b.com/MathsByAmiya ©AMIYA KUMAR . statement-2 is true.. 24 e. NoT 118.. 3rd Floor. 0 d. 8 e. Near Lalpur PS.. 64 c. 275 e. Not 116. 40 c. What is the area of a right angled triangle (in sq. statement-2 is not a correct explanation for statement-1 c. 245 b. Statement-1 is true. if a. nY []Y [^ Y DWln]^ d. 1/3 b. 14 e. If total cost of 3 apples . 7 bananas and 19 mangoes is Rs. n[][^ DWln]^ e. a. b.2.2) b. (3. What is the max value of (a – p) (b – q) (c – r) (a*p + b*q + c*r) . 4 e. There are three flower pots are hanging on a wall such that they are not in straight line and line joining these are making an equilateral triangle.com/MathsByAmiya ©AMIYA KUMAR .3E Learning. Ä ŠnY []Y [^ Y ‹ DWln]^ c. NoT 125. 9 d. if all costs are natural numbers. 2 d. 3rd Floor. T\† T d. T\† † c. . then which would be total cost of 43 apples.(2. NoT 127. 16739 d. 17639 c.5) c. (2. Not T\† Y 124. 126. Fenku is standing on a long straight RTO bridge over a river Yamuna and Pappu is rowing a boat on the river just below Fenku on the bridge.B. What is the probability of having a three digits number whose middle digit is the largest among all. NoT 128. NoT Maths By Amiya. (3. r . NoT What is the coordinates of incenter of triangle whose vertices are (2. 1 c.5.3) e. NoT 130. 0 b. from all three digits number. †Y † †Ä †Æ + T\†Ä + T\†Æ + T\†•Å + ⋯ … … . H. Anand Complex. a. Road Ranchi. 4/15 d.. a. NoT 129.6) a. (5. 15739 b.3) d. √llW m/min d. 256 Sum of last two digits of 1!^1! + 2!^2! + 3!^3! + 4!^4!+. 24/97 c... 095 34 002244 123.. Tol √llW m/min e. 3E Learning.2) & (2. 103 bananas and 131 mangoes.= |C| < 1 T\† Y a. 4/25 e. T b. ALL THE BEST www. Then We have many different places where we can put fourth flower pot so that distance between any two post would be always same. Tlo DW m/min b. Near Lalpur PS. 11 b. a. T\† Y †Y e. (b – q) and (c – r) all are positive quantities a..facebook. =? <. 2219/.and total cost of 47 mangoes. (a – p). p .2). c. 10 c. Tol DW Tlo m/min c. 20000 e. 13 apples and 31 bananas is Rs 5849 /-. b. If Fenku starts walking at constant speed of 4 m/min and the boat moves normal (perpendicular) to the bridge at constant speed of 5 m/min then at what rate are both moving away after 4 minutes if the height of the RTO bridge above the boat is 3 m? a.q. Solution: (d) (6.3) 2C D − 5CF + 3F D − 9C + 12F + 9 = 0 General equation is @C D + 2ℎCF + ¬F D + 2AC + 2<F + ? = 0 Therefore a=2. [b] 1/10 8.e.n)=(132/13 . W=3 & L=6 Three wins and six losses for each player: Total points of each is 0 and total matches =9 Now we just find three wins by three different players in 9 matches.com/MathsByAmiya ©AMIYA KUMAR . Near Lalpur PS. 2. d.. points of one player 2*W + (-1)*L = 0 & W+L=9 So. i. f=6 & c=9 Consider the equations formed by first two rows* of i.Y & Z are win by three different players. its same as arranging X. [d] infinite many 5. [b] 1680 In a game there are one winner and two loosers. b=3. 2C W V − F – = 0 @56 − W C + 3F + 6 = 0 D D D Solving these. 095 34 002244 Solutions 1. 10. 4C − 5F – 9 = 0 .X.e. we get x = 6. g=-9/2. an equation of pair of straight line. [d] None of these 1670 3.Y.B.Y. b.e. Road Ranchi. Note: Š4C − 5F – 9‹(−5C + 6F + 12)  2C D − 5CF + 3F D − 9C + 12F + 9 *It's just a condition to make a general equation.. we get the required point of intersection. h=-5/2.Y.X.Z. According to questions total points after tournament =0 So. 3E Learning. -10/13) www. −5C + 6F + 12 = 0 Solving the above equation. y = 3. 10 4. c. ax + hy + g = 0 and hx + by + f = 0 i. H. 122 (m. and one can get only +2 and -1 point in a game.3E Learning. Anand Complex. [b]4 6. TP l Maths By Amiya.facebook.Z which is 9!/3!*3!*3! =1680 where X. [d] 7. 3rd Floor.Z. 9. W Sol: AXF~BAF and ratio of corresponding sides [check hypotenuses i. 24 13. H. all factors of 144 except 2 19. 095 34 002244 11. 6. Anand Complex. b. None of These. d. 18.B.3E Learning. 3E Learning. c. All . 2.com/MathsByAmiya ©AMIYA KUMAR . Near Lalpur PS. 3rd Floor.8 weight gap then binary base 1&2 T 15. e. 184m 12. d.facebook.. 9 .2^40 mod 9 =6 Maths By Amiya. 150 T 17. Road Ranchi. D 16. c. www.e.None = 2^50 . c. AF and BF] is 1 : rt(5) So required ratio is 1 : 5 or. all are possible except 14. d. www.facebook. ¬ − @? = 0. [a] 30 2sinA + 3cosB =3√2 (i) 3sinB+2cosC=1 (ii) Squaring (i) & (ii) and add. by AM>=GM . √7 ?ŒD 22. @= D respectively. 10 : 26 am 23. taking a . a+b >=12 24. d.B. b. )(1 − = + = D ) is multiple of Only in [d] @D + @D ∗ = D + @o ∗ = o = @D (1 + = D + = o ) = @(1 + = + = D)( denominator @ + @= + @= D = @(1 + = + = D ). 3rd Floor. 3E Learning.3E Learning. Maths By Amiya. c. 12 . a/3=12/b. Near Lalpur PS. a. ab= ab=36 .com/MathsByAmiya ©AMIYA KUMAR . b &c are in GP . Road Ranchi. 3 25. H. c. i. @=. Hence answer is [d] Rationalizing 21. we get sin(A+B) = 1/2 A+B = 30 or 150 But A+B=30 is not a possible case. so C=30. 095 34 002244 20. Anand Complex.e. d. nY []Y [^ Y n[][^ Sol: [d] n√D[] n√D[ [] ]√D\^ ](Dn\^)[√DŠ]Y \n^‹‹ = ∗ = ]√D[^ ]√D[ [^ ]√D\^ D]Y \^ Y D For being rational. since  ≤ 30 & … ≤ 30 wont satisfy the equation (i) Hence A+B=150 . d. C D & C we get . ℎ95?9 Œ@C (@) = 8 − 24/5 = 16/5 5 Maths By Amiya..c. (@> @D + ¬ D +.. S+R=10 . >.facebook. [b] −49 Concept : • If a & b are roots of a quadratic equation then equation is Ëp − (x + Ì)Ë + x ∗ Ì = (Ë − x)(Ë − Ì) = Í • • Sum of all roots taken one at a time = ÕÖ×ØÙÚ×Ù ÙÛÜÝ Product of all roots = ÎÏÐÑÑÒÎÒÐsÓ ÏÑ ËsÔt ÎÏÐÑÑÒÎÒÐsÓ ÏÑ Ës ÎÏÐÑÑÒÎÒÐsÓ ÏÑ Ës Since product of two roots = -45 .(iii) From (i) & (ii) we get. 3rd Floor. 4. so product of rest two roots = − DPoS \oW D = 52 Let S be sum of roots whose product is -45 then equation would be C − ¹C − 45 = 0 and Let R be sum or roots whose product is 62 then equation would be C D − ÞC + 52 = 0 So our give equation can be written as product of above two equations. W Assume rest 4 variables to be equal. @ ≤ Tl W . +9 D = 4D ) now @ + 4C = 8 @D + 4C D = 16 => (8 − 4C)D + 4C D = 16 6 => C = @56 2 5 6 4> >Œ@BB9=. H. a=5.-1 27.(i) 7+SR = m ..b. c. and max(a.e) would obviously be less than 4. TD W (@ + ¬ + ? + 6 + 9) + This value can be realized by putting @ = Alternate solution By Saransh Sood: Tl W Pl W = 16 − Vl W + Pl W = 4...com/MathsByAmiya ©AMIYA KUMAR . Anand Complex. Road Ranchi. www. c.3E Learning. we finally get m=7+14*(-4)=-49 28. C o − 10C P + ŒC D + 838C − 2340 = (C D − ¹C − 45) ∗ (C D − ÞC + 52) C o − 10C P + ŒC D + 838C − 2340 = C o − (¹ + Þ)C P + (7 + ¹Þ)C D + (−52¹ + 45Þ)C − 2340 Equating coefficient of C P ..(ii) -52S + 45R = 838 . S= 14 & R= -4 putting these in (ii) . Near Lalpur PS.B. l and setting the other numbers to . 3E Learning. 16/5 l D l D l D l D l D  @ − W¡ +  ¬ − W¡ +  ? − W¡ +  6 − W¡ +  9 − W¡ = (@D + ¬ D + ? D + 6D + 9 D ) − l Hence ß@ − Wß ≤ 2. . . 095 34 002244 26. B. [c] 2 1 1 @=(∆æç) = » ¼ ∗ 1 ∗ 1 ∗ >4530 30 = 2 4 1 1 ar(∆SRP) = ∗ √2 ∗ √2 ∗ >45 >4530 = 2 2 1 ar(∆PAQ) 2 = =2 ar(∆SRP) 1 4 Maths By Amiya. easy put the value and get answer 35.facebook. a. www. Near Lalpur PS.1 day and night both) ∴ in correct time 8 º 60 = 480 (each minute it will display 1) Remaining 20 hours it will show the incorrect time 16 º 15 = 240 Total incorrect time = 240 + 480 = 720 correct time = 1 – incorrect time = 1 – 720/(24*60) = 1/2 31. [b] 3 30. H. a. 11-12 12 . ^ = á & n = â then X+Y+Z = 11 & XYZ = 1 T T T and ã + ä + å = 8 => XY+YZ+ZX= 8XYZ=8 n˜ ]˜ ^˜ Now]˜ + ^ ˜ + n˜ − 3 = à P + á P + â P − 3àáâ = (à + á + â)(à D + áD + â D − àá − áâ − âà) = (à + á + â)Š(à + á + â â)D − 3(àá + áâ + âà)‹ = 11 ∗ (11D − 3 ∗ 8) = 1067 33. c. c. 3rd Floor. 1067 n ] ^ Let ] = à. 10 -11. 3E Learning. 1/2 The clock will show the incorrect time (between 1 – 2. 12 . 4/9 32. Road Ranchi.3E Learning.com/MathsByAmiya ©AMIYA KUMAR . Anand Complex. 8 34. 095 34 002244 29. x . b. d.. n>25.+ (24C23)*(25C24)+ (24C24)*(25C25) ⇒ 49C24 4 = (24C0)*(25C1)+ (24C1)*(25C2)+ (24C2)*(25C3)+ ..+ 24C23*x^23+24C24*x^24 & (x+1)^25 = 25C0*x^25 + 25C1*x^24 + 25C2*x^23 +....... www. c. (1 + tan 45) =2^23 [(1+tanA)(1+tan(45-A)) A)) = 2] Maths By Amiya. c...... 3.. H.01 => min(n)=26 so digital sum =8 37. [c] 8 .. 3rd Floor... 095 34 002244 36.facebook.. d.01 . 2^23 (1 + cot 45) (1 + cot 46) (1 + cot 47) …. 3E Learning.75 38. None of These 49C24 Sol: We know (1+x)^24 = 24C0 + 24C1*x + 24C2*x^2 +24C3*x^3 +..3E Learning.B... 5/2 3C + 5 ≥ 0 => C ≥ −5/3 & 3C + 5 ≤ 25 => C ≤ 20/3 3 W −P ≤ C ≤ 40.   DS P . (1 + cot 89 89) =(1 + tan 1)) (1 + tan 2) (1 + tan 3) ….com/MathsByAmiya ©AMIYA KUMAR .. d.. Anand Complex.. D ¡ D 41.. then midpoint of this is 5/2 √W\T √W[T .. Road Ranchi.+ 25C24*x+25C25 Multiplying above two and equating coefficient of x^24 coefficient of x^24 in (1+x)^ (1+x)^49 = (24C0)*(25C1)+ (24C1)*(25C2)+ (24C2)*(25C3)+ .. Near Lalpur PS......+ (24C23)*(25C24)+ (24C24)*(25C25) 39. Anand Complex.*Nn) = 9*2^(n-1) here sum of digits (N1*N2*N3*..gl/X7wi3E 46. Non-negative sec o C − 4 tanP C + 4 7@5 C = sec o C − 4 tanD C ∗ 7@5 C + 4 7@5 C = sec o C − 4 (sec D C − 1) ∗ 7@5 C + 4 7@5 C = sec o C − 4 sec D C 7@5 C + 8 7@5 C = (sec D C − 2 7@5 C − 2)D − 4 tanD C − 4 + 4 sec D C = (sec D C − 2 7@5 C − 2)D ≥ 0 48..3E Learning. b. the coefficient is (−1P ) ∗ (7. 1 3 ∗ cosD C ∗ sinD C − sino C − cosD C = 0 ⇒ 3 ∗ (1 − sinD C) ∗ sinD C − sino C − (1 − sinD C) = 0 ⇒ 3 sinD C − 3 sino C − sino C − 1 + sinD C = 0 ⇒ −4 sino C + 4 sinD C − 1 = 0 ⇒ 4 sino C − 4 sinD C + 1 = 0 ⇒ (2sinD C − 1)D = 0 1 sinD C = ⇒ tanD C = 1 2 47.. Road Ranchi... c.facebook. b.644 = pi^2/6 http://goo. Near Lalpur PS.gl/shxyNb Maths By Amiya. positive integral solution http://goo.4608 = 9*2^9 N1= 9 sum of digit = 9 = 9*2^0 N1*N2 = 9*99 = 819 . 3rd Floor.B. sum of digit = 36 = 9*2^2 so sum of digits (N1*N2*N3*. 095 34 002244 T 42.*N10) = 9*2^(10-1)=9*2^9 = 4608 44. =) ∗  C Y ¡ For C o . 3E Learning.. so. d. ¦ If A + B + C = 180° £ ¤ then cos  + cos … + cos  = 1 + 4 ∗ sin ∗ sin ∗ sin £ ¤ ¥ ⇒ 1 + 4 ∗ sin ∗ sin ∗ sin ≤ £ ¤ D ¥ D D T ⇒sin ∗ sin ∗ sin ≤ D D D ¦ P D D D ¥ D P žcos  + cos … + cos  ≤ Ÿ D 43. c... sum of digit = 18 = 9*2^1 N1*N2*N3 = 9*99*999=890109 . DT[é l Y é ∗  C ˜ ¡ = (−1é ) ∗ (7. www. =) ∗ (C = 4 ⇒ = = 3.3) = −35 Y•Èê Å ) 45. H. c. 1. [d] 7 . -35 • Y „ • „\é General term of  C Y − C ˜ ¡ is (−1é ) ∗ (7.com/MathsByAmiya ©AMIYA KUMAR . c(n. Road Ranchi.7> = (−1)”ð€]íé ìî éììï¿ ∗ ^씿ïn”ï Sum of roots = 4 + √3 + b = √3 ⇒b=-4 .it's a right angled triangle . c. www.1)( c(n. • ¹±Œ . all coefficients(including constant) are Rational.1)* c(n. then Total number of Friday party = c(n..com/MathsByAmiya ©AMIYA KUMAR . all coefficients(including constant) are Real.3) . 095 34 002244 49.< @BB =. TSA of ring = pi^2 * (R^2 .10 Let "n" be total number of friend. [b] 148 .2)* c(2n. [d] c (n+2.59 @7 @ 74Œ9 = (−1) ∗ ^ìíîîÀ^Àí”ï ìî † –Ô• • æ=.3) Total number of Friday party with a particular friend = c(n-1 .7> 7@ë95 . Sol: [c] It's a right angled triangle so orthocentre be at right angled vertex of common point of 3x 7y=13 and 14x + 6y = 22 which is (2. ^ìíîîÀ^Àí”ï ìî † – ^ìíîîÀ^Àí”ï ìî † – T T So.-1) How:. [c] 58. check 3x . (5D ) ∗ (45 − 3) Total Number of triangles = all three vertices on different lines + two vertices on same line = c(n. 3rd Floor.1) + c(3. H. equation whose roots are -4 and -1/4 is C D − (−4 + − o)C + (−4) ∗ (− o) = 0 ⇒ CD + T„ o C + 1 = 0 ⇒ 4C D + 17C + 4 = 0 Maths By Amiya.. coefficient interchanges with change in sign. 52. Concepts • Irrational Conjugate Pairs of Roots only Occurs IFF.B. Near Lalpur PS.< @BB =. 2) = 84 => n=10 51. 2) By question.c(n-1 .facebook.c (m+3. Anand Complex.3E Learning. b. 3E Learning.3) m< a+b+c < n ⇒ [0 ≤ @ + ¬ + ? ≤ 5 − 1] − [0 ≤ @ + ¬ + ? ≤ Œ] ⇒ [0 ≤ @ + ¬ + ? + 6 = 5 − 1] − [0 ≤ @ + ¬ + ? + 6 = Œ] 53. [c] 55.7y=13 and 14x + 6y = 22 .6±?7 .1)) = (5D ) ∗ (45 − 3) 50.1)* c(n. • Imaginary Conjugate Pairs of Roots only Occurs IFF.3) . [d] 57. [d] 54.r^2) 56. B. so Let b = 2a + n . You can solve this option elimination method . [c] 864110 61. [c] Hint:. www.22 < @ < 5. so the smallest integral a = 5 for n=2 So smallest b = 2a + n = 2*5 +2 =12 60. for smallest "a" we need smallest "n" which give integral "a" for n=1. 3rd Floor. but the approach is TV o„ < n ] ⇒ 2+ < oP TSP T„ ] oP < n <2+ TSP ⇒ V . Near Lalpur PS. 095 34 002244 59.3E Learning.. we are not getting an integral "a" for n=2 . [b] 12 . LCM=4216=2^3*17*13 .05 . [d].. H. T„ We get (i) ⇒ 2 + oP < ⇒2 + ⇒ T„ ” Dn[” ” <2+ <2+ oP T„ ” V < n < TV oP n n V TV ⇒ V < 2 + TV TV” V <@< oP” T„ Here a and n both are integer . Anand Complex... we have 4. 3E Learning.(i) TV oP < ] n < o„ TV Since b is a natural number and greater than twice of a. so product = (2^ (2^3)^2*17*13 3)^2*17*13 =33728 = Maths By Amiya. Road Ranchi.factors of 320 greater than 32 62. 248=8*31 . so HCF= HCF(Sum.LCM)=2^3 numbers are 136=8*17 . [d] . D<0 63.facebook. Product = 33728 = 2^6*17*31 Sum = 2^7*3 .com/MathsByAmiya ©AMIYA KUMAR . 3E Learning.18} . [c] All even would come in this = 500 All odd perfect square = 16 (from 1D to 31D ) 68. Anand Complex. Murphy's Law According to question = 2n+2 = 142 ⇒ n=70.4.10.com/MathsByAmiya ©AMIYA KUMAR . Near Lalpur PS. Q . So we have to distribute rest three set P.25 + 1) ∗ C for remainder A(C) = C D − 1 = 0 ⇒ C D = 1 (put this in f(x)) We will get remainder.25) ∗ C ”÷S (10. C ∈ ®°95 c@7±=@B c±Œ¬9= & C ≤ 20ó = {2. so smallest number would have five.14. considering all worst conditions. [b] To get all three colors one need to pick at least "n+1" + "n" + 1= 2n + 2 .4. [d] 3l − 1 = 728 ® = ñC│C.gl/C6bL94 67. 095 34 002244 64. If all elements are in set R then  = … =  ∩ … = {2.R &  ∩ …. à − õ = Þ then X is union of disjoint set P. [c] 66. [c] S(M) = 46 . so total pair {A. TS V V <(C.3E Learning.16. www.facebook. for minimum M we need the smallest digit number so we need maximum 9s. we have to exclude this set.B} = 3l − 1 = 728 65. [c] 69.25 + 1) = 2 − 2 ∗ C http://goo.18} .B. 2V − 2V ∗ C D” D”[T − ∑TS <(C) = (C − 1)TS = ∑TS ”÷S (10. (now we cant make number starting from 1) Second Least = 289999 Third least = 298999 Fourth least = 299899 Fifth least = 299989 Maths By Amiya. H. it is sure that we have to arrange rest 6 numbers Take  ∪ … = õ .18. … −  = ç.16.  − … = æ . C D = 1) = ∑TS ”÷S (10. Road Ranchi.8.25) − ∑”÷S (10.4. R in total 3^6 ways. so number of red balls = 69 and total balls = 210 So probability of first ball is red = 69/210 = 23/70 70.20}  ∩ … = {2.16. 3rd Floor. 9s and one 1 So Least M = 199999 .12.Q.6. B. [c] T 1 − 4? > 0 ⇒ ? < o & 9 − 3 + ? < 0 ⇒ ? < −6 so finally ? < −6 From a*b*c = 240 .4) solutions but we have to reduce one value when all digits are 0 so required number = C(13.5 − O9=.4)-1= 714 Alternate: We have to chose 4 numbers .1) + 2^5*10 + 6*(2^5 . . [b] Work in base two and add number or 1s in all base but addition should be in base 2only. [c] N0+N1+N2+N3+N4+N5+N6+N7+N8+N9 = 4 N0 is the number of times 0 occurs N1 is the number of times 1 occurs And so on… so C(13.1. Anand Complex.1) 72. Near Lalpur PS.8.r Let a=w-3.b=x-2. r = (√3-1)/4 Maths By Amiya. 095 34 002244 71. 3E Learning. Œ454Œ±Œ ? = 24 @56 6 = 35 1 + 2 + 3 + 4. .com/MathsByAmiya ©AMIYA KUMAR . .q. p...y. 3rd Floor. . a=a(smallest side) .3.1) + 2^6*20 + 10*(2^6 . Road Ranchi. w.1) + (2^4*4) + 3*(2^4 .facebook. 19 10011 20 10 10 10 6 4 20 19 1 20 19 1 20 18 0 10 6 3 1 20 17 1 10 6 3 1 20 15 1 10 6 3 1 20 10 0 10 6 3 1 20 0 2540 Alternate:(2^4 . + (2@) = ¬ D = (2? + 1)(2? + 2) 2 (2? + 1)(? + 1) = 6D >. total ways = 13c4 = 715 . . + (2? + 1) = ¬ D = 74. c= y-1. Œ454Œ±Œ 5. . H.x.2. . we have to reduce one case when 3. from Pythagoras or 30:60:90 .z strictly descending order out from thirteen numbers 0.3E Learning. d=z . [b] 2@(2@ + 1) 2 D @(2@ + 1) = ¬ >. .9. b= a+2r) => a = 1/2 . www. r(45=@64±>) = n[]\^ D => (where c=1. [c] Use in right angled triangle.2. c is factor (with negative sign) So we have 14 values of c which will satisfy the conditions 73. @ = 4 @56 ¬ = 6 1 + 2 + 3 + 4. .1. .0 so required number = 714 75. [b] 555^555 mod 16 = 3 so last 4 digits in base 2 = 0011 Maths By Amiya. [c] 82. D . [a] DS[WS 20. Since these are collinear points 79. So for circle = 2*pi*r = 440 ⇒ r=70 cm Area of this circle = pi*r^2 = 15400 cm^2 It is the max area . 095 34 002244 76. 3E Learning. …….. T º D º P º …………. T .??±=45A ∗ 10lV ] = 1(428571 =9:9@7> 11 74Œ9>)428 7 Digital Sum (1(428571 repeats 11 times)428) ≡ Digital Sum (1+9*11 + 5) ≡ 6 78.. Road Ranchi. Near Lalpur PS. [d] incentre & excentre 86.??±=45A ∗ 10lV 10„S ø ù = [1. [d] 85. 50 will be an AP with Arithmetic Mean equal to = 35. TSS will also be a AP with Arithmetic Mean equal to 35. [a] (1.428571=9. (here we have perimeter) Perimeter = 40*11 = 440 cm The maximum area of 2-D figure with perimeter = 440 cm is of a circle. Using AM≥GM.facebook..5^2 + (3-x)^2 81.>9? D ¶ = 4(1 + tanD ¶) + 9(1 + cot D ¶) = 13 + 4 ∗ tanD ¶ + 9 ∗ cot D ¶ = 13 + (4 ∗ tanD ¶ + 9 ∗ cot D ¶ − 12) + 12 = 25 + (2 tan ¶ − 3 cot ¶)D So minimum value of above expression is 25 77. D . º TSS =35TSS and its digital sum = 35TSS mod 9 =1 80. [e] Concept: Circle has maximum area for given perimeter.3E Learning.com/MathsByAmiya ©AMIYA KUMAR .B. T . [b] 6 Sol: TS·Z „ = TS „ ∗ 10lV = 1. H. 3rd Floor. [e] NoT Parallelogram is not possible with these three points. D Thus. www. Anand Complex.428571=9.5+x)^2 = 1. ……. [b] 84. [c] 83.. [d] 4>9? D ¶ + 9 ?. º TSS ≤ 35TSS So maximum value ofT º D º P º …………. so area of required fig should be less than this. TSS . and all options are more or equal to this so answer is [e]NoT 87. Near Lalpur PS.com/MathsByAmiya ©AMIYA KUMAR . Road Ranchi. H. [a] 93. Sum of all interior angles of this = 900 a+x+360-a+y+360-a+z+a=900 a+z+a=900 ⇒ x+y+z = 180 89. 0) P (8. [c] x + y ≤ 20 & 360x + 240yy ≤ 5760 Zmax = 22x + 18y & x. Anand Complex. [c] Extend L2 and make a 7 sided polygon.3E Learning. Maths By Amiya. 20) Value of the objective function Z = 22x + 18 18y Z= 22 º 0 + 18 º 0 = 0 Z = 22º16 + 18 º 0 = 352 Z = 22 º 8 + 18 º 12 = 392 Z = 22 º 0 + 20 º 18 = 360 92. 12) B1(0. 3E Learning. 3rd Floor. [c] 91. 095 34 002244 88. y ≥ 0 Points (x.B. 0) A2(16. [a] Till 8th minute from 3rd (B's Start) B would be ahead 90. y) O (0. www.facebook. 3E Learning. H.gl/OyufpL 99. f´(x) = 3*x^2 + 2*a*xx + b . So. 1 ≤ a. T ”\T VV Hence the required probability = ∑TSS ”÷T  TSS¡   TSS ¡ = DSS Maths By Amiya. Near Lalpur PS. [b] obtuse angled 98. [b] 120° Complete the rectangle ABDF. 3.B. so probability = 15/36=5/12 102. Anand Complex.facebook..5) and 8 for rest so to total tal sum is 14.. . [b] http://goo. put the option and check 95.gl/kWdm3d 96. b ≤ 6. AC = BD = AF CB = CD and OB = OD imply CO is the perpendicular bisector of BD and hence of AF as well.5) all product of twins prime have digital sum (remainder by 9) is 8. 3E Learning.gl/zDVQt7 97. [c] best is option elimination method. Follows AC = FC and triangle ACF is equilateral. 3rd Floor. [c] 5/12 Sol. then Rahim can draw any card marked 1. http://goo. 2. So x = supplement of angle CAF = 120 degrees. [b] 99/200 Let ‘Ram’’ draws a card marked with n. y = f(x) is strictly increasing ⇒ f´(x) > 0 ∀ x ⇒ (2a)^2 2 – 4*3*b < 0 This is true for exactly 15 ordered pairs (a. [c] http://goo. www. n–1. we have only two remainders 6 for (3. 095 34 002244 94. Road Ranchi. b). [c] Except (3.com/MathsByAmiya ©AMIYA KUMAR .. 101. [c] 212^2 = 44944 100. Join BF to intersect AD at O. .5 11 3 83⇧ 5 12 1000⇧ 14 6⇧ 13 104.com/MathsByAmiya ©AMIYA KUMAR . we get nY []Y [^ Y [Dn][D]^[D^n ⇒ n[][^ (n[][^)Y n[][^ • • • ≥ (@n ∗ ¬ ] ∗ ? ^ )ೌÈ್È೎ + (@] ∗ ¬ ^ ∗ ? n )ೌÈ್È೎ + (@^ ∗ ¬ n ∗ ? ] )ೌÈ್È೎ • • • ≥ (@n ∗ ¬ ] ∗ ? ^ )ೌÈ್È೎ + (@] ∗ ¬ ^ ∗ ? n )ೌÈ್È೎ + (@^ ∗ ¬ n ∗ ? ] )ೌÈ್È೎ • • • ⇒@ + ¬ + ? ≥ (@n ∗ ¬ ] ∗ ? ^ )ೌÈ್È೎ + (@] ∗ ¬ ^ ∗ ? n )ೌÈ್È೎ + (@^ ∗ ¬ n ∗ ? ] )ೌÈ್È೎ By putting a+b+c =30 (given) • • • ⇒ (@n ∗ ¬ ] ∗ ? ^ )˜Z + (@] ∗ ¬ ^ ∗ ? n )˜Z + (@^ ∗ ¬ n ∗ ? ] )˜Z ≤ 30 • • • So max value of (@n ∗ ¬ ] ∗ ? ^ )˜Z + (@] ∗ ¬ ^ ∗ ? n )˜Z + (@^ ∗ ¬ n ∗ ? ] )˜Z is 30 106... [a] 30 Concept õ ≥ þõ (@ + @ + @+.(A) n∗][]∗^[^∗n ≥ (@] ∗ ¬ ^ ∗ ? n )ೌÈ್È೎ • .19y+ 9z = 0 † ‡ ௭ = − (TW∗V[o∗DT) = (\TW∗TV\o∗DS) ⇒ x:y:z=3:3:5 (DS∗V\TV∗DT) Maths By Amiya. .. . 11003 Use 11003 4 11003⇩ 13 12 60⇩ 0 13 6 Quotient 0⇧ Can take any value 14 785⇩ 11 5 5⇩ 5 Quotient 0 [b] for x=±1 105. . Road Ranchi.21 z = 0 4x . 095 34 002244 103.. . since x:y:z=3:3:5 15x + 20y .3E Learning.0.facebook. . [b] remainders are 13. + ¬ 74Œ9>) + (? + ? + ?+.. .5. . Near Lalpur PS.(B) n∗^[]∗n[^∗] ≥ (@^ ∗ ¬ n ∗ ? ] )ೌÈ್È೎ • . H. Anand Complex. [c] 99 . + @ 74Œ9>) + (¬ + ¬ + ¬+.. www.B. 3rd Floor. . 3E Learning.(C) = n∗n[]∗][^∗^ n[][^ Similarly & n[][^ n[][^ ≥ (@n ∗ ¬ ] ∗ ? ^ )ೌÈ್È೎ By adding all three. + ? 74Œ9>) @+¬+? T ≥ [(@ ∗ @ ∗ @ ∗ … ∗ @)(¬ ∗ ¬ ∗ ¬ ∗ … ∗ ¬)(? ∗ ? ∗ ? ∗ … ∗ ?)]n[][^ • . facebook.gl/1yX6Yq 115. NoT Hint :. then a+b+c =? a.) in base 4.!!! 109.com/MathsByAmiya ©AMIYA KUMAR . Distance(C1toC2) = R1+ R2 (not possible) or Distance(C1toC2) = |R1 . d] 45 . 095 34 002244 • • 107. H. 3E Learning.0).111111 45 ¬@>9 … = http://goo.3E Learning.. 6 b.. 113. [b] 12 + 6√3 112.< C „ = (36 − 300 + 120) = −144 110. [d] (@ ∗ ? ” )–È• + (@” ∗ ?)–È• + ¬ = 0 By option elimination or sum of roots & product of roots concept 108. 5 c. [e] Not 1 45 ¬@>9 10 …−1 ..0). Anand Complex. a=0. Radius R2= c n ßDß = ßD − ?ß ⇒ |a|=|c| but since c is positive so need to write mod with c hence |a|=c 111. [b] | a | = c C D + F D = @C ⇒ C D + F D − @C = 0 CD + FD = ?D For tangential circle.. 3 e. Road Ranchi. www. Near Lalpur PS.Angle bisector theorem and similarity Sol: [d] . First key we just try on 9 locks (max) next on 8 and so on.b=2..c=1 . [c] 240/79.abcabcabc.5?9:7 ∶ 0.Non-positive Maths By Amiya. Radius R1= a/2 Centre C2=(0. solve it and all possible cases. 4 d.R2| n Centre C1=(a/2. [Note method is always useful so try to understand it ] Alternate: l l l Š1 – C – C D + C P ‹ = Š1 – C D ‹ ∗ Š1 – C‹ = [1 – 6C D + 15C o – 20C l ][1 – 6C + 15C D – 20C P + 15C o – 6C W + C l ] . 1/7 in base 10 = (0.B. 3rd Floor..9<<4?4957 . [b] 7 6*x^2 + 5x+1 114. [b] -144 Co-efficient of x^7 = 7!/(a!*b!*c!*d!) with conditions a+b+c+d=7 & b+2c+3d=7 and all are non-negative integers. 5 O+E=59 & O-E=55 .8.. 117.8 in odd places and 9. Anand Complex. [b] 40 Possible digits combinations of numbers with Sum of digits of seven digit is 59 are (i) 9.. 3rd Floor..6 (v) 9..8.9. From 2⇒ O=35 & E= 24 4 ⇒O= 46 & E = 13 6 ⇒O=57 & E = 2 From condition (sum of digits 59 and their combinations) we cant get sum of even place digits as 2 & 13.11...3 O+E=59 & O-E=33 . Road Ranchi.9.. just created question to check and re-check your method or your confidence level. so (1+a)^97=(-a^2)^97 = -a^194= -a^2 = 1 + a =A +Ba 118......7 in even places (4!/3!)*3! C. [d] Statement-1 is false..com/MathsByAmiya ©AMIYA KUMAR ....1 O+E=59 & O-E=11 .... H.7 (iv) 9..6 in even places (4!/3!)*(3!/2!) Total Cases= 40 Maths By Amiya...9.3E Learning..9.5 Rule of divisibility by 11 ..9.. so we have only one condition left O(odd place)=35 & E (even place)= 24 Favourable Cases Number of cases A..8.. 9....9.. [c] 6 Hint : mod 9 121.8....9..9... Difference of sum of odd place and even place digits should be 0...8. [d] 2 .9..8.9.facebook. Let O is sum of odd place digits and E be sum of even place digits.8 in even places 4!/3! B..33.9.9..7 (iii) 9..2 O+E=59 & O-E=22 . 095 34 002244 116..6 We cant get integral values from conditions 1..8 in odd places and 9.gl/WlbehI 120.8 in odd places and 8.9... Sol: [b] 64 http://goo..9.8...3 & 5...... a is cube root of unity ...8.7.. Near Lalpur PS.22.8 (ii) 9. then O+E=59 & O-E=0 .B. 3E Learning. 1 is not true for P=6n-1 119.. statement-2 is true .9..9.9. www....4 O+E=59 & O-E=44 ......9.9.. [e] Not This is not a possible case... 9.9..9.... 9.9.9.9...9. 1>7 79=Œ = CD 1 1 57ℎ 79=Œ = − – = – –Ô• D 1−C 1 − CD 1 − CD ---------------------------------------------------------Adding – 1 1 CD + C ¹5 = − = 1 − C 1 − C D– (1 − C)(1 − C D– ) ” † So...+ n* C(n...1)+2* C(n.2) +3* C(100.100) =100* 2^99 Digital sum = 100* 2^99 mod 9 = 8 123. 3E Learning.n)= n*2^(n By Sambit: (1+x)^n=1+nc1*x+nc2*x^2...2) +3* C(n.....= |C| < 1 –Ô• 124........ Anand Complex.........n)= n*2^(n-1) Now put n=100C(100.facebook..... 095 34 002244 122............+ 100* C(100. H.... put x=1 ..... [c] 4/15 n(s) = 9*10*10 n(E) = ∑V”÷D 5(5 − 1) = 240 P(E) = 240/900 = 4/15 Maths By Amiya..3)+.+ncn*x^n differentiate n*(1+x)^(n-1)=nc1+nc2*x+nc3*x^2. www.....You will get C(n.2) +3* C(n... [d] 8 n*2^(n--1) C(n. .1)+2* C(n. lim”→ஶ ¹” = ¹ஶ = T\† lim”→ஶ C D = 0 <.B....3E Learning.....+ n* C(n.1)+2* C(100. [b] † T\† C 1 1 = − 1 − CD 1 − C 1 − CD 1 1 CD 256 79=Œ = = − o D 1−C 1−C 1 − Co Co 1 1 3=6 79=Œ = = − ¦ o 1−C 1−C 1 − C¦ .3)+...com/MathsByAmiya ©AMIYA KUMAR ... . Near Lalpur PS......3)+.. 3rd Floor. Road Ranchi....... [c] Tlo √llW m/min d. 3rd Floor. b= 4 & c= 3 .B. ∴ AC is perpendicular to each line in the plane ABD. 095 34 002244 125. Tlo √llW ƒ(஽¥) m/min ƒï = ƒŠ√V[oTï Y ‹ ƒï = ¦Dï D∗√V[DWï Y [c] (3. Fenku is at A on the bridge and Pappu is on the boat is at B on the river.2).com/MathsByAmiya ©AMIYA KUMAR . ¸ D = ¸D +  D = 3D + (5 ∗ 7))D + (4 ∗ 7)D [DC is distance between Fenku & Pappy at time t] ¸ = √9 + 417 D ∴ Rate at which they are moving away = So at t=4 .6 @ ∗ à1 + ¬à2 + ?à3 @á1 + ¬á2 + ?á3 ‫ܫ‬5?957=9 = » . Let A (2. AC = 4*t and BD = 5*t and AB = 3 m ∴ from the right-angled ΔABD. Tol √llW m/min e. Road Ranchi. [c] 2.3) 5+4+3 13 12 5+4+3 127. From the question. ¼ = ቆ» ¼ . After "t" minutes. [b] Ä ŠnY []Y [^ Y ‹ DWln]^ Let M = (a – p) (b – q) (c – r) (a*p + b*q + c*r) multiply abc abc*M = abc * (a – p) (b – q)) (c – r) (a*p + b*q + c*r) abc*M = (a^2 – a*p) (b^2 –b* b* q) (c^2 – c*r) (a*p + b*q + c*r) By . AB ⊥ AC and BD ⊥ AC. Near Lalpur PS. ¼ @+¬+? @+¬+? 36 36 5∗2+4∗5+3∗2 5∗2+4∗2+3∗6 =» . DAC. H. a= 5. ∴ AC ⊥ AD. ¸D = … D + …¸D = 3D + (5 ∗ 7))D and from the right-angled ΔDAC.2) & C(2. rate of separation = 126. ∴ AC is perpendicular to the plane of AB and BD. õ ≥ þõ (@D – @ ∗ :) + (¬ ( D – ¬ ∗ ‫ )ݍ‬+ (? D – ? ∗ =) + (@ ∗ : + ¬ ∗ ‫ ݍ‬+ ? ∗ =) 4 T ≥ Š(@ D – @ ∗ :)(¬ D – ¬ ∗ ‫ ?()ݍ‬D – ? ∗ =)(@ ∗ : + ¬ ∗ ‫ ݍ‬+ ? ∗ =)‹o Maths By Amiya. » ¼ቇ = (3.3E Learning.3) 6). Also. www. 3E Learning. B(5. Anand Complex.facebook. NoT In the beginning. ABD. One point above three and one point below three in the line of its circum-centre circum normal to the plane of old pots so that they are in pyramidal shape (upward or downward tetrahedron) 128. Fenku is at C and the boat is at D. we will get 43A+103B+131M = 16739 ..B.facebook..com/kumar..facebook... [c] 16739 3A + 7B+19M = 2219 13A+31B+47M= 5849 4*(2) ..3*(1) . Near Lalpur PS. 3rd Floor. 3rd Floor.com/kumar.com/in/kumaramiya To Join Classes 3E Learning. Near Lalpur PS.amiya http://in...B. Anand Complex.facebook.. 095 34 002244 ⇒ T @D + ¬ D + ? D ≥ Š(@D – @ ∗ :)(¬ D – ¬ ∗ ‫ ?()ݍ‬D – ? ∗ =)(@ ∗ : + ¬ ∗ ‫ ݍ‬+ ? ∗ =)‹o 4 Max value of @¬? õ = Š@ Š D – @ ∗ :‹Š¬ D – ¬ ∗ ‫ ?Š‹ݍ‬D – ? ∗ =‹(@ ∗ : + ¬ ∗ ‫ ݍ‬+ ? ∗ =) o Max Value of M = Ä ŠnY []Y [^ Y ‹ DWln]^ @D + ¬ D + ? D = ቆ ቇ 4 129..3E Learning.. H.... 3E Learning.. Anand Complex.com/MathsByAmiya ©AMIYA KUMAR .. www. 095 34 002244 Maths By Amiya.a https://www...com/MathsByAmiya To Follow Amiya : https://www.(1) . Road Ranchi.linkedin.(2) To get more questions follow www. H. [b] 10 Hint: Have to work till 4!^4! 130. Road Ranchi.facebook. 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