maths by amiya xat 1-130

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Maths By Amiya XAT 1-130 1.1.1.1. 3456 7ℎ9 :;457 ;< 4579=>9?74;5 ;< :@4= ;< >7=@4Aℎ7 B459>

2CD − 5CF + 3FD − 9C + 12F + 9 = 0 a. (0,-1) b. (3,1) c.(9,5) d. (6,3) 2.2.2.2. Total number of integral solution of family of equation C + F = 1 − O, CP + FP = 1 − OD , is a. 3 b. 4 c. 6 d. None of these 3.3.3.3. Total number of real solution pair of (x , y) if C = CD + FD & F = 2CF. a. 3 b. 4 c. 1 d. None of these(???) 4.4.4.4. If (1 + 2C)DS = @S + @TC + @DCD + @PCP + ⋯ + @TVCTV + @DSCDS then 3@S + 2@T + 3@D + 2@P + ⋯ + 2@TV + 3@DS = ? a. W∗PYZ[ P

D b. W∗PYZ\ PD c. W∗PYZ\T

D d. W∗PYZ[TD

e. None of these 5.5.5.5. 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 6.6.6.6. If c = TSTdY

e! , What is the value of integral "k" for maximum value of N a. 9 b. 10 c. 11 d. 101 e. None of these

7.7.7.7. 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 8.8.8.8. 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

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9.9.9.9. 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 isa. l

D[√TP b. √5

10.10.10.10. If 2x+3y=18 intersect y-axis at B. (that AB=AC , where A=(10,10). Then 13*(m+a. 130 b. 132

11.11.11.11. How many positive integral value of m are possible than 1. a. 8 b. 20 c. 22

12.12.12.12. Mr. Sood has three 1 rupee coins and threesame denomination. Then what is the least amount he cannot give to any one with the help of any three coins a. 7 b. 8

13.13.13.13. If we have to fence a 40m*find the minimum length of the required fence.

14.14.14.14. Ten basket, numbered 1-10 are carrying balls. Each balls weighs either 999gm or 1000gm and each basket 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 from tenth basket is 1022870. Then the baskets that have lighter balls are a. 1,3,5 b. 2,4,5

15.15.15.15. If nP = n[]o = n[][

W a.To b.TP

16.16.16.16. If pp[qp[rp[⋯[(ps)p

tp[up[vp[⋯[(ps\t)p a. 99 b. 100

17.17.17.17. If BCKH, HKFA & KDEF are unit area square thenxy(∆{|})xy(∆{~}) =? a. To b.TW c.Tl d.


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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

c. 2 d. TPl e. None of these

axis at B. (m,n) is coordinates of point C(�AB=AC , where A=(10,10). Then 13*(m+n)=?

c. 122 d. 78 e. None of theseHow many positive integral value of m are possible , if TTT

� is a terminating c. 22 d. 24 e. None of These

has three 1 rupee coins and three other coins of any integral (>1) denomination. Then what is the least amount he cannot give to any one with the help of

c. 9 d. 10 e. None of Tm*30m ground to keep people 7 meter away from the ground, then

find the minimum length of the required fence.________________? 10 are carrying balls. Each balls weighs either 999gm or 1000gm

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 from tenth basket is 1022870. Then the baskets that have lighter balls are

b. 2,4,5 c. 1,9 d. 2,8 e. None of these[^ = n[][^[�

l then n][D^[P� =?

c.c.c.c.TD d. 1 e. None of these

))p > 1.01 then maximum (n) = ? c. 125 d. 150 e. None of these

If BCKH, HKFA & KDEF are unit area square then

d.T� e. None of these

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In a triangle. two vertices are (2,3) & (4,0) and its circumcentre is (2,y) then e. None of these

�) on the given line such e. None of these

is a terminating decimal greater e. None of These

other coins of any integral (>1) value but of denomination. Then what is the least amount he cannot give to any one with the help of

e. None of These ground to keep people 7 meter away from the ground, then

10 are carrying balls. Each balls weighs either 999gm or 1000gm 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

e. None of these

e. None of these

e. None of these



18.18.18.18. There are how many integral ordered pair of (x,y) exist for �∗�

TD = TD�\D

a. 5 b. 7 c. 9 d. 29 e. None of these

19.19.19.19. If S ={1,2,3,......,50} , then N is the number of subsets of S, such that it has at least one element divisible by 5, then what would be digital sum of N.

a. 2 b. 4 c. 6 d.7 e. None of these 20.20.20.20. For all positive integers a, b & c , if n√D[]

]√D[^ is a rational number, then among following, which should be always an integer.

a. �nY[]Y\^Y�(n[]\^) b. nY\]Y[^Y

n\][^ c. \nY[]Y[^Y\n[][^

d. nY[]Y[^Yn[][^ e. None of These

21.21.21.21. In an isosceles ∆ABC , AB=AC. D is point on AC such that BD=BC=2cm, then ar(ar(ar(ar(∆ABC)=?∆ABC)=?∆ABC)=?∆ABC)=?

a. 4 cm^2 b. 7 cm^2 c. 2 cm^2 d. √7 ?�D e. None of these

22.22.22.22. Train-A leaves Ranchi at 7 : 30 am and reaches Jamshedpur at 11 : 30 am. Train-B leaves Jamshedpur at 9 : 30 am and reaches Ranchi at 1 :00 pm. If the two trains are travelling with no acceleration then at what time do they cross each other. a. 9: 36 am b. 10 : 16 am c. 10 : 26 am d. Data inadequate e. None of these

23.23.23.23. 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. a. 10 b.11 c. 12 d. 14 e. None of these

24.24.24.24. If line x=k equally divides a triangle whose vertices are (9,1), (1,1) and origin. Then k=? a. 2 b. 3 c. 4 d. 5 e. None of these

25.25.25.25. If 2>45� + 3?;>� = 3√2 & 3>45� + 2?;>� = 1, 7ℎ95 � =? (45 69A=99), where A,B & C are angles of a triangle. a. 30 b. 60 c. 150 d. (a) or (c) e. None f these

26.26.26.26. What would be summation of all possible values of "a" for which quadratic equation

CD + @C + @ + 1 = 0 has integral roots. a. 3 b. 0 c. 4 d. -3 e. None of these 27.27.27.27. If product of two roots of Co − 10CP + �CD + 838C − 2340 = 0 is -45 the find the value of

"m" a. 49 b. -49 c. 52 d. Cannot be determined e. None of these



28.28.28.28. If a+b+c+d+e = 8 & a^2+b^2+c^2+d^2+e^2 = 16, where a,b,c,d& e are real numbers then maximum (a,b,c,d,e)=?

a. 4 b. 2 29.29.29.29. A monkey is at origin (0, 0). It always jumps

point with integer coordinates and always covers a distance of 5 units in each jump. then minimum number of jumps monkey required to go from (0, 0)origin to (0, 1) ?

a. 2 b. 3 30.30.30.30. A 12-hour digital watch displays

7 instead of 1 i.e at 17 past 1 it shows 7:77the correct time ? a. 1/2 b. 5/8

31.31.31.31. Abhishek & Mahajan two players play 5, 6 & Mahajan writes 8, 9, 10 on threepapers randomly. Then AbhishekMahajan adds his two. the onAbhishek wins ? a. 1/3 b. 5/9

32.32.32.32. If n] + ]^ + ^

n = 11 & ]nnon zero real numbers. a. 1967 b. 67

33.33.33.33. A team of harvesters was assigned the task to harvest two fields, one twice the size of the other. Half a day the team worked on the larger field. Then it split into two equal groups: the first remained in the larger field and harvested it by evening; the second group harvested the smaller field, but by evening there still remained a portion to harvest. One member of ta single day’s harvest this portion the next day. How many men were there in the team, assuming all men are equi- a. 4 b. 8

34.34.34.34. If <(C) = (@ − C�)��, where a > 0 and n is

a. x b. C �� c.

35.35.35.35. If ABCDEF is a regular hexagon with side 1 unit and ARQB and ASPF

two squares inside the hexagon then ar(a. 3 b. 4 d. 1 e. None of these



& a^2+b^2+c^2+d^2+e^2 = 16, where a,b,c,d& e are real maximum (a,b,c,d,e)=?

c. 16/5 d. 6/5 e. None of theseA monkey is at origin (0, 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. then minimum number of jumps monkey required to go from (0, 0)origin to (0, 1) ?

c. 4 d. 5 e. None of theseisplays only hour and minute. Due to low battery the watch shows

7 instead of 1 i.e at 17 past 1 it shows 7:77. then what fraction of the day will that c. 3/4 d. 5/6 e. None of theseplayers play a game : on three different papers

writes 8, 9, 10 on three different papers. Then they pick two cards from their bhishek multiplies the two of his paper number

the one who gets larger number will win. What is the probability that c. 4/9 (D)1/9 e. None of these

+ ^] + n

^ = 8 then n�]� + ]�

^� + ^�n� − 3

b. 67 c. 1067 d. Cannot be determined A team of harvesters was assigned the task to harvest two fields, one twice the size of the

eam worked on the larger field. Then it split into two equal groups: the first remained in the larger field and harvested it by evening; the second group harvested the smaller field, but by evening there still remained a portion to harvest. One member of ta single day’s harvest this portion the next day. How many men were there in the team,

-capable of doing work? c.16 d. 32 e. None of these

, where a > 0 and n is a positive integer, then f[f(x)] is equal toc. C� d. @ − C� e. None of These

If ABCDEF is a regular hexagon with side 1 unit and ARQB and ASPF e the hexagon then ar(∆PAQ)/ ar(∆SRP).

c. 2 e. None of these



& a^2+b^2+c^2+d^2+e^2 = 16, where a,b,c,d& e are real e. None of these

from a point having integer coordinates to a point with integer coordinates and always covers a distance of 5 units in each jump. then minimum number of jumps monkey required to go from (0, 0)origin to (0, 1) ?

e. None of these low battery the watch shows

then what fraction of the day will that watch show e. None of these

papers Abhishek writes 3, Then they pick two cards from their

of his paper numbers , he has drawn, and . What is the probability that

e. None of these

=? ; where a, b & c are

d. Cannot be determined e. NoT A team of harvesters was assigned the task to harvest two fields, one twice the size of the

eam worked on the larger field. Then it split into two equal groups: the first remained in the larger field and harvested it by evening; the second group harvested the smaller field, but by evening there still remained a portion to harvest. One member of team in a single day’s harvest this portion the next day. How many men were there in the team,

e. None of these a positive integer, then f[f(x)] is equal to-

e. None of These If ABCDEF is a regular hexagon with side 1 unit and ARQB and ASPF are



36.36.36.36. What is the digital sum of minimum value of positive integer n, for which

√5 + 1 − √5 − 1 < 0.2 a. 6 b. 7 c. 8 d. 9 e. None of these

37.37.37.37. 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, such that PQ=5cm and QR=6 cm then what would be radius of the circle.

a. 2.5 b. 3.5 c. 2.75 d. 3.75 e. None of these

38.38.38.38. (24C0)*(25C1)+ (24C1)*(25C2)+ (24C2)*(25C3)+ ............+ (24C23)*(25C24)+ (24C24)*(25C25) = ? a. 26C13 b. 29C14 c. 29C10 d. None of These

39.39.39.39. What is midpoint of domain of function <(C) = �5 − √3C + 5 for real value of x a. 0 b. -5/2 c. 5/2 d. mid point is not possible since domain is an open interval e. None of these 40.40.40.40. If sides of a triangle is in G.P with common ratio "r" then which defines "r" the best a. �T\√W

D , T[√WD � b. T\√W

D , T[√WD ¡

c. �√W\TD , √W[T

D � d. √W\TD , √W[T

D ¡ e. None of these 41.41.41.41. (1 + cot 45) (1 + cot 46) (1 + cot 47) …. (1 + cot 89)=? a. 2 b. 2^22 c. 2^23 d. 2^24 e. None of these 42.42.42.42. What is the maximum value of sin £

D ∗ sin ¤D ∗ sin ¥

D ; where A,B & C are angles of triangle. a. 3 b. TD c. To d. T¦ e. None of these

43.43.43.43. If N0 = 9, N1 = 99, N2 = 9999,...., Nn=9999..999 (2^n digits number) then sum of all digits of (N1)*(N2)*(N3)*......*(N10)

a. 2304 b.4608 c. 9216 d. 10872 e. None of these 44.44.44.44. What is the coefficient of Co in the expansion of C�

Y − CY�¡�

a. 35 b. 21 c. -35 d. -21 e. None of These 45.45.45.45. T

T + To + T

V + TTl + T

DW + ⋯ … … … . . =? a. 2.44 b. 1.644 c. 1.444 d. 1.555



46.46.46.46. 3 ∗ ?;>DC ∗ sinD C − sino C − cosDC = 0 then tanD C =? a. 0 b. 9/16 c. 1 d. 16/9 e. None of These

47.47.47.47. secoC − 4 tanP C + 4 7@5 C is a. positive b. negative c. Non-negative d. Non-positive e. None of these

48.48.48.48. The number of elements in the set {(C, F) ∶ T�¡ + T

�¡ = 125, C, F ∈ c }, where N is the set of all natural numbers, is- a. 0 b. 3 c. 4 d. 7 e. None of These

49.49.49.49. If there three parallel lines in a same plan and there are "n" points on each of the lines, then how many triangles are possible considering these points as vertices

a. c(n,3) b. 3 ∗ (5D) ∗ (25 − 1) c. (5D) ∗ (45 − 3) d. (5D) ∗ (45 + 3) e. None of these 50.50.50.50. On Friday party night Mr. 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. By this routine if his total Friday party is 84 more than his total Friday party with a particular friend. Then how many friends does he have

a. 8 b.10 c. 12 d. 16 e. None of these 51.51.51.51. If 3x - 7y=13 , 4x- 3y=35 and 14x + 6y = 22 are equation of sides of a triangle the

coordinates of orthocentre is a. (2,1) b. (-2,1) c. (2,-1) d. origin e. None of these 52.52.52.52. Total number of non negative integral ordered solution of (a,b,c) of m< a+b+c < n ; where

0<m<n belongs to integer a. c(n,3) - c(m,3) b. c(n,4) - c(m,3) c. c (n+2 , 3) - c (m+2 ,3) d. c (n+2,3) - c (m+3,3) e. None of these

53.53.53.53. If an unbiased coin is thrown 50 times then what is the probability of getting heads 1,3,5,7,...,21,23 times. a. TD b. To c. T¦ d. None of these

54.54.54.54. If an unbiased coin is thrown 50 times then what is the probability of getting atleast 1 head a. TD b. T

DY© c. DªZ\TDªZ d. None of these



55.55.55.55. What would be total surface area (approx) of uniform circular ring whose internal diameter and external diameters are 14cm and 16 cm respec. a. 47 sq. cm b. 148 sq. cm c. 188 sq. cm d. 592 sq. cm e. None of these

56.56.56.56. (1 − «)(1 − «D) + (2 − «)(2 − «D) + (3 − «)(3 − «D) + ⋯ + (10 − «)(10 − «D) =? ; where « is cube root of unity. a. 420 b. 430 c. 440 d. 450 e. None of These

57.57.57.57. If @ + ¬ + ? = 5; @¬ + ¬? + ?@ = 6 & @¬? = 7; then nY^ + ]Y

^ + ^Yn + ]Y

n + nY] + ^Y

] =? a. 3 b. 23/7 c. 43/7 d. 7 e. None of These

58.58.58.58. If 4 + √3 and "b" are roots of equation CD − √3 C + ? = 0 & then equation whose roots are b and 1/b is − a.a.a.a. 4CD + 17C + 1 = 0 b.b.b.b. 4CD + 17C + 4 = 0 c.c.c.c. Cannot be determined d.d.d.d. No such equation be possible whose roots are b and 1/b since irrational roots always occur in conjugate pair e. e. e. e. None of these

59.59.59.59. What is the minimum value of "b" , if TVo� < n

] < oPTSP for a and b being elements of Natural

number set a. 7 b. 12 c. 17 d. 27 e. None of these

60.60.60.60. If all 6 digits numbers abcdef are taken in where @ ≥ ¬ ≥ ? ≥ 6 ≥ 9 ≥ < (e.g. 998700) and written in ascending order then which number would come at 2005th position from start. a. 864000 b. 864100 c. 864110 d. 864111 e. None of these

61.61.61.61. 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, 11872 & 13152. Then M-m =? a. 320 b. 288 c. 280 d. 160 e. None of these

62.62.62.62. There are how many integral value(s) of K possible, for which f(x) = 3x^2 - 4k*x + k+3 is always positive... a. 0 b. 1 c. 2 d. 3 e. More than 3

63.63.63.63. If sum of two natural numbers is 384 and their LCM is 4216 , then what would be digital sum of their product. a. 2 b. 3 c. 5 d. 9 e. None of these



64.64.64.64. If ® = {C|C, C ∈ ®°95 c@7±=@B c±�¬9= & C ≤ 20} and set A and B are subset of X but � � �. Then how many pairs {A,B} are possible such that � ∩ � = {2,4,16,18} a. 215 b. 242 c. 729 d.728 e. None of these

65.65.65.65. Out of 5 men and 4 women , in how many ways we can select minimum 2 men and any number of women.

a. (5c3)*(2^4) b. (5c3)*(2^6) c. 2^8 d. (5c3)*[(2^6) - 1] e. None of these 66.66.66.66. If <(C) = (C − 1)TS & A(C) = CD − 1 then what would be remainder when f(x) divided by

g(x).

67.67.67.67. For how many values of Natural number 5 ≤ 1000, 5� is a perfect square. a. 500 b. 515 c. 516 d. 539 e. None of These

68.68.68.68. Find the sum of all possible values of "b" for which CD − ¬C + 30 = 0 has integral roots.

a. 72 b. 144 c. 0 d. 36 e. None of these

69.69.69.69. If there are "n" black balls , "n+1" green balls and "n-1" red balls and it is known that one should pick minimum 142 balls (without replacement, in any trial) to get all three color balls then what is the probability of picking one red ball.

a. 70/273 b. 23/70 c. 203/273 d. 47/70 e. NoT 70.70.70.70. If S(N) is denoted as sum of digits (where N is in base 10) , e.g. S(198) = 1+9+8 = 18. If

S(M)= 46 then what would be summation of least value of M and fifth least value of M. a. 799998 b. 597998 c. 499988 d. 588998 e. NoT 71.71.71.71. 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. a. 736 b. 2540 c. 2360 d. 2268 e. NoT 72.72.72.72. For how many integral value of "c". 3 lies in between roots of CD − C + ? = 0 and satisfy

a*b*c = 240 ; where a and b are integers too. a. 6 b. 7 c. 14 d. 15 e. None of these

73.73.73.73. If 1+2+3+4......+ (2a) = b^2 and 1+2+3+4+.....+(2c+1)=d^2 , where a,b,c,d are positive

integers then minimum value of (a+b+c+d) a. 73 b. 69 c. 98 d. Cannot get such values e. NoT 74.74.74.74. There are how many four digits numbers abcd are possible such that @ ≥ ¬ ≥ ? ≥ 6 a. 494 b. 495 c. 714 d. 715 e. NoT



75.75.75.75. If fig is a square of side 1 unit and all five circles are

equal then what is the radius of the circles. a. 1/3 b. 1/4 d. (√3+1)/4

76.76.76.76. What would be minimum value of a. 4 b. 9 77.77.77.77. What is the digital sum of �

sum is continuous summation of digits till single digit. 17 = 1 + 7 ≡ 8 a. 4 b. 6

78.78.78.78. If coordinates of three points are (3,11) , (fourth point such that all four become vertices of a parallelogram. a. (10,32) b. ( e. NoT (???)

79.79.79.79. If �T, �D, …….., �TSS be 100sum of maximum possible value of a. 1 b. 2

80.80.80.80. If fig is a square of side 6 cm, and all four circles are tanother and to respective sides of square. Both smaller circles are congruent and both big circles are congruent too, then find the radius of smaller circle. a. 1 cm b. d. √3 cm e. NoT

81.81.81.81. There are how many intega. 8 b. 9 c. 10



If fig is a square of side 1 unit and all five circles are equal then what is the radius of the circles.

c. (√3-1)/4 e. None of these

What would be minimum value of 4>9?D¶ + 9 ?;>9?D¶ c. 21 d. 25 e. NoT

�TS·Z� � ; where [c] is greatest integer less than equal to N, digital

sum is continuous summation of digits till single digit. ¸4A47@B ¹±�(197 c. 8 d. 9 e. Not

coordinates of three points are (3,11) , (-2,-4) & (5,17) then what would be coordinates of fourth point such that all four become vertices of a parallelogram.

b. (-4, -10) c. (0,2) d. All Three possible

0 arithmetic means between 20 and 50, then what maximum possible value of �T º �D º �P º …………. º �TSS?

c. 4 d. 8 e. NoT If fig is a square of side 6 cm, and all four circles are tangential to each other and to respective sides of square. Both smaller circles are congruent and both big circles are congruent too, then find the radius

b. √2 cm c. 1.5 cm e. NoT

There are how many integral values of x for which �∗(�\W)∗��Y\TW�[WS��Y\Vc. 10 d. infinite e. NoT



is greatest integer less than equal to N, digital (197) ≡ 1 + 9 + 7 ≡

4) & (5,17) then what would be coordinates of d. All Three possible

, then what would be digital

gential to each other and to respective sides of square. Both smaller circles are congruent and both big circles are congruent too, then find the radius

� ≤ 0



82.82.82.82. In how many ways 20 red balls, 20 green balls and 20 black balls are divided in to two boys such that both will get exactly 30 ballsa. 300 b. 330

83.83.83.83. What would be coefficient of »C − 20?1

20?0¼ »C − 2D ∗ 20?220?1 a. -6360 b. -1540

84.84.84.84. There are two towers on a plane.

feet and between them. the angles of elevation of the tops ofand from Q are 60° and 45meters is a. 30 + 15√3 b. e. NoT

85.85.85.85. On a plane there are how many points , from where the shortest distance to three lines (one same plane) in which no two are parallel or collinear are equal. a. 0 b. 1

86.86.86.86. What would be the area of a 40 sided regular figure whose one side is 11 cm. a. 15400 sq. cm b. 16438 sq. cm d. 15840 e. NoT

87.87.87.87. What would be sum of last 4 digits when we convert 555^555 in to base 2 a. 1 b. 2

88.88.88.88. If line L1 || L2 and all angles "a" are equal then x+y+z = ?

a. 120 b. 150 d. 210 e. NoT 89.89.89.89. "A" starts from point A on a certain time and covers 1 m in the first minute and den on

subsequent minutes, he covers 2 m more than the previous minutestarts 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. then "A"? a. 6 b. 7



In how many ways 20 red balls, 20 green balls and 20 black balls are divided in to two boys such that both will get exactly 30 balls

c. 331 d. c(60,30) e. NoWhat would be coefficient of CTV in the expansion of 2

1¼ »C − 3D ∗ 20?320?2¼ ∗ … … … … ∗ »C − 20D ∗ 20

20 c. -2048 d. -1024 e. NoT on a plane. There are two points P and Q are on the line joining

between them. the angles of elevation of the tops of the tower from P are 30and 45°. If PQ is 30m, then what is the distance between the

b. 45 + 15√3 c. 60 – 15√3 d.

On a plane there are how many points , from where the shortest distance to three lines (one same plane) in which no two are parallel or collinear are equal.

c. 3 d. 4 e. NoT d be the area of a 40 sided regular figure whose one side is 11 cm.

b. 16438 sq. cm c. 1432√440 sq. cm e. NoT

What would be sum of last 4 digits when we convert 555^555 in to base 2 c. 3 d. 4 e. NoT

ne L1 || L2 and all angles "a" are equal then c. 180

"A" starts from point A on a certain time and covers 1 m in the first minute and den on subsequent minutes, he covers 2 m more than the previous minute. After 3 minutes, "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. then On how many minutes will the

c. 8 d. 9 e. NoT



In how many ways 20 red balls, 20 green balls and 20 black balls are divided in e. NoT

20?2020?19¼

are on the line joining their tower from P are 30° and 60°

distance between the towers in d. 60 + 15√3

On a plane there are how many points , from where the shortest distance to three lines (one

d be the area of a 40 sided regular figure whose one side is 11 cm.

What would be sum of last 4 digits when we convert 555^555 in to base 2

"A" starts from point A on a certain time and covers 1 m in the first minute and den on . After 3 minutes, "B"

starts from the same place and he covers 12 m in his first minute and on subsequent minutes will the "B" be ahead of



90.90.90.90. Assertion Assertion Assertion Assertion (A) : (A) : (A) : (A) : If m and n are integers and roots of CD + �C + 5 = 0 are rational then these

roots must be integers. Reason (R) : Reason (R) : Reason (R) : Reason (R) : If m and n are integers then roots of CD + �C + 5 = 0 are necessarily integers. a. If both (A) and (R) are true, and (R) is the correct explanation of (A). b. If both (A) and (R) are true but (R) is not the correct explanation of (A). c. If (A) is true but (R) is false. d. If (A) is false but (R) is true. e. NoT

91.91.91.91. DATERAM wishes to date few girls of JLRI and XXM-C. He has only Rs.5760 to spend on date and he can date at most 20 girls. A Date with JLRI would costs him Rs.360 only and that with XXM-C Rs.240 only. He is expected that he could get 22 kisses by a JLRI girl and 18 by XXM-C girls in return in a date. Then find the maximum number of kisses he get in return. a. 352 b. 360 c. 392 d. 412 e. NoT

92.92.92.92. If ?;>¶ + >45¶ = √2 ∗ ?;>¶ then ?;>¶ − >45¶ = ? a. √2 ∗ >45¶ b. ¿À�Á

√D c. √D¿À�Á d. 7@5¶ e. NoT

93.93.93.93. If (1 + C + CD)D� = �S + �T ∗ C + �D ∗ CD+ . . . . . + �D� ∗ CD� then (�S)^2 − (�T)^2 + (�D)^2 − (�P)^2+ . . . . −(�D�\T)^2 + (�D�)^2 = ?

a. 0 b. 1 c. -1 d. �� e.NoT

94.94.94.94. For which value of "a" logDn[P(6@D + 23@ + 21) = 4 – B;APn[�(4@2 + 12@ + 9)

a. 0 b. 2 c.− To d.−4 e. NoT

95.95.95.95. What would be sum of all different remainders when product of a pair or twins prime divided

by 9. (twins prime pair is pair of two primes differ by 2) a. Less than 7 b. More than 6 but less than 14 c. More than 13 but less equal to 21 d. More than 21 e. NoT

96.96.96.96. How many two digits numbers "N" , for which P^N would give reminder 1 when it is divided by 32 , where P can be any prime greater than 32 a. 6 b. 7 c. 11 d. 12 e. NoT

97.97.97.97. 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. 6 d. 9 e. Not



98.98.98.98. If @] + ¬n = ? , where a,b and c are different primes then a+b+c= ?

a. 20 b. 22 e. NoT

99.99.99.99. AABAA is a perfect square then What is the value of A+A+B+A+A=?a. 30 b. 35

100.100.100.100. If m(ABD)=90° , AC=BD and BC=CD then m(AEC)=? a. 115° b. 120° c. 150° d. 130° e. NoT

101.101.101.101. If f(x) = x^3 + a*x^2 + bx + c, where a, b, c aredie three times. Then what is a. 1/4 b. 1/3

102.102.102.102. A bag contains ‘100’ cards marked 1, 2, 3, ......,put back into the bag. Then ‘higher than 'Rahim' is ____ a. 99/100 b. 99/200

103.103.103.103. If a number is successively divided by 11,12,13 & 14 remainders are 3,4,5&6 respectively then what would be sum of remainders when number is successively divided by 14,13,12&11. a. 18 b. 23

104.104.104.104. If <(C) = TS∗��Y�YÄ[D��Y[P��Å[ a. 0 b. 1

105.105.105.105. For positive integer a, b

(@n ∗ ¬] ∗ ?^) ��Z + (@]

a. 30 b. 3^30

106.106.106.106. If 15x + 20y - 21 z = 0 possible value of value of x+y++z.

a. 71 b. 93



, where a,b and c are different primes then a+b+c= ? c. More than one value d. Not Possible

AABAA is a perfect square then What is the value of A+A+B+A+A=? c. 25 d. 20 e. NoT

C=BD and BC=CD then m(AEC)=?

2 + bx + c, where a, b, c are values taken respectivelyThen what is the probability that f(x) is an increasing

c. 5/12 d. 1/2 e. NoT ’ cards marked 1, 2, 3, ......,100. ‘Ram’ draws a card from the bag and

into the bag. Then ‘Rahim’ draws a card. The probability that ‘

b. 99/200 c. 99/10000 c. 99/20000 If a number is successively divided by 11,12,13 & 14 remainders are 3,4,5&6 respectively

then what would be sum of remainders when number is successively divided by c. 32 d. 0 e. NoT

[P�Æ[T then maximum value of f(x) is c. 2 d. 8 e. NoT

b, c and n; a + b + c = 30, then what is the maximum value of ( ] ∗ ¬^ ∗ ?n) �

�Z + (@^ ∗ ¬n ∗ ?]) ��Z

c. 30^3 d 30^(1/30) e. NoT & 4x - 19y+ 9z = 0 then for integral x, y &

value of x+y++z. c. 99 d. 72 e. NoT



, where a,b and c are different primes then a+b+c= ? d. Not Possible

AABAA is a perfect square then What is the value of A+A+B+A+A=?

values taken respectively by throwing a ty that f(x) is an increasing function.

draws a card from the bag and that ‘Ram’ draws a card

d. NoT If a number is successively divided by 11,12,13 & 14 remainders are 3,4,5&6 respectively

then what would be sum of remainders when number is successively divided by

then what is the maximum value of e. NoT

z which option can be



107.107.107.107. If Ç & Ç�are two roots of @CD + ¬C + ? = 0, @ � 0 then which one is correct. a. a+b+c=0 b. @ ∗ ?� + @� ∗ ? + ¬ = 0 c. (@ ∗ ?�)� + (@� ∗ ?)� + ¬ = 0 d. (@ ∗ ?�) �

�È� + (@� ∗ ?) ��È� + ¬ = 0

e. NoT 108.108.108.108. 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, one trail should have unique combination of i and j for (Key-i & Lock-j) ) a. 10! b. 10 c. 55 d. 45 e. NoT

109.109.109.109. What is the coefficient of C� in the expansion of �1 – C – CD + CP�l a. 144 b. -144 c. –132 d. 132 e. NoT

110.110.110.110. Among options which is the correct condition if circles CD + FD = @C and

CD + FD = ?D; (? > 0) are tangential. a. 2|a| = |c| b. | a | = c c. a = 2|c| d. |a| = 2c e. NoT 111.111.111.111. 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. If the height of shahjahan is 6 feet then find the height of the tower in feet.(consider tower is just on the bank of River Yamuna, and 6feet is the height of water level to his eye)

a. 12 − 6√3 b. 12 + 6√3 c. 24 d. Data is inadequate e. NoT 112.112.112.112. In a ∆ABC , AB = 3cm , BC=5cm & CA=4cm. If D and E are points on side AC and AB

respectively such that AD is angle bisector of angle(BAC) and DE||AC, then find the area of quard(ACDE)

a. 9/49 b. 54/49 c. 240/79 d. 40/49 e. NoT 113.113.113.113. Money expensed by Bewkoof Ram on his date is directly proportional to smile of his GF

in quadratic form. If his GF smiles 2 times he spend Rs 35/- , if 5 smiles then Rs 176/- One day his GF was not in mood to smile then he purchases an Alpenliebe cost Rs 1only. Then tell us if Bewkoof Ram is planning to purchase a bouquet cost Rs. 330 then how many time does his GF smile...??? a. 6 b. 7 c. 8 d. Cant expend Rs 330 e. NoT

114.114.114.114. 1/7 in base 10 = (0.abcabcabc.....) in base 4, then a+b+c =? a. 6 b. 5 c. 4 d. 3 e. NoT



115.115.115.115. If set M is containing all values of C − |C| for all real values x, then element(s) of set M is (are) a. 0 b. positive c. negative d. less than x e. Not

116.116.116.116. 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. Then it is found that sum of page numbers on remaining pages is 24000. Then page no on torn page can be ...... a. 245 b. 255 c. 265 d. 275 e. NoT

117.117.117.117. If a is one root of x^2 + x+1=0 and (1+ a)^97 = A + B*a. Then A + B =? a. -2 b.1 c. 0 d. 2 e. NoT

118.118.118.118. StatementStatementStatementStatement----1 : 1 : 1 : 1 : For each prime number P greater than 3, : (P + 1)^(2P+1) – P – 1 is divisible by 12. StatementStatementStatementStatement----2 : 2 : 2 : 2 : For each prime number P greater than 3, P^(2P+1) – P is divisible by 12. a. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1 b. Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 c. Statement-1 is true, statement-2 is false d. Statement-1 is false, statement-2 is true e. NoT

119.119.119.119. What is the area of a right angled triangle (in sq. cm) whose in-radius and circum-radius are 4cm and 6 cm respectively a. 84 b. 64 c. 60 d. 48 e. NoT

120.120.120.120. If 11^11 = 2853bba70abb then a=? a. 2 d. 3 c. 6 d. 1 e. NoT 121.121.121.121. There are how many seven digits numbers are possible whose sum of digits is 59 and are

divisible by 11. a. 32 b. 40 c. 54 d. 24 e. NoT

122.122.122.122. What is the digital sum of C(100,1)+2* C(100,2) +3* C(100,3)+......+ 100* C(100,100) , digital sum is defined as summation till single digit. a. 2 b. 4 c.6 d. 8 e. NoT



123.123.123.123. �T\�Y + �Y

T\�Ä + �ÄT\�Æ + �Æ

T\��Å + ⋯ … … . . =? <;= |C| < 1 a. T

T\� b. �T\� c. T

T\�Y d. �YT\�Y e. Not

124.124.124.124. What is the probability of having a three digits number whose middle digit is the largest among all, from all three digits number.

a. 1/3 b. 24/97 c. 4/15 d. 4/25 e. NoT 125.125.125.125. 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. 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. TloDW m/min b. Tol

DW m/min c. Tlo√llW m/min d. Tol

√llW m/min e. NoT 126.126.126.126. What is the coordinates of incenter of triangle whose vertices are (2,2), (5,2) & (2,6) a. (3,2) b. (2.5,2.5) c. (3,3) d.(2,3) e. NoT 127.127.127.127. 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. Then We have many different places where we can put fourth flower pot so that distance between any two post would be always same.

a. 0 b. 1 c. 2 d. 4 e. NoT 128.128.128.128. What is the max value of (a – p) (b – q) (c – r) (a*p + b*q + c*r) , if a, b, c, p ,q, r ,

(a – p), (b – q) and (c – r) all are positive quantities a. n[][^

DWln]^ b. �nY[]Y[^Y�ÄDWln]^ c. nY[]Y[^Y

DWln]^ d. 256 e. NoT 129.129.129.129. Sum of last two digits of 1!^1! + 2!^2! + 3!^3! + 4!^4!+...... a. 11 b. 10 c. 9 d. 14 e. NoT 130.130.130.130. If total cost of 3 apples , 7 bananas and 19 mangoes is Rs. 2219/- and total cost of 47

mangoes, 13 apples and 31 bananas is Rs 5849 /-, then which would be total cost of 43 apples, 103 bananas and 131 mangoes, if all costs are natural numbers.

a. 15739 b. 17639 c. 16739 d. 20000 e. NoT

ALL THE BEST



Solutions 1.1.1.1. Solution: (d) (6,3)Solution: (d) (6,3)Solution: (d) (6,3)Solution: (d) (6,3) 2CD − 5CF + 3FD − 9C + 12F + 9 = 0 General equation is @CD + 2ℎCF + ¬FD + 2AC + 2<F + ? = 0 Therefore a=2, h=-5/2, b=3, g=-9/2, f=6 & c=9 Consider the equations formed by first two rows* of i.e. ax + hy + g = 0 and hx + by + f = 0 i.e. 2C − W

D F – VD = 0 @56 − W

D C + 3F + 6 = 0 Solving these, we get the required point of intersection. i.e. 4C − 5F – 9 = 0 ; −5C + 6F + 12 = 0 Solving the above equation, we get x = 6, y = 3. Note:Note:Note:Note: �4C − 5F – 9�(−5C + 6F + 12) � 2CD − 5CF + 3FD − 9C + 12F + 9 *It's just a condition to make a general equation, an equation of pair of straight line. 2.2.2.2. [d] infinite many

3.3.3.3. [b]4 4.4.4.4. [d]

5.5.5.5. [d] None of these 1670 6.6.6.6. b. 10

7.7.7.7. [b] 1/10

8.8.8.8. [b] 1680 In a game there are one winner and two loosers... and one can get only +2 and -1

point in a game. According to questions total points after tournament =0 So, points of one player 2*W + (-1)*L = 0 & W+L=9 So, 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. its same as arranging X,X,X,Y,Y,Y,Z,Z,Z which is 9!/3!*3!*3! =1680 where X,Y & Z are win by three different players.

9.9.9.9. d. TP

l

10.10.10.10. c. 122 (m,n)=(132/13 , -10/13)



11.11.11.11. d. 24

12.12.12.12. e. None of These; all are possible except

1 & 2

15.15.15.15. c.c.c.c.TD

16.16.16.16. d. 150

17.17.17.17. b.TW Sol: AXF~BAF and ratio of corresponding sides [check hypotenuses i.e. AF and BF] is 1 : rt(5)So required ratio is 1 : 5 or,

18.18.18.18. c. 9 , all factors of 144 except 2 19.19.19.19. c. 6; All - None = 2^50 - 2^40 mod 9 =6



; all are possible except

13.13.13.13. 184m

14.14.14.14. d. 2,8 weight gap then binary base

AXF~BAF and ratio of corresponding sides [check hypotenuses i.e. AF and BF] is 1 : rt(5)

, all factors of 144 except 2 2^40 mod 9 =6



weight gap then binary base

AXF~BAF and ratio of corresponding sides [check hypotenuses i.e. AF and BF] is 1 : rt(5)



20.20.20.20. d. nY[]Y[^Y

n[][^ Sol: [d]

Rationalizing n√D[]]√D[^ = n√D[

]√D[For being rational, ¬D − @? =Only in [d] @D + @D ∗ =D + @o ∗denominator @ + @= + @=D = 21.21.21.21. d. √7 ?�D

22.22.22.22. c. 10 : 26 am

23.23.23.23. c. 12 ; a/3=12/b, ab=

24.24.24.24. b. 3

25.25.25.25. [a] 30

2sinA + 3cosB =3√2 (i)3sinB+2cosC=1 (ii)Squaring (i) & (ii) and add, we get sin(A+B) = 1/2 A+B = 30 or 150 But A+B=30 is not a possible case, since Hence A+B=150 , so C=30.



[][^ ∗ ]√D\^

]√D\^ = ](Dn\^)[√D�]Y\n^�D]Y\^Y

0, i.e. a, b &c are in GP , taking a , @=, @=D respectively.∗ =o = @D(1 + =D + =o) = @(1 + = + =D)(1@(1 + = + =D); Hence answer is [d]

a/3=12/b, ab=36 , by AM>=GM , a+b >=12

(i) (ii)

Squaring (i) & (ii) and add, we get

But A+B=30 is not a possible case, since � ≤ 30 & � ≤ 30 wont satisfy the equation (i), so C=30.



�respectively.

)( − = + =D) is multiple of

wont satisfy the equation (i)



26.26.26.26. c. 4; a=5,-1 27.27.27.27. [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 = ÎÏÐÑÑÒÎÒÐsÓ ÏÑ ËsÔtÎÏÐÑÑÒÎÒÐsÓ ÏÑ Ës

• Product of all roots = ÕÖ×ØÙÚ×Ù ÙÛÜÝÎÏÐÑÑÒÎÒÐsÓ ÏÑ Ës

Since product of two roots = -45 , so product of rest two roots = − DPoS

\oW = 52 Let S be sum of roots whose product is -45 then equation would be CD − ¹C − 45 = 0 and Let R be sum or roots whose product is 62 then equation would be CD − ÞC + 52 = 0 So our give equation can be written as product of above two equations, Co − 10CP + �CD + 838C − 2340 = (CD − ¹C − 45) ∗ (CD − ÞC + 52) Co − 10CP + �CD + 838C − 2340 = Co − (¹ + Þ)CP + (7 + ¹Þ)CD + (−52¹ + 45Þ)C − 2340 Equating coefficient of CP , CD & C we get , S+R=10 ...(i) 7+SR = m ...(ii) -52S + 45R = 838 ...(iii) From (i) & (ii) we get, S= 14 & R= -4 putting these in (ii) , we finally get m=7+14*(-4)=-49 28.28.28.28. c. 16/5

@ − lW¡D + ¬ − l

W¡D + ? − lW¡D + 6 − l

W¡D + 9 − lW¡D

= (@D + ¬D + ?D + 6D + 9D) − TDW (@ + ¬ + ? + 6 + 9) + Pl

W = 16 − VlW + Pl

W = 4. Hence ß@ − l

Wß ≤ 2, >; @ ≤ TlW .

This value can be realized by putting @ = TlW and setting the other numbers to lW.

Alternate solution By Saransh Sood:Alternate solution By Saransh Sood:Alternate solution By Saransh Sood:Alternate solution By Saransh Sood: Assume rest 4 variables to be equal. and max(a,b,c,d,e) would obviously be less than 4. (@> @D + ¬D+. . . +9D = 4D) now @ + 4C = 8 @D + 4CD = 16 => (8 − 4C)D + 4CD = 16 => C = 6

5 @56 2 65 4> >�@BB9=. ℎ95?9 �@C (@) = 8 − 24/5 = 16/5



29.29.29.29. [b] 3

30.30.30.30. a. 1/2

The clock will show the incorrect time (between 1 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 = 240Total incorrect time = 240 + 480 = 720correct time = 1 – incorrect time= 1 – 720/(24*60) = 1/2

31.31.31.31. c. 4/9

32.32.32.32. c. 1067 Let n] = à, ]

^ = á & ^n = â

then X+Y+Z = 11 & XYZ = 1and T

ã + Tä + T

å = 8 => XY+YZ+ZX= 8XYZ=8 Nown�

]� + ]�^� + ^�

n� − 3 = àP= (à + á + â)�(à + á + â

33.33.33.33. b. 8 34.34.34.34. a. x ; easy put the value and get answer35.35.35.35. [c] 2 @=(∆æ�ç) = »1

2¼ ∗ 1 ∗ 1 ∗ >4530 ar(∆SRP) = 1

2 ∗ √2 ∗ √2 ∗ >45ar(∆PAQ)ar(∆SRP) =

1214= 2



The clock will show the incorrect time (between 1 – 2, 10 -11, 11-12 , 12 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

incorrect time

then X+Y+Z = 11 & XYZ = 1 => XY+YZ+ZX= 8XYZ=8

+ áP + âP − 3àáâ = (à + á + â)(àD + áDâ)D − 3(àá + áâ + âà)� = 11 ∗ (11D − 3

easy put the value and get answer

30 = 14

>4530 = 12



12 , 12 - 1 day and night

D + âD − àá − áâ − âà) ∗ 8) = 1067



36.36.36.36. [c] 8 ; n>25.01 => min(n)=26 so digital sum =8

37.37.37.37. d. 3.75

38.38.38.38. d. None of These 49C24

Sol: We know (1+x)^24 = 24C0 + 24C1*x + 24C2*x^2 +24C3*x^3 +......+ 24C23*x^23+24C24*x^24& (x+1)^25 = 25C0*x^25 + 25C1*x^24 + 25C2*x^23 +......+ 25C24*x+25C25 Multiplying above two and coefficient of x^24 in (1+x)^............+ (24C23)*(25C24)+ ⇒ 49C24 = (24C0)*(25C1)+(24C24)*(25C25)

39.39.39.39. c. 5/2 3C + 5 ≥ 0 => C ≥ −5/3 & 3C + 5 ≤ 25 => C ≤ 20/3− W

P ≤ C ≤ DSP ; then midpoint of this is 5/2

40.40.40.40. d. √W\T

D , √W[TD ¡

41.41.41.41. c. 2^23 (1 + cot 45) (1 + cot 46) (1 + cot 47) …. (1 + cot 89(1 + tan 45) =2^23



.01 => min(n)=26 so digital sum =8

(1+x)^24 = 24C0 + 24C1*x + 24C2*x^2 +24C3*x^3 +......+ 24C23*x^23+24C24*x^24(x+1)^25 = 25C0*x^25 + 25C1*x^24 + 25C2*x^23 +......+ 25C24*x+25C25Multiplying above two and equating coefficient of x^24 coefficient of x^24 in (1+x)^49 = (24C0)*(25C1)+ (24C1)*(25C2)+

(24C23)*(25C24)+ (24C24)*(25C25) 4 = (24C0)*(25C1)+ (24C1)*(25C2)+ (24C2)*(25C3)+ ............+

3 ; then midpoint of this is 5/2

46) (1 + cot 47) …. (1 + cot 89) =(1 + tan 1) (1 + tan 2) (1 + tan 3) …. [(1+tanA)(1+tan(45-A)) = 2]



(1+x)^24 = 24C0 + 24C1*x + 24C2*x^2 +24C3*x^3 +......+ 24C23*x^23+24C24*x^24 (x+1)^25 = 25C0*x^25 + 25C1*x^24 + 25C2*x^23 +......+ 25C24*x+25C25

(24C1)*(25C2)+ (24C2)*(25C3)+

............+ (24C23)*(25C24)+

) (1 + tan 2) (1 + tan 3) ….

A)) = 2]



42.42.42.42. d. T¦ If A + B + C = 180°

then cos � + cos � + cos � = 1 + 4 ∗ sin £D ∗ sin ¤

D ∗ sin ¥D

⇒ 1 + 4 ∗ sin £D ∗ sin ¤

D ∗ sin ¥D ≤ P

D �cos � + cos � + cos � ≤ PD�

⇒sin £D ∗ sin ¤

D ∗ sin ¥D ≤ T

¦ 43.43.43.43. b.4608 = 9*2^9 N1= 9 sum of digit = 9 = 9*2^0 N1*N2 = 9*99 = 819 ; sum of digit = 18 = 9*2^1 N1*N2*N3 = 9*99*999=890109 ; sum of digit = 36 = 9*2^2 so sum of digits (N1*N2*N3*.....*Nn) = 9*2^(n-1) here sum of digits (N1*N2*N3*.....*N10) = 9*2^(10-1)=9*2^9 = 4608 44.44.44.44. c. -35 General term of C�

Y − CY�¡� is (−1é) ∗ �(7, =) ∗ C�

Y¡�\é ∗ CY�¡é = (−1é) ∗ �(7, =) ∗ (CY�Èê

Å ) For Co , DT[é

l = 4 ⇒ = = 3; so, the coefficient is (−1P) ∗ �(7,3) = −35 45.45.45.45. b. 1.644 = pi^2/6

http://goo.gl/X7wi3E

46.46.46.46. c. 1 3 ∗ cosD C ∗ sinDC − sinoC − cosDC = 0 ⇒ 3 ∗ (1 − sinDC) ∗ sinDC − sinoC − (1 − sinDC) = 0 ⇒ 3 sinDC − 3 sinoC − sinoC − 1 + sinDC = 0 ⇒ −4 sinoC + 4 sinDC − 1 = 0 ⇒ 4 sinoC − 4 sinDC + 1 = 0 ⇒ (2sinDC − 1)D = 0 sinD C = 1

2 ⇒ tanD C = 1

47.47.47.47. c. Non-negative seco C − 4 tanP C + 4 7@5 C = seco C − 4 tanD C ∗ 7@5 C + 4 7@5 C

= seco C − 4 (secD C − 1) ∗ 7@5 C + 4 7@5 C = seco C − 4 secD C 7@5 C + 8 7@5 C = (secD C − 2 7@5 C − 2)D − 4 tanD C − 4 + 4 secD C = (secD C − 2 7@5 C − 2)D ≥ 0

48.48.48.48. [d] 7 ; positive integral solution http://goo.gl/shxyNb



49.49.49.49. c. (5D) ∗ (45 − 3) Total Number of triangles = all three vertices on different lines + two vertices on same line

= c(n,1)* c(n,1)* c(n,1) + c(3,1)( c(n,2)* c(2n,1)) = (5D) ∗ (45 − 3) 50.50.50.50. b.10 Let "n" be total number of friend, then Total number of Friday party = c(n,3) Total number of Friday party with a particular friend = c(n-1 , 2) By question, c(n,3) - c(n-1 , 2) = 84 => n=10 51.51.51.51. 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,-1) How:- it's a right angled triangle , check 3x - 7y=13 and 14x + 6y = 22 , coefficient interchanges with change in sign. 52.52.52.52. [d] c (n+2,3) - c (m+3,3)

m< a+b+c < n ⇒ [0 ≤ @ + ¬ + ? ≤ 5 − 1] − [0 ≤ @ + ¬ + ? ≤ �] ⇒ [0 ≤ @ + ¬ + ? + 6 = 5 − 1] − [0 ≤ @ + ¬ + ? + 6 = �]

53.53.53.53. [d] 54.54.54.54. [c] 55.55.55.55. [b] 148 , TSA of ring = pi^2 * (R^2 - r^2)

56.56.56.56. [d]

57.57.57.57. [c] 58.58.58.58. Concepts

• Irrational Conjugate Pairs of Roots only Occurs IFF, all coefficients(including constant) are Rational.

• Imaginary Conjugate Pairs of Roots only Occurs IFF, all coefficients(including constant) are Real.

• ¹±� ;< @BB =;;7> 7@ë95 ;59 @7 @ 74�9 = (−1) ∗ ^ìíîîÀ^Àí�ï ìî ��Ô�^ìíîîÀ^Àí�ï ìî ��

• æ=;6±?7 ;< @BB =;;7> = (−1)�ð�]íé ìî éììï¿ ∗ ^ì�¿ïn�ï

^ìíîîÀ^Àí�ï ìî �� Sum of roots = 4 + √3 + b = √3 ⇒b=-4 , So, equation whose roots are -4 and -1/4 is CD − (−4 + − T

o)C + (−4) ∗ (− To) = 0

⇒ CD + T�o C + 1 = 0 ⇒ 4CD + 17C + 4 = 0



59.59.59.59. [b] 12 , You can solve this option elimination method , but the approach is TVo� < n

] < oPTSP ⇒

⇒ 2 + T�oP < ]

n < 2 + VTV .....(i)

Since b is a natural number and greater than twice of a, so Let b = 2a + n ; We get (i) ⇒ 2 + T�

oP < Dn[�n

⇒2 + T�oP < 2 + �

n < 2 + VTV

⇒ T�oP < �

n < VTV ⇒

Here a and n both are integer , for smallest "a" we need smallest "n" which give integral "a"for n=1, we are not getting an integral "a" for n=2 , we have 4.22 < @so the smallest integral a = 5 So smallest b = 2a + n = 2*5 +2 =12

60.60.60.60. [c] 864110

61.61.61.61. [c] Hint:- factors of 320 greater than 32

62.62.62.62. [d] , D<0

63.63.63.63. [d],

Product = 33728 = 2^6*17*31Sum = 2^7*3 ; LCM=4216=2^3*17*13 , so HCF= HCF(Sum,LCM)=2^3numbers are 136=8*17 , 248=8*31 , so product = (2^



You can solve this option elimination method , but the approach is - ⇒ TSP

oP < ]n < o�

TV .....(i)

Since b is a natural number and greater than twice of a,

� < 2 + VTV

⇒ TV�

V < @ < oP�T�

Here a and n both are integer , for smallest "a" we need smallest "n" which give integral "a"we are not getting an integral "a"

@ < 5.05 , so the smallest integral a = 5 for n=2 So smallest b = 2a + n = 2*5 +2 =12

factors of 320 greater than 32

Product = 33728 = 2^6*17*31 Sum = 2^7*3 ; LCM=4216=2^3*17*13 , so HCF= HCF(Sum,LCM)=2^3numbers are 136=8*17 , 248=8*31 , so product = (2^3)^2*17*13 =



Here a and n both are integer , for smallest "a" we need smallest "n" which give integral "a"

Sum = 2^7*3 ; LCM=4216=2^3*17*13 , so HCF= HCF(Sum,LCM)=2^3 3)^2*17*13 =33728



64.64.64.64. [d] 3l − 1 = 728 ® = ñC│C, C ∈ ®°95 c@7±=@B c±�¬9= & C ≤ 20ó = {2,4,6,8,10,12,14,16,18,20} � ∩ � = {2,4,16,18} , it is sure that we have to arrange rest 6 numbers Take � ∪ � = õ , � − � = æ , � − � = ç, à − õ = Þ then X is union of disjoint set P,Q,R & � ∩ �. So we have to distribute rest three set P, Q , R in total 3^6 ways. If all elements are in set R then � = � = � ∩ � = {2,4,16,18} ; we have to exclude this set. so total pair {A,B} = 3l − 1 = 728 65.65.65.65. [c] 66.66.66.66. 2V − 2V ∗ C

<(C) = (C − 1)TS = ∑ �(10,25) ∗ CD�TS�÷S − ∑ �(10,25 + 1) ∗ CD�[TTS�÷S for remainder A(C) = CD − 1 = 0 ⇒ CD = 1 (put this in f(x)) We will get remainder, <(C, CD = 1) = ∑ �(10,25)TS�÷S − ∑ �(10,25 + 1) =TS�÷S 2V − 2V ∗ C http://goo.gl/C6bL94

67.67.67.67. [c] All even would come in this = 500 All odd perfect square = 16 (from 1D to 31D) 68.68.68.68. [c] 69.69.69.69. [b] To get all three colors one need to pick at least "n+1" + "n" + 1= 2n + 2 ; considering all worst conditions, Murphy's Law According to question = 2n+2 = 142 ⇒ n=70, so number of red balls = 69 and total balls = 210 So probability of first ball is red = 69/210 = 23/70 70.70.70.70. [c] S(M) = 46 , for minimum M we need the smallest digit number so we need maximum 9s, so smallest number would have five, 9s and one 1 So Least M = 199999 , (now we cant make number starting from 1) Second Least = 289999 Third least = 298999 Fourth least = 299899 Fifth least = 299989



71.71.71.71. [b] Work in base two and add number or 1s in all base but addition should be in base 2only.

20 10 10 10 10 10 10

10 6 6 6 6 6

4 3 3 3 3

1 1 1 1

20 20 20 20 20 20 20

19 19 19 18 17 15 10

10011 1 1 0 1 1 0 0 2540 Alternate:- (2^4 - 1) + (2^4*4) + 3*(2^4 - 1) + 2^5*10 + 6*(2^5 - 1) + 2^6*20 + 10*(2^6 - 1) 72.72.72.72. [c] 1 − 4? > 0 ⇒ ? < T

o & 9 − 3 + ? < 0 ⇒ ? < −6 so finally ? < −6 From a*b*c = 240 . c is factor (with negative sign) So we have 14 values of c which will satisfy the conditions

73.73.73.73. [b] 1 + 2 + 3 + 4. . . . . . + (2@) = ¬D = 2@(2@ + 1)

2 @(2@ + 1) = ¬D >; �454�±� 5;5 − O9=; @ = 4 @56 ¬ = 6 1 + 2 + 3 + 4. . . . . . + (2? + 1) = ¬D = (2? + 1)(2? + 2)

2 (2? + 1)(? + 1) = 6D >; �454�±� ? = 24 @56 6 = 35 74.74.74.74. [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(13,4) solutions but we have to reduce one value when all digits are 0 so required number = C(13,4)-1= 714 Alternate: We have to chose 4 numbers , w,x,y,z strictly descending order out from thirteen numbers 0,1,2,3...8,9, p,q,r Let a=w-3,b=x-2, c= y-1, d=z , total ways = 13c4 = 715 , we have to reduce one case when 3,2,1,0 so required number = 714 75.75.75.75. [c] Use in right angled triangle, r(45=@64±>) = n[]\^

D => (where c=1, a=a(smallest side) , b= a+2r) => a = 1/2 ; from Pythagoras or 30:60:90 ; r = (√3-1)/4



76.76.76.76. [d]

4>9?D¶ + 9 ?;>9?D¶ = 4(1 + tanD ¶) + 9(1 + cotD ¶) = 13 + 4 ∗ tanD ¶ + 9 ∗ cotD ¶ = 13 + (4 ∗ tanD ¶ + 9 ∗ cotD ¶ − 12) + 12 = 25 + (2 tan ¶ − 3 cot ¶)D So minimum value of above expression is 25

77.77.77.77. [b] 6 Sol: TS·Z

� = TS� ∗ 10lV = 1.428571=9;??±=45A ∗ 10lV

ø10�S7 ù = [1.428571=9;??±=45A ∗ 10lV] = 1(428571 =9:9@7> 11 74�9>)428

Digital Sum (1(428571 repeats 11 times)428) ≡ Digital Sum (1+9*11 + 5) ≡ 6

78.78.78.78. [e] NoT Parallelogram is not possible with these three points, Since these are collinear points

79.79.79.79. [a] 20, �T, �D, …….., �TSS, 50 will be an AP with Arithmetic Mean equal to

DS[WSD = 35.

Thus, �T, �D, …….., �TSS will also be a AP with Arithmetic Mean equal to 35.

Using AM≥GM, �T º �D º �P º …………. º �TSS ≤ 35TSS So maximum value of�T º �D º �P º …………. º �TSS =35TSS and its digital sum = 35TSS mod 9 =1 80. [a] (1.5+x)^2 = 1.5^2 +

(3-x)^2

81.81.81.81. [c] 82.82.82.82. [c] 83.83.83.83. [b]

84.84.84.84. [d] 85.85.85.85. [d] incentre & excentre

86.86.86.86. [e]

Concept: Circle has maximum area for given perimeter, (here we have perimeter) Perimeter = 40*11 = 440 cm The maximum area of 2-D figure with perimeter = 440 cm is of a circle, 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 , so area of required fig should be less than this, and all options are more or equal to this so answer is [e]NoT

87.87.87.87. [b] 555^555 mod 16 = 3 so last 4 digits in base 2 = 0011



88.88.88.88. [c] Extend L2 and make a 7 sided polygon.Sum of all interior angles of this = 900a+x+360-a+y+360-a+z+a=900 89.89.89.89. [a] Till 8th minute from 3rd (B's Start) B would be ahead

90.90.90.90. [c]

91.91.91.91. [c]

x + y ≤ 20 & 360x + 240y Zmax = 22x + 18y & x, y ≥ Points (x, y) Value of the objective function Z = 22x + 18O (0, 0) Z= 22 º 0 + 18 º 0 = 0A2(16, 0) Z = 22º16 + 18 º 0 = 352P (8, 12) Z = 22 º 8 + 18 º 12 = 392B1(0, 20) Z = 22 º 0 + 20 º 18 = 360

92.92.92.92. [a]

93.93.93.93.



[c] Extend L2 and make a 7 sided polygon. Sum of all interior angles of this = 900

a+z+a=900 ⇒ x+y+z = 180 [a] Till 8th minute from 3rd (B's Start) B would be ahead

y ≤ 5760 ≥ 0

Value of the objective function + 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





94.94.94.94. [c] best is option elimination method, put the option and check

95.95.95.95. [c]

Except (3,5) all product of twins prime have digital sum (remainder by 9) is 8.So, we have only two remainders 6 for (3,5) and 8 for rest so to http://goo.gl/kWdm3d

96.96.96.96. [c] http://goo.gl/zDVQt7

97.97.97.97. [b] obtuse angled

100.100.100.100. [b] 120° Complete the rectangle ABDF.AC = BD = AF CB = CD and OB = OD imply CO is theand hence of AF as well. Follows AC = FC and triangle ACF is equilateral.degrees.

101.101.101.101. [c] 5/12 Sol. Sol. Sol. Sol. f´(x) = 3*x^2 + 2*a*x + b⇒ f´(x) > 0 ∀ x ⇒ (2a)^2 This is true for exactly 15 ordered pairs (a, b); 1

102.102.102.102. [b] 99/200 Let ‘Ram’ draws a card marked with Hence the required probability =



[c] best is option elimination method, put the option and check

Except (3,5) all product of twins prime have digital sum (remainder by 9) is 8.So, we have only two remainders 6 for (3,5) and 8 for rest so total sum is 14.

98.98.98.98. [b]

http://goo.gl/OyufpL99.99.99.99. [c] 212^2 = 44944

Complete the rectangle ABDF. Join BF to intersect AD at O. CB = CD and OB = OD imply CO is the perpendicular bisector of BD Follows AC = FC and triangle ACF is equilateral. So x = supplement of angle CAF = 120

x + b ; y = f(x) is strictly increasing 2 – 4*3*b < 0

This is true for exactly 15 ordered pairs (a, b); 1 ≤ a, b ≤ 6, so probability =

’ draws a card marked with n, then Rahim can draw any card marked 1, 2, 3, .... required probability = ∑ T

TSS¡ �\TTSS ¡ = VV

DSSTSS�÷T



Except (3,5) all product of twins prime have digital sum (remainder by 9) is 8. tal sum is 14.

http://goo.gl/OyufpL [c] 212^2 = 44944

So x = supplement of angle CAF = 120

6, so probability = 15/36=5/12

draw any card marked 1, 2, 3, .... n–1.



14 11003⇩ 13 13 785⇩ 5 12 60⇩ 0 11 5⇩ 5

Quotient 0

103.103.103.103. [b] remainders are 13,5,0,5

Use 11003

104.104.104.104. [b] for x=±1

105.105.105.105. [a] 30 Concept �õ ≥ þõ (@ + @ + @+. . . . + @ 74�9>) + (¬ + ¬ + ¬+. . . . + ¬ 74�9>) + (? + ? + ?+. . . . + ? 74�9>)

@ + ¬ + ?≥ [(@ ∗ @ ∗ @ ∗ … ∗ @)(¬ ∗ ¬ ∗ ¬ ∗ … ∗ ¬)(? ∗ ? ∗ ? ∗ … ∗ ?)] Tn[][^

=n∗n[]∗][^∗^n[][^ ≥ (@n ∗ ¬] ∗ ?^) �

�È�È� ...(A) Similarly n∗][]∗^[^∗n

n[][^ ≥ (@] ∗ ¬^ ∗ ?n) ��È�È� ...(B)

& n∗^[]∗n[^∗]

n[][^ ≥ (@^ ∗ ¬n ∗ ?]) ��È�È� ...(C)

By adding all three, we get nY[]Y[^Y[Dn][D]^[D^n

n[][^ ≥ (@n ∗ ¬] ∗ ?^) ��È�È� + (@] ∗ ¬^ ∗ ?n) �

�È�È� + (@^ ∗ ¬n ∗ ?]) ��È�È�

⇒(n[][^)Yn[][^ ≥ (@n ∗ ¬] ∗ ?^) �

�È�È� + (@] ∗ ¬^ ∗ ?n) ��È�È� + (@^ ∗ ¬n ∗ ?]) �

�È�È� ⇒@ + ¬ + ? ≥ (@n ∗ ¬] ∗ ?^) �

�È�È� + (@] ∗ ¬^ ∗ ?n) ��È�È� + (@^ ∗ ¬n ∗ ?]) �

�È�È� 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

106.106.106.106. [c] 99 ; since x:y:z=3:3:5 15x + 20y - 21 z = 0 4x - 19y+ 9z = 0 �

(DS∗V\TV∗DT) = − �(TW∗V[o∗DT) = �

(\TW∗TV\o∗DS) ⇒ x:y:z=3:3:5

11 11003 3 12 1000⇧ 4 13 83⇧ 5 14 6⇧ 6

Quotient 0⇧ Can take any value



107.107.107.107. [d] (@ ∗ ?�) �

�È� + (@� ∗ ?) ��È� + ¬ = 0

By option elimination or sum of roots & product of roots concept 108.108.108.108. d] 45 ; First key we just try on 9 locks (max) next on 8 and so on...!!! 109.109.109.109. [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, solve it and all possible cases. [Note method is always useful so try to understand it ]

Alternate:Alternate:Alternate:Alternate: �1 – C – CD + CP�l = �1 – CD�l ∗ �1 – C�l = [1 – 6CD + 15Co– 20Cl][1 – 6C + 15CD– 20CP + 15Co– 6CW + Cl] �;9<<4?4957 ;< C� = (36 − 300 + 120) = −144 110.110.110.110. [b] | a | = c

CD + FD = @C ⇒ CD + FD − @C = 0 Centre C1=(a/2,0), Radius R1= a/2

CD + FD = ?D Centre C2=(0,0), Radius R2= c For tangential circle, Distance(C1toC2) = R1+ R2 (not possible) or Distance(C1toC2) = |R1 - R2| ßn

Dß = ßnD − ?ß ⇒ |a|=|c| but since c is positive so need to write mod with c hence |a|=c

111.111.111.111. [b] 12 + 6√3

112.112.112.112. [c] 240/79; Hint :- Angle bisector theorem and similarity 113.113.113.113. [b] 7 6*x^2 + 5x+1

114.114.114.114. 1/7 in base 10 = (0.abcabcabc.....) in base 4, then a+b+c =?

a. 6 b. 5 c. 4 d. 3 e. NoT Sol: [d] , a=0,b=2,c=1 �;5?9:7 ∶ 0.111111 45 ¬@>9 � = 1

� − 1 45 ¬@>9 10 http://goo.gl/1yX6Yq

115.115.115.115. [e] Not - Non-positive



116.116.116.116. [e] Not This is not a possible case, just created question to check and re-check your method or your confidence level.

117.117.117.117. [d] 2 ; a is cube root of unity , so (1+a)^97=(-a^2)^97 = -a^194= -a^2 = 1 + a =A +Ba

118.118.118.118. [[[[d] Statement-1 is false, statement-2 is true ; 1 is not true for P=6n-1 119.119.119.119. Sol: [b] 64

http://goo.gl/WlbehI

120.120.120.120. [c] 6 Hint : mod 9 121.121.121.121. [b] 40

Possible digits combinations of numbers with Sum of digits of seven digit is 59 are (i) 9,9,9,8,8,8,8 (ii) 9,9,9,9,8,8,7 (iii) 9,9,9,9,9,7,7 (iv) 9,9,9,9,9,8,6 (v) 9,9,9,9,9,9,5 Rule of divisibility by 11 , Difference of sum of odd place and even place digits should be 0,11,22,33...... Let O is sum of odd place digits and E be sum of even place digits. then O+E=59 & O-E=0 .............1 O+E=59 & O-E=11 .............2 O+E=59 & O-E=22 .............3 O+E=59 & O-E=33 .............4 O+E=59 & O-E=44 .............5 O+E=59 & O-E=55 .............6 We cant get integral values from conditions 1,3 & 5. 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, so we have only one condition left O(odd place)=35 & E (even place)= 24 Favourable Cases Number of cases A. 9,9,9,8 in odd places and 8,8,8 in even places 4!/3! B. 9,9,9,8 in odd places and 9,8,7 in even places (4!/3!)*3! C. 9,9,9,8 in odd places and 9,9,6 in even places (4!/3!)*(3!/2!) Total Cases= 40



122.122.122.122. [d] 8 C(n,1)+2* C(n,2) +3* C(n,3)+......+ n* C(n,n)= n*2^(nC(n,1)+2* C(n,2) +3* C(n,3)+......+ n* C(n,n)= n*2^(nC(n,1)+2* C(n,2) +3* C(n,3)+......+ n* C(n,n)= n*2^(nC(n,1)+2* C(n,2) +3* C(n,3)+......+ n* C(n,n)= n*2^(n----1)1)1)1) By Sambit:By Sambit:By Sambit:By Sambit: (1+x)^n=1+nc1*x+nc2*x^2..............+ncn*x^n differentiate n*(1+x)^(n-1)=nc1+nc2*x+nc3*x^2........... put x=1 ,You will get C(n,1)+2* C(n,2) +3* C(n,3)+......+ n* C(n,n)= n*2^(n-1) Now put n=100C(100,1)+2* C(100,2) +3* C(100,3)+......+ 100* C(100,100) =100* 2^99 Digital sum = 100* 2^99 mod 9 = 8

123.123.123.123. [b] �T\�

1>7 79=� = C1 − CD = 1

1 − C − 11 − CD

256 79=� = CD1 − Co = 1

1 − CD − 11 − Co

3=6 79=� = Co1 − C¦ = 1

1 − Co − 11 − C¦

............... ............ ............... 57ℎ 79=� = CD�Ô�

1 − CD� = 11 − CD�Ô� − 1

1 − CD� ---------------------------------------------------------- Adding ¹5 = 1

1 − C − 11 − CD� = CD� + C

(1 − C)(1 − CD�) lim�→� CD� = 0 <;= |C| < 1 So, lim�→� ¹� = ¹� = �

T\�

124.124.124.124. [c] 4/15 n(s) = 9*10*10 n(E) = ∑ 5(5 − 1)V�÷D = 240 P(E) = 240/900 = 4/15



125.125.125.125. [c] Tlo

√llW m/min In the beginning, Fenku is at A on the bridge and"t" minutes, Fenku is at C and the boat is at D.From the question, AB ⊥ AC and BD ∴ AC is perpendicular to the plane of AB and BD.∴ AC is perpendicular to each line in the plane∴ AC ⊥ AD. Also, AC = 4*t and BD = 5*t ∴ from the right-angled ΔABD,and from the right-angled ΔDAC,¸� = √9 + 417D [DC is distance between Fenku & Pappy at time ∴ Rate at which they are moving away = So at t=4 ; rate of separation = 126.126.126.126. [c] (3,3)

Let A (2,2), B(5,2) & C(2,6

�5?957=9 = »@ ∗ à1 + ¬à2 +@ + ¬ + ?

= »5 ∗ 2 + 4 ∗ 55 + 4 +

127.127.127.127. [c] 2; One point above three and one point below three in the line of its circum

normal to the plane of old pots so that they are in pyramidal shape (upward or downward tetrahedron)

128.128.128.128. [b] �nY[]Y[^Y�Ä

DWln]^ Let M = (a – p) (b – q) (c – r) (multiply abc abc*M = abc * (a – p) (b – q) (abc*M = (a^2 – a*p) (b^2 –b* By �. õ ≥ þõ

(@D– @ ∗ :) + (¬D– ¬

≥ �(@



d. Tol√llW m/min e. NoT

is at A on the bridge and Pappu is on the boat is at B on the river. After is at C and the boat is at D.

AC and BD ⊥ AC. is perpendicular to the plane of AB and BD.

AC is perpendicular to each line in the plane ABD. and AB = 3 m

ABD, �¸D = ��D + �¸D = 3D + (5 ∗ 7)DDAC, ¸�D = �¸D + ��D = 3D + (5 ∗ 7)D[DC is distance between Fenku & Pappy at time

Rate at which they are moving away = �(¥)�ï = ��√V[oTïY�

�ï = ¦DïD∗√V[DWïY

rate of separation = Tlo√llW m/min

6), a= 5, b= 4 & c= 3 , ?à3 , @á1 + ¬á2 + ?á3@ + ¬ + ? ¼

5 + 3 ∗ 2+ 3 , 5 ∗ 2 + 4 ∗ 2 + 3 ∗ 6

5 + 4 + 3 ¼ = »3613¼ , »36

12One point above three and one point below three in the line of its circum

normal to the plane of old pots so that they are in pyramidal shape (upward or downward

) (a*p + b*q + c*r)

) (c – r) (a*p + b*q + c*r) b* q) (c^2 – c*r) (a*p + b*q + c*r)

( ¬ ∗ �) + (?D– ? ∗ =) + (@ ∗ : + ¬ ∗ � + ?4

�(@D– @ ∗ :)(¬D– ¬ ∗ �)(?D– ? ∗ =)(@ ∗ : + ¬



the boat is at B on the river. After

)D ) + (4 ∗ 7)D

[DC is distance between Fenku & Pappy at time t]

¼ »3612¼� = (3,3)

One point above three and one point below three in the line of its circum-centre normal to the plane of old pots so that they are in pyramidal shape (upward or downward

? ∗ =)

¬ ∗ � + ? ∗ =)�To



⇒ @D + ¬D + ?D4 ≥ �(@D– @

Max value of @¬? õ = �@D– @

Max Value of M =�nY[]Y[^Y�ÄDWln]^

129.129.129.129. [b] 10 Hint: Have to work till 4!^4!

130.130.130.130. [c] 16739

3A + 7B+19M = 2219 13A+31B+47M= 5849 4*(2) - 3*(1) , we will get 43A+103B+131M = 16739 To get more questions follow www.facebook.com/MathsByAmiya

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@ ∗ :)(¬D– ¬ ∗ �)(?D– ? ∗ =)(@ ∗ : + ¬ ∗ �

� @ ∗ :��¬D– ¬ ∗ ��?D– ? ∗ =�(@ ∗ : + ¬ ∗ �= @D + ¬D + ?D

4 �o

�

Hint: Have to work till 4!^4!

..........(1) ..........(2)

To get more questions follow www.facebook.com/MathsByAmiya

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+ ? ∗ =)�To

� + ? ∗ =)

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maths by amiya xat 1-130

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