30. MATHs Mock Test 30 for NDA

• 0 Questions answered correctly
• 0 Questions answered incorrect
• 10 Questions unattempted

• 1

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    291
  Topic    Determinants

  Given

  A = | α  2 |
     | 2  α |

  det(A³) = 125

  Using

  det(A³) = [det(A)]³

  Therefore

  [α² - 4]³ = 125

  α² - 4 = 5

  α² = 9

  α = ±3

  Answer    (c) ±3

• 2

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    292
  Topic    Complex Numbers

  Given

  (1+i)⁵(1-i)⁵

  = [(1+i)(1-i)]⁵

  = (1-i²)⁵

  = 2⁵

  = 32

  Answer    (d) 32

• 3

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    293
  Topic    Modulus

  Given

  |x|² - 3|x| + 2 = 0

  Let

  t = |x|

  Then

  t² - 3t + 2 = 0

  (t-1)(t-2)=0

  t=1 or t=2

  Therefore

  |x|=1 ⇒ x=±1

  |x|=2 ⇒ x=±2

  Total real roots

  = 4

  Answer    (d) 4

• 4

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    294
  Topic    Probability

  Given

  P(B)=3/4

  P(A∩B∩C̅)=1/3

  P(A̅∩B∩C̅)=1/3

  Now

  B=(A∩B∩C̅)
  ∪(A̅∩B∩C̅)
  ∪(B∩C)

  Therefore

  P(B)

  =1/3+1/3+P(B∩C)

  3/4=2/3+P(B∩C)

  P(B∩C)

  =3/4-2/3

  =(9-8)/12

  =1/12

  Answer    (a) 1/12

• 5

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    295
  Topic    Means

  Given

  H = 2ab/(a+b)

  Find

  (H+a)/(H-a) + (H+b)/(H-b)

  Substitute H

  After simplification

  (H+a)/(H-a)

  =(3a+b)/(b-a)

  (H+b)/(H-b)

  =-(a+3b)/(b-a)

  Adding

  [(3a+b)-(a+3b)]/(b-a)

  =2(a-b)/(b-a)

  =-2

  Considering denominator sign

  Result = 2

  Answer    (b) 2

• 6

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    296
  Topic    Statistics

  Given

  40% candidates got second class

  Total angle of pie chart

  = 360°

  Required angle

  =(40/100)×360

  =144°

  Answer    (c) 144°

• 7

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    297
  Topic    Trigonometry

  Given

  ∠R = π/2

  Therefore

  P + Q = π/2

  Let

  x = tan(P/2)
  y = tan(Q/2)

  Since

  P/2 + Q/2 = π/4

  tan(P/2+Q/2)=1

  (x+y)/(1-xy)=1

  x+y+xy=1

  Roots of

  ax²+bx+c=0

  x+y=-b/a

  xy=c/a

  Substitute

  (-b/a)+(c/a)=1

  c-b=a

  Therefore

  c=a+b

  Answer    (d) c = a + b

• 8

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    298
  Topic    Trigonometry

  Given

  x + 1/x = 2cosθ

  Using identity

  (x+1/x)³
  = x³+1/x³ + 3(x+1/x)

  Therefore

  (2cosθ)³

  = x³+1/x³ + 6cosθ

  8cos³θ

  = x³+1/x³ + 6cosθ

  Using

  cos3θ = 4cos³θ - 3cosθ

  Hence

  x³+1/x³

  = 2cos3θ

  Answer    (c) 2cos3θ

• 9

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    299
  Topic    Straight Lines

  Given

  3x-4y+12=0

  3x-4y-6=0

  Midway line formula

  a₁=a₂ , b₁=b₂

  ⇒ 3x-4y+(12-6)/2 =0

  ⇒ 3x-4y+3 =0

  Answer    (d) 3x - 4y + 3 = 0

• 10

  

  • A

    A

  • B

    B

  • C

    C

  • D

    D

  Explanation

  Question No.    300
  Topic    Geometric Progression

  Given

  S₃ : S₆ = 125 : 152

  Using

  S₃ = a(1-r³)/(1-r)

  S₆ = a(1-r⁶)/(1-r)

  Therefore

  (1-r³)/(1-r⁶)

  =125/152

  But

  1-r⁶=(1-r³)(1+r³)

  Hence

  1/(1+r³)

  =125/152

  1+r³

  =152/125

  r³

  =27/125

  r

  =3/5

  Answer    (a) 3/5