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Aptitude miscellaneous
  1. If x = y + z then x³ – y³ – z³ is








  1. View Hint View Answer Discuss in Forum

    x = y + z ⇒ x – y – z = 0
    If a + b + c = 0 then a³ + b³ + c³ = 3abc
    ∴ (x)³ + (–y)³ + (–z)³ = 3x (–y) (–z) = 3xyz

    Correct Option: B

    x = y + z ⇒ x – y – z = 0
    If a + b + c = 0 then a³ + b³ + c³ = 3abc
    ∴ (x)³ + (–y)³ + (–z)³ = 3x (–y) (–z) = 3xyz

  1. If x = 11, the value of x5 – 12x4 + 12x³ – 12x² + 12x - 1 is








  1. View Hint View Answer Discuss in Forum

    x5 – 12x4 + 12x³ – 12x² + 12x – 1
    = x5 – (11 + 1) x4 + (11 + 1)x³ – (11 + 1)x² + (11 + 1) x – 1
    = x5 – 11x4 – x4 + 11x³ + x³ – 11x² – x² + 11x + x – 1
    = x – 1 = 11 – 1 = 10
    [∵ x = 11]

    Correct Option: B

    x5 – 12x4 + 12x³ – 12x² + 12x – 1
    = x5 – (11 + 1) x4 + (11 + 1)x³ – (11 + 1)x² + (11 + 1) x – 1
    = x5 – 11x4 – x4 + 11x³ + x³ – 11x² – x² + 11x + x – 1
    = x – 1 = 11 – 1 = 10
    [∵ x = 11]

  1. If a = 101, then the value of a (a² – 3a + 3) is :








  1. View Hint View Answer Discuss in Forum

    a = 101 (Given)
    ∴ a (a² – 3a + 3)
    = a³ – 3a² + 3a – 1 + 1
    = (a – 1)³ + 1 = (100)³ + 1 = 1000001

    Correct Option: C

    a = 101 (Given)
    ∴ a (a² – 3a + 3)
    = a³ – 3a² + 3a – 1 + 1
    = (a – 1)³ + 1 = (100)³ + 1 = 1000001

  1. If a +
    1
    		² = 3, then the value of a³ +
    1
    		is :
    aa³








  1. View Hint View Answer Discuss in Forum

    Givena +
    1
    		² = 3
    a

    ⇒ a +
    1
    		 = √3
    a

    On cubing both sides,
    a +
    1
    		³ = (√3)³
    a

    ⇒ a³ +
    1
    		 + 3a +
    1
    		 = 3√3
    a³

    ⇒ a³ +
    1
    		 = 3√3 = 3√3
    a³

    ⇒ a³ +
    1
    		 = 3√3 - 3√3 = 0
    a³

    Correct Option: A

    Givena +
    1
    		² = 3
    a

    ⇒ a +
    1
    		 = √3
    a

    On cubing both sides,
    a +
    1
    		³ = (√3)³
    a

    ⇒ a³ +
    1
    		 + 3a +
    1
    		 = 3√3
    a³

    ⇒ a³ +
    1
    		 = 3√3 = 3√3
    a³

    ⇒ a³ +
    1
    		 = 3√3 - 3√3 = 0
    a³

  1. If
    a² + b²
    		 =
    b² + c²
    		 =
    c² + a²
    		 =
    1
    		 (k ≠ 0) then k = ?
    c²a²b²k








  1. View Hint View Answer Discuss in Forum

    a² + b²
    		 =
    b² + c²
    		 =
    c² + a²
    		 =
    1
    c²a²b²k

    ⇒ c² = k (a² + b²);
    a² = k (b² + c²);
    b² = k (c² + a²)
    ∴ a² + b² + c² = k (b² + c² + c² + a² + a² + b²)
    ⇒ a² + b² + c² = 2k (a² + b² + c²)
    ⇒ 2k = 1
    ⇒ k =
    1
    2

    Correct Option: D

    a² + b²
    		 =
    b² + c²
    		 =
    c² + a²
    		 =
    1
    c²a²b²k

    ⇒ c² = k (a² + b²);
    a² = k (b² + c²);
    b² = k (c² + a²)
    ∴ a² + b² + c² = k (b² + c² + c² + a² + a² + b²)
    ⇒ a² + b² + c² = 2k (a² + b² + c²)
    ⇒ 2k = 1
    ⇒ k =
    1
    2