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









  1. View Hint View Answer Discuss in Forum

    Using Rule 8,

    ⇒ a +
    1
    		2 = 3
    a

    ⇒ a +
    1
    		 = √3
    a

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

    ⇒ a3 +
    1
    		 + 3a +
    1
    		 = 3√3
    a3a

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

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

    Correct Option: A

    Using Rule 8,

    ⇒ a +
    1
    		2 = 3
    a

    ⇒ a +
    1
    		 = √3
    a

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

    ⇒ a3 +
    1
    		 + 3a +
    1
    		 = 3√3
    a3a

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

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

  1. The numerical value of
    (a - b)2
    		 +
    (b - c)2
    		 +
    (c - a)2
    		 is (a ≠ b ≠ c)
    (b - c)(c - a)(c - a)(a - b)(a - b)(b - c)









  1. View Hint View Answer Discuss in Forum

    Expression =
    (a - b)2
    		 +
    (b - c)2
    		 +
    (c - a)2
    (b - c)(c - a)(c - a)(a - b)(a - b)(b - c)

    Expression =
    (a - b)3 + (b - c)3 + (c - a)3
    (a - b)(b - c)(c - a)

    Expression =
    3(a - b)(b - c)(c - a)
    		 = 3
    (a - b)(b - c)(c - a)

    [Here, a – b + b – c + c – a = 0.
    If x + y + z = 0, x3 + y3 + z3 = 3xyz ]

    Correct Option: D

    Expression =
    (a - b)2
    		 +
    (b - c)2
    		 +
    (c - a)2
    (b - c)(c - a)(c - a)(a - b)(a - b)(b - c)

    Expression =
    (a - b)3 + (b - c)3 + (c - a)3
    (a - b)(b - c)(c - a)

    Expression =
    3(a - b)(b - c)(c - a)
    		 = 3
    (a - b)(b - c)(c - a)

    [Here, a – b + b – c + c – a = 0.
    If x + y + z = 0, x3 + y3 + z3 = 3xyz ]

  1. An example of an equality relation of two expressions in x, which is not an identity is








  1. View Hint View Answer Discuss in Forum

    According to equality relation (x + 2)2 = x2 + 4x + 4 is not an identity

    Correct Option: C

    According to equality relation (x + 2)2 = x2 + 4x + 4 is not an identity

  1. If p = 99, then the value of p(p2 + 3p + 3) is








  1. View Hint View Answer Discuss in Forum

    Using Rule 8,
    p = 99 (Given)
    ∴ p(p2 + 3p + 3) = p3 + 3p2 + 3p
    = p3 + 3p2 + 3p + 1 - 1
    = (p + 1)3 - 1 = (99 + 1)3 - 1
    ∴ p(p2 + 3p + 3) = 1003 – 1 = 999999

    Correct Option: C

    Using Rule 8,
    p = 99 (Given)
    ∴ p(p2 + 3p + 3) = p3 + 3p2 + 3p
    = p3 + 3p2 + 3p + 1 - 1
    = (p + 1)3 - 1 = (99 + 1)3 - 1
    ∴ p(p2 + 3p + 3) = 1003 – 1 = 999999

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








  1. View Hint View Answer Discuss in Forum

    x = 11 (Given)
    ∴ 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
    When x = 11,
    = 115 – 115 - 114 + 114 + 113 – 113 - 112 + 112 + 11 – 1
    = 0 - 0 + 0 - 0 + 10 = 10

    Correct Option: B

    x = 11 (Given)
    ∴ 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
    When x = 11,
    = 115 – 115 - 114 + 114 + 113 – 113 - 112 + 112 + 11 – 1
    = 0 - 0 + 0 - 0 + 10 = 10