Algebra


  1. If x varies inversely as (y2 – 1) and is equal to 24 when y = 10, then the value of x when y = 5 is









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    x ∝
    1
    y2 − 1

    ⇒  x =
    k
    y2 − 1

    Where k is a constant.
    When y = 10, x = 24, then
    ∴  24 =
    k
    ⇒ 24 =
    k
    y2 − 199

    ⇒  k = 24 × 99
    When y = 5, then
    ∴  x =
    k
    =
    24 × 99
    =
    24 × 99
    = 99
    y2 − 152 − 124

    Correct Option: A

    x ∝
    1
    y2 − 1

    ⇒  x =
    k
    y2 − 1

    Where k is a constant.
    When y = 10, x = 24, then
    ∴  24 =
    k
    ⇒ 24 =
    k
    y2 − 199

    ⇒  k = 24 × 99
    When y = 5, then
    ∴  x =
    k
    =
    24 × 99
    =
    24 × 99
    = 99
    y2 − 152 − 124


  1. If   2x +
    1
    = 5,   find the value of  
    5x
    .
    3x6x2 + 20x + 1









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    2x +
    1
    = 5
    3x

    ⇒  6x2 + 1 = 15x
    ⇒  6x2 + 20x + 1 = 15x + 20x = 35x
    ⇒ 
    5x
    =
    5x
    =
    1
    6x2 + 20x + 135x7

    Correct Option: D

    2x +
    1
    = 5
    3x

    ⇒  6x2 + 1 = 15x
    ⇒  6x2 + 20x + 1 = 15x + 20x = 35x
    ⇒ 
    5x
    =
    5x
    =
    1
    6x2 + 20x + 135x7



  1. If (a – 1)2 + (b + 2)2 + (c + 1)2 = 0, then the value of 2a – 3b + 7c is









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    (a – 1)2 + (b + 2)2 + (c + 1)2 = 0
    ⇒  a – 1 = 0 ⇒ a = 1;
    b + 2 = 0 ⇒ b = –2
    c + 1 = 0 ⇒ c = –1
    ∴  2a – 3b + 7c
    = 2 – 3 (– 2) + 7 (–1)
    = 2 + 6 – 7 = 1

    Correct Option: D

    (a – 1)2 + (b + 2)2 + (c + 1)2 = 0
    ⇒  a – 1 = 0 ⇒ a = 1;
    b + 2 = 0 ⇒ b = –2
    c + 1 = 0 ⇒ c = –1
    ∴  2a – 3b + 7c
    = 2 – 3 (– 2) + 7 (–1)
    = 2 + 6 – 7 = 1


  1. If   x = 7 – 4√3 , then √x +
    1
      is equal to :
    x









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    x = 7 – 4√3
    ∴  √x = √7 – 4√3
    = √7 – 2 × 2 × √3
    = √4 + 3 − 2 × 2 × √3
    = √(2 − √3)2 = 2 − √3

    ∴ 
    1
    =
    1
    x2 − √3

    =
    1
    ×
    2 + √3
    =
    2 + √3
    2 − √32 + √34 − 3

    = 2 + √3
    ∴  √x +
    1
    = 2 − √3 + 2 + √3 = 4
    x

    Correct Option: D

    x = 7 – 4√3
    ∴  √x = √7 – 4√3
    = √7 – 2 × 2 × √3
    = √4 + 3 − 2 × 2 × √3
    = √(2 − √3)2 = 2 − √3

    ∴ 
    1
    =
    1
    x2 − √3

    =
    1
    ×
    2 + √3
    =
    2 + √3
    2 − √32 + √34 − 3

    = 2 + √3
    ∴  √x +
    1
    = 2 − √3 + 2 + √3 = 4
    x



  1. If x, y and z are real numbers such that (x– 3)2 + (y – 4)2 + (z – 5)2 = 0 then (x + y + z) is equal to









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    (x – 3)2 + (y – 4)2 + (z – 5)2 = 0
    ⇒  x – 3 = 0, y – 4 = 0 and z – 5 = 0
    ⇒  x = 3, y = 4 and z = 5
    ∴  x + y + z = 3 + 4 + 5 = 12

    Correct Option: D

    (x – 3)2 + (y – 4)2 + (z – 5)2 = 0
    ⇒  x – 3 = 0, y – 4 = 0 and z – 5 = 0
    ⇒  x = 3, y = 4 and z = 5
    ∴  x + y + z = 3 + 4 + 5 = 12