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Quant Quiz On Quadratic Equation Questions Day 5 Bag

Quant Quiz On Quadratic Equation Questions Day 5 Bag


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  • For which of the following equations the value of X is less than Y(X<Y)
    I.5x + 2y = 31; 3x + 7y = 36
    II.2x² -15x + 27 = 0; 5y² – 26y + 33 = 0
    III.25/√x + 9/√x = 17√x; √y/3 + 5√y/6 = 3/√y
    1.Only I
    2.Only II
    3.Only III
    4.Both II and III
    5.None

    Answer & Explanation
    Answer – 3.Only III
    Explanation :
    25/√x + 9/√x = 17√x
    34 = 17x
    x = 2
    21√y/18 = 3/√y
    y = 18/7 = 2.57
  • For which of the following equations the value of X is less than or equal to Y (X≤ Y)
    I.X2– 4X + 3= 0; Y2 – 8Y + 15 = 0
    II. 3X2 – 19X + 28= 0; 4Y2 – 29Y + 45 = 0
    III.x2 – (16)2 = (23)2 – 56; y1/3 – 55 + 376 = (18)2
    1.Only I
    2.Only II
    3.Both I and III
    4.Both II and III
    5.All follow

    Answer & Explanation
    Answer – 3.Both I and III
    Explanation :
    From I, (x-3)(x-1) = 0 X=1,3
    (y-5)(y-3) = 0 Y = 5,3
    From III, x2 – (16)2 = (23)2 – 56
    x2 = 729
    x = ± 27
    y1/3 – 55 + 376 = (18)2
    y = 33 = 27
  • For which of the following equations the value of X is greater than or equal to Y (X ≥ Y)
    I.X2 -3X – 4 = 0; 3Y2 – 10Y+8 = 0
    II.√(x + 6) = √121 – √36; y2 + 112 = 473
    III.5x2– 7x – 6 = 0; 5y2 + 23y + 12 = 0
    1.Only I
    2.Only II
    3.Both I and III
    4.Both II and III
    5.None follow

    Answer & Explanation
    Answer – 4.Both II and III
    Explanation :
    From II, √(x + 6) = √121 – √36
    x + 6 = 25; x = 19
    y2 + 112 = 473
    y2 = 361; y = ± 19
    From III, 5x2 – 7x – 6 = 0; x = -3/5, 2
    5y2 + 23y + 12 = 0; y = -4, -3/5
  • For which of the following equations the value of X is greater than Y(X>Y)
    I.3X2+23X + 44 = 0; 3Y2 + 20Y +33 = 0
    II.3X2+29 X +56 = 0; 2Y+ 15Y + 25 = 0
    III.3X2– 16X + 21 = 0; 3Y2 – 28Y + 65 = 0
    1.Only I
    2.Only II
    3.Both I and III
    4.Both II and III
    5.None follow

    Answer & Explanation
    Answer – 5.None follow
    Explanation :
    From I, X = -4, -11/3; Y = -3, -11/3
    From II, X=-7, -8/3; Y = -5, -5/2
    From III,X= 3,7/3; Y = 5,13/3
  • For which of the following equations the value of X=Y or relationship cannot be established.
    I.1/x + 1/(x-10) = 8/75; 132/y – 132/(y + 11) = 1
    II.(3x – 2)/y = (3x + 6)/(y + 16); (x + 2)/(y + 4) = (x + 5)/(y + 10)
    III. x² – 4x – 21 = 0; y² – 35y + 306 = 0
    1.Only I
    2.Only II
    3.Both I and III
    4.Both II and III
    5.All follow

    Answer & Explanation
    Answer – 1.Only I
    Explanation :
    From I, x = 25, 3.75; y = -44, 33
  • For which of the following equations the value of Y is greater than X(Y>X)
    I.7x + 6y + 4z = 122; 4x + 5y + 3z = 88; 9x + 2y + z = 78
    II.7x + 6y = 110; 4x + 3y = 59
    III.2x + 5y = 23.5, 5x+ 2y = 22
    1.Only I
    2.Only II
    3.Both I and II
    4.Both II and III
    5.All follow

    Answer & Explanation
    Answer – 5.All follow
    Explanation :
    From I, x = 6, y = 8
    From II, x = 8; y = 9
    From III, x = 3, y = 3.5
  • For which of the following equations the value of X is less than or equal to Y (X≤ Y)
    I.[48 / x4/7] – [12 / x4/7] = x10/7; y³ + 783 = 999
    II.9/√x + 19/√x = √x; y– [(28)11/2 /√y] = 0
    III.6X2 – 7X + 2 = 0; 12Y2 – 7Y+1 = 0
    1.Only I
    2.Only II
    3.Both I and II
    4.Both I and III
    5.All follow

    Answer & Explanation
    Answer – 1.Only I
    Explanation :
    From I, x = ± 6; y = 6
  • For which of the following equations the value of X is greater than or equal to Y (X ≥ Y)
    I.3X2 +17X + 10=0; 10Y2 + 9Y+2 = 0
    II.X2 +X – 56 = 0; Y2 – 17Y+72 = 0
    III.X2– 12X + 35= 0; Y2 + Y – 30 = 0
    1.Only I
    2.Only II
    3.Both I and II
    4.Both I and III
    5.None of these

    Answer & Explanation
    Answer – 5.None of these
    Explanation :
    Only III follows, From III, X = 7,5; Y = 5, -6
  • For which of the following equations the value of X is greater than Y(X>Y)
    I.2X+ 15X + 25= 0; 3Y2 + 29Y + 56 = 0
    II.3X2 – 19X + 28= 0; 4Y2 – 29Y + 45 = 0
    III.X2 – 13X + 42= 0; Y– 9Y + 20 = 0
    1.Only I
    2.Only II
    3.Only III
    4.Both II and III
    5.None of these

    Answer & Explanation
    Answer – 3.Only III
    Explanation :
    From III, X=7,6; Y =5,4
  • For which of the following equations the value of X=Y or relationship cannot be established.
    I.x2 – 18x + 72= 0; 3y2 + 7y + 4 = 0
    II.2x2 + x – 36 = 0; 2y2 – 13y + 20 = 0
    III.4X2 – 19X + 12= 0; 3Y2 + 8Y + 4 = 0
    1.Only I
    2.Only II
    3.Both I and II
    4.Both II and III
    5.None of these

    Answer & Explanation
    Answer – 2.Only II
    Explanation :
    From II, x = 4 , -9/2; y = 4, 5/2

 

  • 5/√x + 7/√x = √x 
    4/√y + 6/√y = √y
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    A. X > Y
    Explanation:

    5/√x + 7/√x = √x
    1/√x (5 + 7) = √x => x = 12
    4/√x + 6/√x = √x => y = 10
  • x² – 300 = 325
    y – √144 = √169
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    D. X ≤ Y
    Explanation:

    x² – 300 = 325
    x = ± 25
    y – √144 = √169
    y = 12 + 13 = 25
  • x² – 115/2/√x = 0
    y² – 135/2/√y = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    B. X < Y
    Explanation:

    x² – 115/2/√x = 0
    x5/2 = 115/2
    x = 11
    y² – 135/2/√y = 0 => y = 13
  • √(x + 20) = √256 – √121
    y² + 576 = 697
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    E. X = Y or relation cannot be established
    Explanation:

    √(x + 20) = √256 – √121
    x + 20 = 25 => x = 5
    y² + 576 = 697
    y = ± 11
  • y² – x² = 32
    y – x = 4
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    B. X < Y
    Explanation:
    y² – x² = 32
    y – x = 4
    (y + x) (y – x) = 32
    (y + x) 4 = 32
    y + x = 8
    y – x = 4
    y = 16 and x = 12
  • x² + 12x + 32 = 0
    y² + 19y + 90 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    A. X > Y
    Explanation:

    x² + 12x + 32 = 0
    x = – 4, – 8
    y² + 19y + 90 = 0
    y = -9, -10
  • x² – 15x + 56 = 0
    y² + 17y + 72 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    A. X > Y
    Explanation:

    x² – 15x + 56 = 0
    x = 8, 7
    y² + 17y + 72 = 0
    y = -8, -9
  • x² – 23x + 132 = 0
    y² + 13y + 42 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    A. X > Y
    Explanation:

    x² – 23x + 132 = 0
    x = 11, 12
    y² + 13y + 42 = 0
    y = -6, -7
  • x² – 32x + 255 = 0
    y² – 35y + 306 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    D. X ≤ Y
    Explanation:

    x² – 32x + 255 = 0
    x = 17, 15
    y² – 35y + 306 = 0
    y = 17, 18
  • x² – 21x + 108 = 0
    y² – 17y + 72 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    C. X ≥ Y
    Explanation:

    x² – 21x + 108 = 0
    x = 11, 9
    y² – 17y + 72 = 0
    y = 9, 8

Here I am giving you some examples of quadratic equations.

Ex 1: x2 + 3x – 10 = 0 and y2 – 7y + 12 = 0

By solving we get x = -5, 2 and y = 3, 4
Put these values as on a number line as:
-5(x)           2(x)     3(y)      4(y)
By this, it is clear that x is always less than y. So, we will say x < y.

Ex 2: x2 – 8x + 12 = 0 and y2 + 3y – 10 = 0

By solving we get x = 2, 6 and y = -5, 2
Put these values as on a number line as:
-5(y)             2(y,x)         6(x)
When x = 2, x = y=2 and x > y = -5, so x ≥ y.
When x = 6, x > y = 2 and x > y = -5.
By this, it is clear that either x is greater than y always or is equal to y at value 2. So, we will say x ≥ y.

Ex 3: x2 + 3x – 4 = 0 and y2 – y – 6 = 0

By solving we get x = -4, 1 and y = -2, 3
Put these values as on a number line as:
-4(x)             -2(y)         1(x)       3(y)
When there are values in between, one cannot find the relationship. So, we will say that no relationship exists between x and y.

Ex 4: x2 – 5x + 6 = 0 and y2 – 7y + 10 = 0

By solving we get x = 2, 3 and y = 2, 5
Put these values as on a number line as:
2(x,y)        3(x)            5(y)
When x = 2, x = y=2 and x < y = 5, so x ≤ y.
When x = 3, x > y = 2 and x < y = 5.
By these values, one cannot find the exact relationship. So, we will say that no relationship exists between x and y.

 

 

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