Set-3 Quadratic Equations (Inequalities) For SBI PO and SBI Clerk 2019 | Must Go Through These Questions

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We are providing the most important Quadratic Equations (Inequalities) for SBI PO 2019, SBI Clerk 2019 and all other competitive bank and insurance exams. These questions have very high chances to be asked in SBI PO 2019, SBI Clerk 2019.
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Directions:1-10) In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

1.

I) 4x-3y-2=0

II) 6x+y-36=0

2.

I) x2-4x-621=0

II) y2-35y+276=0

3.

I) 3x2+25x-18=0

II) 18y2-41y+21=0

4.

I) 6x2+x-2=0

II) 30y2+11y+1=0

5.

I) x3=1728

II) y2=144

6.

  1. 5x2– 18x + 9 = 0
  2. 20y2– 13y + 2 = 0

7.

  1. x3– 878 = 453
  2. y2– 82 = 39

8.

  1. 3/√x + 4/√x = √x
  2. y3– (7)7/2/√y = 0

9.

  1. 9x – 15.45 = 54.55 + 4x
  2. √(y + 155) – √36 = √49

10.

  1. x2+ 11x + 30 = 0
  2. y2+ 7y + 12 = 0

 

Check your Answers below:

 

 

  • Directions:1-10) In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

    1. Question

    I) 4x-3y-2=0

    II) 6x+y-36=0

    Ans:2
    I) 4x-3y-2=0 …………..(1)II) 6x+y-36=0 …………………. (2)

    After solving (i) and (ii), we have

    x=5 and y=6

    Hence x<y 

  • 2. Question

    I) x2-4x-621=0

    II) y2-35y+276=0

    Ans:5
    I) x2-4x-621=0x2-27x+23x-621=0

    (x-27)(x+23)=0

    x=27 or -23

    II) y2-35y+276=0

    y2-23y-12y+276=0

    (y-23)(y-12)=0

    y=23 or 12

    Hence the relationship cannot be determined.

  • 3. Question

    I) 3x2+25x-18=0

    II) 18y2-41y+21=0

    Ans:2
    I) 3x2+25x-18=03x2+27x-2x-18=0

    (x+9) (3x-2) =0

    x=2/3 or -9

    II) 18y2-41y+21=0

    18y2-27y-14y+21=0

    (2y-3)(9y-7)=0

    y=3/2 or 7/9

    Hence x<y

  • 4. Question

    I) 6x2+x-2=0

    II) 30y2+11y+1=0

    Ans:5
    I) 6x2+x-2=06x2-3x +4x -2 =0

    (2x-1)(3x+2)=0

    x=1/2 or -2/3

    II) 30y2+11y+1=0

    30y2+6y+5y+1=0

    (5y+1)(6y+1)=0

    y=-1/5 or -1/6

    Hence the relationship cannot be determined.

  • 5. Question

    I) x3=1728

    II) y2=144

    Ans:3
    I) x3=1728x=12

    II) y2=144

    y=12 or -12

    Hence x ≥ y

  • 6. Question
    1. 5x2– 18x + 9 = 0
    2. 20y2– 13y + 2 = 0
    Ans:1
    I. 5x2– 18x + 9 = 05x2 – 15x – 3x + 9 = 0

    5x (x -3) – 3(x – 3) = 0

    (x – 3) (5x – 3) = 0

    x = 3 or 3/5

    II. 20y2– 13y + 2 = 0

    20y2 – 8y – 5y + 2 = 0

    4y(5y – 2) -1(5y – 2) = 0

    (4y – 1) (5y – 2) = 0

    y = 1/4 or 2/5

    Clearly x > y

  • 7. Question
    1. x3– 878 = 453
    2. y2– 82 = 39
    Ans:2
    I. x3– 878 = 453x = 3√1331 = 11

    x = 11

    II. y2– 82 = 39

    y2 = 82 + 39 = 121

    y = ±11

    Hence x ≥ y

  • 8. Question
    1. 3/√x + 4/√x = √x
    2. y3– (7)7/2/√y = 0
    Ans:5
    I. 3/√x + 4/√x = √x3 +4 = x

    x = 7

    II. y3– (7)7/2/√y = 0

    y3+1/2 – (7)7/2 = 0

    y7/2 = 77/2

    y = 7

    Clearly, x = y

  • 9. Question
    1. 9x – 15.45 = 54.55 + 4x
    2. √(y + 155) – √36 = √49
      Ans:5
    I. 9x – 15.45 = 54.55 + 4x
    9x – 4x = 705x = 70

    x = 14

    II. √(y + 155) – √36 = √49

    √(y + 155) = 6 + 7

    √(y + 155) = 13

    y + 155 = 169

    y = 169 – 155 = 14

    Clearly, x = y

  • 10. Question
    1. x2+ 11x + 30 = 0
    2. y2+ 7y + 12 = 0
    Ans:3
    I. x2+ 11x + 30 = 0x2 + 6x + 5x + 30 = 0

    x(x +6) + 5(x + 6) = 0

    (x + 5) (x + 6) = 0
    x = -5 (or) -6

    II. y2+ 7y + 12 = 0

    y2 + 4y + 3y + 12 = 0

    y(y+4) + 3(y+4) = 0

    (y+3) (y+4) = 0

    y = -3 (or) -4
    Hence x<y

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