Quadratic Equation Mock: IBPS 27717


Directions: In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer.

1.  I. x2-32x+256=0

    II. y2-33y+272=0

A. x < y

B. x ≤ y

C. x > y

D. x ≥ y

E. If either x=y or the relationship can’t be
established

2.   I. 3x-4y+9=0

      II.7x-7y-7=0

A. x < y

B. x ≤ y

C. x > y

D. x ≥ y

E. If either x=y or the relationship can’t be
established

3.   I. x2-2x-15=0

      II. y2-9y+14=0

A. x < y

B. x ≤ y

C. x > y

D. x ≥ y

E. If either x=y or the relationship can’t be
established

4.      I. 4x2-8x+3=0

        II. 4y2+8y+3=0

A. x < y

B. x ≤ y

C. x > y

D. x ≥ y

E. If either x=y or the relationship can’t be
established

5.   I. 2x2-3x+1=0

      II. 2y2-4y+2=0

A. x < y

B. x ≤ y

C. x > y

D. x ≥ y

E. If either x=y or the relationship can’t be
established

Directions: In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer.

6.  I. 3x2+12x-180=0

     II. 2y2+4y-96=0

A. x < y

B. x ≤ y

C. If either x=y or the relationship can’t be
established

D. x > y

E.  x ≥ y

7.   I. 36x2+30x+6=0

      II.45y2+24y+3=0

A. x < y

B. x ≤ y

C. If either x=y or the relationship can’t be
established

D. x > y

E.  x ≥ y

8.   I. 2x2-9x+9=0

      II. y2-11y+24=0

A. x < y

B. x ≤ y

C. If either x=y or the relationship can’t be
established

D. x > y

E.  x ≥ y

9.      I. x2-13x+40=0

         II. y2+9y+18=0

A. x < y

B. x ≤ y

C. If either x=y or the relationship can’t be
established

D. x > y

E.  x ≥ y

10.   I. 42x2-162x-24=0

      II. 12y2+24y-288=0

A. x < y

B. x ≤ y

C. If either x=y or the relationship can’t be
established

D. x > y

E.  x ≥ y


Answer-

  1. I. x2-32x +256 =0
    => (x-16)2 =0
    => x = 16,16
    and II. y2– 33y + 272 =0
    => (y-16)(y-17)=0
    => y =16,17
    So, x ≤ y
    Hence, option B

 

  1. Solving I and II, we get
    x = 13 and y =12
    So, x > y. Hence, option C

  2. I.   x2-2x -15 =0
    =>x2-5x +3x -15 =0
    => (x-5)(x+3) =0
    => x= 5, -3
    II. y2 – 9y +14 =0
    =>y2 – 2y -7y +14 =0
    =>(y -2)(y-7) =0
    =>y = 2,7
    So, relation between x and y is not determined.
    Hence, option e.

4. C

 

  1. B

 

6.

From I,
x2 + 4x – 60 = 0
x2– 6x + 10x – 60 = 0
x(x – 6)+10(x – 6) = 0
(x – 6)(x + 10) = 0
x = 6 or -10
From II,
y2 + 2y – 48 = 0
y2– 6y + 8y – 48 = 0
y(y – 6)+8(y – 6) = 0
(y – 6)(y + 8) = 0
y = -8 or 6

So, no relationship can be established between x
and y. Hence, option c.

7.

From I,
6x2 + 5x + 1 = 0
6x2 + 3x + 2x + 1 = 0
3x(2x + 1) + 1(2x + 1) = 0
(3x + 1)(2x + 1) = 0
x = -1/3 or -1/2
From II,
15y2 + 8y + 1 = 0
15y2 + 5y + 3y + 1 = 0
5y(3y + 1) + 1(3y + 1) = 0
(5y + 1)(3y + 1) = 0
y = -1/5 or -1/3

So, y≥x. Hence, option b.

8.

From I,
2x2– 6x – 3x + 9 = 0
2x(x – 3)-3(x – 3) = 0
(2x – 3)(x – 3) = 0
x = 3/2 or 3
From II,
y2– 3y – 8y + 24 = 0
y(y – 3)-8(y – 3) = 0
(y – 3)(y – 8) = 0
y = 3 or 8
So, y ≥ x. Hence, option b

9.

From I,
x2– 5x – 8x + 40 = 0
x(x – 5)-8(x – 5) = 0
(x – 5)(x – 8) = 0
x = 5 or 8
From II,
y2 + 3y + 6y + 18 = 0
y(y + 3)+6(y + 3) = 0
(y + 3)(y + 6) = 0
y = -3 or -6
So, x > y. Hence, option d.

10.

From I,
7x2– 27x – 4 = 0
7x2– 28x + x – 4 = 0
7x(x – 4) + 1(x – 4) = 0
(x – 4)(7x + 1) = 0
x = -1/7 or 4
From II,
y2 + 2y – 24 = 0
y2– 4y + 6y – 24 = 0
y(y – 4)+6(y – 4) = 0
(y – 4)(y + 6) = 0
y = 4 or -6

So, no relationship can be established between x
and y. Hence, option c.

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