# Quadratic Equation Questions Quizzes with PDF for CET 2021, NRA CET, Bank Clerk and Insurance Exams

Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
Important for :
1
I. x2 – 19x + 88 = 0

II. y2 – 12y + 35 = 0
» Explain it
A
According to the given equations :

I. x– 19x + 88 = 0

x– 11x – 8x + 88 = 0

x (x – 11) – 8 (x – 11) = 0

(x – 8) (x – 11) = 0

x = 8, 11

II. y– 12y + 35 = 0

y– 7y – 5y + 35 = 0

y (y – 7) – 5 (y – 7) = 0

(y – 7) (y – 5) = 0

y = 7, 5

After comparison of both equations, the conclusion is x > y

Hence, option A is correct.
2
I. x2 – 11x + 24 = 0

II. y2 – 16y + 63 = 0
» Explain it
E
According to the given equations :

I. x2 – 11x + 24 = 0

x2 – 3x – 8x + 24 = 0

x (x – 3) – 8 (x – 3) = 0

(x – 3) (x – 8) = 0

x = 3, 8

II. y2 – 16y + 63 = 0

y2 – 7y – 9y + 63 = 0

y (y – 7) – 9 (y – 7) = 0

(y – 7) (y – 9) = 0

y = 7, 9

While comparing the root values of x and y, we find that one root value of x is lies between the root values of y. Hence, the relation between x and y can't be established.

Hence, option E is correct.
3
I. 2x2 – 24x + 70 = 0

II. y2 – 20y + 91 = 0
» Explain it
B
According to the given equations :

I. 2x2 – 24x + 70 = 0

 2x2 – 24x + 70 = 0 2

x2 – 12x + 35 = 0

x2 – 5x – 7x + 35 = 0

x (x – 5) – 7 (x – 5) = 0

(x – 5) (x – 7) = 0

x = 5, 7

II.  y2 – 20y + 91 = 0

y2 – 7y – 13y + 91 = 0

y (y – 7) – 13 (y – 7) = 0

(y – 7) (y – 13) = 0

y = 7, 13

While comparing the root values of x and y, we find that one root value of y is equal to x's and another one is greater than x's root values. Hence, the relation between x and y will be x ≤ y.

Hence, option B is correct.
4
I. x3 = 73 – 127

II. y = 182 – 315
» Explain it
D
According to the given equations :

I. x3 = 73 – 127

x3 = 343 – 127

x3 = 216

x = 6

II. y = 182 – 315

y = 324 – 315

y = 9

After comparison of both equations, the conclusion is x < y

Hence, option D is correct.
5
I. 3x + 5y = 76

II. x + 3y = 36
» Explain it
A
According to the given equations :

I. 3x + 5y = 76

Applying x's value from equation (ii), we get

3(36 – 3y) + 5y = 76

108 – 9y + 5y = 76

108 – 76 = 4y

32 = 4y ; y = 8

II. x + 3y = 36

x = 36 – 3y

x = 36 – 3 × 8

x = 36 – 24 = 12

While comparing the root values of x and y we find that x > y.

Hence, option A is correct.