Hello Aspirants, As we all know that Quadratic Equation Questions is a important part of quantitative aptitude section for every competitive exams.Different kind of Quadratic Equation Questions asked in every competitative exam.So here, In this article we will provide Quadratic Equation Quiz.These Quadratic Equation Questions are important for Bank, SSC, SEBI, NABARD, RBI, LIC, and Other state exams. You can attempt these questions & boost your preparation for your examination.

In the Banking exams Quadratic Equation Questions asked in the Prelims as well as Mains exam.There are 5 Quadratic Equation Questions asked in the Prelims exam (Bank).You want to score more in the Quadratic Equation section then you should practice more and more Quadratic Equation.

## Quadratic Equation Questions Quiz -11

Directions-(1-5) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2– 16x + 63 = 0
II. y2– 8y + 15 = 0
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

A. x > y

From I,

x2– 16x + 63 = 0
x2 – 7x – 9x + 63 = 0
x(x – 7) – 9(x – 7) = 0
(x – 7)(x – 9) = 0
x = 7, 9
From II, y2– 8y + 15 = 0
y2– 3y – 5y + 15 = 0
y(y – 3) – 5(y – 3) = 0
(y – 3)(y – 5) = 0
y = 3, 5
Therefore, x > y

2.In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2– 13x + 42 = 0
II. y2– 17y + 70 = 0
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

E. x ≤ y

From I, x2– 13x + 42 = 0
x2 – 6x – 7x + 42 = 0
x(x – 6) – 7(x – 6) = 0
(x – 6)(x – 7) = 0
x = 6, 7
From II, y2– 17y + 70 = 0
y2– 7y – 10y + 70 = 0
y(y – 7) – 10(y – 7) = 0
(y – 7)(y – 10) = 0
y = 7, 10
Therefore, x ≤ y

3. I. x3 = 133– 866
II. y3= 431 + 92

A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

A. x > y

From I, x3 = 133 – 866
x3 = 2197 – 866 = 1331
x = 11
From II, y3 = 431 + 92
y3 = 431 + 81 = 512
y = 8
Therefore, x > y

4.I. x2– 15x + 54 = 0
II. y2– 9y + 18 = 0
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

D. x ≥ y.

From I, x2– 15x + 54 = 0
x2 – 6x – 9x + 54 = 0
x(x – 6) – 9(x – 6) = 0
(x – 6)(x – 9) = 0
x = 6, 9
From II, y2– 9y + 18 = 0
y2– 3y – 6y + 18 = 0
y(y – 3) – 6(y – 3) = 0
(y – 3)(y – 6) = 0
y = 3, 6
Therefore, x ≥ y

5. I. 7x + 4y = 63
II. 9x + y = 52
A  x > y
B  x ≥ y
C  x < y
D  x ≤ y
E  x = y or the relationship cannot be established

C. x < y

From II, we get y = 52 – 9x
So,
7x + 4 × (52 – 9x) = 63
7x + 208 – 36x = 63
x = 5
So, y = 52 – 9 × 5 = 7
Therefore, x < y

6.In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. 45x – 14 = 25x2
II. 6y2+ 15 = 19y
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

B. x < y

7. In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. 10x– 33x + 20 = 0
II. 9y2 = 14 – 15y
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

A. x > y

8.

A. x > y
B. x < y
C. x = y or the relationship cannot be established
D. x ≥ y
E. x ≤ y

C. x = y or the relationship cannot be established

9. I. 9x – 2y = 4
II. 4x + 3y = 29
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

B. x < y

10. I. x2 = 2x + 15
II. y2 – 12y + 35 = 0
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

E. x ≤ y

From I:
x2 = 2x + 15
=>x2 – 2x – 15 = 0
=>x2 + 3x – 5x – 15 = 0
=>x(x + 3) – 5(x + 3) = 0
=>(x + 3)(x – 5) = 0
=>x = -3, 5
From II:
y2– 12y + 35 = 0
=>y2– 7y – 5y + 35 = 0
=>y(y – 7) – 5(y – 7) = 0
=>(y – 7)(y – 5) = 0
=>y = 7, 5
So, x ≤ y