 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 -7

1. In the following questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x2 +14x -207 = 0
II. y2-33y +266 = 0
A. x < y
B. x > y
C. x ≤ y
D. x ≥ y
E. x = y or relationship cannot be established

A. x < y

From equation I:
x2 +14x -207 = (x + 23)(x -9)= 0
=> x = -23, 9
From equation II:
y2-33y +266 = (y -19)(y -14) = 0
=> y = 19, 14

 X = -23 X = 9 Y =19 x < y x < y Y = 14 x < y x < y

So, x < y

2.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x2-44x +483 = 0
II. 25y2 +15y -54 = 0

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

B. x > y

From equation I:
x2-44x +483 = (x -21)(x -23)= 0
=> x = 21, 23
From equation II:
25y2 +15y -54 = (5y -6)(5y + 9) = 0
=> y = 6/5, -9/5

 X = 21 X = 23 Y = 6/5 x > y x > y Y = -9/5 x > y x > y

So, x > y

3.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x2-31x +184 = 0
II. y2-22y +96 = 0

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

E. x = y or relationship cannot be established

From equation I:
x2-31x +184 = (x -8)(x -23)= 0
=> x = 8, 23
From equation II:
y2-22y +96 = (y -6)(y -16) = 0
=> y = 6, 16

 X = 8 X = 23 Y =6 x > y x > y Y = 16 x < y x > y

So, relationship cannot be established between x and y

4.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 15x2 +22x +8 = 0
II. y2 +y -462 = 0

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

E. x = y or relationship cannot be established

From equation I:
15x2 +22x +8 = (5x + 4)(3x + 2)= 0
=> x = -4/5, -2/3
From equation II:
y2 +y -462 = (y + 22)(y -21) = 0
=> y = -22, 21

 X = -4/5 X = -2/3 Y =-22 x > y x > y Y = 21 x < y x < y

So, relationship cannot be established between x and y

5.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x2-5x -300 = 0
II. y2 +46y +528 = 0

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

B. x > y

From equation I:
x2-5x -300 = (x + 15)(x -20)= 0
=> x = -15, 20
From equation II:
y2 +46y +528 = (y + 24)(y + 22) = 0
=> y = -24, -22

 X = -15 X = 20 Y =-24 x > y x > y Y = -22 x > y x > y

So, x > y

6.In the following questions, two equations numbered I and II are given.You have to solve both the equations and choose the correct option.
187x2 + 6x – 1 = 0
153y2– 26y + 1 = 0

A. If x ≤ y
B. If x ≥ y
C. If x < y
D. If x > y
E. If x = y or no specific relation cannot be established.

A. If x ≤ y

First equation is –
187x2 + 6x – 1 = 0
or, 187x2– 11x + 17x – 1= 0
or, 11x (17x – 1) + 1 (17x – 1) = 0
or, (11x + 1) (17x – 1) = 0
x = – 1/11, 1/17
Second equation is –
153y2– 26y + 1 = 0
or, 153y2– 17y – 9y + 1 = 0
or, 17y (9y – 1) – 1 (9y – 1) = 0
or, (17y – 1) (9y – 1) = 0
y = 1/17, 1/9
y is always more than x or equal to x.

7.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 10x2 +5x -5 = 0
II. y2-16y +15 = 0

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

A. x < y

From equation I-
10x2 +5x -5 = (5x + 5)(2x -1)= 0
=> x = -5/5, 1/2
From equation II:
y2-16y +15 = (y -1)(y -15) = 0
=> y = 1, 15
So, x < y

8.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x = √484
II. 7y – 4x = 45

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

C. x > y

From I, x=22
From II, 7y – 88 = 45
=> 7y = 133
=> y = 19
Hence x > y

9.In the following questions, two equations numbered I and II are given.You have to solve both the equations and choose the correct option.
I. 3.5x2– 28x + 42 = 0
II. 1.5y2 + 3y – 12 = 0

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

B. x ≥ y

From I, 3.5(x-2)(x-6) = 0
=> x = 2, 6
From II, 1.5(y-2)(y+4) = 0
=> y = 2, -4
Hence x ≥ y

10.In the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I- y6 = 729
II- x5 = 243

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