# Quantitative Aptitude: Quadratic Equations Set 12

Type of Quadratic Equations Questions Asked in All Exams – Bank PO, Insurance and Other related Exams.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

1. I. 20x2 – 31x + 12 = 0
II. 3y2 – 16y + 16 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option B
Solution:

20x2 – 31x + 12 = 0
20x2 – 16x – 15x + 12 = 0
So x = 3/4, 4/5
3y2 – 16y + 16 = 0
3y2 – 14y – 4y + 16 = 0
Gives y = 4, 4/3
2. I. 3x2 + 22 x + 24 = 0
II. 2y2 – 5y – 12 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option E
Solution:

3x2 + 22 x + 24 = 0
3x2 + 18x + 4x + 24 = 0
So x = -4/3, -6
2y2 – 5y – 12 = 0
2y2 – 8y + 3y – 12 = 0
Gives y = -3/2, 4
3. I. 2x2 – 9x + 4 = 0
II. 4y2 – 13y – 12 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option E
Solution:

Solution:
2x2 – 9x + 4 = 0
2x2 – 8x – x + 4 = 0
So x = 4 , 1/2
4y2 – 13y – 12 = 0
4y2 – 16y + 3y – 12 = 0
Gives y = -3/4, 4
4. I. 5x2 + 23x + 12 = 0
II. 5y2 – 7y – 6 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option D
Solution:

5x2 + 23x + 12 = 0
5x2 + 20x + 3x + 12 = 0
So x = -4, -3/5
5y2 – 7y – 6 = 0
5y2 – 10y + 3y – 6 = 0
So y = -3/5, 2
Put all values on number line and analyze the relationship
-4….. -3/5…. 2
5. I. 7x2 + 19x – 6 = 0,
II. 2y2 – 7y + 3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option B
Solution:

7x2 + 19x – 6 = 0
7x2 + 21x – 2x – 6 = 0
Gives x = -3, 2/7
2y2 – 7y + 3 = 0
2y2 – 6y – y + 3 = 0
So y = 1/2, 3
6. I. 4x2 – 12x + 5 = 0,
II. 2y2 – 19y + 35 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option D
4x2 – 12x + 5 = 0
4x2 – 2x – 10x + 5 = 0
x = 1/2, 5/2
2y2 – 19y + 35 = 0
2y2 – 14y – 5y + 35 = 0
So y = 5/2, 7
7. I. 2x2 + 5x ¬– 12 = 0,
II. 4y2 + 13y – 12 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option E
2x2 + 5x ¬– 12 = 0
2x2 + 8x ¬– 3x – 12 = 0
So x = -4 , 3/2
4y2 + 13y – 12 = 0
4y2 + 16y – 3y – 12 = 0
y = -4, 3/4
8. I. 3x2 + 22x + 24 = 0,
II. 4y2 – 9y – 9 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option B
3x2 + 22 x + 24 = 0
3x2 + 18x + 4x + 24 = 0
Gives x = -4/3, -6
4y2 – 9y – 9 = 0
4y2 – 12y + 3y – 9 = 0
y = -3/4, 3
9. I. 20x2 – 31x + 12 = 0,
II. 4y2 + 9y – 9 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option C
20x2 – 31x + 12 = 0
20x2 – 16x – 15x + 12 = 0
Gives x = 3/4, 4/5
4y2 + 9y – 9 = 0
4y2 + 12y – 3y – 9 = 0
y = 3/4, -3
10. I. 6x2 – 7x – 3 = 0,
II. 4y2 + 5y – 6 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established