Quantitative Aptitude: Quadratic Equations Set 27

Quadratic Equations Practice Sets for IBPS PO, IBPS RRB, LIC, UIICL, OICL, Bank of Baroda and other competitive 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. 4x2 – x – 14 = 0,
II. 2y2 – 13y + 20 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option B
Solution:

4x2 – x – 14 = 0
4x2 – 8x + 7x – 14 = 0
Gives x = -7/4, 2
2y2 – 13y + 20 = 0
2y2 – 8y – 5y + 20 = 0
Gives y = 5/2, 4
2. I. 2x2 – 11x + 14 = 0,
II. 3y2 + 13y + 14 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option A
Solution:

2x2 – 11x + 14 = 0
2x2 – 4x – 7x + 14 = 0
Gives x = 2, 7/2
3y2 + 13y + 14 = 0
3y2 + 6y + 7y+ 14 = 0
Gives y = -7/3, -2
3. I. 3x2 + 14x + 15 = 0,
II. 3y2 – 13y – 30 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option D
Solution:

3x2 + 14x + 15 = 0
3x2 + 9x + 5x + 15 = 0
Gives x = -5/3, 6
3y2 – 13y – 30 = 0
3y2 – 18y + 5y – 30 = 0
Gives y = -3, -5/3
4. I. 3x2 + 28x + 60 = 0,
II. 2y2 – 3y – 20 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option B
Solution:

3x2 + 28x + 60 = 0
3x2 + 18x + 10x + 60 = 0
Gives x = -6, -10/3
2y2 – 3y – 20 = 0
2y2 – 8y + 5y – 20 = 0
Gives y= -5/2, 4
5. I. 3x2 – 8x – 35 = 0,
II. 3y2 + 37y + 104 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option A
Solution:

3x2 – 8x – 35 = 0
3x2 – 15x + 7x – 35 = 0
Gives x = 5, -7/3
3y2 + 37y + 104 = 0
3y2 + 24y + 13y + 104 = 0
Gives y= -8, -13/3
6. I. 3x2 – 5x – 78 = 0,
II. 3y2 + 28y + 65 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option C
Solution:

3x2 – 5x – 78 = 0
3x2 – 18x + 13x – 78 = 0
Gives x = -13/3, 6
3y2 + 28y + 65 = 0
3y2 + 15y + 13y + 65 = 0
Gives y = -5, -13/3
7. I. 3x2 – 7x – 40 = 0,
II. 3y2 + 26y + 48 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option C
Solution:

3x2 – 7x – 40 = 0
3x2 – 15x + 8x – 40 = 0
Gives x = -8/3, 5
3y2 + 26y + 48 = 0
3y2 + 18y + 8y + 48 = 0
Gives y = -6, -8/3
8. I. x2 + (4 + 2√2)x + 8√2 = 0
II. 3y2 + (3 + 3√2)y + 3√2 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option B
Solution:

x2 + (4 + 2√2)x + 8√2 = 0
(x2 + 4x) + (2√2x + 8√2) = 0
x (x + 4) + 2√2 (x + 4) = 0
So x = -4, -2√2 (-2.8)
3y2 + (3 + 3√2)y + 3√2 = 0
(3y2 + 3y) + (3√2y + 3√2) = 0
3y (y + 1) + 3√2 (y + 1) = 0
So, y = -1, -√2 (-1.41)
9. I. 6x2 – (3 + 4√3)x + 2√3 = 0,
II. 4y2 – (2 + 4√3)y + 2√3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
Option E
Solution:

6x2 – (3 + 4√3)x + 2√3 = 0
(6x2 – 3x) – (4√3x – 2√3) = 0
3x (2x- 1) – 2√3 (2x – 1) = 0,
So x = 1/2 (0.5), 2√3/3 (1.16)
4y2 – (2 + 4√3)y + 2√3 = 0
(4y2 – 2y) – (4√3y – 2√3) = 0
2y (2y – 1) – 2√3 (2y – 1) = 0
So, y = 1/2 (0.5), √3 (1.73)
10. I. x2 + (4 + 2√2)x + 8√2 = 0
II. y2 – (2 + 3√3)y + 6√3 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
Option B
Solution:

x2 + (4 + 2√2)x + 8√2 = 0
(x2 + 4x) + (2√2x + 8√2) = 0
x (x + 4) + 2√2 (x + 4) = 0
So x = -4, -2√2 (-2.82)
y2 – (2 + 3√3)y + 6√3 = 0
(y2 – 2y) – (3√3y – 6√3) = 0
y (y – 2) – 3√3 (y – 2) = 0
So y = 2, 3√3 (5.2)

14 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 27”

1. Sandy

@MOD
qsn 8
2nd equation mai y ki value -1 & -root2 ayegi….
So overall X<Y ans hoga…
Do check

2. ?Jab Harry Met Sejal?

sir
1st Ques me
2*14 how we get factors 8 /2 or 5/2
may be instead of 14 there is 20
or instead of 13 there is 11
2y^2 -13Y +14 = 0

1. yes its 20

3. ครђ

Thanks ma’am nice set :))

4. sonali

ty AZ…nice ques..

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tq mam

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