Quantitative Aptitude: Quadratic Equations Set 17

Quadratic Equations SBI PO, IBPS PO, NIACL, NICl, LIC, Dena Bank PO PGDBF, BOI, 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. 3x2 â€“ 25x + 52 = 0,
II. 3y2 â€“ 16y + 16 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option C
Solution:

3x2 – 25x + 52 = 0
3x2 – 12x â€“ 13x + 52 = 0
Gives x = 4, 13/3
3y2 â€“ 16y + 16 = 0
3y2 â€“ 14y â€“ 4y + 16 = 0
Gives y = 4, 4/3
2. I. 3x2 â€“ 8x â€“ 16 = 0,
II. 2y2 â€“ 5y â€“ 12 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option E
Solution:

3x2 – 8x – 16 = 0
3x2 – 12x + 4x – 16 = 0
Gives x = -4/3, 4
2y2 â€“ 5y â€“ 12 = 0
2y2 â€“ 8y + 3y â€“ 12 = 0
Gives y = -3/2, 4
3. I. 3x2 + 22x + 24 = 0,
II. 2y2 + 7y â€“ 4 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If 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 + 7y – 4 = 0
2y2 + 8y – y – 4 = 0
Gives y = -4, 1/2
4. I. 2x2 â€“ 9x + 4 = 0,
II. 2y2 â€“ 17y + 36 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option D
Solution:

2x2 – 9x + 4 = 0
2x2 – 8x – x + 4 = 0
So x = 4 , 1/2
2y2 – 17y + 36 = 0
2y2 – 8y – 9y + 36 = 0
Gives y= 4, 9/2
5. I. 3x2 + 7x â€“ 6 = 0,
II. 3y2 â€“ 19y + 20 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option B
Solution:

Explanation:
3x2 + 7x â€“ 6 = 0
3x2 + 9x â€“ 2x â€“ 6 = 0
Gives x = -3, 2/3
3y2 â€“ 19y + 20 = 0
3y2 â€“ 15y â€“ 4y + 20 = 0
Gives y = 4/3, 5
6. I. 3x2 â€“ 4x â€“ 4 = 0,
II. 3y2 + 16y + 20 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option A
Solution:

3x2 – 4x â€“ 4 = 0
3x2 – 6x + 2x â€“ 4 = 0
Gives x = -2/3, 2
3y2 + 16y + 20 = 0
3y2 + 6y + 10y + 20 = 0
Gives y = -10/3, -2
7. I. 7x2 + 19x â€“ 6 = 0,
II. 2y2 + 13y + 21 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relationship cannot be determined
Option C
Solution:

7x2 + 19x â€“ 6 = 0
7x2 + 21x – 2x â€“ 6 = 0
Gives x = -3, 2/7
2y2 + 13y + 21 = 0
2y2 + 6y + 7y + 21 = 0
So y = -7/2, -3
8. I. x2 + (4 + âˆš2)x + 4âˆš2 = 0
II. 3y2 â€“ (1 + 3âˆš3)y + âˆš3 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option B
Solution:

x2 + (4 + âˆš2)x + 4âˆš2 = 0
(x2 + 4x) + (âˆš2x + 4âˆš2) = 0
x (x + 4) + âˆš2 (x + 4) = 0
So x = -4, -âˆš2 (-1.4)
3y2 – (1 + 3âˆš3)y + âˆš3 = 0
(3y2 – y) – (3âˆš3y – âˆš3) = 0
y (3y â€“ 1) – âˆš3 (3y â€“ 1) = 0
So, y = 1/3, âˆš3 (1.7)
9. I. 3x2 + (3 + 2âˆš2)x + 2âˆš2 = 0
II. 5y2 + (2 + 5âˆš2)y + 2âˆš2 = 0
A) If x > y
B) If x < y
C) If x â‰¥ y
D) If x â‰¤ y
E) If x = y or relation cannot be established
Option E
Solution:

3x2 + (3 + 2âˆš2)x + 3âˆš2 = 0
(3x2 + 3x) + (2âˆš2x + 2âˆš2) = 0
3x (x + 1) + 2âˆš2 (x + 1) = 0
So x = -1, -2âˆš2/3
5y2 + (2 + 5âˆš2)y + 2âˆš2 = 0
(5y2 + 2y) + (5âˆš2y + 2âˆš2) = 0
y (5y + 2) + âˆš2 (5y + 2) = 0
So, y = -2/5, -âˆš2
10. I. 4x2 â€“ 12x + 5 = 0,
II. 2y2 + 3y â€“ 20 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option E
Solution:

4x2 – 12x + 5 = 0
4x2 – 2x – 10x + 5 = 0
x = 1/2, 5/2
2y2 + 3y – 20 = 0
2y2 + 8y â€“ 5y – 20 = 0
So y = -3, 5/2

35 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 17”

1. ?@shh...ishhh..â›·?

thnx..

2. Pawan

1. M@nish...

yes

3. AVIâ„¢ (Sinchen loveR)

8/8
Q8 ans may be E ?

1. Both values of x are negative. And that if y positive. So why E ??

4. sachin shukla@ my turn 2017

ty…

5. Ayushi

thanku mam

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