Quantitative Aptitude Set -623
Jun 16 2025
Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.
In the following questions, two equations I and II are given. You have to solve both the equations and give an answer as,
1) I) 3x2 –
II) 4y2+ 9y – 55 = 0
A.x > y
B.x = y or the relation cannot be established
C.x < y
D.x ≤ y
E.x ≥ y
2) I) 4x2 – 25x + 36 = 0
II) 2y2– 25y + 72 = 0
A.x < y
B.x ≥ y
C.x > y
D.x ≤ y
E.x = y or the relation cannot be established
3) I). 6x² + 77x + 121 = 0
II). y² + 30y + 225 = 0
A.x > y
B.x < y
C.x ≥ y
D.x ≤ y
E.x = y or relation between x and y cannot be determined.
4) I). 4x² – 13x + 9 = 0
II). 4y2 + 13y + 3 = 0
A.x = y or relation between x and y cannot be determined.
B.x < y
C.x ≥ y
D.x ≤ y
E.x > y
5) I). 3x2 + 8x – 35 = 0
II). y2 + 2y – 195 = 0
A.x > y
B.x < y
C.x ≥ y
D.x ≤ y
E.x = y or relation between x and y cannot be determined.
Answers :
1) Answer: B
3x2 – 5x – 42 = 0
3x2 + 9x – 14x – 42 = 0
3x (x + 3) – 14 (x + 3) = 0
(3x – 14) (x + 3) = 0
x = 14/3, -3 = 4.667, -3
4y2 + 9y – 55 = 0
4y2 + 20y – 11y – 55 = 0
4y (y + 5) – 11 (y + 5) = 0
(4y – 11) (y + 5) = 0
y = 11/4, -5 = 2.75, -5
Hence, the relationship between x and y cannot be determined
2) Answer: A
4x2 – 25x + 36 = 0
4x2 – 16x – 9x + 36 = 0
4x (x – 4) – 9 (x – 4) = 0
(4x – 9) (x – 4) = 0
x = 2.25, 4
2y2 – 25y + 72 = 0
2y2 – 16y – 9y + 72 = 0
2y (y – 8) – 9 (y – 8) = 0
(2y – 9) (y – 8) = 0
y = 4.5, 8
Hence, x < y
3) Answer: A
6x² + 77x + 121 = 0
=> 6x2 + 66x + 11x + 121 = 0
=> 6x(x + 11) + 11(x + 11) = 0
=> (6x + 11)(x + 11) = 0
=> x = -11/6, -11
y² + 30y + 225 = 0
=> y2 + 15x + 15x + 225 = 0
=>y(y + 15) + 15(y + 15) = 0
=> (y + 15)(y + 15) = 0
=> y = -15, -15
Hence, x > y
4) Answer: E
4x² – 13x + 9 = 0
=> 4x2 – 4x – 9x + 9 = 0
=> 4x(x – 1) – 9(x – 1) = 0
=> (4x – 9)(x – 1) = 0
=> x = 9/4, 1
4y2 + 13y + 3 = 0
=> 4y2 + 12y + y + 3 = 0
=> 4y(y + 3) + 1(y + 3) = 0
=> (4y + 1)(y + 3) = 0
=> y = -1/4, -3
Hence, x > y
5) Answer: E
3x2 + 8x – 35 = 0
=>3x2 + 15x – 7x – 35 = 0
=> 3x(x + 5) – 7(x + 5) = 0
=> (3x – 7)(x + 5) = 0
=> x = 7/3, -5
y2 + 2y – 195 = 0
=> y2 + 15y – 13y – 195 = 0
=>y(y + 15) – 13(y + 15) = 0
=> (y + 15)(y – 13) = 0
=> y = -15, 13
Hence, the relationship between x and y cannot be determined.
