Quantitative Aptitude Set -386
Oct 22 2024
Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.
Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,
1)
II)y2– y – 6 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
2) I) x2 + 39x + 380 = 0
II)y2+ y – 342 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
3) I) x2 + 3x – 108 = 0
II)y2+ 25y + 156 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
4) I) 2x + y = 11
II)4x – y = 7
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
5) I) x2 – 30x + 221=0
II)y2– 23y + 132=0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
Answers :
1) Answer: C
x2 – x – 12 = 0
x2 – 4x + 3x – 12 = 0
x(x – 4) + 3(x – 4) = 0
(x + 3)(x – 4) = 0
x = 4, -3
y2 – y – 6 = 0
y2 – 3y + 2y – 6 = 0
y(y – 3) + 2(y – 3) = 0
(y + 2)(y – 3) = 0
y = -2, 3
Relationship between x and y cannot established.
2) Answer: E
x2 + 39x + 380 = 0
x2 + 20x + 19x + 380 = 0
x(x + 20) + 19(x + 20) = 0
(x + 19)(x + 20) = 0
x = -19, -20
y2 + y – 342 = 0
y2 + 19y – 18y – 342 = 0
y(y + 19) – 18(y + 19) = 0
(y – 18)(y + 19) = 0
y = 18, -19
Hence , x ≤ y
3) Answer: B
x2 + 3x – 108 = 0
x2 + 12x -9x – 108 = 0
x(x + 12) – 9(x + 12) = 0
(x – 9)(x + 12) = 0
x = 9, -12
y2 + 25y + 156 = 0
y2 + 12y + 13y + 156 = 0
y(y + 12)+ 13(y + 12) = 0
(y + 13)(y + 12) = 0
y = -13, -12
Hence, x ≥ y
4) Answer: D
2x + y = 11 -----(1)
4x – y = 7 -----(2)
From (1) and (2)
6x = 18
x = 3
y = 11 – 6 = 5
Hence, x < y
5) Answer: A
x2 – 30x + 221=0
x2 – 13x – 17x + 221 = 0
x(x – 13) – 17(x – 13) = 0
(x – 17)(x – 13) = 0
x = 17, 13
y2 – 23y + 132=0
y2 – 11y – 12y + 132 = 0
y(y -11) – 12(y – 11) = 0
(y – 12)(y – 11) = 0
y = 12, 11
Hence, x > y
