Dear Aspirants, Our IBPS Guide team is providing new series of Quants Questions for IBPS PO Prelims 2023 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
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Directions (01-10): 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) I)Â x2Â + 19x + 88 = 0
II)y2+ 24y + 143 = 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 – 2x – 48 = 0
II)y2– 17y + 72 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
3) I)Â x2Â + 45 = 81
II)y2– 15y + 56 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
4) I)Â 2x + y = 10
II)3x + 5y = 29
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
5) I) x2 – 19x + 34 = 0
II)y2+ 21y – 46 =0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
6) I) x2 – 15x + 54 = 0
II)y2– 10y + 24 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
7) I) 2x2 + 12x – 54 = 0
II)2y2– 14y + 12 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
8) I)Â 18x + 3y = 24
II)x + 5y = 11
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
9) I) x2 – 32x + 240 = 0
II)y2– 27y + 180 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
10) I)Â x2Â + 19x + 70 = 0
II)2y2+ 26y + 80 = 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: B
x2Â + 19x + 88 = 0
x2Â + 11x + 8x + 88 = 0
x(x + 11) + 8(x + 11) = 0
(x + 8)(x + 11) = 0
x = -8, -11
y2Â + 24y + 143 = 0
y2Â + 11y + 13y + 143 = 0
y(y + 11) + 13(y + 11) = 0
(y + 13)(y + 11) = 0
y = -13, -11
x ≥ y
2) Answer: E
x2 – 2x – 48 = 0
x2 – 8x + 6x – 48 = 0
x(x – 8) + 6(x – 8) = 0
(x + 6)(x – 8) = 0
x = -6, 8
y2 – 17y + 72 = 0
y2 – 8y – 9y + 72 = 0
y(y – 8) – 9(y – 8) = 0
(y – 9)(y – 8) = 0
y = 9, 8
x ≤ y
3) Answer: D
x2Â + 45 = 81
x2Â = 36
x = 6, -6
y2Â – 15y + 56 = 0
y2 – 7y – 8y + 56 = 0
y(y – 7) – 8(y – 7) = 0
(y – 7)(y – 8) = 0
y = 7, 8
x < y
4) Answer: D
2x + y = 10 —–(1)
3x + 5y = 29 —–(2)
From (1) and (2)
7x = 21
x = 3
y = 4
x < y
5) Answer: B
x2 – 19x + 34 = 0
x2 – 17x – 2x + 34 = 0
x(x – 17) – 2(x – 17) = 0
(x – 2)(x – 17) = 0
x = 2, 17
y2Â + 21y – 46 =0
y2 + 23y – 2y – 46 = 0
y(y + 23) – 2(y + 23) = 0
(y – 2)(y + 23) = 0
y = 2, -23
x ≥ y
6) Answer: B
x2 – 15x + 54 = 0
x2 – 6x – 9x + 54 = 0
x(x – 6) – 9(x – 6) = 0
(x – 9)(x – 6) = 0
x = 6, 9
y2 – 10y + 24 = 0
y2 – 6y – 4y + 24 = 0
y(y – 6) – 4(y – 6) = 0
(y – 4)(y – 6) = 0
y = 4, 6
x ≥ y
7) Answer: C
2x2 + 12x – 54 = 0
2x2 + 18x – 6x – 54 = 0
2x(x + 9) – 6(x + 9) = 0
(2x – 6)(x + 9) = 0
x = 3, -9
2y2 – 14y + 12 = 0
2y2 – 12y – 2y + 12 = 0
2y(y – 6) – 2(y – 6) = 0
(2y – 2)(y – 6) = 0
y = 1, 6
Relationship between x and y cannot be established.
8) Answer: D
18x + 3y = 24 ——(1)
x + 5y = 11 ——(2)
From (1) and (2)
x = 1
y = 2
x < y
9) Answer: C
x2 – 32x + 240 = 0
x2 – 20x – 12x + 240 = 0
x(x – 20) – 12(x – 20) = 0
(x – 20)(x – 12) = 0
x = 20, 12
y2 – 27y + 180 = 0
y2 – 15y – 12y + 180 = 0
y(y – 15) – 12(y – 15) = 0
(y – 12)(y – 15) = 0
y = 12, 15
Relationship between x and y cannot be established.
10) Answer: C
x2Â + 19x + 70 = 0
x2Â + 14x + 5x + 70 = 0
x(x + 14) + 5(x + 14) = 0
(x + 5)(x + 14) = 0
x = -5, -14
2y2Â + 26y + 80 = 0
2y2Â + 10y + 16y + 80 = 0
2y(y + 5) + 16(y + 5) = 0
(2y + 16)(y + 5) = 0
y = -8, -5
Relationship between x and y cannot be established.
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Directions (11-20): In the following questions, two equations I and II are given. You have to solve both the equations and give an answer as,
11) I) 3x2 – 5x – 42 = 0
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
12) 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
13) 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.
14) 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
15) 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.
16) I). 2x² + 13x + 21 = 0
II). 5y² + 11y – 36 = 0
A.x > y
B.x < y
C.x = y or relation between x and y cannot be determined.
D.x ≤ y
E.x ≥ y
17) I) x2 – 20x + 75 = 0
II) 2y2– 14y + 24 = 0
A.x > y
B.x ≥ y
C.x < y
D.x ≤ y
E.x = y or the relation cannot be established
18) I) 2x – 3y = -1
II)3x – 5y = 0
A.x > y
B.x ≥ y
C.x = y or the relation cannot be established
D.x ≤ y
E.x < y
19) I) x2Â + 13x + 40 = 0
II) y2– 7y – 60 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
20) I) 2x + 2y = 30
II)Â 2x + y = 22
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
Answers :
11) 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
12) 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
13) Answer: A
6x² + 77x + 121 = 0
=> 6x2Â + 66x + 11x + 121 = 0
=> 6x(x + 11) + 11(x + 11) = 0
=> (6x + 11)(x + 11) = 0
=> x = -11/6, -11
y² + 30y + 225 = 0
=> y2Â + 15x + 15x + 225 = 0
=>y(y + 15) + 15(y + 15) = 0
=> (y + 15)(y + 15) = 0
=> y = -15, -15
Hence, x > y
14) 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
4y2Â + 13y + 3 = 0
=> 4y2Â + 12y + y + 3 = 0
=> 4y(y + 3) + 1(y + 3) = 0
=> (4y + 1)(y + 3) = 0
=> y = -1/4, -3
Hence, x > y
15) 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.
16) Answer: C
2x² + 13x + 21 = 0
=> 2x2Â + 6x + 7x + 21 = 0
=> 2x(x + 3) + 7(x + 3) = 0
=> (2x + 7)(x + 3) = 0
=> x = -7/2, -3
5y² + 11y – 36 = 0
=>5y2 + 20y – 9y – 36 = 0
=> 5y(y + 4) – 9(y + 4) = 0
=> (5y – 9)(y + 4) = 0
=> y = 9/5, -4
Hence, the relationship between x and y cannot be determined.
17) Answer: A
x2 – 20x + 75 = 0
x2 – 15x -5x + 75 = 0
(x – 15) (x – 5) = 0
x = 15, 5
2y2 – 14y + 24 = 0
2y2 – 8y – 6y + 24 = 0
2y (y – 4) – 6 (y – 4) = 0
(2y – 6) (y – 4) = 0
y = 3, 4
Hence, x > y
18) Answer: E
2x – 3y = -1 –> (1)
3x – 5y = 0 —> (2)
(1) * 3 = > 6x – 9y – 3 –> (3)
(2) * 2 = > 6x – 10y = 0 –> (4)
Subtract the equations (3) and (4), we get
= > y = – 3
Substitute the y value in equation (1), we get
2x – 3 * (-3) = -1
2x = -10
= > x = -5
By solving the equation (1) and (2), we get,
x = -5, y = -3
Hence, x < y
19) Answer: E
x2Â + 13x + 40 = 0
x2Â + 8x + 5x + 40 = 0
x(x + 8) + 5(x + 8) = 0
(x + 5)(x + 8) = 0
x = -5, -8
y2 – 7y – 60 = 0
y2 – 12y + 5y – 60 = 0
y(y – 12) + 5(y – 12) = 0
(y + 5)(y – 12) = 0
y = -5, 12
Hence, x ≤ y
20) Answer: D
2x + 2y = 30 ——–(1)
2x + y = 22 ———-(2)
(1) – (2)
y = 8
2x = 22 – 8 = 14
x = 7
Hence x < y
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