Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for IBPS PO Prelims 2020 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.
Missing number series
Direction (1-5): What value should come in the place of (?) in the following number series?
1) 4080, 3360, 2730, 2184, 1716, ?
A) 1320
B) 1850
C) 1560
D) 1210
E) 1080
2) 3, 21, 33, 39, 39, ?
A) 52
B) 48
C) 33
D) 28
E) 22
3) 19, 17, 30, 82, 312, ?
A) 1528
B) 1236
C) 1328
D) 1124
E) 1456
4) 94, 95, 98, 104, 114, ?
A) 110
B) 116
C) 119
D) 124
E) 129
5) 2430, 1620, 1080, 720, ?, 320
A) 640
B) 560
C) 480
D) 680
E) 520
Direction (6-10): 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,
a) If x > y
b) If x ≥ y
c) If x = y or relationship can’t be determined.
d) If x < y
e) If x ≤ y
6)
I) x2+ 2x – 195 = 0
II)y2+ y – 182 = 0
7)
I)x2– 25x + 136 = 0
II)y2– y – 56 = 0
8)
I)Â 2x2+ 16x + 30 = 0
II)y2+ 6y + 8 = 0
9)
I)x2+ 5x + 6 = 0
II)y2– 2y + 1 = 0
10)
I) x2– 11x + 30 = 0
II)y2– 21y + 90 = 0
Answers :
Directions (1-5) :
1) Answer: A
163 – 16 = 4080
153 – 15 = 3360
143 – 14 = 2730
133 – 13 = 2184
123 – 12 = 1716
113 – 11 = 1320
2) Answer: C
3 + (3 * 6) = 21
21 + (2 * 6) = 33
33 + (1 * 6) = 39
39 + (0 * 6) = 39
39 + (-1 * 6) = 33
3) Answer: A
19 * 1 – 2 = 17
17 * 2 – 4 = 30
30 * 3 – 8 = 82
82 * 4 – 16 = 312
312 * 5 – 32 = 1528
4) Answer: E
94 + 1 = 95
95 + 1 + 2 = 98
98 + 1 + 2 + 3 = 104
104 + 1 + 2 + 3 + 4 = 114
114 + 1 + 2 + 3 + 4 + 5 =Â 129
5) Answer: C
2430 ÷ 1.5 = 1620
1620 ÷ 1.5 = 1080
1080 ÷ 1.5 = 720
720 ÷ 1.5 = 480
480 ÷ 1.5 = 320
Directions (6-10):
6) Answer: C
x2 + 2x – 195 = 0
x2 + 15x – 13x – 195 = 0
x(x + 15) – 13(x + 15) = 0
(x – 13)(x + 15) = 0
X = 13, -15
y2 + y – 182 = 0
y2 + 14y – 13y – 182 = 0
y(y + 14) – 13(y + 14) = 0
(y – 13)(y + 14) = 0
Y = 13, -14
Relationship between x and y cannot be established.
7) Answer: B
x2 – 25x + 136 = 0
x2 – 17x – 8x + 136 = 0
x(x – 17) – 8(x – 17) = 0
(x – 8)(x – 17) = 0
x = 8, 17
y2 – y – 56 = 0
y2 – 8y + 7y – 56 = 0
y(y – 8) + 7(y – 8) = 0
(y + 7)(y – 8) = 0
y = -7, 8
Hence, x ≥ y
8) Answer: C
2x2Â + 16x + 30 = 0
2x2Â + 10x + 6x + 30 = 0
2x(x + 5) + 6(x + 5) = 0
(2x + 6)(x + 5) = 0
x = -3, -5
y2Â + 6y + 8 = 0
y2Â + 4y + 2y + 8 = 0
y(y + 4) + 2(y + 4) = 0
(y + 2)(y + 4) = 0
y = -2, -4
Relationship between x and y cannot be established.
9) Answer: D
x2Â + 5x + 6 = 0
x2Â + 3x + 2x + 6 = 0
x(x + 3) + 2(x + 3) = 0
(x + 2)(x + 3) = 0
x = -2, -3
y2 – 2y + 1 = 0
y2 – y – y + 1 = 0
y(y – 1) – 1(y – 1) = 0
(y – 1)(y – 1) = 0
y = 1, 1
Hence, x < Y
10) Answer: E
x2 – 11x + 30 = 0
x2 – 6x – 5x + 30 = 0
x(x – 6) – 5(x – 6) = 0
(x – 5)(x – 6) = 0
x = 5, 6
y2 – 21y + 90 = 0
y2 – 15y – 6y + 90 = 0
y(y – 15) – 6(y – 15) = 0
(y – 6)(y – 15) = 0
y = 6, 15
Hence, x ≤ y
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