Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk 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.
Quadratic equation
Directions (01-05): 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 cannot be determined.
d)If x < y
e)If x ≤ y
1) I) 2x2Â + 24x + 54 =0
II) 3y2– 21y+ 36 = 0
2) I) 3x2Â + 12x – 63 =0
II) y2– 16y + 63 = 0
3) I) (x + 2)2Â = 9
II) y2 =Â 25
4) I) 1/x2 × √x = 1/(2)3
II) y2 – 7y + 12 = 0
5) I) x = √841
II) y2= 961
missing number series
Directions (6-10): Find out the missing number in the following number series.
6) 1, 9, 21, ?, 63, 93Â Â Â Â
A.39
B.42
C.37
D.41
E.None of these
7) 342, ?, 210, 156, 110
A.306
B.272
C.240
D.281
E.None of these
8) 1, 5/2, 5, ?, 13, 37/2, 25
A.21/2
B.19/2
C.17/2
D.23/2
E.None of these
9) 3, 3, 4.5, 11.25, 45, 270, ?
A.2250
B.2285
C.2295
D.2025
E.None of these
10) 1.42, 1.94, 2.54, ?, 3.98, 4.82
A.2.23
B.3.22
C.2.77
D.3.24
E.None of these
Answers :
1) Answer: D
2x2 + 18x + 6x + 54 =0
2x(x + 9)Â +Â 6 (x + 9) = 0
(2x + 6) (x + 9) = 0
Roots, x = – 9, -3
3y2Â – 21y + 36 = 0
3y2Â – 12y – 9y + 36 = 0
3y (y – 4) – 9(y- 4) = 0
(3y – 9) (y – 4) = 0
Roots, y = Â 4, 3
Hence, x < y
2) Answer: D
3x2 + 12x – 63 = 0
3x2Â + 21x – 9x – 63 = 0
3x(x + 7) – 9(x + 7) = 0
(x + 7) (3x – 9) = 0
Roots, x = -7, 3
y2Â – 16y + 63 = 0
y2Â – 9y – 7y + 63 = 0
y(y – 9) – 7(y – 9) = 0
(y – 9) (y – 7) = 0
y = 9, 7
Hence, x < y
3) Answer: C
x2Â + 4x + 4 = 9
x2Â + 4x – 5 = 0
x2 + 5x – x – 5 = 0
x(x + 5) – 1(x + 5) =0
(x + 5) (x – 1) = 0
Roots, x= -5, 1
y2 =Â 25
Roots, y = 5, – 5
Hence, relationship cannot be determined.
4) Answer: B
1/x2 × √x = 1/(2)3
x-2 × x1/2 = 1/(√4)3
x-3/2Â = 1/(41/2)3
x-3/2Â = 1/(4)3/2
x-3/2Â = (4)-3/2
x = 4
 y2 – 7y + 12 = 0
y2Â – 4y – 3y + 12 = 0
y(y – 4) – 3 (y – 4) =0
(y – 4)(y – 3) = 0
Roots, y = 4, 3
Hence, x ≥ y
5) Answer: C
x = √841
x = 29
y2Â = 961
y = +31, – 31
Hence, relationship can’t be determined
6) Answer: A
3 + 5 = 8 (sum of prime numbers)
5 + 7 = 12
7 + 11 = 18
11 + 13 = 24
13 + 17 = 30
Therefore, 21 + 18 = 39
7) Answer: B
192 – 19 = 342
172 – 17 = 272
152 – 15 = 210
132 – 13 = 156
112 – 11 = 110
8) Answer: C
1 + 3/2 = 5/2
5/2 + 5/2 = 5
5 +Â 7/2 =Â 17/2
17/2 + 9/2 = 13
13 + 11/2 = 37/2
37/2 + 13/2 = 25
9) Answer: C
(3 × 1) = 3
(3 × 1.5) = 4.5
(4.5 × 2.5) = 11.25
(11.25 × 4) = 45
(45 × 6) = 270
(270 × 8.5) = 2295
10) Answer: B
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