Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for IBPS Clerk Prelims 2021 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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Simplification
Directions (01-05): What value should come in the place of (?) in the following questions.
1) 30% of 170 + 25% of 120 =? * 9
A.11
B.12
C.10
D.13
E.9
2) ? * 29 = 117 * √169 + 51.5 * 2
A.49
B.56
C.47
D.43
E.61
3) (0.5)12 * (0.0625)10 ÷ (0.25)8 = (0.5)?  Â
A.28
B.32
C.36
D.40
E.42
4) 80% of 875 – 200 = ? + 25 % of 1740
A.65
B.55
C.45
D.70
E.40
5) 48 * 25 – 35 * 25 = ?2Â + 1
A.11
B.13
C.18
D.19
E.20
Quadratic equation
Directions (06-10): 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,
6) I: 5x2 – 26x + 33 = 0
II: 2y2 – 15y + 27 = 0
A.If x < y
B.If x > y
C.If x ≤ y
D.If x ≥ y
E.If relationship between x and y cannot be determined
7) I: 3x2Â – 8x + 4 = 0
II: 5y2Â – 18y + 16 = 0
A.If x < y
B.If x > y
C.If x ≤ y
D.If x ≥ y
E.If relationship between x and y cannot be determined
8) I: x2 – 13x + 42 = 0
II: y2Â – 9y + 20 = 0
A.If x < y
B.If x > y
C.If x ≤ y
D.If x ≥ y
E.If relationship between x and y cannot be determined
9) I: 3x2 – 23x + 40 = 0        Â
II: 6y2Â + y – 2 = 0
A.If x < y
B.If x > y
C.If x ≤ y
D.If x ≥ y
E.If relationship between x and y cannot be determined
10) I: 3x2Â +14x + 8 = 0
II: 2y2 – 17y + 36 = 0
A.If x < y
B.If x > y
C.If x ≤ y
D.If x ≥ y
E.If relationship between x and y cannot be determined
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Answers :
1) Answer: E
30% of 170 + 25% of 120 =? * 9
51 + 30 = ? * 9
? = 9
2) Answer: B
? * 29 = 117 * √169 + 51.5 * 2
? * 29 = 1521 + 103
? = 56
3) Answer: C
(0.5)12 * (0.0625)10 ÷ (0.25)8 = (0.5)?
(0.5)12 * (0.5)40 ÷ (0.5)16 = (0.5)?
(0.5)12 + (40 – 16) = (0.5)?
(0.5)36Â = (0.5)?
? = 36
4) Answer: A
80% of 875 – 200 = ? + 25 % of 1740
700 – 200 = ? + 435
? = 65
5) Answer: C
48 * 25 – 35 * 25 = ?2Â + 1
1200 – 875 = ?2 + 1
? = 18
6) Answer: C
From I =>5x2 – 26x + 33 = 0
5x2 – 11x -15x + 33 = 0
x(5x-11)-3(5x-11) = 0
=> (5x – 11) (x – 3) = 0
=> x = 2.2, 3
From II =>2y2 – 15y + 27 = 0
2y2 – 9y – 6y +27 = 0
y(2y-9)-3(2y-9) = 0
=> (y – 3) (2y – 9) = 0
=> y = 3, 4.5
Hence, x ≤ y
7) Answer: E
From I =>3x2Â – 8x + 4 = 0
3x2 – 6x – 2x + 4 = 0
3x(x-2) -2(x-2) = 0
=>(x – 2) (3x – 2) = 0
=> x = 2, 2/3
From II =>5y2Â – 18y + 16 = 0
5y2Â – 10y -8y + 16 = 0
5y(y-2)-8(y-2) =0
=> (5y – 8) (y – 2) = 0
=> y = 2, 8/5
Hence, relationship between x and y cannot be determined
8) Answer: B
From I =>x2 – 13x + 42 = 0
x2 – 7x -6x + 42 = 0
x(x-7)-6(x-7) = 0
=> (x – 7) (x – 6) = 0
=> x = 6, 7
From II =>y2Â – 9y + 20 = 0
y2Â – 5y -4y + 20 = 0
y(y-5)-4(y-5) = 0
=> (y – 5) (y – 4) = 0
=> y = 4, 5
Hence, x > y
9) Answer: B
From I =>3x2 – 23x + 40 = 0
3x2 – 15x -8x + 40 = 0
3x(x-5)-8(x-5) = 0
=> (3x – 8) (x – 5) = 0
=> x = 8/3, 5
From II =>6y2Â + y – 2 = 0
6y2Â + 4y -3y – 2 = 0
2y(3y+2)-1(3y+2) = 0
=> (2y – 1) (3y + 2) = 0
=> y = 1/2, -2/3
Hence, x > y
10) Answer: A
From I =>3x2Â + 14x + 8 = 0
3x2Â + 12x +2x + 8 = 0
3x(x+4)+2(x+4) = 0
=> (3x + 2) (x + 4) = 0
=> x = -2/3, -4
From II =>2y2 – 17y + 36 = 0
2y2 – 9y -8y + 36 = 0
y(2y-9)-4(2y-9) = 0
=> (2y – 9) (y – 4) = 0
=> y = 9/2, 4
Hence, x < y
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