Quantitative Aptitude Quiz for IBPS Clerk Prelims: 10th November 2018

Dear Students,


Quantitative Aptitude Quiz For IBPS PO Mains: 10th November 2018
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.



Directions (1-5): What should come in place of question mark (?) in the following questions? 

Q1.
50
60
55
75
80
Solution:


Q2.
137.5
125.5
117.5
112.5
107.5
Solution:


Q3. 22.5 × 12 + (11)² – √(?) = (19)²
800
750
825
900
950
Solution:
270 + 121 – √(?) = 361 √(?)=30 ? = 900

Q4. 624/?+3.5×6+27=(6)²+440/2
5
3
4
8
7
Solution:


Q5.
44
46
56
54
64
Solution:


Directions (6-10): What will come in the place of the question mark (?) in the following number series? 

Q6. 125, 189, 314, 530, 873, ?
1385
1265
1525
1375
1785
Solution:


Q7. 150, 300, 60, 120, 24, 48, ?
16.4
7.6
9.6
12
30
Solution:
Pattern is
150 × 2 = 300
300 ÷ 5 = 60
60 × 2 = 120
120 ÷ 5 = 24
24 × 2 = 48
48 ÷ 5 = 9.6

Q8. 106, 184, 267, 357, 456, ?
486
566
626
766
546
Solution:


Q9. 142, 119, 100, 83, 70, ?
69
49
42
59
53
Solution:
Pattern is
142 – 23 = 119
119 – 19 = 100
100 – 17 = 83
83 – 13 = 70
70 – 11 = 59

Q10. 120, 60, 90, 225, ?
782.5
787.5
752.5
875
855.5
Solution:
Pattern is
 ×0.5, ×1.5, ×2.5, ×3.5, ×4.5 ….
∴ 225 × 3.5 = 787.5

Directions (11-15): Solve the given quadratic equations and mark the correct option based on your answer— 

Q11. (i) x² – x – 132 = 0 
         (ii) y² – 22y + 112 = 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
X² – 12x + 11x – 132 = 0
x(x – 12) + 11 (x – 12) = 0
(x + 11) (x – 12) = 0
x = –11, 12
y² – 14y – 8y + 112 = 0
y(y – 14) – 8(y – 14) = 0
(y – 8) (y – 14) = 0
y = 8, 14
No relation can be established between x & y.

Q12. (i) x² – 8x + 15 = 0 
         (ii) y² – 2y – 3 = 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
x² – 3x – 5x + 15 = 0
(x – 3) (x – 5) = 0
x = 3, 5 y² – 3y + y – 3 = 0
y(y – 3) + 1 (y – 3) = 0
(y + 1) (y – 3) =0
y = –1, 3
x ≥ y

Q13. (i) x² + 5√3 x+18=0 
         (ii) y²-√3 y-18=0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
x²+2√3 x+3√3 x+18=0
x(x+2√3)+3√3 (x+2√3)=0
(x+3√3)(x+2√3)=0
x=-3√3,-2√3
y²+2√(3) y-3√(3 ) y-18=0
(y+2√3)(y-3√3)=0
y= -2√(3,) 3√3
y ≥ x

Q14. (i) 2x² – x – 3 = 0 
         (ii) 4y² – 24y + 35 = 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
2x² – x – 3 = 0
(x + 1) (2x – 3) = 0
x= -1, 3/2 4y² – 24y + 35 = 0
4y² – 14y – 10y + 35 = 0
y=5/2, 7/2
y > x

Q15. (i) 2x² – 7x + 6 = 0 
         (ii) 2y² – y– 1= 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
2x²– 3x – 4x + 6 = 0
x(2x – 3) – 2 (2x – 3) = 0
(x – 2) (2x – 3) = 0
x=2, 3/2
2y² – 2y + y – 1 = 0
2y(y – 1) + 1 (y – 1) = 0
(2y + 1) (y – 1) = 0
y = -1/2 , 1
x > y

               




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