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

**Dear Students,**

**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?**

50

60

55

75

80

137.5

125.5

117.5

112.5

107.5

**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

44

46

56

54

64

**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

**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

**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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