Dear Aspirants,
Directions (1-5): What will come at the place of question mark in the following questions? (You are not expected to find exact value)
Q1.


Q6. 148, 152, 161, 177, ? , 238
Q7. 339, 355, 323, 371, 307, ?
Q8. 5, 14, 40, 117, 347, ?
Q9. 12, 24, 96, 576, ? , 46080
Q10. 156, 468, 780, ? , 1404, 1716
Directions (11-15): In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
Q11. I. x² = 144
II. y² - 24y + 144 = 0
Q12. I. 2x² - 9x + 10 = 0
II. 2y² - 13y + 20 = 0
Q13. I. 2x² + 15x + 27 = 0
II. 2y² + 7y + 6 = 0
Q14. I. 3x² - 13x + 12 = 0
II. 3y² - 13y + 14 = 0
Q15. I. 5x² + 8x + 3 = 0
II. 3y² + 7y + 4 = 0
Quantitative Aptitude Quiz For LIC AAO
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 will come at the place of question mark in the following questions? (You are not expected to find exact value)
Q1.
71
81
86
75
91
Q2. ∛2197 × 861.08 ÷ 90.98 - 57.98 =? 55
52
65
61
59
Q3. 69.98% of 259.98 – 29.98% of 529.98 = ? 19
23
20
27
18
Q4. 23
19
28
24
35
Q5. 489
476
550
525
500
Directions (6-10): What will come in place of question mark (?) in the following number series? Q6. 148, 152, 161, 177, ? , 238
208
214
202
198
192
Solution:
Pattern is +2², +3², +4², +5², +6²
∴ ? = 177 + 25 = 202
∴ ? = 177 + 25 = 202
Q7. 339, 355, 323, 371, 307, ?
407
354
328
387
388
Solution:
Pattern is +16, –32, +48, –64, +80
∴ ? = 307 + 80 = 387
∴ ? = 307 + 80 = 387
Q8. 5, 14, 40, 117, 347, ?
920
745
1124
1036
694
Solution:
Pattern is ×3–1, ×3–2, ×3–3, ×3–4, ×3–5,
∴ ? = 347 × 3 – 5 = 1036
∴ ? = 347 × 3 – 5 = 1036
Q9. 12, 24, 96, 576, ? , 46080
3542
3890
4248
4608
3246
Solution:
Patter is ×2, ×4, ×6, ×8, ×10
∴ ? = 576 × 8 = 4608
∴ ? = 576 × 8 = 4608
Q10. 156, 468, 780, ? , 1404, 1716
1096
1092
1290
9610
1910
Solution:
Patter is +312, +312, +312, +312, ….
∴ ? = 780 + 312 = 1092
∴ ? = 780 + 312 = 1092
Directions (11-15): In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
Q11. I. x² = 144
II. y² - 24y + 144 = 0
If x < y
If x ≤ y
If y < x
If y ≤ x
If x =y or if no relationship can be established.
Solution:
I. x² = 144
⇒x= ±12
II. y² - 24y + 144 = 0
⇒ (y – 12)² = 0
⇒ y – 12 = 0
⇒ y = 12
So, x≤y
⇒x= ±12
II. y² - 24y + 144 = 0
⇒ (y – 12)² = 0
⇒ y – 12 = 0
⇒ y = 12
So, x≤y
Q12. I. 2x² - 9x + 10 = 0
II. 2y² - 13y + 20 = 0
If x < y
If x ≤ y
If y < x
If y ≤ x
If x =y or if no relationship can be established.
Solution:
I. 2x² - 9x + 10 = 0
⇒ (x – 2) (2x - 5) = 0
⇒ x=2 or 5/2
II. 2y² - 13y + 20 = 0
⇒ (y - 4) (2y - 5) = 0
⇒ y=4 or 5/2
∴ y ≥ x
⇒ (x – 2) (2x - 5) = 0
⇒ x=2 or 5/2
II. 2y² - 13y + 20 = 0
⇒ (y - 4) (2y - 5) = 0
⇒ y=4 or 5/2
∴ y ≥ x
Q13. I. 2x² + 15x + 27 = 0
II. 2y² + 7y + 6 = 0
If x < y
If x ≤ y
If y < x
If y ≤ x
If x =y or if no relationship can be established.
Solution:
I. 2x² + 15x + 27 = 0
⇒ (2x + 9) (x + 3) = 0
⇒ x=-9/2 or-3
II. 2y² + 7y + 6 = 0
⇒ (2y + 3) (y + 2) = 0
⇒ y= -3/2 or-2
∴ x < y
⇒ (2x + 9) (x + 3) = 0
⇒ x=-9/2 or-3
II. 2y² + 7y + 6 = 0
⇒ (2y + 3) (y + 2) = 0
⇒ y= -3/2 or-2
∴ x < y
Q14. I. 3x² - 13x + 12 = 0
II. 3y² - 13y + 14 = 0
If x < y
If x ≤ y
If y < x
If y ≤ x
If x =y or if no relationship can be established.
Solution:
I. 3x² - 13x + 12 = 0
⇒ (3x – 4) (x - 3) = 0
⇒ x=4/3 or 3
II. 3y² - 13y + 14 = 0
⇒ (3y - 7) (y - 2) = 0
⇒y=7/3 or 2
So, no relation exists between x and y
⇒ (3x – 4) (x - 3) = 0
⇒ x=4/3 or 3
II. 3y² - 13y + 14 = 0
⇒ (3y - 7) (y - 2) = 0
⇒y=7/3 or 2
So, no relation exists between x and y
Q15. I. 5x² + 8x + 3 = 0
II. 3y² + 7y + 4 = 0
If x < y
If x ≤ y
If y < x
If y ≤ x
If x =y or if no relationship can be established.
Solution:
I. 5x² + 8x + 3 = 0
⇒ (5x + 3) (x + 1) = 0
⇒ x= -3/5 or-1
II. 3y² + 7y+ 4= 0
⇒ (y + 1) (3y + 4) = 0
⇒y= -1 or -4/3
So, x≥y
⇒ (5x + 3) (x + 1) = 0
⇒ x= -3/5 or-1
II. 3y² + 7y+ 4= 0
⇒ (y + 1) (3y + 4) = 0
⇒y= -1 or -4/3
So, x≥y

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