Dear Aspirants,

### Quantitative Aptitude Quiz For NIACL AO

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.

Direction (1-5): - Table given below shows total number of employees (in % out of total employees) working in given companies. Study the data carefully and answer the following questions

Total employees in A = 1200
Total employees in B = 800
Total employees in C = 900

Q1. Total number of employees whose age is 30 and above in company ‘C’ is what percent more than total number of employees who age is 40 and above in company ‘A’.
50%
37.5%
25%
12.5%
10%
Solution:

Q2. Total number of persons in company ‘B’ whose age is 30 and above but less than 40 is how much more than total number of persons in company ‘A’ whose age is 40 and above but less than 50.
60
80
100
120
140
Solution:

Q3. Find the ratio between total number of employees in company ‘C’ whose age is 30 and above but less than 50 to total number of employees in company ‘A’ whose age is 30 and above but less than 50.
9 : 8
1 : 2
3 : 4
4 : 3
1 : 4
Solution:

Q4. Total number of person in company ‘B’ whose age is 40 and above is what percent of the total number of person in company ‘A’ whose age is 50 and above.
50%
0%
200%
100%
150%
Solution:

Q5. Total number of person in company ‘A’ whose age is less than 40 is how much more/less than total number of person in company ‘B’ whose age is less than 50.
80
160
40
120
200
Solution:

Directions (6-10): Find the wrong number in the following number series:

Q6. 31, 53, 105, 182, 280, 391
391
31
280
53
105
Solution:

Q7. 1, 1, 3, 23, 367, 11745
11745
1
3
23
367
Solution:

Q8. 125, 127, 137, 163, 213, 296
125
127
163
296
213
Solution:

Q9. 675, 338, 170, 86, 44, 23
23
338
170
44
675
Solution:

Q10. 48, 62, 96, 224, 992, 7136
48
62
224
992
7136
Solution:

Directions (11-15): In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer.

Q11. I. 25x² - 90x + 72 = 0
II. 5y² - 27y + 36 = 0
If x = y or no relation can be established
If x > y
If x < y
If x ≥ y
If x ≤ y
Solution:
I. 25x² – 90x + 72 = 0
⇒ 25x² – 30x – 60x + 72 = 0
⇒ 5x (5x – 6) – 12 (5x – 6) = 0
⇒ x = 6/5 or 12/5
II. 5y² – 27y + 36 = 0
⇒ 5y² – 15y – 12y + 36 = 0
⇒ 5y (y – 3) – 12 (y – 3) = 0
⇒ y = 3 or 12/5
y ≥ x

Q12. I. 12x² + 46x + 42 = 0
II. 3y² - 16y + 21 = 0
If x = y or no relation can be established
If x > y
If x < y
If x ≥ y
If x ≤ y
Solution:
I. 12x² + 46x + 42 = 0
⇒ 12x² + 18x + 28x + 42 = 0
⇒ 6x (2x + 3) + 14 (2x + 3) = 0
⇒ x = (–3)/2 or (–14)/6
II. 3y² – 16y + 21 = 0
⇒ 3y² – 9y – 7y + 21 = 0
⇒ 3y (y – 3) – 7 (y – 3) = 0
⇒ y = 3 or 7/3
y > x

Q13. I. 4x² + 10x = 14
II. 15 = 16y – 4y²
If x = y or no relation can be established
If x > y
If x < y
If x ≥ y
If x ≤ y
Solution:
I. 4x² + 10x – 14 = 0
⇒ 4x² + 14x – 4x – 14 = 0
⇒ 2x (2x + 7) – 2 (2x + 7) = 0
⇒ x = 1 or (–7)/2
II. 4y² – 16y + 15 = 0
⇒ 4y² – 6y – 10y + 15 = 0
⇒ 2y (2y – 3) – 5 (2y – 3) = 0
⇒ y = 3/2 or 5/2
y > x

Q14. I. 6x² + 15x – 36 = 0
II. 4y² - 2y – 10 = -8
If x = y or no relation can be established
If x > y
If x < y
If x ≥ y
If x ≤ y
Solution:
I. 6x² + 15x – 36 = 0
⇒ 6x² + 24x – 9x – 36 = 0
⇒ 6x (x + 4) – 9 (x + 4) = 0
⇒ x = –4 or 9/6
II. 4y² – 2y – 2 = 0
⇒ 4y² – 4y + 2y – 2 = 0
⇒ 4y (y – 1) + 2 (y – 1) = 0
⇒ y = 1 or (–1)/2
Relationship can’t be established

Q15. I. 2x² - 19x + 44 = 0
II. 3y² - 22y + 40 = 0
If x = y or no relation can be established
If x > y
If x < y
If x ≥ y
If x ≤ y
Solution:
I. 2x² – 19x + 44 = 0
⇒ 2x² – 8x – 11x + 44 = 0
⇒ 2x (x – 4) – 11 (x – 4) = 0
⇒ x = 4 or 11/2
II. 3y² – 22y + 40 = 0
⇒ 3y² – 12y – 10y + 40 = 0
⇒ 3y (y – 4) – 10 (y – 4) = 0
⇒ y = 4 or 10/3
x ≥ y

You May also like to Read: