# Quantitative Aptitude For NIACL AO Prelims: 26th January 2019

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**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%

**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

**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

**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%

**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

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

**Q6. 31, 53, 105, 182, 280, 391**

391

31

280

53

105

**Q7. 1, 1, 3, 23, 367, 11745**

11745

1

3

23

367

**Q8. 125, 127, 137, 163, 213, 296**

125

127

163

296

213

**Q9. 675, 338, 170, 86, 44, 23**

23

338

170

44

675

**Q10. 48, 62, 96, 224, 992, 7136**

48

62

224

992

7136

**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

⇒ 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

⇒ 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

⇒ 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

⇒ 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

⇒ 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

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