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

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

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_3.1

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:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_4.1

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.

100

120

140

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_5.1

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:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_6.1

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%

200%

100%

150%

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_7.1

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.

160

120

200

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_8.1

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

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

391

280

105

Solution:

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

11745

367

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_10.1

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

125

127

163

296

213

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_11.1

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

338

170

675

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_12.1

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

224

992

7136

Solution:

Quantitative Aptitude For NIACL AO Prelims: 26th January 2019 |_13.1

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