Quantitative Aptitude Quiz For NIACL AO
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’.
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.
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.
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.
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.
Directions (6-10): Find the wrong number in the following number series:
Q6. 31, 53, 105, 182, 280, 391
Q7. 1, 1, 3, 23, 367, 11745
Q8. 125, 127, 137, 163, 213, 296
Q9. 675, 338, 170, 86, 44, 23
Q10. 48, 62, 96, 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
⇒ 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
⇒ 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
⇒ 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
⇒ 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
⇒ 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