# IBPS RRB PO Prelims Quantitative Aptitude Questions 2019 (Day-04)

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## IBPS RRB PO Mains Free Mock Test 2019

[WpProQuiz 6568]

Directions (Q. 1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

a) If x > y

b) If x â‰¥ y

c) If x < y

d) If x â‰¤ y

e) If x = y or the relation cannot be established

1)

I) 2x2 â€“ 16x + 32 = 0

II) 3y2 â€“ 15y + 18 = 0

2)

I) x2 â€“ 4x â€“ 77 = 0

II) y2 â€“ 27y + 92 = 0

3)

I) 4x + 5y + 26 = 0

II) 3x â€“ 4y + 4 = 0

4)

I) 3x2 + 14x + 15 = 0

II) 3y2 – 13y + 14 = 0

5)

I) 4x2 â€“ 6x â€“ 18 = 0

II) 5y2 + 6y â€“ 27 = 0

Directions (Q. 6 â€“ 10): Study the following information carefully and answer the given questions:

The following table shows the total number of employees working in 5 different departments and the percentage of females among them.

6) Find the ratio between the total number of males working in department A to that of total number of females working in department D?

a) 13 : 18

b) 27 : 35

c) 22 : 29

d) 11 : 15

e) None of these

7) Find the total number of unmarried males in department B and E together, if the ratio between the total number of married males to that of unmarried males is 19 : 13 and 2 : 1 respectively?

a) 584

b) 632

c) 678

d) 556

e) None of these

8) Find the average number of females working in department B, C and D together?

a) 812

b) 844

c) 878

d) 926

e) None of these

9) Total number of females working in department A and E together is approximately what percentage of total number of males working in department C and D together?

a) 109 %

b) 125 %

c) 142 %

d) 95 %

e) 156 %

10) Total number of employees working in department A and B together is approximately what percentage more than the total number of employees working in department E?

a) 6 %

b) 32 %

c) 17 %

d) 45 %

e) 58 %

Directions (1-5):

I) 2x2 â€“ 16x + 32 = 0

2x2 â€“ 8x â€“ 8x + 32 = 0

2x (x â€“ 4) – 8 (x â€“ 4) = 0

(2x â€“ 8) (x â€“ 4) = 0

x = 8/2, 4 = 4, 4

II) 3y2 â€“ 15y + 18 = 0

3y2 â€“ 9y – 6y + 18 = 0

3y (y â€“ 3) â€“ 6 (y â€“ 3) = 0

(3y â€“ 6) (y â€“ 3) = 0

y = 6/3, 3 = 2, 3

x > y

I) x2 â€“ 4x â€“ 77 = 0

(x â€“ 11) (x + 7) = 0

x = 11, -7

II) y2 â€“ 27y + 92 = 0

(y â€“ 23) (y â€“ 4) = 0

y = 23, 4

Canâ€™t be determined

4x + 5y = – 26 –> (1)

3x â€“ 4y = – 4 –> (2)

By solving the equation (1) and (2), we get,

x = -4, y = -2

x < y

I) 3x2 + 14x + 15 = 0

3x2 + 9x + 5x + 15 = 0

3x (x + 3) + 5 (x + 3) = 0

(3x + 5) (x + 3) = 0

x = -5/3, -3

II) 3y2 – 13y + 14 = 0

3y2 – 6y – 7y + 14 = 0

3y (y – 2) – 7 (y – 2) = 0

(3y – 7) (y – 2) = 0

y = 7/3, 2

x < y

I) 4x2 â€“ 6x â€“ 18 = 0

4x2 â€“ 12x + 6x â€“ 18 = 0

4x(x â€“ 3) + 6(x â€“ 3) = 0

(4x + 6) (x â€“ 3) = 0

x = -6/4, 3 = -3/2, 3 = -1.5, 3

II) 5y2 + 6y â€“ 27 = 0

5y2 + 15y â€“ 9y â€“ 27 = 0

5y(y + 3) – 9(y + 3) = 0

(5y â€“ 9) (y + 3) = 0

y = 9/5, -3 = 1.8, -3

Canâ€™t be determined

Directions (6-10):

The total number of males working in department A

= > 1500 * (52/100)

The total number of females working in department D

= > 2000 * (54/100)

Required ratio = [1500 * (52/100)] : [2000 * (54/100)]

= > 13 : 18

The total number of unmarried males in department B and E together

= > 1200 * (48/100) * (13/32) + 2300 * (42/100) * (1/3)

= > 234 + 322 = 556

The average number of females working in department B, C and D together

= > [1200 * (52/100) + 1800 * (46/100) + 2000 * (54/100)]/3

= > [624 + 828 + 1080]/3

= > 2532/3 = 844

Total number of females working in department A and E together

= > 1500 * (48/100) + 2300 * (58/100)

= > 720 + 1334 = 2054

Total number of males working in department C and D together

= > 1800 * (54/100) + 2000 * (46/100)

= > 972 + 920 = 1892

Required % = (2054/1892) * 100 = 108.56 % = 109 %

Total number of employees working in department A and B together

= > 1500 + 1200 = 2700

The total number of employees working in department E = 2300

Required % = [(2700 â€“ 2300) / 2300] * 100 = 17.39 % = 17 % more