IBPS RRB Clerk Prelims Quantitative Aptitude Questions 2019 (Day-16)

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IBPS RRB Clerk Prelims Quantitative Aptitude Questions 2019 (Day-16)

maximum of 10 points
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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 ≥ y

c) If x < y

d) If x ≤ y

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

 

1) I) x3 =1331

II) 2y2 -21y +55 =0

 

2) I) 5x = 7y+21

II) 11x +4y+109 =0

 

3) I) 2x2 -11x +12 =0

II) 2y2 -17y +36 =0

 

4) I) 6x2 – 32x + 42 = 0

II) y2 + 7y + 12 = 0

 

5) I) x2 + 6x + 8 = 0

II) 4y2 – 13y + 9 = 0

 

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

The following table shows the total number of students from various schools playing various different games. Each person is playing only one game 

6) Total number of students playing various sports in School A is 32 % of total number of students in School A. Find the total number of students not playing any games in School A?

a) 1480

b) 1620

c) 1360

d) 1570

e) None of these

7) Find the ratio between the total number of students playing Badminton from School A, C and E together to that of total number of students playing Football from School B, C and D together?

a) 56 : 59

b) 41 : 47

c) 33 : 35

d) 101 : 105

e) None of these

8) Find the average number of students playing Kabaddi and Volleyball together from all the Schools together?

a) 710

b) 655

c) 680

d) 620

e) 735

9) Total number of students playing Cricket from School C is approximately what percentage of total number of students playing Football from School C and E together?

a) 45 %

b) 56 %

c) 28 %

d) 33 %

e) 60 %

10) Find the difference between the total number of students playing all the given games together from School B to that of School C?

a) 205

b) 275

c) 310

d) 380

e) None of these

Answers :

Direction (1-5) :

1) Answer: a)

I) x3 =1331

x = ∛1331 =11

II) 2y2 -21y + 55 =0

2y2 -10y -11y +55 =0

2y(y – 5)-11 (y – 5) =0

(2y – 11) (y – 5) =0

y = 11/2, 5

x > y

2) Answer: a)

5x – 7y = 21 —> (1)

11x +4y = -109 –> (2)

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

x = -7, y = -8

x > y

3) Answer: d)

I) 2x2 -11x +12 = 0

2x2 – 8x – 3x +12 = 0

2x (x – 4) -3 (x – 4) = 0

(2x – 3) (x – 4) =0

X = 3/2, 4

II) 2y2 -17y +36 = 0

2y2 – 9y – 8y + 36 = 0

y (2y – 9) – 4 (2y – 9) = 0

(y – 4) (2y – 9) =0

y = 4, 9/2

x y

4) Answer: a)

I) 6x2 – 32x + 42 = 0

6x2 – 18x – 14x + 42 = 0

6x(x – 3)-14 (x – 3) =0

(6x – 14)(x – 3) = 0

x = 3, 7/3

II) y2 + 7y + 12 = 0

(y + 4) (y + 3) = 0

y = -4, -3

x > y

5) Answer: c)

I) x2 + 6x + 8 = 0

(x + 4)(x + 2) = 0

x = -4, -2

II) 4y2 – 13y + 9 = 0

4y2 – 4y – 9y + 9 = 0

4y(y – 1) – 9 (y – 1) =0

(4y – 9) (y – 1) = 0

y = 1, 9/4

x < y

Direction (6-10) :

6) Answer: c)

Total number of students playing various sports in School A

= > 150 + 120 + 180 + 100 + 90 = 640

(32/100)*Total students in School A = 640

Total students in School A = 640*(100/32) = 2000

The total number of students not playing any games in School A

= > 2000*(68/100) = 1360

7) Answer: d)

The total number of students playing Badminton from School A, C and E together

= > 150 + 225 + 130 = 505

The total number of students playing Football from School B, C and D together

= > 145 + 250 + 130 = 525

Required ratio = 505 : 525 = 101 : 105

8) Answer: b)

The total number of students playing Kabaddi and Volleyball together from all the Schools together

= > 100 + 150 + 110 + 115 + 125 + 120 + 130 + 200 + 150 + 110

= > 1310

Required average = 1310/2 = 655

9) Answer: d)

Total number of students playing Cricket from School C = 130

Total number of students playing Football from School C and E together

= > 250 + 150 = 400

Required % = (130/400)*100 = 32.5 % = 33 %

10) Answer: a)

Total number of students playing all the given games together from School B

= > 175 + 130 + 145 + 150 + 110 = 710

Total number of students playing all the given games together from School C

= > 225 + 200 + 250 + 110 + 130 = 915

Required difference = 915 – 710 = 205

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