RRB PO Prelims Quantitative Aptitude (Day-14)

Quantitative Aptitude for RRB PO 2020

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for RRB PO 2020 Prelims so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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IBPS RRB PO Prelims

Data Interpretation

Direction (1-5): Study the following information carefully and answer the given questions.

The following table shows the total number of people bought books from 6 different stores in May 2019 and the ratio of male and female among them.

1) Find the difference between the total number of males bought books from store A and B together to that of total number of females bought books from store E and F together?

A) 108

B) 115

C) 136

D) 127

E) None of these

2) Find the ratio between the total number of people bought books from store C to that of total number of females bought books from store D?

A) 47: 55

B) 60: 49

C) 52: 43

D) 75: 67

E) None of these

3) Total number of males bought books from store A and F together is approximately what percentage of total number of females bought books from store B?

A) 223 %

B) 192 %

C) 256 %

D) 177 %

E) 165 %

4) Find the total number of females bought books from store B, C, D and E together?

A) 844

B) 916

C) 872

D) 998

E) None of these

5) If the percentage of horror books bought from store B and C is 45 % and 60 % respectively, then find the total number of story books bought from store B and C together (The available books from all the stores are Horror and Story books only)?

A) 420

B) 390

C) 360

D) 440

E) None of these

Quadratic Equation

Direction (6-10): 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

6)

I) 2x2– 3x – 44 = 0

II) 3y2– 20y + 25 = 0

7)

I) 3x – 2y = 2

II) 5x – 4y = 0 

8)

I) x2+ 7x – 78 = 0

II) y2+ 27y + 182 = 0 

9)

I) 2x2– 10x – 48 = 0

II) y2+ 9y + 20 = 0

10)

I) x = 65536 % of 450 – 122+ 80

II) y3= 1728

Answers :

Directions (1-5):

1) Answer: A

The total number of males bought books from store A and B together

= > 2400 * (18/100) * (5/8) + 2400 * (15/100) * (4/9)

= > 270 + 160 = 430

The total number of females bought books from store E and F together

= > 2400 * (14/100) * (5/8) + 2400 * (12/100) * (7/18)

= > 210 + 112 = 322

Required difference = 430 – 322 = 108

2) Answer: B

The total number of people bought books from store C

= > 2400 * (20/100)

The total number of females bought books from store D

= > 2400 * (21/100) * (7/9)

Required ratio = [2400 * (20/100)]: [2400 * (21/100) * (7/9)]

= > 60: 49

3) Answer: A

Total number of males bought books from store A and F together

= > 2400 * (18/100) * (5/8) + 2400 * (12/100) * (11/18)

= > 270 + 176 = 446

Total number of females bought books from store B

= > 2400 * (15/100) * (5/9) = 200

Required % = (446/200) * 100 = 223 %

4) Answer: E

The total number of females bought books from store B, C, D and E together

= > [2400 * (15/100) * (5/9) + 2400 * (20/100) * (1/3) + 2400 * (21/100) * (7/9) + 2400 * (14/100) * (5/8)]

= > (200 + 160 + 392 + 210)

= > 962

5) Answer: B

The total number of story books bought from store B and C together

= > 2400 * (15/100) * (55/100) + 2400 * (20/100) * (40/100)

= > 198 + 192 = 390

Directions (6-10) :

6) Answer: E

I) 2x2– 3x – 44 = 0

2x2 + 8x – 11x – 44 = 0

2x (x + 4) – 11 (x + 4) = 0

(2x – 11) (x + 4) = 0

x = 11/2, -4 = 5.5, -4

II) 3y2– 20y + 25 = 0

3y2 – 15y – 5y + 25 = 0

3y (y – 5) – 5 (y – 5) = 0

(3y – 5) (y – 5) = 0

y = 5/3, 5 = 1.66, 5

Can’t be determined

7) Answer: C

3x – 2y = 2 –> (1)

5x – 4y = 0 –> (2)

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

x = 4, y = 5

x < y

8) Answer: B

I) x2+ 7x – 78 = 0

(x + 13) (x – 6) = 0

x = -13, 6

II)y2+ 27y + 182 = 0

(y + 14) (y + 13) = 0

y = – 14, -13

x ≥ y

9) Answer: A

I) 2x2– 10x – 48 = 0

2x2 – 16x + 6x – 48 = 0

2x (x – 8) + 6 (x – 8) = 0

(2x + 6) (x – 8) = 0

x = -3, 8

II) y2+ 9y + 20 = 0

(y + 5) (y + 4) = 0

y = -5, -4

x > y

10) Answer: C

I) x = ∜65536 % of 450 – 122+ 80

x = 16 % of 450 – 144 + 80

x = (16/100) * 450 – 144 + 80

x = 72 – 144 + 80 = 8

II) y3= 1728

y = 12

x < y

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