Quant Test for IBPS RRB 2018 Prelim Exam Set – 9

Institute of Banking Personnel Selection (IBPS) had released the official notification for the Common Recruitment Process for RRBs (CRP RRBs VII) for the recruitment of Group “A”-Officers (Scale-I, II & III) and Group “B”-Office Assistant (Multipurpose)

Click here to know the details of the Examination

The examination will be held in two phases i.e. Preliminary Examination and Main Examination. The RRB Scale I Preliminary Exam is scheduled on 11th, 12th & 18th of August 2018. And RRB Assistant Preliminary Exam is scheduled on 19th, 25th August & 1st September 2018. Details of the exam are as under:

Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam. 

Directions(1-5): Find the relation between x and y and choose a correct option.

1. I. 15x² – 34x + 15 = 0
   II. 15y² + 22y + 8 = 0
   y>x
   x>=y
   x>y
   y>=x
   x=y or relation cannot be established.
   Option C
   I. 15x²- (9+25)x + 15 =0
   15x² – 9x – 25x + 15 = 0
   3x (5x-3)-5 (5x-3)=0
   (3x-5) (5x-3)=0
   II. 15y² +(10+12)y + 8 = 0
   15y² + 10y + 12y + 8 = 0
   5y (3y+2) +4 (3y+2)=0
   (3y+2) (5y+4)=0
   x > y

    

2. I. 4x^4 – 5x^2 + 1 = 0
   II. 2y^2 + 11y + 14 = 0
   y>x
   x=y or relation cannot be established.
   x>=y
   x>y
   y>=x
   Option D
   I.4x^4 – 5x^2 + 1 = 0
   (x^2 – 1)(4^2 – 1) = 0
   x = +1, -1, 1/2 or -1/2
   II.2y^2 + 11y + 14 = 0
   (y + 2)(2y + 7) = 0
   y = -2, -7/2
   x > y

    

3. I. x^2 – 7 √3 x + 36 = 0
   II. y^2 – 11 √3 y + 84 = 0
   x=y or relation cannot be established.
   x>y
   y>=x
   y>x
   x>=y
   Option C
   I. x^2 – 7 √3 x + 36 = 0
   => x^2 – 4 √3 x – 3 √3 x + 36 = 0
   => x (x – 4 √3) – 3√3 (x – 4√3) = 0
   => (x – 3√3) (x – 4√3) = 0
   x = 3√3, 4√3
   II. y^2 – 11√3 y + 84 = 0
   => y^2 – 4√3 y – 7√3 y + 84 = 0
   => y (y – 4√3)– 7√3 (y – 4√3) = 0
   => (y – 7 √3) (y – 4√3) = 0
   y = 7 √3, 4√3
   y>=x

    

4. I. x^2 = 361
   II. y^3 = 7269 + 731
   x=y or relation cannot be established.
   y>x
   y>=x
   x>=y
   x>y
   Option B
   I. x^2 = 361
   x = ± 19
   II. y^3 = 7269 + 731
   y^3 = 8000
   y = 20
   x < y

    

5. I. x^2 – 6x + 135 = 0
   II. y^2–30y+225 = 0
   x>y
   y>=x
   y>x
   x>=y
   x=y or relation cannot be established.
   Option B
   I.x^ 2–6x + 135 = 0
   x^2 – 15x + 9x + 135 = 0
   x = +15, –9
   II.y^2–30y+225 = 0
   y^2–15y–15y+225 = 0
   y = +15, +15
   y>=x

    


Directions(6-10): Study the following graph carefully and answer the questions given below.
Percentage growth in the population of six villages A, B, C, D, E and F from 2011 to 2012 and 2012 to 2013.

Actual Population of these villages in three different years:


6. Which of the following villages had maximum population in the year 2011?
   B
   A
   D
   C
   E
   Option C
   Population in 20011 A = 3000 × 100/120 = 2500
   B = 1980 × 100/110 = 1800
   C = 2400 × 100/120 = 2000
   D = 3000
   E = 1250
   F = 1200

    

7. What was the population of village ‘F’ in 2013?
   2025
   2525
   2020
   3025
   2000
   Option A
   F’s population in 2012 = 1620
   In 2013 = 1620 × 125/100 = 2025

    

8. Population of village ‘E’ in 2011 was approximately what per cent of its population in 2013?
   63%
   66%
   70%
   64%
   60%
   Option D
   E’s population in 2011 = 1250
   2012 = 1500
   2013 = 1500 × 130/100 = 1950
   Required % = 1250/1950*100 = 64.1% == 64%

    

9. If the populations of villages ‘C’ and ‘D’ in 2012 are in the ratio of 2 : 3 respectively, what was the population of village ‘D’ in 2011?
   4000
   5000
   2000
   3000
   1000
   Option D
   C’s population in 2013 = 2760
   C’s population in 2012 = 2760 × 100/115 = 2400
   D’s population in 2012 = 2400 × 3/2 = 3600
   D’s population in 2011 = 3600 × 100/120 = 3000

    

10. Which of the following villages had minimum population in the year 2012?(Refer previous question)
    E
    B
    A
    C
    D
    Option A
    A’s population = 3750 × 100/125 = 3000
    B’s population = 1980
    C’s population = 2400
    D’s population = 3600
    E’s population = 1250 × 120/100 = 1500
    F’s population = 1200 × 135/100 = 1620
    In village E the population is minimum.