Quant Test for IBPS RRB 2018 Prelim Exam Set – 18

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 2018Details 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. 


 

  1. I. 4x^2–25x+25 = 0
    II. 2y^2–13y+21 = 0

    x>=y
    y>x
    x>y
    x=y or relation cannot be established.
    y>=x
    Option D
    I. 4x^2–25x+25 = 0
    4x^2–25x+25 = 0
    4x^2–20x–5x+25 = 0
    x = 20/4 = 5, x = 5/4 = 1.25
    II. 2y^2–13y+21 = 0
    2y^2–13y+21 = 0
    2y^2–7y–6y+21 = 0
    y = 7/2 = 3.5, y = 6/2 = 3
    Hence, there is no relation.

     

  2. I. 2x^2+21x+34 = 0
    II. 3y^2+23y+42 = 0

    y>x
    y>=x
    x>y
    x>=y
    x=y or relation cannot be established.
    Option E
    I. 2x^2+21x+34 = 0
    2x^2+17x+4x+34 = 0
    x = – 17/2 = –8.5 , – 4/2 = –2
    II. 3y^2+23y+42 = 0
    3y^2+14y+9y+42 = 0
    y = – 14/3 = – 4.66, – 9/3 = –3
    Hence, no relation.

     

  3. I. 2x^2–6x–48 = 0
    II. y^2–13y+42 = 0

    x>=y
    x>y
    x=y or relation cannot be established.
    y>x
    y>=x
    Option E
    I. 2x^2–6x–48 = 0
    2x^2–4x–48 = 0
    2(x^2–2x–24) = 0
    x^2–2x–24 = 0
    x^2–6x+4x–24 = 0
    x = +6, –4
    II. y^2–13y+42 = 0
    y^2–13y+42 = 0
    y^2–7y–6y+42 = 0
    y = +7, +6
    Hence, y>=x

     

  4. I. 3x^2–27x+54 = 0
    II. 2y^2–9y+10 = 0

    x>y
    y>=x
    x=y or relation cannot be established.
    y>x
    x>=y
    Option A
    I. 3x^2–27x+54 = 0
    3x^2–27x+54 = 0
    3x^2–18x–9x+54 = 0
    x = 18/3 = 6, 9/3 = 3
    II. 2y^2–9y+10 = 0
    2y^2–9y+10 = 0
    2y^2–5y–4y+10 = 0
    y = 5/2 = 2.5 y = +, 4/2 = 2
    Hence, x>y

     

  5. I. x^2 + 6x + 8 = 0
    II. y^2 + 7y + 12 = 0

    x=y or relation cannot be established.
    x>=y
    y>=x
    y>x
    x>y
    Option A
    x = 4, 2
    y = 4, 3
    No relation

     

  6. The simple interest on a sum of money is equal to the principal and the number of years is equal to the rate percent per annum. Find the rate percent.
    50%
    40%
    30%
    10%
    20%
    Option D
    Let SI = Principal = x
    Years = rate = y
    R = SI×100/P×N
    y = x×100/x×y
    y= 100/y
    y^2 = 100
    y = 10%

     

  7. Two trains start from same station in different times at a speed of 60 km/hr and 72 km/hr respectively and the ratio of length of faster and slower trains is 3:2. If fast train overtakes the slower train in 150 sec, find the length of the faster train.
    600 m
    700 m
    400 m
    200 m
    300 m
    Option E
    72 − 60 = total length/150 × 3600
    Total length = 12 × 150/3600 = 0.5 km
    Length of the 1st train = 0.5/5 × 3 = 0.3km = 300 m

     

  8. The ratio of speed of boat in still water to speed of stream is 5:3. If the boat goes 24 km distance in downstream and returns to the starting point in 15 hours. Find the speed of stream.
    6 km/hr.
    4 km/hr.
    1 km/hr.
    3 km/hr.
    5 km/hr.
    Option D
    24/(5x + 3x) + 24/(5x − 3x) = 15
    24/8x + 24/2x = 15
    24 + 96 = 120𝑥
    120x = 120
    x = 1
    Speed of stream = 3 km/hr.

     

  9. Four years ago the ratio of father’s age to son’s age was 8:1 and ratio of present age of mother and son is 4:1. If the sum of present ages of mother and son is 4 years more than the father’s present age, find father’s present age.
    44 years
    50 years
    42 years
    36 years
    30 years
    Option D
    Let present age of father, mother and son be x, y and z
    (x – 4)/(z – 4) = 8/1
    => 𝑥 − 8𝑧 = −28 − − − −(1)
    y/z = 4/1
    => 𝑦 = 4𝑧 − − − −(2)
    y + z = 4 + x
    => y + z − x = 4 − − − −(3)
    Solve the above three equation,
    x = 36 years,
    y = 32 years,
    z = 8 years
    Father’s age = 36 years

     

  10. A basket contains 8 Red and 6 Pink toys. There is another basket which contains 7 Red and 8 Pink toys. One toy is to drawn from either of the two baskets. What is the probability of drawing a Pink toys?
    91/210
    111/211
    101/210
    99/200
    103/211
    Option C
    Probability of one basket = 1/2
    1st Basket Pink toy probability = 1/2* (6c1/14c1)
    2nd Basket Pink toy probability = 1/2* (8c1/15c1)
    Adding both cases (1/2*6/14) + (1/2*8/15)
    = 3/14+4/15 = 101/210

     


Related posts

Leave a Comment