Quant Test for SBI PO 2018 Prelim Exam Set – 36

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

Click here to know the details of the Examination

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 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): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.

  1. I. 21×2 – 2x – 8 = 0
    II. 24y2 – 10y – 25 = 0

    x=y or relation cannot be established.
    x>=y
    x>=y
    x>y
    y>x
    Option A
    I. 21x2 – 2x – 8 = 0
    II. 24y2 – 10y – 25 = 0
    I. 21x2 – 2x – 8 = 0
    =>21x2 – 14x + 12x – 8 = 0
    =>7x (3x – 2) + 4 (3x – 2) = 0
    =>x = 2/3, -4/7
    II. 24y2 – 10y – 25 = 0
    =>24y2 + 20y – 30y – 25 = 0
    =>4y (6y + 5) – 5(6y + 5) = 0
    =>(4y – 5)(6y + 5) = 0
    =>y = 5/4, -5/6
    Hence, relationship cannot be established between x and y.

     

  2. I. 18y2 + 63y + 49 = 0
    II. 20x2 + 7x – 6 = 0

    x>=y
    y>x
    x=y or relation cannot be established.
    x>y
    x>=y
    Option D
    I. 18y2 + 63y + 49 = 0
    II. 20x2 + 7x – 6 = 0
    I. 18y2 + 63y + 49 = 0
    =>18y2 + 21y + 42y + 49 = 0
    =>3y (6y + 7) + 7 (6y + 7) = 0
    =>(3y + 7)(6y + 7) = 0
    =>y = -7/3 , -7/6
    II. 20x2 + 7x – 6 = 0
    =>20x2 -8x + 15x – 6 = 0
    =>4x (5x – 2) + 3 (5x – 2) = 0
    =>(4x + 3)(5x – 2) = 0
    =>x = – 3/4, 2/5
    Hence, x > y

     

  3. I. 6y2 – 29y + 35 = 0
    II. 20x2 – 57x + 40 = 0

    x>=y
    y>x
    x>y
    x>=y
    x=y or relation cannot be established.
    Option B
    I. 6y2 – 29y + 35 = 0
    II. 20x2 – 57x + 40 = 0
    I. 6y2 – 29y + 35 = 0
    => 6y2 – 14y – 15y + 35 = 0
    =>2y (3y – 7) – 5(3y – 7) = 0
    =>(2y – 5) (3y – 7) = 0
    =>y = 5/2, 7/3
    II. 20x2 – 57x + 40 = 0
    => 20x2 – 32x – 25x + 40 = 0
    =>4x (5x – 8) – 5(5x – 8) = 0
    =>(4x – 5)(5x – 8) = 0
    =>x = 5/4, 8/5
    Hence, x < y

     

  4. I. 21x2 + 40x – 21 = 0
    II. 15y2 – 8y – 16 = 0

    x>=y
    x=y or relation cannot be established.
    y>x
    x>y
    x>=y
    Option B
    I. 21x2 + 40x – 21 = 0
    II. 15y2 – 8y – 16 = 0
    I. 21x2 + 40x – 21 = 0
    => 21x2 – 9x + 49x – 21 = 0
    =>3x (7x – 3) + 7(7x – 3) = 0
    =>(3x + 7)(7x -3) = 0
    =>x = 7/3, 3/7
    II. 15y2 – 8y – 16 = 0
    =>15y2 – 20y + 12y – 16 = 0
    =>5y (3y – 4) + 4(3y – 4) = 0
    =>(5y + 4)(3y – 4) = 0
    => y = – 4/5, 4/3
    Hence, relationship cannot be established between x and y

     

  5. I. 2y2 – 11y – 21 = 0
    II. 6x2 – 25x + 14 = 0

    x>=y
    x>y
    x=y or relation cannot be established.
    y>x
    x>=y
    Option C
    I. 2y2 – 11y – 21 = 0
    II. 6x2 – 25x + 14 = 0
    I. 2y2 – 11y – 21 = 0
    =>2y2 +3y – 14y – 21 = 0
    =>y (2y + 3) -7(2y + 3) = 0
    =>(y -7) (2y + 3) = 0
    =>y = 7, -3/2
    II. 6x2 – 25x + 14 = 0
    =>6x2 – 21x – 4x + 14 = 0
    =>3x (2x – 7) – 2(2x – 7) = 0
    =>(3x – 2)(2x – 7) = 0
    =>x = 2/3, 7/2
    Hence, relationship cannot be established between x and y.

     

  6. Points S and T are 200 km apart on a highway. One car starts from S and another one from T at same time. If they travel in same direction, they meet in 5 hours. But if they travel towards each other, they meet in 2 hours. Find the speeds of the two cars.
    60km/hr. and 50km/hr.
    70km/hr. and 40km/hr.
    50km/hr. and 10km/hr.
    70km/hr. and 30km/hr.
    80km/hr. and 20km/hr.
    Option D
    Let speeds of two cars be x km/hr and y km/hr.
    Relative speed when they are travelling in same direction x – y = 200/5 = 40 km/hr
    And relative speed when they are travelling in opposite direction x + y = 200/2 =100 km/hr
    x + y = 100
    x – y = 40
    By solving both equations, we get
    x = 70km/hr and y = 30km/hr

     

  7. A sum of money lent at compound interest at the rate of 20% per annum is paid back in three equal instalments of Rs 54,600. Find the sum of money?
    Rs. 1,63,500
    Rs. 4,22,400
    Rs. 3,32,500
    Rs. 1,58,000
    Rs. 2,38,493
    Option E
    Sum =
    Instalment [(1+20/100) + (1+20/100)^2 + (1+20/100)^3 ] = 54600 (6/5 + 36/25 + 216/125)
    = 54600 [(150 + 180 + 216)/125] Total sum = Rs. 2,38,493

     

  8. There are two vessels P and Q. Vessel P contains 150 litres of pure milk and vessel Q contains 70 litres of pure water. From vessel P, 30 litres of milk is taken out and poured into vessel Q. Then 20 litres of mixture (milk and water) is taken out from vessel Q and poured into vessel P. What is the ratio of the quantity of pure milk in vessel P to that of pure water in vessel Q?
    5:2
    7:3
    4:3
    9:4
    2:5
    Option D
    Initially milk in vessel P = 150 lit
    water in vessel Q = 70 lit
    After 1st operation Milk in vessel P = 150 – 30 = 120 lit
    Water in vessel Q = 70 lit
    Milk in vessel Q = 30 lit
    Mixture in vessel Q = 70 + 30 = 100 lit
    after 2nd operation (when 20 lit or 20/100 = 1/5 of the mixture is taken out from Q it means 70/5 litres water and 30/5 litres of milk is taken out)
    Milk in vessel P = 120 + 30/5 = 126 lit water in vessel Q = 70 – 70/5 = 56 lit
    required ratio = 126:56 = 9:4

     

  9. The base radius and height of roller are 0.14 cm and 10 cm respectively. When it revolves 2400 times then it levels only 44% area. If the cost of levelling is Rs 1.50/cm2 , then find the total cost of levelling the whole field.
    Rs. 64000
    Rs. 72000
    Rs. 55000
    Rs. 87000
    Rs. 50000
    Option B
    Area covered by roller in one revolution = 2 (22/7) r h
    = 2 (22/7) × 0.14 × 10 = 8.8 cm^2
    let the total area be A 44% of A = 2400 × 8.8 A
    = (2400 × 8.8)/44 × 100 = 48000 cm^2
    Total cost of levelling = 1.50 × 48000
    = Rs. 72000

     

  10. There are 8 members in a delegation which is to be sent abroad. The total no. of members is 18. In how many ways can the selection be made so that 2 members are always included?
    8008
    5000
    7520
    8000
    4088
    Option A
    Required ways = 16C6 = 8008

     


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