Quant Test for SBI PO 2018 Prelim Exam Set – 27

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): Find the relation x and y ,and choose a correct option.

  1. I. 12x2 – 5x – 3 = 0
    II. 4y2 – 11y + 6 = 0

    x>y
    x=y or relation cannot be established.
    y>x
    x>=y
    y>=x
    Option E
    I. 12x2 – 5x – 3 = 0
    II. 4y2 – 11y + 6 = 0
    I. 12x2 – 5x – 3 = 0
    =>12x2 – 5x – 3 = 0
    =>12x2 + 4x – 9x – 3 = 0
    x = -1/3, 3/4
    II. 4y2 – 11y + 6 = 0
    => 4y2 – 11y + 6 = 0
    =>4y2 – 8y – 3y + 6 = 0
    y= 3/4, 2
    x ≤ y

     

  2. I. 20x2 – 31x + 12 = 0
    II. 3y2 – 16y + 16 = 0

    x>=y
    y>x
    x>y
    y>=x
    x=y or relation cannot be established.
    Option B
    I. 20x2 – 31x + 12 = 0
    II. 3y2 – 16y + 16 = 0
    I. 20x2 – 31x + 12 = 0
    =>20x2 – 31x + 12 = 0
    =>20x2 – 16x – 15x + 12 = 0
    x = 3/4, 4/5
    II. 3y2 – 16y + 16 = 0
    => 3y2 – 16y + 16 = 0
    => 3y2 – 14y – 4y + 16 = 0
    y = 4, 4/3
    x < y

     

  3. I. x2 – 2x – √5x + 2√5 = 0
    II. y2 – 3y – √6y + 3√6 = 0

    y>=x
    y>x
    x=y or relation cannot be established.
    x>y
    x>=y
    Option B
    I. x2 – 2x – √5x + 2√5 = 0
    II. y2 – 3y – √6y + 3√6 = 0
    I. x2 – 2x – √5x + 2√5 = 0
    => x2 – 2x – √5x + 2√5 = 0
    =>x (x – 2) – √5 (x – 2) = 0
    x = 2, √5 (2.23)
    II. y2 – 3y – √6y + 3√6 = 0
    => y2 – 3y – √6y + 3√6 = 0
    =>y (y – 3) – √6 (y – 3) = 0
    y = 3, √6 (2.44)
    x < y

     

  4. I. 2x2 – 15√3x + 84 = 0
    II. 3y2 – 2y – 8 = 0

    x=y or relation cannot be established.
    x>=y
    y>x
    y>=x
    x>y
    Option E
    I. 2x2 – 15√3x + 84 = 0
    II. 3y2 – 2y – 8 = 0
    I. 2x2 – 15√3x + 84 = 0
    =>2x2 – 8√3x – 7√3x + 84 = 0
    => 2x (x – 4√3) – 7√3 (x – 4√3x) = 0
    x = 3.5√3, 4√3
    II. 3y2 – 2y – 8 = 0
    =>3y2 – 2y – 8 = 0
    => 3y2 – 6y + 4y – 8 = 0
    y = -4/3, 1
    x > y

     

  5. I. 2x2 – (6 + √3)x + 3√3 = 0
    II. 3y2 – (9 + √3)y + 3√3 = 0

    x>y
    y>=x
    y>x
    x=y or relation cannot be established.
    x>=y
    Option D
    I. 2x2 – (6 + √3)x + 3√3 = 0
    II. 3y2 – (9 + √3)y + 3√3 = 0
    I. 2x2 – (6 + √3)x + 3√3 = 0
    =>2x2 – 6x – √3x + 3√3 = 0
    =>2x (x- 3) – √3 (x – 3) = 0
    x = 3, √3/2
    II. 3y2 – (9 + √3)y + 3√3 = 0
    =>3y2 – 9y – √3y + 3√3 = 0
    =>3y (y – 3) – √3 (y – 3) = 0
    x = 3, √3/3
    x=y or relation cannot be established.

     

  6. Mr. A lends 40% of sum at 15% p.a. 50% of rest sum at 10% p.a. and the rest at 18% p.a. rate of interest. What would be the rate of interest if the interest is calculated on the whole sum ?
    13.3%
    10.2%
    15.5%
    14.4%
    11.5%
    Option D
    Let the amount be Rs. x
    Investment is done as given below.
    Amount left = x – 40/100 x = 60x/100
    Now, 40x/100 at 15% p.a
    50/100 of 60x/100 = 30x/100 at 10% p.a
    Rest amount = x – 40x/100 – 30x/100 = 30x/100 at 18% p.a
    Interest earned by each at end of 1 year By 1st
    => 15/100 × 40x/100 = 60x/1000
    By 2nd 10/100 × 30x/100 = 30x/1000
    By 3rd 18/100 × 30x/100 = 54x/1000
    Total interest = 144x/1000
    Rate% = [144x/1000]/x * 100
    = 14.4%

     

  7. A man starts going for morning walk every day. The distance walked by him on the first day was 2 kms. Everyday he walks half of the distance walked on the previous day. What can be the maximum total distance walked by him in his life time.
    200km
    Inadequate data.
    100km
    80km
    150km
    Option B
    When a train crosses a platform,
    the distance covered = Length of platform and the train.
    Speed = Length/Time taken
    Thus, we have inadequate data.

     

  8. The area of a rectangle gets reduced by 9 sq. metre if its length is reduced by 5 m and breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 67sq. metre. Find the length of the rectangle.
    22m
    17m
    25m
    10m
    30m
    Option B
    Let the length of the rectangle be x metre and breadth be y metre.
    =>xy – (x – 5)(y + 3) = 9
    => xy – (xy – 5y + 3x – 15) = 9
    => 5y – 3x + 15 = 9
    => 3x – 5y – 6 = 0 ….(i)
    Again, (x + 3) (y + 2) – xy = 67
    =>3y + 2x + 6 = 67
    => 2x + 3y – 61 = 0 ….(ii)
    By equation (i) × 3 + (ii) × 5
    9x – 15y – 18 = 0
    10x + 15y -305 = 0
    19x = 323
    => x = 17 m

     

  9. A shopkeepers announce the same price of Rs. 700 for a shirt. The first offers successive discounts of 30% and 6% while the second offers successive discounts of 20% and 16%. How much more discount offered by the shopkeeper .
    Rs. 8.80
    Rs. 7.80
    Rs. 5.80
    Rs. 6.80
    Rs. 9.80
    Option E
    According to the question,
    Selling price of first shopkeeper. = 700 × 70/100 × 94/100
    = Rs. 460.60
    Selling price of second shopkeeper = 700 × 80/100 × 84/100
    = Rs. 470.40
    Required difference = 470.40 – 460.60
    = Rs. 9.80

     

  10. A grocer purchased 2 kg. of rice at the rate of Rs. 15 per kg. and 3 kg of rice at the rate of Rs. 13 per kg. At what price per kg should he sell the mixture to earn 33(1/3) % profit on the cost price ?
    Rs.12.6
    Rs.15.5
    Rs.18.4
    Rs.20.5
    Rs.17.2
    Option C
    Mixture : 2 kg of rice at Rs. 15/kg + 3 kg of rice at Rs. 13/kg
    Total weight = 2 + 3 = 5 kg
    Total cost price = (2 × 15) + (3 × 13)
    = 30 + 39 = Rs. 69
    Cost price per kg of the mixture = 69/5
    = Rs.13.80
    Selling price to get 33(1/3)% = [10+ 33( 1/3)]/100 × Rs. 13.80
    = [400 × Rs. 13.80]/[3*100] = 4/3 × Rs 13.80
    = Rs. 18.40

     


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