Quant Test for SBI PO 2018 Prelim Exam Set – 42

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 values of x and y and choose a correct answer .

  1. I. 5x2 – 87x + 378 = 0
    II. 3y2 – 49y + 200 = 0

    y>=x
    x>=y
    y>x
    x>y
    x=y or no relation cannot be established.
    Option D
    I. 5x2 – 87x + 378 = 0
    5x2 – 45x – 42x + 378 = 0
    5x(x – 9) – 42(x – 9) = 0
    (5x – 42) (x – 9) = 0
    x = 9, 42/5
    II. 3y2 – 49y + 200 = 0
    3y2 – 24y – 25y + 200 = 0
    3y(y – 8) –25(y – 8) = 0
    (y – 8) (3y – 25) = 0
    y = 8,25/3
    x > y

     

  2. I. 10x2 – x – 24 = 0
    II. y2 – 2y = 0

    x=y or no relation cannot be established.
    x>y
    x>=y
    y>x
    y>=x
    Option A
    I. 10x2 – x – 24 = 0
    10x2 – 16x + 15x – 24 = 0
    2x(5x – 8) + 3(5x – 8) = 0
    (2x + 3) (5x – 8) = 0
    x = -3/2 ,8/5
    II. y2 – 2y = 0
    y2 – 2y = 0
    y(y – 2) = 0
    y = 0, 2
    No relationship exists between x and y.

     

  3. I. x2 – 5x + 6 = 0
    II. 2y2 – 15y + 27 = 0

    y>=x
    x=y or no relation cannot be established.
    y>x
    x>=y
    x>y
    Option A
    I. x2 – 5x + 6 = 0
    x2 – 2x – 3x + 6 = 0
    x(x – 2) – 3(x – 2) = 0
    (x – 2) (x – 3) = 0
    x = 2, 3
    II. 2y2 – 15y + 27 = 0
    2y2 – 6y – 9y + 27 = 0
    2y(y – 3) – 9(y – 3) = 0
    (y – 3) (2y – 9) = 0
    y = 3, 9/2
    y>=x

     

  4. I. 3x + 2y = 301
    II. 7x – 5y = 74

    x>y
    x>=y
    x=y or no relation cannot be established.
    y>x
    y>=x
    Option D
    eqn (I) × 5 + eqn (II) × 2
    15x + 10y = 1505
    14x – 10y = 148
    => 29x = 1653
    => x = 1653/29 = 57
    and y = 65
    y>x

     

  5. I. 14x2 – 37x + 24 = 0
    II. 28y2 – 53y + 24 = 0

    x=y or no relation cannot be established.
    x>y
    y>=x
    y>x
    x>=y
    Option E
    I. 14x2 – 37x + 24 = 0
    14x2 – 37x + 24 = 0
    14x2 – 21x – 16x + 24 = 0
    7x(2x – 3) –8(2x – 3) = 0
    (2x – 3) (7x – 8) = 0
    x = 3/2, 8/7
    II. 28y2 – 53y + 24 = 0
    28y2 – 53y + 24 = 0
    28y2 – 21y – 32y + 24 = 0
    7y(4y – 3) –8(4y – 3) = 0
    (7y – 8) (4y – 3) = 0
    y = 8/7, ¾
    x>= y

     

  6. In a class, the average age of student is 8 years, and average age of 16 teachers is 28 years. If the average age of the combined group of all the teachers and students is 10, then find the number of students.
    110
    120
    100
    144
    122
    Option D
    Let the number of students be x
    =>10 = [8x + 16*28]/ [x+16] => x = 144

     

  7. The perimeter of a rectangle, whose length is 8 m more than its breadth, is 92 m. What would be the area of a triangle whose base is equal to the half of the diagonal of the rectangle and whose height is equal to the length of the rectangle?(Answer is in approximate value)
    223 sq. m
    200 sq.. m
    150 sq. m
    198 sq. m
    188 sq. m
    Option A
    2 (b + 8 + b) = 92
    => 2 (2b + 8) = 92
    => b = 19m and l = 19 + 8 = 27 m
    Diagonal of the rectangle = [(27)2 + (19)2 ]1/2 = 33 (approx.)
    Area of the triangle = ½*33/2*27 = 223 sq.m

     

  8. In a container, there is 480 ltr of pure milk from which 24 ltr of milk is replaced with 24 ltr of water, again 24 ltr milk is replaced by same amount of water, as this process is done once more. Now, what is the amount of pure milk?
    405.5 litre
    355 litre
    411.54 litre
    333.20 litre
    412.11 litre
    Option C
    Amount of pure milk = a [ 1 – b/a]^3
    Where, a = pure milk in starting
    b = amount replaced
    = 480 [1- 24/480]^3
    = 480[1 – 1/20]^3
    = 480 * 19/20 * 19/20 * 19/20
    = 411.54 litre

     

  9. A trader marks his goods at such a price that after allowing a discount of 18%, he makes a profit of 28%. What is the marked price of an article whose cost price is Rs.1804?
    Rs.2055
    Rs.2816
    Rs.2060
    Rs.3500
    Rs.2007
    Option B
    CP : SP : MP = 100 : 128 : 128(100/82)
    The marked price of an article = 128*1804/82 = Rs.2816

     

  10. A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked random. Find the probability that the ball will be white.
    3/7
    11/15
    9/14
    6/13
    7/15
    Option E
    The probability of selecting one bag =1/2
    Now, probability of getting a white ball from bag A =1/2×3/5=3/10
    and probability of getting a white ball from bag B =1/2×2/6=1/6
    Hence, Probability that white ball is drawn either first or second bag =3/10+1/6
    =7/15

     


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