Quant Test for SBI PO 2018 Prelim Exam Set – 66

Banya

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 between x and y and choose a correct option.

  1. I. x2 + x – 30 = 0
    II. y2 – 11y + 30 = 0

    x>=y
    y>=x
    x>y
    x=y or relation cannot be established.
    y>x
    Option B
    x2 + x – 30 = 0
    x2 + 6x – 5x – 30 = 0
    x = – 6, 5
    y2 – 11y + 30 = 0
    y2 – 5y – 6y + 30 = 0
    y = 5, 6
    x ≤ y

     

  2. I. x2 + 5x – 14 = 0
    II. y2 + 24y + 128 = 0

    x=y or relation cannot be established.
    x>=y
    x>y
    y>x
    y>=x
    Option C
    x2 + 5x – 14 = 0
    x2 + 7x – 2x – 14 = 0
    x = -7, 2
    y2 + 24y + 128 = 0
    y2 + 16y + 8y + 128 = 0
    y = -16, -8
    x > y

     

  3. I. x2 – 6x – 91 = 0
    II. y2 – 32y + 247 = 0

    y>x
    y>=x
    x>=y
    x=y or relation cannot be established.
    x>y
    Option B
    x2 – 6x – 91 = 0
    x2 – 13x + 7x – 91 = 0
    x = 13, -7
    y2 – 32y + 247 = 0
    y2 – 19y -13y + 247 = 0
    y = 19, 13
    x ≤ y

     

  4. I. x2 – 2x – 15 = 0
    II. y2 + 22 = 122

    y>x
    x=y or relation cannot be established.
    y>=x
    x>y
    x>=y
    Option B
    x2 – 2x – 15 = 0
    x2 – 5x + 3x – 15 = 0
    x = 5, -3
    y2 + 22 = 122
    y2 = 100
    y = + 10, -10
    x = y or relationship cannot be determined .

     

  5. I. x2 – 7x + 12 = 0
    II. y2 – 5y + 6 = 0

    y>=x
    x>y
    y>x
    x>=y
    x=y or relation cannot be established.
    Option D
    x2 – 7x + 12 = 0
    x2 – 3x – 4x + 12 = 0
    x = 3, 4
    y2 – 5y + 6 = 0
    y2 – 3y – 2y + 6 = 0
    y = 2, 3
    x ≥ y

     

  6. The average weight of four men A,B,C and D is 67 kg. The fifth man E is included and the average weight decreases by 2 kg. A is replaced by F. The weight of F is 4 kg more than E. Average weight decreases because of the replacement of A and now the average weight is 64 kg. Find the weight of A
    98 kg
    75 kg
    86 kg
    62 kg
    66 kg
    Option E
    Weight of E = (67-2)- 4*2 = 57 kg
    Weight of F = 61 kg
    Average weight decreased by 1 kg of replacing A.
    So, weight of A = 61 + 5*1 = 66 kg

     

  7. Train A starts its journey from P to Q while B starts from Q to P. After crossing each other they finish their journey in 81 hours and 121 hours respectively. Then what will be the speed of train B if train A speed is 44 km/hr.?
    45 km/hr.
    36 km/hr.
    25 km/hr.
    22 km/hr.
    40 km/hr.
    Option B
    Ratio of their speeds = (121)^1/2 : (81)^1/2
    = 11:9
    B’s speed = 36 km/hr.

     

  8. A, B and C are partners in a business partnership. A invested Rs.4000 for whole year. B invested Rs. 6000 initially but increased this investment upto Rs. 8000 at the end of 4 months, while C invested Rs. 8000 initially, but withdraw Rs. 2000 at the end of 9 months. At the end of year total earned profit is Rs. 16950 , find the share of profit C.
    6750
    6550
    7750
    5750
    6050
    Option A
    A : B : C = 4000*12 : 6000*4 + 8000*8 : 8000*9+6000*3
    = 48 : 88 : 90
    = 24 : 44 : 45 == total 113
    A = 3600 B = 6600 C = 6750

     

  9. Murugan invests Rs. 4500 as fixed deposit at a bank at the rate of 3(1/2)% per annum SI. But due to some needs he has to withdraw the entire money after 2(1/2) years, for which the bank allowed him a lower rate of interest. If he gets Rs 281.25 less than what he would have got at the end of 5 years, find the rate of interest allowed by the bank .
    6.0%
    3.3%
    1.5%
    4.5%
    5.2%
    Option D
    P=4500
    R=3.5%
    T=5
    [4500 × 3.5 × 5]/ 100 – [4500 × 5 × 𝑅]/ [100 × 2] = 281.25
    112.5R=787.5-281.25
    112.5R=506.25
    R=4.5%

     

  10. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
    9/20
    9/20
    7/20
    11/21
    7/12
    Option B
    S = {1, 2, 3, 4, …., 19, 20}.
    Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
    P(E) = n(E)/n(S) = 9/20.

     


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