Quant Test for SBI PO 2018 Prelim Exam Set – 72

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.𝑥^2−3481=0
    II.3𝑦^2=(216000)^1/3

    x>=y
    x>=y
    x>y
    y>x
    x=y or relation cannot be established.
    Option E
    I.x=±59
    II.3y^2=60
    ⇒y=±√20
    No relation exists.

     

  2. I. 5x^2+2x−3=0
    II. 2y^2+7y+6=0

    x=y or relation cannot be established.
    x>y
    x>=y
    x>=y
    y>x
    Option B
    I.5x^2+5x−3x−3=0
    5x(x+1)−3(x+1)=0
    x=35,−1
    II.2y^2+4y+3y+6=0
    2y(y+2)+3(y+2)=0
    y=−32,−2
    x>𝑦

     

  3. I. (17)^2 +144 ÷18 = x
    II. (26)^2 − 18 ×21 = y

    x>=y
    y>x
    x=y or relation cannot be established.
    x>=y
    x>y
    Option B
    I.x=289+14418=297
    II. y=298
    x<𝑦

     

  4. I. x2−5x−14=0
    II.y2+7y+10 = 0

    x>=y
    x=y or relation cannot be established.
    y>x
    x>=y
    x>y
    Option D
    I. x2−7x+2x−14=0
    x(x−7)+2(x−7)=0
    x=7,−2
    II.y2+5y+2y+10=0
    y=−2,−5
    x≥y

     

  5. I.√𝑥−(18)^15/2/𝑥2=0
    II.√𝑦=(19)^9/2/𝑦

    x>y
    x>=y
    y>=x
    y>x
    x=y or relation cannot be established.
    Option D
    I.x^5/2=(18)^15/2
    ⇒x=(18)^3
    II.y^3/2=(19)^9/2
    ⇒y=19^3
    𝑥<𝑦

     

  6. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:
    2
    8
    3
    4
    7
    Option D
    P(1+20/100)^n>2𝑃
    =>(6/5)^>2 By simple observation, if n = 3, then (6/5)^𝑛<2, but At n = 4, (6/5)^𝑛>2.
    Thus minimum value of n is 4 years.

     

  7. A contract is to be completed in 50 days and 105 men were set to work, each working 8 h a day. After 25 days, 2/5th of the work is finished. How many additional men be employed, so that the work may be completed on time, each man now working 9 h a day?
    15 men
    65 men
    35 men
    25 men
    55 men
    Option C
    𝑀1𝐷1𝑇1/𝑊1=𝑀2𝐷2𝑇2/𝑊2 [by technique 1] Given,
    𝑀1=105,
    𝐷1=25,
    𝑇1=8,
    𝑊1=25
    Now, let the additional men be x.
    Then, 𝑀2=105+𝑥 𝑇2=9 𝐷2=25 And 𝑊2=1−25=35
    On putting these values in the above formula.
    105 × 25 × 82/5=(105 + 𝑥) × 25 × 93/5
    =105 × 82=(105 + 𝑥) × 93
    =105×4=(105+𝑥)×3
    =105×4=105×3+3𝑥 = 3x = 105
    => x = 35 men

     

  8. 20 litres of a mixture contains milk and water in the ratio 5 : 3. If 4 litres of this mixture be replaced by 4 litres of milk, the ratio of milk to water in the new mixture would be.
    3 : 1
    7 : 5
    11 : 6
    7 : 3
    5 : 4
    Option D
    Quantity of milk in 16 litres of mix. = (16×58) litres
    = 10 litres.
    Quantity of milk in 20 litres of new mix = (10 + 4) litres.
    Quantity of water in it = (20 – 14) litres = 6 litres.
    Ratio of milk and water in the new mix = 14 : 6
    = 7 : 3

     

  9. In an examination paper of five questions, 5 % of the candidates answered all of them and 5% answered none,of the rest, 25% candidates answered only one question and 20% answered 4 questions. If 396 candidates answered either 2 questions or 3 questions,What is the number of candidates that appeared for the examination?
    600
    800
    900
    700
    500
    Option B
    5% answered all and 5% answered none Remaining percentage = 90%
    Candidates percentage answered 1 question = 90×25100=22.5%
    Candidates percentage answered 4 questions = 90×20100=18%
    All these makes (18 + 22.5 + 10) = 50.5%
    Remaining percentage of candidates = 49.5%
    Let the total number of candidates = x 𝑥×49.5/100=396 𝑥=[396 × 100]/49.5
    Hence, x = 800

     

  10. If 60 engineers or 120 doctors or 100 teachers can finish a work in 360 days then in how many days will 10 engineers, 40 doctors and 50 teachers working together will complete the same work?
    360 days
    420 days
    500 days
    760 days
    660 days
    Option A
    60E = 120D = 100T = 1 unit can finish a work in 360 days. Also 10E + 40D + 50T =16+13+12 unit = 1 unit
    Therefore, 10 engineers, 40 doctors and 50 teachers working together will complete the same work in same time i.e. 360 days

     


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