Quant Test for SBI PO 2018 Prelim Exam Set – 15

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 relationship between x and y and choose a required option.

1. I. x^2 – 87x -270 = 0
   II. 7y^2 – 11y -18 = 0
   y>x
   x>y
   x=y or relation cannot be established.
   y>=x
   x>=y
   Option C
   I. x^2 – 87x -270 = 0
   x^2 -90x + 3x – 270 =0
   => x = 90,-3
   II. 7y^2 – 11y -18 = 0
   7y^2 -18y +7y -18=0
   => y = 18/7, -1
   x=y or relation cannot be established.

    

2. I. 4x^2 + 23x + 15 = 0
   II. 4y^2 – 11y + 6 = 0
   y>x
   y>=x
   x>y
   x>=y
   x=y or relation cannot be established.
   Option A
   I. 4×2 + 23x + 15 = 0
   4x^2 + 20x + 3x + 15 = 0
   => x = -5, -3/4
   II. 4y2 – 11y + 6 = 0
   4y^2 – 8y – 3y + 6 = 0
   => y= 3/4, 2
   y>x

    

3. I. 55x^2 – 495x +1100 = 0
   II. 5y^2 + 10y -120 = 0
   x=y or relation cannot be established.
   y>=x
   x>=y
   x>y
   y>x
   Option C
   I. 55x^2 – 495x +1100 = 0
   55x^2 – (220+275)x – 1100 = 0
   => x = 4,5
   II. 5y^2 + 10y -120 = 0
   5y^2 +30y – 20y – 120 = 0
   => y = -6,4
   x>=y

    

4. I. x^2 – 87x -270 = 0
   II. 7y^2 – 11y -18 = 0
   x>=y
   y>x
   x=y or relation cannot be established.
   x>y
   y>=x
   Option C
   I. x^2 – 87x -270 = 0
   x^2 -90x + 3x – 270 =0
   => x = 90,-3
   II. 7y^2 – 11y -18 = 0
   7y^2 -18y +7y -18=0
   => y = 18/7, -1
   x=y or relation cannot be established.

    

5. I. 9x^2 – 94.5x + 243 = 0
   II. 4.5 y^2 – 13.5y – 486=0
   y>x
   y>=x
   x>y
   x=y or relation cannot be established.
   x>=y
   Option D
   I. 9x^2 -94.5x + 243 = 0
   9x^2 – (54+40.5)x + 243 =0
   => x = 6,4.5
   II. 4.5 y^2 – 13.5y -486=0
   4.5y^2 – 54y + 40.5y – 486=0
   => y = 12, -9
   x=y or relation cannot be established.

    

Directions(6-10): Following line-graph shows the population of seven cities (in lakh) and the table shows the percentage of literate population in these cities.



6. What is the percentage rise in the population of City C from 2010 to 2012?
   41.2%
   30.5%
   25.12%
   33.33%
   50.12%
   Option D
   Required %
   = [(6.4-4.8)/4.8]*100 = 33.33%

    

7. The literate population of City E in the year 2012 is approximately what percentage more than its literate population in 2010?
   15.5%
   22.5%
   40.2%
   37.3%
   20.5%
   Option D
   City E (2010) = 5.5*(67.7/100) = 3.7235 lakh
   City E(2012) = 7.2*(71/100) = 5.112 lakh
   Required % = [(5.112 – 3.7235)/3.7235]*100 = 37.3%

    

8. What is the difference between the Literate population and illiterate population of City D in the year 2010? (in lakh)
   1.500 lakh
   1.250 lakh
   2.115 lakh
   1.508 lakh
   5.002 lakh
   Option D
   Total population = 5.2 lakh
   Percentage of literate population = 64.5%
   Percentage of illiterates = 100-64.5 = 35.5%
   Difference = 64.5 – 35.5 = 29%
   Required answer = 5.2*(29/100) = 1.508 lakh

    

9. What is the difference between the total illiterate population of City G and City F in the year 2010? (in lakh)
   0.1700 lakh
   0.1437 lakh
   0.4300 lakh
   0.7500 lakh
   0.1220 lakh
   Option B
   Required difference
   = 7.5*(100-68.9)/100 – 6.4*(100-65.8)/100 = 0.1437 lakh

    

10. What is the total literate population of City A in the year 2010 and 2012 together (in lakh)?
    5.445 lakh
    3.450 lakh
    5.000 lakh
    6.444 lakh
    5.005 lakh
    Option A
    Literate population of City A in 2010
    = 3.6*(57.8/100) = 2.0808 lakh
    Literate population of City A in 2012 = 5.4*(62.3/100)
    = 3.3642 lakh
    Total = 2.0808 + 3.3642 = 5.445 lakh