Quant Test for SBI PO 2018 Prelim Exam Set – 17

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 accordingly.

1. I. (17)^2 + 144 /18 = x
   II. (26)^2 – 18*21 = y
   x>=y
   x>y
   y>=x
   y>x
   x=y or relation cannot be established.
   Option D
   I. (17)² + 144 /18 = x
   => x = 297
   II. (26)² – 18*21 = y
   =>y = 298
   y>x

    

2. I. 16x² -106x +13=0
   II. 8y² + 47y – 6 = 0
   y>x
   y>=x
   x=y or relation cannot be established.
   x>y
   x>=y
   Option E
   I. 16x² -106x +13=0
   16x² – 104x – 2x + 13 =0
   => (2x – 13)(8x – 1)=0
   => x= 13/2 , 1/8
   II. 8y² + 47y – 6 = 0
   8y² + 48y -1y – 6 =0
   => (y+6)(8y – 1) = 0
   => y = -6, 1/8
   x>=y

    

3. I. x² – 529 = 0
   II. y³-19683 = 0
   x>y
   x=y or relation cannot be established.
   x>=y
   y>=x
   y>x
   Option E
   I. x² – 529 = 0
   => x = +23, -23
   II. y³ -19683 = 0
   => y = 27
   y>x

    

4. I. 40x² -101x +12 =0
   II. 40y² – 13y + 1 =0
   x>=y
   y>=x
   y>x
   x>y
   x=y or relation cannot be established.
   Option A
   I. 40x² -101x +12 =0
   40x² – 96x – 5x + 12 =0
   => (5x – 12)(8x – 1) = 0
   => x = 12/5, 1/8
   II. 40y² – 13y + 1 =0
   40y² – 8y – 5y + 1 =0
   => (8y – 1)(5y – 1) = 0
   => y = 1/8, 1/5
   x>=y

    

5. I. 55x² – 495x + 1100 =0
   II. 5y² + 10y – 120 = 0
   y>=x
   y>x
   x>=y
   x=y or relation cannot be established.
   x>y
   Option C
   I. 55x² – 495x + 1100 =0
   55x² – 220x – 275x + 1100 =0
   => x = 4,5
   II. 5y² + 10y – 120 = 0
   5y² + 30y – 20y – 120 =0
   => y = -6,4
   x>=y

    

Directions(6-10): Study the following table carefully and answer the questions that follow.
Few values are missing in the table. A candidate is expected to calculate the missing value, if it is required to answer the given questions on the basis of the given information.


6. The number of appeared candidates from State B increased by 100% from 2010 to 2011. If the total number of qualified candidates from State B in 2010 and 2011 together is 408, what is the number of appeared candidates from State B in 2010?
   300
   280
   340
   330
   310
   Option C
   Let the number of appeared candidates form State B in 2010 be x.
   Then number of appeared candidates in 2012 = x + (x*100)/100 = 2x
   Now, (x*30)/100 + (2x*45)/100 = 408
   =>12x = 408*10
   => x= 340

    

7. If the ratio of the number of qualified candidates from State A in 2013 to that in 2014 is 14:9, what is the number of qualified candidates from State A in 2014?
   150
   200
   188
   216
   110
   Option D
   In 2013, number of qualified candidates
   = (480*70)/100 = 336
   Let the number of qualified candidates from State A in 2013 be 14x and that in 2014 be 9x.
   Then,
   14x = 336
   => x = 24
   So, the number of qualified candidates from State A in 2014
   = 9*24 = 216

    

8. What is the difference between the number of qualified candidates from State A in 2010 and that in 2011?
   12
   10
   8
   15
   20
   Option A
   Required difference
   = (450*60)/100 – (600*43)/100
   =270-258 = 12

    

9. If the average number of qualified candidates from State B in 2012, 2013 and 2014 is 210, what is the number of qualified candidates from State B in 2014?
   125
   187
   170
   154
   136
   Option B
   Let the percentage of candidates who qualify from State B in 2014 be x%.
   Average
   => [(280*60)/100 + (550*50)/100 + 400*x%]/3= 210
   => x = 46.75%
   Number of qualified candidates from State B in 2014 = (400*46.75)/100
   = 187

    

10. Out of the number of qualified candidates from State A in 2012, the ratio of male to female candidates is 11:7. If the number of female qualified candidates from State A in 2012 is 126, what is the number of appeared candidates (both male and female) from State A in 2012?
    400
    480
    380
    520
    540
    Option E
    Let the number of male and female qualified candidates be 11x and 7x respectively.
    Then, 7x = 126
    => x = 18
    Males = 18*11 = 198
    Total number of qualified candidates = 198+126 = 324
    Let the number of appeared candidates be y.
    Then , y * 60% = 324
    => y = (324*100)/60 = 540