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.

- I. 20x
^{2}– 31x + 12 = 0

II. 6y^{2}– 7y + 2 = 0x>yx=y or relation cannot be established.x>=yy>=xy>xOption A

I. 20x^{2}– 31x + 12 = 0

20x^{2}– 31x + 12 = 0

20x^{2}– 16x – 15x + 12 = 0

x = 3/4, 4/5

II. 6y^{2}– 7y + 2 = 0

6y^{2}– 7y + 2 = 0

6y^{2}– 3y – 4y + 2 = 0

y = 1/2, 2/3

x>y - I. 2x
^{2}+ 5x – 12 = 0

II. 2y^{2}– 19y + 35 = 0x=y or relation cannot be established.y>xx>=yx>yy>=xOption B

I. 2x^{2}+ 5x – 12 = 0

2x^{2}+ 5x – 12 = 0

2x^{2}+ 8x – 3x – 12 = 0

x = -4 , 3/2

II. 2y^{2}– 19y + 35 = 0

2y^{2}– 19y + 35 = 0

2y^{2}– 14y – 5y + 35 = 0

y = 5/2, 7

x < y - I. 5x
^{2}– 13x + 6 = 0

II. 3y2 – 22y – 35 = 0y>=xx=y or relation cannot be established.x>=yx>yy>xOption B

I. 5x^{2}– 13x + 6 = 0

5x^{2}– 13x + 6 = 0

5x^{2}– 10x – 3x + 6 = 0

x = 3/5, 2

II. 3y^{2}– 22y – 35 = 0

3y^{2}– 22y – 35 = 0

3y^{2}– 15y – 7y – 35 = 0

y = -7/3, 5

Relation cannot be established. - I. 50x
^{2}– 95x + 42 = 0

II. 50y^{2}– 65y + 21 = 0y>xx=y or relation cannot be established.x>=yx>yy>=xOption C

I. 50x^{2}– 95x + 42 = 0

50x^{2}– 95x + 42 = 0

50x^{2}– 60x – 35x + 42 = 0

x = 7/10, 6/5

II. 50y^{2}– 65y + 21 = 0

50y^{2}– 65y + 21 = 0

50y^{2}– 65y + 21 = 0

y = 3/5, 7/10

x ≥ y - I. 3x
^{2}– 4x – 15 = 0

II. 5y^{2}– 9y – 18 = 0x>=yx=y or relation cannot be established.y>=xy>xx>yOption B

I. 3x^{2}– 4x – 15 = 0

3x^{2}– 4x – 15 = 0

3x^{2}– 9x + 5x – 15 = 0

x = -5/3, 3

II. 5y^{2}– 9y – 18 = 0

5y^{2}– 9y – 18 = 0

5y^{2}– 15y + 6y – 18 = 0

y = -6/5, 3

x = y or relation cannot be established - What is the total number of vacancies available in Chennai in 2000 and in Mumbai in the year 2010?
3.50 lakh1.02 lakh3.01 lakh2.238 lakh4.15 lakhOption D

Sum = 5.4* 8/100 + 8.6* 21/100

= 0.432 + 1.806

= 2.238 lakh - What is the difference between the number of vacancies available in Bangalore in the year 2010 and 2000?
3.250 lakh5.080 lakh4.303 lakh2.112 lakh1.426 lakhOption E

Difference = 8.6*26/100 – 5.4*15/100

= 2.236 – 0.81 = 1.426 lakh - What is the average number of vacancies available in Hyderabad in the year 2000 and 2010?
7000052420600003542045500Option A

Hyderabad(2000) = 5.4*10/100 = 0.54 lakh

Hyderabad(2010) = 8.6*10/100 = 0.688

Average = 0.54 + 0.86/2 = 0.7 lakh = 70,000 - What is the percentage rise in vacancies available in Hyderabad from year 2000 to 2010? (Give approximate value only).
50%45%59%61%48%Option C

Hyderabad(20OO) = 5.4*10/100 = 0.54 lakh

Hyderabad(2010) = 8.6*10/100 = 0.688 lakh

% rise =(0.86-0.54)/0.54*100 ==59% - If the number of vacancies in Pune is 48000 in the year 2010 and the percentage distribution is the same as given in the chart, what is number of vacancies available in NCR in 2010?
130000150000120000164571110000Option D

Total number of vacancies in 2010= 48000*100/7

=685714.3

Vacancies in NCR = 24% of 685714.3 = 164571

**Directions(6-10):** Following pie-charts show the percentage distribution of job vacancies

in IT industries in the year 2000 and 2010. In the year 2000, the total number of vacancies was 5.4 lakh, and in the year 2010, it was 8.6 lakh.