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.𝑥^2−3481=0

II.3𝑦^2=(216000)^1/3x>=yx>=yx>yy>xx=y or relation cannot be established.Option E

I.x=±59

II.3y^2=60

⇒y=±√20

No relation exists. - I. 5x^2+2x−3=0

II. 2y^2+7y+6=0x=y or relation cannot be established.x>yx>=yx>=yy>xOption 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 - I. (17)^2 +144 ÷18 = x

II. (26)^2 − 18 ×21 = yx>=yy>xx=y or relation cannot be established.x>=yx>yOption B

I.x=289+14418=297

II. y=298

x<𝑦 - I. x2−5x−14=0

II.y2+7y+10 = 0x>=yx=y or relation cannot be established.y>xx>=yx>yOption 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 - I.√𝑥−(18)^15/2/𝑥2=0

II.√𝑦=(19)^9/2/𝑦x>yx>=yy>=xy>xx=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 - The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:
28347Option 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. - 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 men65 men35 men25 men55 menOption 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 - 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 : 17 : 511 : 67 : 35 : 4Option 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 - 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?
600800900700500Option 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 - 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 days420 days500 days760 days660 daysOption 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