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. 6x
^{2}+ 77x + 121 = 0

II. y^{2}+ 9y – 22 = 0y>xx=y or relation cannot be established.y>=xx>yx>=yOption B

I. 6x^{2}+ 77x + 121 = 0

=> 6x^{2}+ 66x + 11x + 121 = 0

=> 6x (x + 11) + 11 (x + 11) = 0

=> (x + 11) (6x + 11) = 0

=> x = – 11 or – 11/6

II. y^{2}+ 11y – 2y – 22 = 0

=>y (y + 11) – 2 (y + 11) = 0

=>(y – 2) (y + 11) = 0

=>y = 2 or -11

x=y or relation cannot be established. - I. 16x
^{2}+ 8x – 15 = 0

II. 4y^{2}+ 29y + 30 = 0x=y or relation cannot be established.y>=xx>yx>=yy>xOption D

I. 16x^{2}+ 8x – 15 = 0

=>16x^{2}+ 8x – 15 = 0

=>16x^{2}+ 20x – 12x – 15 = 0

=>x = -5/4, 3/4

II. 4y^{2}+ 29y + 30 = 0

=>4y^{2}+ 29y + 30 = 0

=>4y^{2}+ 24y + 5y + 30 = 0

=>y = -6, -5/4

x ≥ y - I. 3x
^{2}+ 23x + 30 = 0

II. 3y^{2}– 4y – 4 = 0x>yx>=yx=y or relation cannot be established.y>xy>=xOption D

I. 3x^{2}+ 23x + 30 = 0

=> 3x^{2}+ 23x + 30 = 0

=> 3x^{2}+ 18x + 5x + 30 = 0

=> x = -5/3, -6

II. 3y^{2}– 4y – 4 = 0

=> 3y^{2}– 4y – 4 = 0

=> 3y^{2}– 6y + 2y – 4 = 0

=> y = 2, -2/3

x < y - I. 5x
^{2}– 36x – 32 = 0

II. 3y^{2}– 17y – 6 = 0x>=yx>yy>=xy>xx=y or relation cannot be established.Option E

I. 5x^{2}– 36x – 32 = 0

=>5x^{2}– 36x – 32 = 0

=>5x^{2}+ 4x – 40x – 32 = 0

=> x = -4/5, 8

II. 3y^{2}– 17y – 6 = 0

=>3y^{2}– 17y – 6 = 0

=> 3y^{2}+ y – 18y – 6 = 0

=> y= -1/3, 6

x = y or relation cannot be established - I. 3x
^{2}+ 20x + 32 = 0

II. 3y^{2}– 4y – 4 = 0y>xx=y or relation cannot be established.y>=xx>=yx>yOption A

I. 3x^{2}+ 20x + 32 = 0

=>3x^{2}+ 20x + 32 = 0

=>3x^{2}+ 12x + 8x + 32 = 0

=>x = -4, -8/3

II. 3y^{2}– 4y – 4 = 0

=>3y^{2}– 4y – 4 = 0

=>3y^{2}– 6y + 2y – 4 = 0

=> y = -2/3, 2

x < y - A sum was put at simple interest at a certain rate for 3 yr. Had it been put at 1% higher rate, it would have fetched Rs 5100 more. Find the sum.
Rs 1,10,000Rs 1,20,000Rs 1,70,000Rs 1,30,000Rs 1,50,000Option C

Simple interest for 1 yr = 5100/3 = Rs 1700

1% of sum 1700

Sum = 1700 * 100 = Rs 1,70,000 - A shopkeeper buys 144 items at 90 paise each. On the way 20 items broken. He sells the remainder at Rs.1.20 each. His gain percent correct to one decimal place.
14.81%14.61%17.21%13.20%15.11%Option A

20 items are broken out of 144.

CP of 124 items = 144*90/100 = Rs.129.6

Total SP = 1.2*124 = Rs.148.8

Gain% = [148.8 – 129.6]/129.6 *100 = 14.81% - A man gave 50% of his savings of Rs. 84,100 to his wife and divided the remaining sum among his two sons A and B of 15 and 13 years old of age resp. He divided it in such a way that each of his sons, when they attain the age of 18 years , would receive the same amount at 5% compound interest per annum. Find the share of B.
Rs.10,000Rs.80,000Rs.20,000Rs.50,000Rs.60,000Option C

50% of savings = 50/100*84100 = Rs.42050

Difference of their ages = 2 years

Rate = 5%

Ratio of shares = A:B = (21/20)^2 = 441/400

B’s share = 42050/841*400 = Rs.20,000 - A trader buys goods at 20% discount on marked price. If he wants to make a profit of 25% after allowing a discount of 20% , by what percent should his marked price be greater than the original marked price?
35%10%15%25%40%Option D

Let the original price be 100x.

cost price = 80% of 100x = 80x

Profit 25% selling price = 125/100*80x = 100x

Allowed discount = 20%

Trader’s market price = 100x/80*100 = 125

Required % = [125x-100x]/100x*100 = 25% - Average age of A,B and C is 84 years. When D joins them the average age becomes 80 years. A new person E, whose age is 4 years more than D, replaces A and the average of B,C,D and E becomes 78 years. What is the age of A?
60 years40 years80 years90 years50 yearsOption C

A+B+C = 3*84 = 252 years

A+B+C+D = 80*4 = 320 years

D’s age = 320 – 252 = 68 years

E’s age = 68 + 4 = 72 years

B+C+D+E = 78*4 = 312 years

B+C = 312 – 68 – 72 =172 years

A’s age = 252 – 172 = 80 years