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 x and y ,and choose a correct option.

- I. 12x
^{2}– 5x – 3 = 0

II. 4y^{2}– 11y + 6 = 0x>yx=y or relation cannot be established.y>xx>=yy>=xOption E

I. 12x^{2}– 5x – 3 = 0

II. 4y^{2}– 11y + 6 = 0

I. 12x^{2}– 5x – 3 = 0

=>12x^{2}– 5x – 3 = 0

=>12x^{2}+ 4x – 9x – 3 = 0

x = -1/3, 3/4

II. 4y^{2}– 11y + 6 = 0

=> 4y^{2}– 11y + 6 = 0

=>4y2 – 8y – 3y + 6 = 0

y= 3/4, 2

x ≤ y - I. 20x
^{2}– 31x + 12 = 0

II. 3y^{2}– 16y + 16 = 0x>=yy>xx>yy>=xx=y or relation cannot be established.Option B

I. 20x^{2}– 31x + 12 = 0

II. 3y^{2}– 16y + 16 = 0

I. 20x^{2}– 31x + 12 = 0

=>20x^{2}– 31x + 12 = 0

=>20x^{2}– 16x – 15x + 12 = 0

x = 3/4, 4/5

II. 3y^{2}– 16y + 16 = 0

=> 3y^{2}– 16y + 16 = 0

=> 3y^{2}– 14y – 4y + 16 = 0

y = 4, 4/3

x < y - I. x
^{2}– 2x – √5x + 2√5 = 0

II. y^{2}– 3y – √6y + 3√6 = 0y>=xy>xx=y or relation cannot be established.x>yx>=yOption B

I. x^{2}– 2x – √5x + 2√5 = 0

II. y^{2}– 3y – √6y + 3√6 = 0

I. x^{2}– 2x – √5x + 2√5 = 0

=> x^{2}– 2x – √5x + 2√5 = 0

=>x (x – 2) – √5 (x – 2) = 0

x = 2, √5 (2.23)

II. y^{2}– 3y – √6y + 3√6 = 0

=> y^{2}– 3y – √6y + 3√6 = 0

=>y (y – 3) – √6 (y – 3) = 0

y = 3, √6 (2.44)

x < y - I. 2x
^{2}– 15√3x + 84 = 0

II. 3y^{2}– 2y – 8 = 0x=y or relation cannot be established.x>=yy>xy>=xx>yOption E

I. 2x^{2}– 15√3x + 84 = 0

II. 3y^{2}– 2y – 8 = 0

I. 2x^{2}– 15√3x + 84 = 0

=>2x^{2}– 8√3x – 7√3x + 84 = 0

=> 2x (x – 4√3) – 7√3 (x – 4√3x) = 0

x = 3.5√3, 4√3

II. 3y^{2}– 2y – 8 = 0

=>3y^{2}– 2y – 8 = 0

=> 3y^{2}– 6y + 4y – 8 = 0

y = -4/3, 1

x > y - I. 2x
^{2}– (6 + √3)x + 3√3 = 0

II. 3y^{2}– (9 + √3)y + 3√3 = 0x>yy>=xy>xx=y or relation cannot be established.x>=yOption D

I. 2x^{2}– (6 + √3)x + 3√3 = 0

II. 3y^{2}– (9 + √3)y + 3√3 = 0

I. 2x^{2}– (6 + √3)x + 3√3 = 0

=>2x^{2}– 6x – √3x + 3√3 = 0

=>2x (x- 3) – √3 (x – 3) = 0

x = 3, √3/2

II. 3y^{2}– (9 + √3)y + 3√3 = 0

=>3y^{2}– 9y – √3y + 3√3 = 0

=>3y (y – 3) – √3 (y – 3) = 0

x = 3, √3/3

x=y or relation cannot be established.

- Mr. A lends 40% of sum at 15% p.a. 50% of rest sum at 10% p.a. and the rest at 18% p.a. rate of interest. What would be the rate of interest if the interest is calculated on the whole sum ?
13.3%10.2%15.5%14.4%11.5%Option D

Let the amount be Rs. x

Investment is done as given below.

Amount left = x – 40/100 x = 60x/100

Now, 40x/100 at 15% p.a

50/100 of 60x/100 = 30x/100 at 10% p.a

Rest amount = x – 40x/100 – 30x/100 = 30x/100 at 18% p.a

Interest earned by each at end of 1 year By 1st

=> 15/100 × 40x/100 = 60x/1000

By 2nd 10/100 × 30x/100 = 30x/1000

By 3rd 18/100 × 30x/100 = 54x/1000

Total interest = 144x/1000

Rate% = [144x/1000]/x * 100

= 14.4% - A man starts going for morning walk every day. The distance walked by him on the first day was 2 kms. Everyday he walks half of the distance walked on the previous day. What can be the maximum total distance walked by him in his life time.
200kmInadequate data.100km80km150kmOption B

When a train crosses a platform,

the distance covered = Length of platform and the train.

Speed = Length/Time taken

Thus, we have inadequate data. - The area of a rectangle gets reduced by 9 sq. metre if its length is reduced by 5 m and breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 67sq. metre. Find the length of the rectangle.
22m17m25m10m30mOption B

Let the length of the rectangle be x metre and breadth be y metre.

=>xy – (x – 5)(y + 3) = 9

=> xy – (xy – 5y + 3x – 15) = 9

=> 5y – 3x + 15 = 9

=> 3x – 5y – 6 = 0 ….(i)

Again, (x + 3) (y + 2) – xy = 67

=>3y + 2x + 6 = 67

=> 2x + 3y – 61 = 0 ….(ii)

By equation (i) × 3 + (ii) × 5

9x – 15y – 18 = 0

10x + 15y -305 = 0

19x = 323

=> x = 17 m - A shopkeepers announce the same price of Rs. 700 for a shirt. The first offers successive discounts of 30% and 6% while the second offers successive discounts of 20% and 16%. How much more discount offered by the shopkeeper .
Rs. 8.80Rs. 7.80Rs. 5.80Rs. 6.80Rs. 9.80Option E

According to the question,

Selling price of first shopkeeper. = 700 × 70/100 × 94/100

= Rs. 460.60

Selling price of second shopkeeper = 700 × 80/100 × 84/100

= Rs. 470.40

Required difference = 470.40 – 460.60

= Rs. 9.80 - A grocer purchased 2 kg. of rice at the rate of Rs. 15 per kg. and 3 kg of rice at the rate of Rs. 13 per kg. At what price per kg should he sell the mixture to earn 33(1/3) % profit on the cost price ?
Rs.12.6Rs.15.5Rs.18.4Rs.20.5Rs.17.2Option C

Mixture : 2 kg of rice at Rs. 15/kg + 3 kg of rice at Rs. 13/kg

Total weight = 2 + 3 = 5 kg

Total cost price = (2 × 15) + (3 × 13)

= 30 + 39 = Rs. 69

Cost price per kg of the mixture = 69/5

= Rs.13.80

Selling price to get 33(1/3)% = [10+ 33( 1/3)]/100 × Rs. 13.80

= [400 × Rs. 13.80]/[3*100] = 4/3 × Rs 13.80

= Rs. 18.40