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 onÂ **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.Â

**Directions(1-5):** Find the relationship between x and y and choose a correct option accordingly.

- I. 2x
^{2}â€“ (2 + 2âˆš5)x + 2âˆš5 = 0

II. 4y^{2}â€“ (6 + 2âˆš2)y + 3âˆš2 = 0x>yx>=yy>=xy>xx=y or relation cannot be established.Option E

I. 2x^{2}â€“ (2 + 2âˆš5)x + 2âˆš5 = 0

=>2x^{2}â€“ 2x â€“ 2âˆš5x + 2âˆš5 = 0

=>2x (x â€“ 1) â€“ 2âˆš5 (x â€“ 1) = 0

x = 1, âˆš5 (2.2)

II. 4y^{2}â€“ (6 + 2âˆš2)y + 3âˆš2 = 0

=>4y^{2}â€“ 6y â€“ 2âˆš2y + 3âˆš2 = 0

=> 2y (2y â€“ 3) â€“ âˆš2 (2y â€“ 3) = 0

y = 3/2, 1/âˆš2 (0.7)

x = y or relation cannot be established - I. 3x
^{2}+ 22x + 24 = 0

II. 3y^{2}â€“ 8y â€“ 16 = 0y>=xy>xx>=yx>yx=y or relation cannot be established.Option A

I. 3x^{2}+ 22x + 24 = 0

=>3x^{2}+ 22 x + 24 = 0

=> 3x^{2}+ 18x + 4x + 24 = 0

x = -4/3, -6

II. 3y^{2}â€“ 8y â€“ 16 = 0

=>3y^{2}â€“ 8y â€“ 16 = 0

=> 3y^{2}â€“ 12y + 4y â€“ 16 = 0

y = -4/3, 4 x â‰¤ y - I. 20x
^{2}â€“ 31x + 12 = 0

II. 3y^{2}â€“ 16y + 16 = 0y>xy>=xx>yx>=yx=y or relation cannot be established.Option A

I. 20x^{2}â€“ 31x + 12 = 0

=>20x^{2}â€“ 31x + 12 = 0

=>20x^{2}â€“ 16x â€“ 15x + 12 = 0

x = 3/4, 4/5

II. 3y^{2}â€“ 16y + 16 = 0

=>3y^{2}â€“ 16y + 16 = 0

=>3y^{2}â€“ 14y â€“ 4y + 16 = 0

y = 4, 4/3

x < y - I. x
^{2}+ âˆš5x â€“ 10 = 0

II. 2y^{2}+ 9âˆš5y + 50 = 0x>=yy>xy>=xx=y or relation cannot be established.x>yOption A

I. x^{2}+ âˆš5x â€“ 10 = 0

=>x^{2}+ âˆš5x â€“ 10 = 0

=>x^{2}+ 2âˆš5x â€“ âˆš5x â€“ 10 = 0

x = -2âˆš5, âˆš5

II. 2y^{2}+ 9âˆš5y + 50 = 0

=>2y^{2}+ 9âˆš5y + 50 = 0

=> 2y^{2}+ 4âˆš5y + 5âˆš5y + 50=0

y = -2âˆš5, -5âˆš5/2

x â‰¥ y - I. 8x
^{2}+ 5x â€“ 13 = 0

II. 2y^{2}+ 23y + 63 = 0y>=xx>yx>=yy>xx=y or relation cannot be established.Option B

I. 8x^{2}+ 5x â€“ 13 = 0

=>8x^{2}+ 5x â€“ 13 = 0

=>8x^{2}+ 13x â€“ 8x â€“ 13 = 0

x = -1.625, 1

II. 2y^{2}+ 23y + 63 = 0

=>2y^{2}+ 23y + 63 = 0

=> 2y^{2}+ 14y + 9y + 63 = 0

y = -7, -4.5

x > y - What is the average number of A items sold by all six companies together?
8850075080896806995095000Option C

A items sold by P = 16*(24/100)*(5/8)*(65/100)

= 1.56 lakh

Similarly,

Total A items = 1.56 + 0.896 + 0.6912 + 1.44 + 0.4096 + 0.384

= 5.3808 lakh

Average = 5.3808/6

= 0.8968 lakh

= 89680 - The number of items sold by Company P is what percentage of the number of A items sold by Company U?
320.11%415.02%520.20%406.25%300.05%Option D

A sold by P = 156000,

A sold by U = 38400

Reqd % = (156000/38400)*100

= 406.25% - What is the difference between the total items produced by Company P and Q together and the total items produced by Company S?
2.24 lakh3.23 lakh4.00 lakh3.52 lakh1.20 lakhOption A

Total items = [(24+18) – 28]*16/100 = 2.24 lakh - What is the difference between the total number of A items and the total number of B items produced by Company U?
1580034500155002560026000Option D

Total = 16 Ã— (8/100) = 1.28 lakh,

Item A = (1.28/5)* 3= 0.768

Item B = (1.28/5)*2 = 0.512

Diff = 0.768 â€“ 0.512 = 0.256 lakh

= 25600 - What is the difference between the number of A items sold and the number of B items sold by Company T?
12350 lakhs15600 lakhs32560 lakhs20000 lakhs14560 lakhsOption E

Item A = 16*( 7/100)*( 4/7)* (64/100) = 0.4096

Similary, Item B = 0.2640

Diff = 0.4096 – 0.2640 = 0.1456 lakh

= 14560 lakhs

**Directions(6-10):** Following pie-chart shows the percentage distribution of total items (Item A and Item B) produced by six companies (P,Q,R,S,T and U) and the table shows the ratio of A to B and percentage sale of A and B.

Total items = 16 lakh

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