**Institute of Banking Personnel Selection (IBPS)** had released the official notification for the Common Recruitment Process for RRBs (CRP RRBs VII) for the recruitment of Group “A”-Officers (Scale-I, II & III) and Group “B”-Office Assistant (Multipurpose)

Click here to know the details of the Examination

The examination will be held in two phases i.e. Preliminary Examination and Main Examination. The **RRB Scale I** Preliminary Exam is scheduled on 11th, 12th & 18th of August 2018. And **RRB Assistant** Preliminary Exam is scheduled on 19th, 25th August & 1st September 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. 2x² – 25x + 77 = 0

II. 2y² – 21y + 55 = 0x>=yx>yy>=xx=y or relation cannot be established.y>xOption A

I. 2x² – 25x + 77 = 0

=>2x² – 14x – 11x + 77 = 0

=>2x(x- 7) – 11 (x- 7) = 0

=>x = 7, 11/2

II. 2y² – 21y + 55 = 0

=>2y² – 10y – 11y + 55 = 0

=>2y(y – 5) – 11 (y – 5) = 0

=>y = 11/2, 5

x ≥ y - I. x^2 + (5 + 3(3)^1/2)x +15(3)^1/2 = 0

II. y^2 + 7(3)^1/2y + 36 = 0x=y or relation cannot be established.y>=xx>yx>=yy>xOption D

I. x^2 + (5 + 3(3)^1/2)x + 15(3)^1/2 = 0

=> x^2 + 5x + 3(3)^1/2x + 15(3)^1/2 = 0

=> x(x+5) + 3(3)^1/2(x+5) = 0

=> (x+5) (x+3(3)^1/2) = 0

=> x = -5 , -3(3)^1/2

II. y^2 + 7(3)^1/2y + 36 = 0

=> y^2 + 4(3)^1/2 + 3(3)^1/2 + 36 = 0

=> y(y + 4(3)^1/2) + 3(3)^1/2(y + 4(3)^1/2) = 0

=> (y + 4(3)^1/2) (y + 3(3)^1/2) = 0

=> y = – 3(3)^1/2, -4(3)^1/2

x>=y - I. 9x² – 33x + 28 = 0

II. 6y² – 25y + 25 = 0y>xx>=yx=y or relation cannot be established.y>=xx>yOption C

I. 9x² – 33x + 28 = 0 =>9x² – 12x – 21x + 28 = 0

=> 3x (3x – 4) – 7 (3x – 4) = 0

=> x = 4/3, 7/3

II. 6y² – 25y + 25 = 0

=> 6y² – 15y – 10y + 25= 0

=>3y(2y – 5) – 5 (2y -5) = 0

=>y = 5/2, 5/3

No relation - I. 9x² + 3x – 2 = 0

II. 8y² + 6y + 1 = 0x=y or relation cannot be established.x>=yy>xx>yy>=xOption A

I. 9x² + 3x – 2 = 0

=> 9x² + 6x – 3x – 2 = 0

=>3x(3x+ 2) -1 (3x + 2)= 0

=> x = −2/3,1/3

II. 8y² + 6y +1 = 0

=>8y² + 4y + 2y + 1 = 0

=> 4y(2y + 1) + 1 (2y + 1) = 0

=> y = −1/4,−1/2

No relation - I. 2x² + 9x + 7 = 0

II. 2y² + 9y + 10 = 0y>xx>yy>=xx>=yx=y or relation cannot be established.Option E

I. 2x² + 9x + 7 = 0

=> 2x² + 7x+ 2x + 7 = 0

=>x(2x + 7) + 1(2x + 7) = 0

=>x = −1, −7/2

II. 2y² +9y + 10 = 0

=> 2y² + 5y + 4y +10 = 0

=>y(2y+ 5) +2(2y +5) = 0

=>y = −2, −5/2

No relation - Wheels of radius 7 cm and 14 cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make same number of revolutions per second. If both of them meet after 10 seconds, find the speed of the smaller wheel
70 cm/sec86 cm/sec52 cm/sec80 cm/sec66 cm/secOption E

Distance travelled by smaller wheel in one revolution = 2× 22/7 ×7 = 44 cm

And by larger wheel = 2× 22 7 ×14 = 88 cm

Now, if speed of smaller wheel is x cm/sec.

Then, speed of larger wheel = 2x cm/sec 10x + 10 × 2x

= 1980

=> x = 66 cm/sec - 18 litres of pure water was added to a vessel containing 80 litres of pure milk. 49 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 2 : 1. If the resultant respective ratio of milk and waterin the vessel was 4 : 1, what was the quantity of pure milk added in the vessel ?(in litres)
14 litres11 litres8 litres4 litres2 litresOption D

Initial ratio of milk and water = 80 : 18 = 40 : 9

Let 2x and x amount of pure milk and pure water were added respectively.

[80 – 49*40/49] + 2x / [18 – 49*9/49] + x = 4/1

=> x = 2

Required quantity of milk added = 2 × 2 = 4 litres - Train A, travelling at 84 kmph, overtook train B, traveling in the same direction, in 10 seconds. If train B had been traveling at twice its speed, then train A would have taken 22.5 seconds to overtake it. Find the length of train B, given that it is half the length of train A.
70 m40 m50 m80 m60 mOption C

Let speed of train B be x m/s And length of train B be y m.

Then length of train A is 2y m

Speed of train A = 84*5/18 = 210/9 = 70/3 m/s

Now, 2y+y/10 =70/3 – x —-(1)

2y+y/22.5 = 70/ – x —–(2) F

rom (1) and (2), we get

y = 50 m - Ram invested his money in two schemes A and B in the ratio 3 : 5 for the same time. If schemes A and B offer 15% and 10% interest rates to Ram respectively then find after how much time total income of Ram will be Rs 4560 if his capital amount was Rs. 12800?
7 years5 years6 years3 years2 yearsOption D

Let Ram invested Rs. 3x and Rs. 5x amounts in schemes A and B respectively.

x = 12800/8

=>x= 1600 rupees

Let total time was t years.

4800×15×t 100 + 8000×10×t 100 = 4560

=> 720t + 800t = 4560

=> t = 4560/1520

=> t = 3 years - Two cards are drawn in succession from a pack of 52 cards, without replacement. What is the probability, that the first is a Queen and the second is a Jack of a different suit?
1/2211/1171/1211/1191/220Option A

The probability of first Queen = 4 / 52

The probability of Second Jack of different suit = 3 / 51

Required Probability = (4/52) x (3/51) = (1/13) x (1/17)

= 1/221

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