Quant Test for SBI PO 2018 Prelim Exam Set – 63

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.

1. I. 3x² + 16x + 20 = 0
   II. 3y² – 14y – 5 = 0
   x>=y
   y>x
   x=y or relation cannot be established.
   x>y
   y>=x
   Option B
   3x² + 16x + 20 = 0
   3x² + 6x + 10x + 20 = 0
   x = -2, -10/3
   3y² – 14y – 5 = 0
   3y² – 15y + y – 5 = 0
   y = -1/3, 5
   x < y

    

2. I. 2x² – 15x + 7 = 0
   II. 8y² + 6y – 5 = 0
   y>=x
   x>y
   x>=y
   y>x
   x=y or relation cannot be established.
   Option C
   2x² – 15x + 7 = 0
   2x² – 14x – x + 7 = 0
   x = 1/2, 7
   8y² + 6y – 5 = 0
   8y² – 4y + 10y – 5 = 0
   y = -5/4, ½
   x ≥ y

    

3. I. 3x² + 14x – 5 = 0
   II. 4y² + 5y – 6 = 0
   x>y
   y>=x
   x>=y
   x=y or relation cannot be established.
   y>x
   Option D
   3x² + 14x – 5 = 0
   3x² + 15x – x – 5 = 0
   x = -5, 1/3
   4y² + 5y – 6 = 0
   4y² + 5y – 6 = 0
   y = -2, ¾
   x = y or relation cannot be established.

    

4. I. 20x² – 31x + 12 = 0
   II. 4y² + 5y – 6 = 0
   x=y or relation cannot be established.
   y>x
   x>=y
   y>=x
   x>y
   Option C
   20x² – 31x + 12 = 0
   20x² – 16x – 15x + 12 = 0
   x = 3/4, 4/5
   4y² + 5y – 6 = 0
   4y² + 5y – 6 = 0
   y = -2, ¾
   x ≥ y

    

5. I. 3x² + 22 x + 24 = 0
   II. 2y² + 11y + 12 = 0
   x>y
   y>=x
   y>x
   x>=y
   x=y or relation cannot be established.
   Option E
   3x² + 22 x + 24 = 0
   3x² + 18x + 4x + 24 = 0
   x = -4/3, -6
   2y² + 11y + 12 = 0
   2y² + 8y + 3y + 12 = 0
   y = -4, -3/2
   x = y or relation cannot be established .

    


Directions(6-10): Study the following graph and answer accordingly
The following line graph shows the Time taken by four pipes to full an empty tank.

The following table shows the comparison of efficiency between A,B,C ,D and to P,Q,R,S(either
inlet or outlet).


6. Four pipes C,D,P and Q can fill a tank . When initially the tank is empty pipes C, D, P and Q start to fill the tank .After two hours pipe C is closed, after another 1hour pipe D is closed and Pipe P is closed 2 hours before the tank is full and the remaining part is filled by pipe Q alone .Find the time taken by all the pipes together to fill the tank.
   4
   12
   8
   9
   10
   Option D
   2/ 60 + 3 /15 + (𝑥 – 2)/ 15 + 𝑥/ 30 = 1
   [2 + 12 + 4𝑥 − 8 + 2𝑥 ]/60 = 1
   => 6x + 6=60
   => 6x=60-6=54
   => X=54/6 = 9 hrs

    

7. Initially two inlet pipes A and D are opened for two hours and the remaining part is filled by pipe Q alone .Find the time taken by pipe Q to fill the remaining tank?
   30
   24
   20
   15
   22
   Option B
   (A+D)’s one hour= 1 /30 + 1 /15 = 3/ 30
   = 1/ 10
   For 2 hours = 1/ 5
   Remaining part of the tank=1 – 1/ 5 = 4 /5
   Remaining tank is filled by Q alone in 4/ 5 × 30 = 24 ℎ𝑜𝑢𝑟

    

8. Two inlet pipes A and B fill the tank and two outlet pipes R and S empty the tank. All the four pipes are opened together. Find the time taken to full the tank.
   30
   40
   20
   50
   10
   Option A
   1/ 30 + 1/ 15 – 1/ 20 – 1/ 60
   = 1 [2 + 4 − 3 – 1]/ 60
   = 1 [ 6 – 4]/ 60 = 1
   = 1/ 30 ℎ𝑜𝑢𝑟𝑠 = 30ℎ𝑜𝑢𝑟

    

9. Find the time taken to fill the tank when all the inlet pipes A,B,C and D are opened together and also when two outlet pipes P and Q are opened together.
   10
   18
   14
   12
   16
   Option D
   A ,B ,C and D one hour work
   = 1 /30 + 1 /60 + 1 /15 + 1/ 15 = 11 /60 𝑝𝑎𝑟𝑡
   => 60/ 11 ℎ𝑜𝑢𝑟𝑠
   P=2A
   Efficiency ratio of P:A=2:1
   Time taken ratio=1:2
   P=15days
   Q=(1/2)×B
   Time taken ratio of B and Q=2:1
   Q=30 hours
   P+Q’s one hour work= 1/ 15 + 1/ 30 = 1/ 10 𝑝𝑎𝑟𝑡
   One hour work of A,B,C,D,and P and Q
   = 11 /60 – 1/ 10 = 5/ 60 = 12 ℎ𝑜𝑢𝑟s

    

10. Initially both the pipes A and C are opened together to fill a tank and when the tank was 1/3rd full ,a leak developed, through which 1/ 4 𝑡ℎ of the water supply by both the pipes leaked out .Find after how many hours the tank be full.
    17 (3 /8 )
    16 (5 /9 )
    17 (7 /9 )
    10 (2 /7 )
    11 (5/9 )
    Option C
    Let total work be 60 units
    A=2units
    C=1units
    A+C=3units
    Leakage= (1+2)*1/4=-3/4
    A+C-leak=2 + 1 − 3 /4 = 9 /4
    A+C=60 × 1/ 3 = 20 units
    Remaining = 40 units
    𝑇𝑖𝑚𝑒 = 40 /9/ 4 = 160/ 9 ℎ𝑜𝑢𝑟𝑠
    = 17 (7 /9 ) ℎ𝑜𝑢𝑟𝑠