Post Date
|
Last Date
|
Bank Name
|
Post Name
|
No. Post
|
Bank Website
|
07/08/2015
|
21/08/2015
|
Jharkhand Gramin Bank
|
Officer Scale I, II, Office Assistant
|
42
|
|
06/08/2015
|
20/08/2015
|
Sarva U.P. Gramin Bank
|
Officer Scale I, II, III, Office Assistant
|
275
|
|
31/07/2015
|
13/08/2015
|
Aryapuram Co-Operative Urban Bank Ltd.
|
Officer, Clerk
|
20
|
|
30/07/2015
|
13/08/2015
|
Madhyanchal Gramin Bank
|
Office Assistant
|
207
|
|
30/07/2015
|
13/08/2015
|
Paschim Banga Gramin Bank
|
Officer Scale I, II, III, Office Assistant
|
103
|
|
28/07/2015
|
12/08/2015
|
Uttarakhand Gramin Bank
|
Officer Scale II & III
|
43
|
Saturday 8 August 2015
Bank Jobs Qualify in RRB CWE 3
Thursday 6 August 2015
Syllabus of IBPS CWE 5 online examination
Online test questions except the English Language
Test will be available in English and Hindi.
I] Preliminary
Examination:
Sr. No.
|
Test Name
|
No. of Questions
|
Maximum Marks
|
Duration (Time)
|
1
|
English Language
|
30
|
30
|
1 Hour (Composite time of 60 Minutes)
|
2
|
Numerical Ability
|
35
|
35
|
|
3
|
Reasoning Ability
|
35
|
35
|
|
Total
|
100
|
100
|
||
II] Main Examination:
Sr. No.
|
Test Name
|
No. of Questions
|
Maximum Marks
|
Duration (Time)
|
1
|
Reasoning
|
40
|
40
|
2 Hour (Composite time of 120 Minutes)
|
2
|
English Language
|
40
|
40
|
|
3
|
Quantitative Aptitude
|
40
|
40
|
|
4
|
General Awareness(Related to Banking Industry) |
40
|
40
|
|
5
|
Computer Knowledge |
40
|
40
|
|
Total
|
200
|
200
|
||
Tuesday 4 August 2015
Bank clerks exam in 2015
Apply Online
for Clerks CWE V in 2015: ibps.in has released notification to conduct
online examination for Clerks CWE V examination. Any graduate candidate can
apply online from 11-08-2015 on ibps.in site. Candidates have to fulfill the criteria
of Age Limit and Education Qualification.
Age limit: 20 years minimum age
limit and 28 years is maximum age limit as on 01-08-2015, Age relaxation is
applicable 5 years for SC/ ST, 3 years for OBC, 10 years for PWD & Others
as per rules.
Educational Qualification:
Candidate should possess Degree (Graduation) in any discipline from a recognized
University by the Government of India.
Application Fee:Rs.100/- for SC/
ST/ PWD category candidates and Rs. 600/- for all others. Payment mode is online
only through payment gateway on or before 01-09-2015.
How to Apply: Eligible
candidates may apply online through IBPS’s website www.ibps.in from 11-08-2015
to 01-09-2015. After submission of online application form, candidates have to
take print out of it for future purpose. System generated printout of the
online application form, original and self attested photocopies of all relevant
documents must be produced at the time of interview.
Very Important Dates candidates have to
remember:
Starting Date to Apply Online and to Payment of
Fee :11-08-2015
Last Date to Apply Online & Payment of Fee :01-09-2015
Download of Call Letter for Pre-Examination
Training :03-11-2015to17-11-2015
Download of Call letter for Online Preliminary Examination : 18-11-2015 onwards
Online
Examination-Preliminary :05-12-2015,
06-12-2015,12-12-2015, 13-12-2015
Declaration
of Result of Preliminary Examination : December 2015
Download of
Call letter for Online Main Examination : December 2015
Date of
Online Main Examination :02-01-2016&03-01-2016
Declaration
of Result of Main Examination : January 2016
Sunday 2 February 2014
Solved examples on Time and Distance
You have to understand some basic
concepts before we start to solve problems on Time and Distance.
*Speed Definition: The time covered
per unit time is called as speed.
Basic formula: Speed = (Distance/Time)
Manipulation of
formula: Time = (Distance/Speed)
Distance
= (Speed * Time)
*Unit Conversion: If you want to convert speed from
KMPH (Kilometres Per Hour) into m/s (metres per second) follow
some basic tricks.
1KMPH = (1*1000)/(60*60) metres/seconds
=(1000/3600)
m/s
=10/36 m/s
=5/18
m/s
so 1 KMPH is 5/18 m/s.
so 1 KMPH is 5/18 m/s.
Problem I] if you wants to convert speed from KMPH to m/s, multiply that number with (5/18)
i.e. 54 KMPH speed can be
converted into m/s as (54)*(5/18)
=3*5 (as
18 cancel 54)
=15
m/s
Problem II] if you want to convert speed
from m/s to KMPH, multiply that number with (18/5)
i.e. 25 m/s speed can be converted
into KMPH as 25*(18/5)
=5*18 (as
5 cancel 25)
=90KMPH
*Average Speed = (Total distance
traveled/ Total time taken)
Consider a bus traveling from A to
B of distance d with the constant speed p and same bus travel back from B to A
of same distance d with different constant speed q.
As per the speed formula Time
= (Distance/Speed)
So Total
time taken = (Time while going)+(Time while coming back)
=(d/p)+(d/q)
=d{(p+q)/(pq)}
Total
distance traveled:(d+d)=2d
By the formula
Average
Speed = (Total distance traveled/ Total timer taken)
Average
Speed =
(2d/d{(p+q)/(pq)})
Average
Speed = 2pq/(p+q)
*If a bus covers part of there
journey at a constant speed p and remaining part of journey at
a different constant speed q, and the distance traveled of the two
part is in the ratio as m:n, then the Average Speed for
entire journey is given as below
Average
Speed = {(m+n)*(pq)}/(mq+np)
Previously asked in Bank
Examinations, most important and Solved questions on Time and Distance
1] Express 72 KMPH in metres/second?
Sol: 72
KMPH
=
72*(5/18)
=
4*5 m/s
=20
m/s
2] Express 35 m/s in KMPH?
Sol: 35
m/s
=35*(18/5)
=7*18
=126
KMPH
3] A car can travel 350KM in 4 hours.
If its speed is decreased by 12.5 KMPH, How much time does the can takes to
caver a distance of 450KM?
Sol: Speed of
the car
Speed =
(Distance/Time)
Speed =(350/4)
Speed
=87.5 KMPH
Now speed of the is
reduced by 12.5
So new speed of the becomes
(87.5-12.5) = 75 KMPH
If car travels with new speed of
75 KMPH, the time taken to cover distance of 450 is given by
Time = (Distance/ Speed)
Time = (450/75)
Time = 6 Hours
So if car travels with speed 75
KMPH, 6 hours will be required to cover 450 KM of distance.
4] A bus can travel a certain
distance travelling at a speed 60 KMPH and return with the speed 40 KMPH to the
starting point. Find the average speed for entire journey?
Sol: Average
Speed = (Total distance traveled/ Total timer taken)
Average
Speed = 2pq/(p+q)
Here
no distance is mentioned, so do not about distance, only place the values of p
and q
Where
p is speed while going
And q
is speed while coming back to the original point.
So,
Average
Speed = (2*60*40)/(60+40)
=(4800)/100
=48
KMPH
Average
Speed of the bus is 48 KMPH
5] A boy wins 200m race in 24
seconds, then his speed is?
Sol: Speed
= (Distance/Time)
Speed = (200/24)
Speed
= 30 m/s
Speed of the is 30 metres per
second.
6] If a train cover distance of
350km in one hour, What is the speed of the train in m/s?
Sol: Speed = (Distance/Time)
Speed = (350km/1hr)
Speed = 350 KMPH
To convert
this speed from KMPH to m/s multiply it with 5/18
Speed in m/s = 350*(5/18)
= 97.22 m/s
7] A bus can travel a certain
distance travelling at average speed 48 KMPH and it takes 3 hour to come back
at starting point. Find the distance covered
by bus in one side?
Sol: Let total distance covered by bus be x;
one
side distance covered by bus = x/2;
Average Speed = (Total distance
traveled/ Total timer taken)
48 = x/3;
X=144
Here,
total distance covered by bus is 144 KM;
so,
one
side distance covered by bus = x/2;
=144/2
one side distance covered by bus is 72 KM.
Thursday 30 January 2014
computer storage devices and memory units
Area where we store Data or Information is called as computer data
storage, in other way we can call it as storage
or memory, this is a main functionality
of computer. All type of manipulations and computations performed in the central
processing unit (CPU) of a computer. As per the Von Neumann architecture, the
CPU is made up of two main parts: control unit(CU) and arithmetic logic unit
(ALU). The control unit controls the flow of data between the CPU and memory,
while ALU performing any arithmetic or logical operations on data. The most
common unit of storage is the byte, and 1 byte is equal to 8 bits. 1 bit is
nothing but alphabet, numbers, pictures pixel, audio, video, or any other form
of data or information can be converted into a string of bit, or binary number
form, which is a value of 1 or 0. Below table will specify the units of memory
storage.
|
Sr. No.
|
Unit
|
Description
|
|
1
|
1 Bit
|
A binary digits as 0 & 1
|
|
2
|
1 Nibble
|
A group of 4 bits
|
|
3
|
1 Byte
|
A group of 8 bits, and 2 nibbles
|
|
4
|
1 Word
|
It is group of byte, which varies from computer to computer, but is
fixed for each computer
|
|
5
|
1 Kilobyte (KB)
|
1024 bytes
|
|
6
|
1 Megabyte (MB)
|
1024 KB
|
|
7
|
1 Gigabyte (GB)
|
1024 MB
|
|
8
|
1 Terabyte (TB)
|
1024 GB
|
|
9
|
1 Petabyte (PB)
|
1024 TB
|
Memory storage types:
Memory storage is again divided in three sections, which are
primary storage, secondary storage and Tertiary storage.
I]Primary storage:
CPU has direct access to Primary storage or it is referred simply as main memory
or internal memory. As CPU is directly connected with main memory by memory bus
so CPU continuously read the instruction from the memory and stored data to the
main memory. This is volatile storage means it lose the data when there is
power supply interrupt at any time. i.e. RAM(Random Access Memory)
II]Secondary storage:
Secondary storage is known as external memory or auxiliary storage, which is differs
in functionality from primary storage as it is not directly accessible by the
CPU. There is not direct connection from CPU. The computer usually uses its input/output
channels to access secondary storage. Secondary storage does not lose the data
when there is power supply interrupt at any time, so it is known as non-volatile
storage. i.e. ROM(Read Only Memory), CD or DVD
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