SBI JOBS

State Bank of India invited online applications for recruitment of 514 Specialist Officer in Associate Banks of SBI.
Interested candidates can apply online between 08/10/2012 and 28/10/2012 through bank’s website www.sbi.co.in OR www.statebankofindia.com.

Important Dates:

• Opening date for online registration: 08/10/2012

• Last date for online registration : 28/10/2012

Total No of Vacancies: 514

Details of vacancies:

Sl No.	Name of post/post code	Bank	Grade	Max. Age (as on 01/10/2012)	No of Vacancies
1	Asst. General manager (Chief Economist)/ AGEC	SBBJ	SMGS-V	40	1
2	Manager (Economist)/ MGEC	SBI	MMGS-III	35	2
3	Chief Manager (CA)/CMCA	SBBJ	SMGS-IV	35	1
4	Manager (CA)/MGCA	SBM, SBBJ	MMGS-III	30	(1+1) 2
5	Manager (Hardware)/ MGHW	SBH	MMGS-III	40	2
6	Dy. Manager (Hardware)/ DMHW	SBBJ, SBH, SBP	MMGS-II	35	(10+4+2) 16
7	Asst. Manager (Hardware) / AMHW	SBP	JMGS-I	30	9
8	Manager (Networking)/ MGNW	SBH	MMGS-III	40	1
9	Dy. Manager (Networking)/ DMNW	SBH	MMGS-II	35	3
10	Dy. Manager (Systems)/ DMSY	SBBJ,SBP, SBT	MMGS-II	30	(10+8+7) 25
11	Asst. Manager (Systems)/ AMSY	SBBJ,SBH,SBP, SBM,SBT	JMGS-I	30	(80+84+49+ 14+4) 231
12	Asst. Manager (Comp. & Comm.) /AMCC	SBI	JMGS-I	30	5
13	Dy. Manager (Civil Engg.)/ DMCE	SBBJ,SBH,SBM,SBT	MMGS-II	35	(8+2+1+2) 13
14	Asst. Manager (Civil Engg.)/ AMCE	SBI, SBM,SBP,SBT	JMGS-I	30	(20+3+6+3) 32
15	Dy. Manager (Electrical Engg.)/ DMEE	SBH,SBM	MMGS-II	33	(3+1) 4
16	Asst. Manager (Electrical Engg.)/ AMEE	SBI,SBP,SBT	JMGS-I	30	(17+1+3) 21
17	Dy. Manager (Official Lang.)/ DMOL	SBM,SBP	MMGS-II	30	(3+5) 8
18	Manager (Official Lang.)/MGOL	SBM,SBP	MMGS-III	35	2
19	Asst. Manager (Official Lang.)/ AMOL	SBT	JMGS-I	30	1
20	Manager (Law)/MGLA	SBI	MMGS-III	35	42
21	Dy. Manager (Law)/ DMLA	SBH,SBP	MMGS-II	35	(1+3) 4
22	Asst. Manager (Law)/ AMLA	SBI, SBH, SBT	JMGS-I	30	(42+23+13) 78
23	Dy. Manager (Security)/ DMSE	SBH, SBT	MMGS-II	35	(6+4) 10
24	Dy. Manager (Fire)/ DMFS	SBBJ	MMGS-II	35	1

[Minimum age limit is 21 years, i.e. those born on or after 01/10/1991 are not eligible to apply.]
[A candidate can apply for only one post.]

Pay Scale

• JMGS-I: Rs.14500-600/7-18700-700/2 20100-800/7-25700
• MMGS-II:  Rs.19,400-700/1-20, 100-800/10-28100
• MMGS-III: Rs. 25700-800/5-29, 700-900/2-31500
• SMGS-IV: Rs. 30600-900/4-34, 200-1000/2-36200
• SMGS-V: Rs. 36200-1000/2-38200-1100/2-40400

Educational Qualification: Candidates must possess the prescribed qualification and relevant full time post qualification experience for each post as on 01/10/2012.

Application Fee and Intimation Charges: (Non-Refundable)

Category                  Fee         Intimation Charges        Total

a. SC/ST/PWD/XS      NIL             Rs. 50                      Rs.50
b. Others                 Rs.450         Rs. 50                      Rs.500

Probation: 

1) For JMGS-I & MMGS-II posts:    Two years
2) For remaining posts:                One year

SELECTION PROCEDURE:

The selection of the candidates will be on the basis of either
(i) Written tests and interviews         OR
(ii) Short-listing and interview.
Written test (WT) will be held on-line. If the number of applications is less, Bank reserves the right to consider selection of the candidate(s) through short listing and interview, instead of written test & interview.

How to Apply:

Candidates should first register themselves online through Bank's websitewww.statebankofindia.com / www.sbi.co.in after which they should pay the requisite application fee through one of the following mode-

(i) Offline Mode: Through a system generated cash voucher/challan form pre-printed with the candidate's details to be presented at any State Bank of India counter with the requisite fee. On payment of the requisite fee through this challan form, registration of the online application will be complete.

Fee payment through System generated fee payment challan will be accepted from 2nd working day after registration and can be made within three working days at any branch of State Bank of India.

(ii) Online mode: Using debit card/credit card/Internet Banking. Transaction charges for online payment, if any, will be borne by the candidates.

Application Procedure: Visit Bank’s website www.statebankofindia.com or www.sbi.co.in and open the appropriate Online Application Format, available in the 'Recruitment link'. Fill the application carefully and submit the data.
In case, the candidate is not able to fill the data in one go, he can save the entered data and note down the provisional registration number and password generated by the system. Candidate can re-open the saved data using registration number and password and edit the particulars, if needed. This facility will be available for three times only. 

For offline fee payment: Once the application is filled in completely, candidate should submit the data and take a printout of the system generated fee payment challan immediately and deposit the fee as mentioned above.

For online fee payment: candidates are required to pay fees through the payment gateway integrated with the application, after completely and correctly filling in the application form, and then submitting the data.
On successful completion of the transaction, e-receipt and application form will be generated by the system. Take a print out of both the documents for record.

Call Letter: The candidates should download their call letter and an “acquaint yourself” booklet by entering their registration number and password / date of birth, after 19/11/2012 from the Bank’s website. Call letter/acquaint yourself booklet will be not sent by post

NOTE: Candidates should have valid active email ID for getting call letter/interview advices etc. by email.

For detailed notification, Click Here

The sum of 1/4,1/9,1/16....1/n2 is :
(1)	less than 1	(2)	greater than 1
(3)	equal to 1 or less than 1	(4)	equal to 1/n2 (n + 2)
The right answer is Option (1).
¼ + 1/9 + 1/16 +…+ 1/n2 Consider ¼: ¼ = 1/(2 x 2) < 1/(1 x 2) because ½ < 1 Similarly, 1/9 = 1/(3 x 3) < 1/(2 x 3) Since 1/3 < ½ 1/16 = 1/(4 x 4) < 1/(3 x 4) Because ¼ < 1/3 …………………………………………………….. …………………………………………………….. 1/n2 = 1/(nxn) < 1/(n – 1)n Since 1/n < 1/(n – 1) ¼ + 1/9 + …+ 1/n2 < 1/(1 x 2) + 1/(2 x 3) + … + 1/[(n – 1)n] i.e. ¼ + 1/9 + 1/16 + … + 1/n2 < (1/1 – ½) + (1/2 – 1/3) +…+ [1/(n – 1) – 1/n] i.e. ¼ + 1/9 + 1/16 +… + 1/n2 < 1/1 – 1/n which is less than 1. Hence, the given sum is less than 1 only. Answer: (1) What is the digit in the unit place of  3946252 + 734733 ? (1) 8 (2)  7 (3) 6 (4) 4 The right answer is Option (1). 3946252 + 734733 First we will find the unit digit of 3946252. Divide 46, by 4, we get 2 as remainder. Actual remainder (2)252 Again dividing (2)252  by 4, we get 0 as remainder. And ‘9’ is an odd digit number. So, the last digit of 3946252 will be 1. Last digit of 734733 Þ By dividing 47 by 4, we get 3 or – 1 as a remainder. Now, (– 1) odd, when divided by 4 will give – 1 or 3 as remainder. So, the last digit of 734733 will be. Last digit of 3 ´ 3 ´ 3 = i.e. 7. So, the last digit is 1 + 7 = 8. Answer: (1) The expression 333555 + 555333 is divisible by: (1) 2 (2)  3 (3) 37 (4) All of these The right answer is Option (4). 333555 + 555333 Check it by option. The last digit of 333555 will be odd, and the last digit of 555333 will be odd. odd + odd Þ even So, the last digit is even, which is divisible by 2. Both the numbers are individually divisible by 3. So 333555 + 555333 is divisible by 3. And similarly, both the numbers are divisible by 37, so, 333555 + 555333 is divisible by 37. Hence this expression is divisible by all three numbers.   Answer : (4) Note : If the number consists of same in the multiples of 3, then it is always a multiple of 37 � l ; p� �ߡ ight:150%;font-family:"Arial","sans-serif";color:#333333'> So, the last digit is 1 + 7 = 8. Answer: (1)

divisiable rule

Number	Method	Example
7	Subtract 2 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 7, the original number is also divisible by 7	Check for 945: : 94-(2*5)=84. Since 84 is divisible by 7, the original no. 945 is also divisible
13	Add 4 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 13, the original number is also divisible by 13	Check for 3146:: 314+ (4*6) = 338:: 33+(4*8) = 65. Since 65 is divisible by 13, the original no. 3146 is also divisible
17	Subtract 5 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 17, the original number is also divisible by 17	Check for 2278:: 227-(5*8)=187. Since 187 is divisible by 17, the original number 2278 is also divisible.
19	Add 2 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 19, the original number is also divisible by 19	Check for 11343:: 1134+(2*3)= 1140. (Ignore the 0):: 11+(2*4) = 19. Since 19 is divisible by 19, original no. 11343 is also divisible
23	Add 7 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 23, the original number is also divisible by 23	Check for 53935:: 5393+(7*5) = 5428 :: 542+(7*8)= 598:: 59+ (7*8)=115, which is 5 times 23. Hence 53935 is divisible by 23
29	Add 3 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 29, the original number is also divisible by 29	Check for 12528:: 1252+(3*8)= 1276 :: 127+(3*6)= 145:: 14+ (3*5)=29, which is divisible by 29. So 12528 is divisible by 29
31	Subtract 3 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 31, the original number is also divisible by 31	Check for 49507:: 4950-(3*7)=4929 :: 492-(3*9) :: 465:: 46-(3*5)=31. Hence 49507 is divisible by 31
37	Subtract 11 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 37, the original number is also divisible by 37	Check for 11026:: 1102 - (11*6) =1036. Since 103 - (11*6) =37 is divisible by 37. Hence 11026 is divisible by 37
41	Subtract 4 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 41, the original number is also divisible by 41	Check for 14145:: 1414 - (4*5) =1394. Since 139 - (4*4) =123 is divisible by 41. Hence 14145 is divisible by 41
43	Add 13 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 43, the original number is also divisible by 43.*This process becomes difficult for most of the people because of multiplication with 13.	Check for 11739:: 1173+(13*9)= 1290:: 129 is divisible by 43. 0 is ignored. So 11739 is divisible by 43
47	Subtract 14 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 47, the original number is also divisible by 47. This too is difficult to operate for people who are not comfortable with table of 14.	Check for 45026:: 4502 - (14*6) =4418. Since 441 - (14*8) =329, which is 7 times 47. Hence 45026 is divisible by 47

Notes:

Isro's 100th mission: PSLV-C21 puts 2 foreign satellites in orbit
CHENNAI: The Indian space odyssey crossed a historic landmark on Sunday when a PolarSatellite Launch Vehicle (PSLV-C21) put in orbit two foreign satellites.

It marked the 100th space mission of the Indian Space Research Organisation (Isro) which started the journey in 1975 with the launch of its first satellite 'Aryabhata.'

Among those who watched the majestic rocket lift off from the Satish Dhawan Space Centre in Sriharikota, 100km north of Chennai, was Prime Minister Manmohan Singh.

At the end of a 51-hour countdown that started on Friday, PSLV-C21 lifted off at 9.53am. After 17 minutes and 49 seconds, it injected the first satellite, the French SPOT-6, into orbit. Soon, the second satellite, the Japenese Proiteres was put in orbit. There were, however, some anxious moments when the trajectory on the giant screens at the mission control room showed the rocket's blip deviate slightly midflight, which appeared to have been corrected later.

Scientists cheered and shook hands as the mission was pronounced a success and the Prime Minister congratulated the scientists. "India is justly proud of its space scientists who have overcome many odds to develop our own technology for space missions," he said.

It was PSLV's 21st consecutively successful flight. PSLV-C21 is India's 38th satellite launch vehicle to lift off from Sriharikota. India has so far put in orbit 62 Indian satellites. The two numbers add up the century that Isro celebrated on Sunday.

Besides buttressing its technological mastery, Isro, with its latest launch, has cemented its place in among the space-faring nations as a sought-after commercial launcher. Isro's commercial arm Antrix Corporation has received several requests from foreign countries to launch their satellites for a price.