Name: SSC KKR
Post Name :
| Sr.No. | Post Name | No. of Posts |
| 1 | Senior Technical Assistant | 32 |
| 2 | Assistant Epigraphist (Dravidian Inscriptions) | 01 |
| 3 | Data Entry Operator | 02 |
| 4 | Technical Clerk | 03 |
No.Of Posts :38
Qualification :
| Post Name | Qualifications |
| Senior Technical Assistant | Diploma in Civil Engineering with Two years certificate course of Draftsman from ITI. |
| Assistant Epigraphist (Dravidian Inscriptions) | Master Degree in Telugu/Tamil/Malayalam/Kannada language orMaster Degree in History with Ancient Indian History. |
| Data Entry Operator | Graduation in any discipline and 8000 key depression /Hour of Data Entry Work. |
| Technical Clerk | 03 |
Age Limit :
| Post Name | Age Limit |
| Senior Technical Assistant | 30 years |
| Assistant Epigraphist (Dravidian Inscriptions) | 30 years |
| Data Entry Operator | 18-25 years |
| Technical Clerk | 18-27 years |
Pay Scale :
| Post Name | Pay Scale | Grade pay |
| Senior Technical Assistant | Rs.9300-34800/- | Rs.4200/- |
| Assistant Epigraphist (Dravidian Inscriptions) | Rs.9300-34800/- | Rs.4200/- |
| Data Entry Operator | Rs.5200-20200/- | Rs.2800/- |
| Technical Clerk | Rs.5200-20200/- | Rs.2400/- |
Selection Process :Selection made will be Education qualifications, Proficiency Test & Interview marks merit.
Application Fee :
General/OBC Categories – Rs.50/-
SC/ST/PWD/Women/Ex.Serviceman candidates – Nil
Payment Mode – Central Recruitment Fee Stamp, available at any Postal Offices.
General/OBC Categories – Rs.50/-
SC/ST/PWD/Women/Ex.Serviceman candidates – Nil
Payment Mode – Central Recruitment Fee Stamp, available at any Postal Offices.
How To Apply :Eligible Applicants can download the prescribed Application form from official website or SSC Regional Offices and Dully filled application, attested copies of SSLC/HSC, Graduation, PG, Diploma, Ph.D, Experience, Caste, Disability certificates, DD slip, recent passport size color photograph send to The Regional Director (KKR), Staff Selection Commission, 1st Floor, E Wing , Kendriya Sadan, Koramangala, Bangalore-560034.
NOTE : The Notification will be coming soon and still will be available at Employment News paper up dated from 13th to 19th December 2014.Stay connected with gurujobalert.
Important Dates :
Last date for submission of filled Application : 09th January 2015
For Official Notification : Click Here
SSC CHSL Recruitment 2015
Name :SSC
1. Combined Higher Secondary Level (10+2) Examination => DEO / LDC - Various posts
No.Of Posts :Various posts
Qualification :Candidates should have done 12th Pass or its equivalent qualification from a recognized university.
Age Limit :Candidates age should be between 18 - 27 Years As On 01-08-2015. Age relaxations will be applicable as per the rules
Selection Process :All Eligible Candidates will Be Selected Based on Their Performance In Written Exam, Interview .
Application Fee:For General/OBC Candidates Application Fee is - 100/- & For All Other Candidates (PH/ST/SC) Application Fee is - Nil
How To Apply :All Eligible and Interested candidates may fill the online application through official website http://ssc.nic.in from 11-04-2015 To 10-05-2015.
Important Dates :
Advertisement Date => 11-04-2015
Date of Exam => 19-7-2015 (Sunday) / 02-08-2015 (Sunday) / 09-08-2015 (Sunday)
Last Date for Registration of Online Application Form Is: 10-05-2015
For Official Notification : Click Here
SSC Model Test Paper - Aptitude, analytical reasoning
If a = 5 + 2 √6, the value of √a - 1/√a is —
(A) 2 √2 (Ans)
(B) 2 √3
(C) 3 - √2
(D) 1 + √5
Explanation : a = 5 + 2 √6
= (√3)2 + 2 (√3) * (√2) + (√2)2
= (√3 + √2)2
∴ √a = √3 + √2
∴ √a - 1/√a = (√3 + √2) - (1/√3 + √2)
= (√3 + √2)2 - 1 / (√3 + √2) = 5 + 2 √6 - 1 / √3 + √2
= 4 + 2 √6 / √3 + √2 = 2 √2 (√2 + √3) / √3 + √2
= 2 √2
The arrangement of rational numbers -7/10, 5/-8, 2/-3 in ascending order is —
(A) -7/10, 5/-8, 2/-3
(B) -7/10, 2/-3, 5/-8 (Ans)
(C) 2/-3, 5/-8, -7/10
(D) 5/-8, -7/10, 2/-3
Explanation : -7/10 = -0.7
-5/8 = - 0.625
and 2/-3 = - 0.667
On writing these numbers in ascending order, we get
-7/10, 2/-3, and -5/8
Given that log10 2 = 0.3010, then log2 10 is equal to —
(A) 0.3010
(B) 0.6990
(C) 1000/301 (Ans)
(D) 699/301
Explanation : log2 10 = 1/log10 2
= 1/0.3010 = 1000/301
_ _
3.9 + 5.7 is equal to —
_
(A) 9.6
_
(B) 8.6
_
(C) 7.6 (Ans)
_
(D) 1.6
_ _
Explanation : 3.9 + 5.7 = - 3 + .9 - 5 + .7
_
= - 8 + 1.6 = 7.6
If 1/6.198 = 0.16134, then the value of 1/0.0006198 is —
(A) 16134
(B) 1613.4 (Ans)
(C) 0.16134
(D) 0.016134
Explanation : 1/6.198 = 0.16134
∴ 1/0.0006198 = 1/6.198 * 10-4
= 1 * 104 / 6.198 = 1/6.198 * 10,000
= 0.16134 * 10,000
= 1613.4
Square root of 0.081/0.0064 * 0.484/6.25 * 2.5/12.1 is —
(A) 0.45 (Ans)
(B) 0.75
(C) 0.95
(D) 0.99
Explanation : √0.081/0.0064 * 0.484/6.25 * 2.5/12.1
= √810/64 * 484/6250 * 25/121
= √81/64 * 484/625 * 25/121
= 9/8 * 22/25 * 5/11 = 0.45
If xy = yx, then (x/y)x/y is equal to —
(A) xx/y
(B) xx/y -1 (Ans)
(C) xy/x
(D) xy/x -1
Explanation : xy = yx
or xy/x = y
∴ x/y = x/xy/x = x1-y/x = xx-y/x
∴ (x/y)x/y = (xx-y/x)x/y = xx-y/y = xx/y-1
Which is the greatest out of the following ?
(i) 3√1.728
(ii) √3 - 1 / √3 + 1
(iii) (1/2)-2
(iv) 17/8
(A) (i)
(B) (ii)
(C) (iii) (Ans)
(D) (iv)
Explanation : (i) 3√1.728 = 1.2
(ii) √3 - 1 / √3 + 1 = (√3 - 1)(√3 + 1)
(√3 + 1)(√3 - 1)
= 3 + 1 - 2 √3
3 - 1
= 2 - √3 = 0.26
(iii) (1/2)-2 = (2/1)2 = 4
(iv) 17/8 = 2.125
∴ (iii) is the greatest.
If (a + b = 3), then what is the value of —
(a3 + b3 + 9ab) ?
(A) 18
(B) 27 (Ans)
(C) 81
(D) Cannot be determined due to insufficient data
Explanation : Given,
a + b = 3
∴ a3 + b3 = (a + b) (a2 + b2 - ab)
= (a + b) [(a + b)2 - 3ab]
= 3 [9 - 3ab]
= 27 - 9ab
∴ a3 + b3 + 9ab = 27 - 9ab + 9ab = 27
Simplify —
√[(12.1)2 - (8.1)2 ÷ [(0.25)2 + (0.25) (19.95)]
(A) 1
(B) 2
(C) 3
(D) 4 (Ans)
Explanation : √[(12.1)2 - (8.1)2 ÷ [(0.25)2 + (0.25) (19.95)]
= √[(12.1 + 8.1) (12.1 - 8.1)] ÷ [0.25 (0.25 + 19.95)]
= √(20.2 * 4) ÷ (0.25 * 20.2)
= √20.2 * 4 / 0.25 * 20.2 = √16 = 4
Simplify —
(2.3)3 - 0.027 / (2.3)2 + 0.69 + 0.09
(A) 0
(B) 1.6
(C) 2 (Ans)
(D) 3.4
On simplification of —
1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/100 we get —
(A) 2/27
(B) 1/9
(C) 5/27
(D) 6/55 (Ans)
If 0 < a < 1, then the value of a + 1/a is —
(A) Greater than 2 (Ans)
(B) Less than 2
(C) Greater than 4
(D) Less than 4
Explanation : a + 1/a - 2 = (√a - 1/√a)2 = positive
∴ a + 1/a > 2
The simplification of
2.002 + 7.9 {2.8 - 6.3 (3.6 - 1.5) + 15.6} yields —
(A) 2.002
(B) 4.2845
(C) 40.843
(D) 42.845 (Ans)
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