JSSC Kakshpal Mains admit card Jharkhand warder download exam call letter 2016

JSSC Kakshpal Mains admit card Jharkhand warder download exam call letter 2016

Qualified in JSSC Kakshpal Prelims??? Then what’s next, yes hope your preparation is going well for next round of selection process i.e. Main written exam. As exam going to be held on few days, make yourself ready to download your hall ticket at any time and keep on checking the website for Hall ticket updates.

Well- known JSSC

The Jharkhand Staff Selection Commission is formerly known as JSSC. It had been constituted by the Jharkhand Staff Selection Commission Act 2008. It is the State government organization and works under the Staff Selection Commission of Jharkhand. JSSC is the authorized body that governed by the government of India to hire the candidates or staff for various posts as per their requirement. JSSC organization will frequently release the notification for different sectors and complete the selection process as per the schedule. 1394 Kakshpal vacancies are available earlier in JSSC website. Interested candidates appeared and few succeed in preliminary written exam can now download their call letter from official link for their Main written exam.

Kakshpal Question Pattern

Check out whether your preparation are based on question paper pattern as follows, alike Prelims exam you will also get main exam with Objective typed Multiple choice questions. It consists of two papers such as Paper I and Paper II. The Paper I will covers General Hindi, General Studies and Numerical ability. It comes under total of 100 marks with time duration of 2hours. The Paper II questions will be asked from local language for 100marks. It also take place for 2 hours as time duration. The complete syllabus for Kakshpal can be checked on their official website. Let us check how prepared are you.

1. Which one of the following is not a prime number?

1. 31  
2. 61
3. 71
4. 91

Answer: Option D

Explanation:  91 is divisible by 7. So, it is not a prime number.

1. It is being given that (232 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?

1. 216 + 1
2. 216 – 1
3. 296 + 1
4. 7x 223

Answer: Option C

Explanation:  Let 232 =x , Then it will be x+1

This ( x+1 ) be completely divisible by natural number N, then

296 + 1 = [(232 )3 + 1]

= x3 +1 = (x+1)( x2 – x+1) , thus (x+1) is divisible by N.

1. What least number must be added to 1056, so that the sum is completely divisible by 23 ?

1. 2
2. 3
3. 18
4. 21

Answer: Option A

Explanation:  If the number 1056 is completely divisible by 23 means, remainder should come zero.

But if we divide 1056 by 23, the remainder is 2.

So if 2 is added to the 1056, we get remainder 0.

Therefore solution is 2

Can download now 

The call letter can be downloaded from their official link from 28th June 2016 onwards. Go to official link given below, find the correct link to download call letter for JSSC Kakshpal Mains written exam 2016 on homepage of newsfeed or recent notification. Thus it will ask for some particular details to be entered. Just give your details and click on the Submit button to download your call letter. The Call letter will be downloaded in form of pdf format, where you can’t make changes. Take print out, which is mandatory and need to be carried out to enter the exam hall which is 10th July 2016. Put your effort surely you can crack this exam. This exam is followed by Physical Test and Medical Examination. U have to complete all these level to became JSSC kakshpal.

Important Dates

Exam date : 10th July 2016

Download call letter: 28th June 2016 onwards.

Previous important dates for your info, Prelims exam date: 20th March 2016.

Official Link: >>> Click here <<<

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