IndianOil Educational Scholarship Scheme 2013 - Engineering, MBBS & MBA Courses

IndianOil Educational Scholarship Scheme 2013 -Last Date:10 Sep 2013-IndianOil offers 2600 Scholarships for Students of 10+/ITI, Engineering, MBBS & MBA Courses on merit.

---

To Apply this -Click here 

Source:http://www.iocl.com/Aboutus/Scholarships.aspx

IndianOil Educational Scholarship Scheme 2013

IndianOil offers 2600 Scholarships for Students of 10+/ITI, Engineering, MBBS & MBA Courses on merit-cum-means basis as under:

Sl.No.	Stream	Duration for which Scholarship will be awarded and minimum duration of course	No. of Scholarships	Qualifying Examination	Minimum Eligibility marks in Qualifying Exam			Scholarship amount per month
					GEN	SC/ST/OBC/Girls	PwDs
1	10+/ITI	2 years	2000	10th	65%	60%	50%	Rs. 1000
2	Engineering	4 years	300	12th	65%	60%	50%	Rs. 3000
3	MBBS	4 years	200	12th	65%	60%	50%	Rs. 3000
4	MBA	2 years	100	Graduation	65%	60%	50%	Rs. 3000

Abbreviations: SC: Scheduled Caste, ST: Scheduled Tribe, OBC: Other Backward Class, GEN: General, PwDs: Persons with Disabilities.

SCHOLARSHIPS OFFERED :

1."IndianOil Merit Scholarships Scheme" for 10+/ITI studies

2000 Scholarships are offered for 10+/ITI students. The equitable distribution of scholarships will be State-wise, based on “Number of students who appeared for 10th Standard examination during 2012-13”.

2. "IndianOil Scholarships Scheme" for Graduate and Post-Graduate studies

600 scholarships are offered to students pursuing professional courses (Engineering–300 Nos., MBBS-200 Nos. & MBA-100 Nos.). The scholarships will be equitably distributed Zone-wise viz. North, South, East & West, based on “Number of students who appeared for 12th Standard examination during 2012-13”.

ELIGIBILITY: Students pursuing full time/regular courses (not correspondence/distance/online mode) in these streams & studying in Schools / Colleges / Institutions / Universities recognized by MCI / AICTE / State Education Boards / State Govt. / ICSE / ISC / CBSE / Central Govt. / Association of Indian universities, shall be eligible to apply for the different streams as under :

1. 10+/ITI :

(For 10+) – Bonafide students (full time / regular) in 11th standard / Ist Year of Intermediate course or equivalent course in 2013-14 are eligible to apply.

(For ITI) - Bonafide students (full time / regular) in 1st year of 2 years ITI course in 2013-14 are eligible to apply.

2. Engineering & MBBS : Bonafide students (full time/regular) in the first year of Engineering/MBBS degree courses in 2013-14 (minimum 4 years course duration) are eligible to apply.

3. MBA : Bonafide students (full time/regular) in the first year of 2 years post graduate course in Business Administration/Management in 2013-14 are eligible to apply.

The marks obtained in the qualifying examination making the students eligible to seek admission in the respective first year of these courses will be considered for selection. Minimum eligibility percentage of marks required from the students of various categories in qualifying examination is given in the table.

Percentage of marks obtained in qualifying examination shall be calculated on total aggregate marks of all subjects. In case of Graduates, percentage of marks obtained shall be calculated on total aggregate of marks of all subjects for all the academic years of graduation.

Normalization of marks for all courses shall be done by using percentile method so as to rationalize the variation of marks provided by different Boards and Universities/Institutions.

INCOME : Gross Joint income of the family of the student from all sources during financial year 2012-13 should not exceed Rs. 1,00,000/- (Rupees One Lakh only) per annum. However, in case of tie, preference will be given to the students whose family income (gross joint income from all sources in 2012-13) is upto Rs. 60,000/- (Rs. Sixty thousand only). Income certificate should be got issued by appropriate civil authorities of the State/UTs/Central Government only, details of which have been specified in General Instructions.Any other Affidavit submitted in support of Income Proof shall not be accepted.

AGE LIMIT:Minimum 15 years and maximum 30 years as on 01.07.2013. Persons born between 01.07.1983 and 01.07.1998 (both days inclusive) are eligible to apply. However, relaxation in the upper age limit by 3 years for OBC candidates, 5 years for SC/ST candidates and 10 years for Persons with Disabilities (PwDs) is permissible

RESERVATION:49.5% scholarships are reserved i.e. SC(15%), ST(7.5%) & OBC (27%).Only those listed in the Central Govt’s OBC list will be considered against OBC quota. OBC candidates of State list but not covered by Central Govt’s OBC list may apply as general candidates. 25% of scholarships for girl students and 10% of scholarships for persons with Disabilities (PwDs) (minimum 40% disability as per definition under Disabilities Act 1952) are reserved in each category/sub-category. The above reservations shall be maintained on overall All India Basis.

APPLICATION:Students meeting above eligibility criteria can apply online by logging on to http://ioclscholar.sifyitest.com or athttp://www.iocl.com/Aboutus/Scholarships.aspx. This site will be kept open to receive online applications from 05.07.2013 to 30.09.2013. Once the online application is submitted by an applicant, the system will generate a unique Registration Number. The applicants should take a print out of the same for future reference/correspondence till the Scholarships are awarded.

SUBMISSION OF DOCUMENTS:All the applicants are required to submit the documents along with a print out of online application form duly signed by the applicants for scrutiny/verification. The detail of such forms/documents to be submitted is given in the above website.

If the detail provided in the online application is found to be incorrect & does not match with the submitted documents, the application of such applicants shall be summarily rejected and no correspondence shall be made / entertained on the subject.

Above documents in a sealed cover should reach at the following address on or before 15.10.2013 :

Sify Technologies Limited
(IOCL ESS I-TEST DIVISION)
Plot No B 7, Sector-132,
Noida-201301 (Uttar Pradesh)
(Landmark : Next to Naukri.com office)

Documents received after 15.10.2013 shall not be accepted. The applicants are advised to send their documents preferably by Speed post or Registered Post.

GENERAL INSTRUCTIONS:

1.Criterion for selection will be merit and family’s income

2.A scholarship holder under this scheme should not be receiving any other scholarship/stipend/financial aid from any other source during the period of this scholarship.

3.There will be no obligation on part of IOCL to provide employment to any Scholar at any point of time.

4.IOCL reserves the right to reject any application as well as discontinue/suspend the scholarship without giving any notice/assigning any reason.

5.Wards of the IOCL employees, its Joint Venture and Subsidiary companies and Consultant managing the Scheme are not eligible to apply.

6.Applicants will have to provide valid & active e-mail address, mobile number & Bank account no. in their online application form & keep checking their personal e Mail / Mobile SMS, website of Administrator of the Scheme http://ioclscholar.sifyitest.com as well as the web site of IOCLhttp://www.iocl.com/Aboutus/Scholarships.aspx regularly for instructions specific for them from Administrator of the Scheme/IOCL

7.IOCL will not be responsible for any postal delay or loss/damage of documents in transit or whatsoever reason.

8.Court of jurisdiction for any dispute will be at Delhi.

9.Any query/information or correspondence be made with the Administrator of the Scheme at the following address : 

Click here to Apply

Sify Technologies Limited
(IOCL ESS I-TEST DIVISION)
Plot No B 7, Sector-132,
Noida-201301 (Uttar Pradesh)
(Landmark : Next to Naukri.com office)
Phone: 9810421827 / 9810422305
e-Mail : iocless@sifyitest.com 

Scholarship

Pals Tamil e-Dictonary Free Download

PALS Tamil e-DICTIONARY

PALS e-DICTIONARY

ஆங்கிலம் - ஆங்கிலம் - தமிழ்

Pals e-Dictionary என்பது ஒரு குறுவட்டிலுள்ள ஆங்கிலம் - ஆங்கிலம் - தமிழ் அகராதி.

இந்த அகராதி சுமார் 22,000 முக்கிய சொற்களையும் மற்றும் 35,000 வழி சொற்களையும் கொண்டுள்ளது. இந்த ஒவ்வொரு சொற்களின் பொருளும் முதலில் ஆங்கிலத்திலும் பின் தமிழிலும் கொடுக்கப்பட்டுள்ளது. அனைத்து பொதுவாக பயன்படுத்தப்படும் சொற்களும் அதன் ஆங்கிலம் மற்றும் தமிழ் பொருளுடன் அதற்கான சொற்களின் கீழ் கொடுக்கப்பட்டுள்ளது. மேலும் இந்த அகராதியில் சில பயனுள்ள தகவல்களான சுருக்கங்கள், இயல்பில்லா வினைச்சொற்கள், கோணங்களின் ஒப்பீடு மற்றும் கிரேக்க எழுத்துக்கள் போன்றவை உள்ளன.

e-Dictionary ஒரு உலாவும் நிரலையும் தேடும் நிரலையும் கொண்டுள்ளது.

உலாவும் நிரல் ஒருவரை அகராதியின் பக்கங்களை புத்தக வடிவினை போல உலாவ அனுமதிக்கிறது. பக்கங்கள் முதல் இரண்டு எழுத்தில் அகரவரிசைப்படுத்தப்பட்டுள்ளது. அகராதியை உலாவும் போது ஒருவர் புல்லாங்குழல் வித்வான் ஷாஷன்க்கின் இசையைக் கேட்டுக் கேட்டே உலாவலாம்.

தேடும் நிரல் ஒரு பயனருக்கு ஒரு குறிப்பிட்ட சொல்லின் பொருளை விரைவாக தேடிப் பெற அனுமதிக்கிறது. தேடும் சொல் ஆங்கிலம் அல்லது தமிழாக இருக்கிறது. மேலும் விரிவாக தேடும் முறையில் ஒரு குறிப்பிட்ட சொல்லுக்கான அனைத்து பொருளையும் கொடுக்கிறது. அனைத்து சொற்களும் ஹைபர்லிங்க் செய்யப்பட்டுள்ளது. எனவே ஒரு சொல்லை கிளிக் செய்தே அதன் பொருளை பெறலாம்.

தமிழக அரசால் அங்கீகரிக்கப்பட்ட புதிய தமிழ்99 விசைப்பலகை மற்றும் TAM எழுத்துரு குறிமுறை தரப்படுத்தல் தமிழுக்குக் கொடுக்கப்பட்டுள்ளது.

PALS Tamil e-DICTIONARY

விண்டோஸுக்கு:

• நிறுவல் தகவல்கள்
-(73 எம்பி)
• இந்த தகவல்களை பின்பற்றி மென்பொருளை நிறுவவும் :
• Pals-Tamil-e-Dictionary.zipஐ கீழேயுள்ள இணைப்பிலிருந்து பதிவிறக்கி உங்கள் கணினியில் ஒரு அடைவில் சேமிக்கவும்.
• zip கோப்பு சேமிக்கப்பட்ட அடைவுக்குச் செல்லவும்.
• zip கோப்பினை பிரித்தெடுக்கவும்
• பால்ஸ் தமிழ் e-அகராதி கோப்புறையை திறக்கவும்
• பால்ஸ் தமிழ் e-அகராதி கோப்புறையை திறக்கவும்
• Setup.exe நிரலை அதன் சின்னத்தை இரட்டை சொடுக்குவது மூலம் இயக்கவும்.
• நிறுவலுக்கு பின் கணினியை மீண்டும் துவக்கவும்.
• பணிப்பட்டையில் Start பொத்தானை கிளிக் செய்யவும்.
• Programs -> Pals Tamil e-Dictionary -> Tamil e-Dictionary என தேர்ந்தெடுக்கவும்
• அகராதியைப் பயன்படுத்த தேவையான அனைத்து தகவல்களும் மென்பொருளிலேயே உள்ளது
•  (73 எம்பி)

பகுதிகளாக பதிவிறக்க இங்கே கிளிக் செய்யவும். அனைத்து பகுதிகளையும் பதிவிறக்கி ஒரு அடைவில் சேமிக்கவும். பதிவிறக்கிய பின் இங்கிருந்து பதிவிறக்கவும்: 1 2 3 4 5 6 7 8 (ஒவ்வொன்றும் 10 எம்பி).

- தமிழ் – ஆங்கிலம்:

Pals e-Dictionary என்பது ஒரு குறுவட்டிலுள்ள தமிழ் - தமிழ் - ஆங்கிலம் அகராதி.

இந்த அகராதில் 49000க்கும் அதிகமான சொற்கள் உள்ளது, இதிலும் ஒரு உலாவும் முறைமை மற்றும் ஒரு தேடும் முறைமையைக் கொண்டுள்ளது

உலாவும் முறைமை ஒருவரை அகராதியின் பக்கங்களை புத்தக வடிவினை போல உலாவ அனுமதிக்கிறது. பக்கங்கள் முதல் இரண்டு எழுத்தில் அகரவரிசைப்படுத்தப்பட்டுள்ளது.

தேடும் நிரல் ஒரு பயனருக்கு ஒரு குறிப்பிட்ட சொல்லின் பொருளை விரைவாக தேடிப் பெற அனுமதிக்கிறது. தேடும் சொல் ஆங்கிலம் அல்லது தமிழாக இருக்கிறது. மேலும் விரிவாக தேடும் முறையில் ஒரு குறிப்பிட்ட சொல்லுக்கான அனைத்து பொருளையும் கொடுக்கிறது. அனைத்து சொற்களும் ஹைபர்லிங்க் செய்யப்பட்டுள்ளது. எனவே ஒரு சொல்லை கிளிக் செய்தே அதன் பொருளை பெறலாம்.

தேடும் முறைமையில் இன்னொரு வசதி என்னவென்றால் வேர் சொற்களின் பொருளை ஒருவர் பெறலாம்.

ஒருவர் ஒரு முறையிலிருந்து மற்றொரு முறைமைக்கு தேவையான பொத்தானை கிளிக் செய்வது மூலம் மாற்றலாம்.

தமிழக அரசால் அங்கீகரிக்கப்பட்ட புதிய தமிழ்99 விசைப்பலகை மற்றும் TAM எழுத்துரு குறிமுறை தரப்படுத்தல் தமிழுக்குக் கொடுக்கப்பட்டுள்ளது.

பழனியப்பா பிரதர்ஸ் நியூ 25, 
பீட்டர்ஸ் ரோடு சென்னை - 600 014.தொலைபேசி: 91-44-28132863. மின்னஞ்சல்: sales@palbrothers.comஇணையதளம் :www.palbrothers.com

தமிழ் எழுத்துருக்கள் :

TTF (ட்ரூ டைப் எழுத்துரு) எழுத்துருக்கள் வின்டோஸ் 95/ வின்டோஸ் 98/ வின்டோஸ்  NT  ப்ளாட்பாரம் உள்ள கணினிகளில் மிக அதிகமாக பயன்படுத்தப்படுகின்றன. வின்டோஸ்  2000/XP/2003 மற்றும் லினக்ஸ் ப்ளாட்பாரம் உள்ள கணினிகளுக்காக OTF (ஓப்பன் டைப் எழுத்துருக்கள்) வெளியிடப்பட்டுள்ளன. இந்த முயற்சி முதல்முறையாக எடுத்துக்கொள்ளப்பட்டுள்ளது மற்றும் பல வர்ட் பிராசஸர்களால்  (உள்ளவை மற்றும் புதியவை) இந்த எழுத்துருக்களை பயன்படுத்தமுடியும்.  தமிழ்நெட்/தமிழ் 99 மற்றும் தட்டச்சு விசைப்பலகைகள் உள்ள பயனர்களுக்கு தட்டச்சு செய்வதற்கு பயன்படுத்த இதன் மூலம் பலவித எழுத்துருக்கள் கிடைக்கும்.

ஓப்பன்டைப் எழுத்துரு என்பது ஓப்பன் டைப் ஸ்பெசிபிகேஷனுக்கு ஏற்ப இருக்கும் டிஜிட்டல் எழுத்துருக்கள். ஓப்பன்டைப் எழுத்துருக்கள் சிக்கலான எழுத்துமுறைமைகளுக்கு தேவைப்பட்ட உயர்தர தட்டச்சு அம்சங்களை அளிக்கின்றது – பிராம்மி மற்றும் செமான்டிக் மொழி போன்ற ஸ்க்ரிப்டுகளுக்கு.  ஓப்பன்டைப் ஸ்பெசிபிகேஷன் எழுத்துக்களின் இணைப்புருகளுக்கு,  எழுத்துகளின் இட அமைப்பு, மாற்று எழுத்துக்கள் மற்றும் மற்ற மாற்றமைப்புகளை ஆதரிக்கிறது. ஓப்பன்டைப் எழுத்துருக்களில் இரு-பரிமாண க்லிஃப் இட அமைப்பு மற்றும் க்லிஃப் இணைப்புகளை ஆதரிக்கும் விவரங்களும் இருக்கலாம்.

.ஓப்பன்டைப் எழுத்துருக்கள் ஸ்கிரிப்ட் பிராசஸிங்க் லைபரரி மற்றும் மற்றும் ஓப்பன் டைப் லைப்ரரி சர்வீசஸ்களுடன்< (OTLS) சேர்ந்து பயன்படுத்தப்படுகிறது. OTLS என்பது உரை-பிராசஸிங் உதவி செயற்நிரல்களின் ஒரு கோவை.  வின்டோஸ் ஆபரேட்டிங் ஸிஸ்டத்தில், Uniscribe  மற்றும் OLTS  ஆபரேட்டிங் ஸிஸ்டம் லைப்ரரிகளின் பாகங்கள்.

GIST-DVOT/GIST-TMOT என்பவை சி-டாக்கின் ஜிஸ்ட் க்ரூப் தயாரித்த தேவ்நகரிதமிழ் ஸ்கிரிப்டு வகைகள். தேவ்நக்ரி ஸ்கிரிப்டை பயன்படுத்தும் மொழிகள் - ஹிந்தி, மராத்தி ஆகியவை இந்த எழுத்துருவை பயன்படுத்தலாம். தேவ்நக்ரி ஸ்கிரிப்டின் விசேஷ அம்சங்களான , ‘akhands’ (கூட்டெழுத்துக்கள்),  ‘nukta’, எழுத்துக்களை மறுவரிசைப்படுத்துவது மற்றும் இடம் அமைப்பது, கீழே வரும் வடிவங்கள், அரை வடிவங்கள் ஆகியவை இந்த எழுத்துருக்களில் தரப்பட்டுளன.  இந்த எழுத்துருவின் க்லிஃப் கோவைகள் தலைசிறந்த பலனை அளிக்குமாறு அமைக்கப்பட்டுள்ளன.

எழுத்துருக்கள் பொருளடக்கங்களை உருவாக்க மற்றும் காண்பிப்பதற்கு என்றே விசேஷமாக கலைநயத்துடன் தயாரிக்கப்பட்டுள்ளன.  எழுத்துருவின் க்லிஃப்களின் அளவு ஒரே அளவில் அமைக்கப்பட்டிருப்பதால் எழுத்துருவை படிப்பது மேம்பட்டு உள்ளது. எழுத்துருவை எம்பெட் செய்யலாம் என்பதால், இந்த எழுத்துரு மூலம் நிலையான வலைய பொருளடக்கத்தை தயாரித்து, எந்த விதமான பதிவிறக்கமும் செய்யாமலேயே உலாவியில் காண்பிக்கலாம்.

Target application areas:

பயன்படக்கூடிய இடங்கள்:

• இந்திய மொழி வலைய பொருளக்கத்தின் தயாரிப்பதில்
• இந்திய மொழியை ஆதரிக்கும் டெஸ்க்டாப் அப்ளிகேஷன்களை தயாரிப்பதில்
• யூனிகோட் ஆதாரத்தில் இந்திய மொழி தேடல் எஞ்சின்களை தயாரிப்பதில்

பல மொழி கருவிகளுக்கு இந்திய மொழி தொகுப்புகளை அதாவது கார்பஸை தயாரிப்பதில்.

Blogger Read More Option

Create 'After the jump' summaries

After the Jump is a feature which lets you create expandable post summaries in your blog posts, so longer posts appear as an intro with a link to Read More.

Creating jump breaks in your blog posts can be easily done right from the Post Editor, without the need for any HTML changes. First, decide where in the post you want to create the jump break, and place your cursor in that position:

Once your mouse cursor is placed at the jump point, simply click the Insert Jump Breaktoolbar icon:

Clicking the icon will insert a grey bar at the cursor point, illustrating where in the post your break will appear. The bar can be dragged though, so you can always re-position it after insertion.

Once you decide on the jump break's location within the post, you're ready to publish your post. After publishing, you'll notice that the Read More link is placed where you set the jump break. Clicking on the Read More link will then display the full text of the post.

If you feel like changing the Read More text to your own custom phrase, you can easily do this from the Layout tab. Click Edit on the Blog Post widget, and then change the Post page link text to whatever you'd like.

The Jump Break feature doesn't change how your post appears in your feed. You can configure post feed options by going to Settings | Basic | Other, and editing Allow Blog Feed.

HCL Time & Work Problems

Solved Time & Work Problems For HCL Aptitude Tests

Dear Reader,

Below are three problems on time and work which you can expect in HCL and other companies placement tests.

Question 1

In a factory, there are two machines A and B. A machine can produce 1000 watts power in 5 hours and by running both the machines together can produce same watts power in 3 hours. Find the time taken by B machine alone to produce same watts of power.

a) 5 hrs b) 7 hrs & 30 mins c) 6 hrs d) 6 hrs & 30 mins

Answer : b) 7 hrs & 30 mins

Solution :

A type machine can produce 1000 watts power in 5 hours.
Then A's 1 hour work = 1/5.
Both A and B can produce in 3 hours.
And 1 hour work of (A + B)'s = 1/3.
B's 1 hour work = 1/3 = 1/5 = 2/15.
Therefore, machine B can produce 1000 watts power in 15/2 hours.
i.e., 15/2 hours = 7.5 hours = 7 hours and 30 minutes.

Question 2

There is a building with 4 floors and each floor has equal area. There are three sweepers P, Q and R. P can clean 1 floor in 10 hours, Q can clean 8/5 of the floors in 40 hours and R can clean 4/3 of the floors in 13 hours. Then who will clean all the 4 floors at first?

a) R b) R & P c) Q d) Q & P

Answer : a) R

Solution :

P can sweep 1 floor in 10 hours.
Then 4 floors will be cleaned by P in = 10 x 4 = 40 hours.

Q can sweep 8/5 part in 40 hours.
Then 1 floor will be done by Q in 40/(8/5) = 25 hours.
And 4 floors will be done by Q in 25 x 4 = 100 hours.

R can clean 4/3 part of the floors in 13 hours.
Then 1 floor will be cleaned by R in 13/(4/3) = 13 x 3/4 hours.
And 4 floors will be done by R in 13 x 3/4 x 4 = 13 x 3 = 39 hours.

Thus P, Q and R takes 40, 100 and 39 hours respectively.
Hence the sweeper R will complete the work at first.

Question 3

A man and his son planned to paint their own house. The man alone can do in 20 days. He paints for 4 days and then his son completed in 8 days. Find how many days will they take working together?

a) 12 days b) 11 days c) 7 days d) 8 days

Answer : c) 7 days

Solution :

Man's 1 day work = 1/20
Then 4 days work = 4/20 = 1/5
Remaining work = 1 - 1/5 = 4/5

Now, 4/5 work is done by the son in 8 days.
Then, whole work will be done by him in (8 x 5/4) days = 10 days.
Then his 1 day work = 1/10

The man and his son's 1 day work = 1/10 + 1/10 = 3/10
They will complete the work together in 20/3 days = 6 2/3 days = 7 days(nearly).
Hence the answer is 7 days.

Infosys Sample Clock Problems

Clock Problems for infosys placement

Dear Reader,

Below are three problems based on clocks, which you can expect in infosys and other companies placement test.

Question 1

Three analog clocks were set to correct time. First one run with the exact time. Second one loss one hour per day. Third one gains one hour per day. After how many days will they show the same time?

a)180 days & 10 hours b)287 days & 12 hours c)212 days d)199 days

Answer : b) 287 days & 12 hours

Solution :

Let a day be a cycle of 24 hours.
First one completes a cycle within a day
Second one completes a cycle in 25 hours
And third one completes a cycle in 23 hours.
Now, the LCM of 23,24,25 is 23 x 24 x 25 = 13800.

Therefore, after 13,800 hours (least period) they will show the correct time together. (for 24 hours cycle)
i.e., 13,800/24 = 575
On 575th day they will show the correct time.
Since the clocks are analog (12 hours cycle), after 287 days and 12 hours (575/2 = 287.5)they will show the same time.

Question 2

Three clocks were set to true time and starts to run together, makes a beep sound at intervals of 8,10 and 12 minutes respectively. In 2 days, how many times do they will make the sound together?

a)12 b)15 c)25 d)30

Answer : c)25

Solution:

The clocks make sound at intervals of 8, 10 and 12 minutes respectively.
The LCM of 8,10 and 12 is 120.
So the clocks will make sound together after every 120 minutes.
i.e., after every 2 hours.

In 2 days(48 hours), they will make the sound together (48/2) = 24 times.
At the end of 2 day,they will make one more beep together.
Therefore, totally (24+1) 25 times beep sound is made together.
Hence the answer is 25 times.

Question 3

Three analog clocks were set to the exact time at 12 am and they run together and gain 3, 6 and 9 minutes per hour respectively. At what time, they will show the same time again?

a)2.42am b)12.02am c)1.58pm d)12.02pm

Answer : a)2.42am

Solution :

Let an hour(60 minutes) be a cycle.
The clocks complete a cycle early by 3, 6 and 9 minutes respectively.
i.e., a cycle is completed in 57, 54 and 51 minutes respectively.
LCM of 57,54 and 51 is 17442 minutes.
The clocks show the same time after 17442 minutes

To convert 17442 minutes, we have to divide by 60 :

17442/60 = 17400/60 + 42/60 = 290 hours + 42/60 hours = 290 hours and 42 minutes
(42/60 hours = 42/60 x 60 minutes = 42 minutes)

After 290 hours and 42 minutes from 12 am, an analog clock shows 2.42am. (since 290 = 24 x 12 + 2).

Hence, they show the exact time again at 2.42 am .

HCL Distance Puzzles

Solved Distance Puzzles For HCL Placement Tests

Dear Reader,

Below are three problems based on distance and time calculations.

Question 1

A boy drops a bouncing toy from 1st floor to ground floor. Each time, it bounces vertically to a height of 1/2 of the preceding bounce's height. Find the distance the toy will cross if it totally takes 6 bounces and the distance between each floor is 10m.

a)39 m b)30 m c)40 m d)15 m

Answer : b)30 m

Solution :

The distance between two floors is 10m.
The boy dropped the bouncing toy from a height of 10 m.
As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and falls back down 5 meters. This process will repeat for 5 more times.

Then:

Distance that the toy travels down = 10 + 10x1/2 + 10x 1/2 x 1/2 + ...+ 10x(1/2)⁶
= 10 + 10x(1/2)¹ + 10(1/2)² + 10(1/2)³ + 10x(1/2)⁴ + 10(1/2)⁵ + 10(1/2)⁶ which is a g.p series with 7 terms and first term = 10 common term = 1/2.

we know that, sum of n terms in a g.p series is
sum = a(1 - rⁿ)/1 - r when r is less than one
Here, distance = 10 x [1 - (1/2)⁷]/(1-1/2) = 10 x 127/128 / 1/2
= 20x127/128 = 19.84 meters.
Distance that the toy travels up = 10x(1/2)¹ + 10(1/2)² + 10(1/2)³ + 10x(1/2)⁴ + 10(1/2)⁵ + 10(1/2)⁶
= 19.84 - 10 = 9.84 meters.

Thus, the total distance is 19.84 + 9.84 = 29.68 meters = 30 meters (approximately)

Question 2

A spider starts jumping up from the bottom of a ladder with 32 steps. Each minute, it jumps 4 steps and slips back 2 steps. How much time would the spider take to reach the top of the ladder?

a)15 minutes b)14 minutes c)16 minutes d)18 minutes

Answer : a)15minutes.

Solution :

Think of it like a number line: In the first minute, the spider goes 4 steps up, and slips back 2 steps.

0...1...2...3...4...5...6
|---|---|---|---|---|---|
---------------->
........<-------

(Imagine that's going up instead of horizontal.)

The spider effectively climbs 4 steps - 2 steps = 2 steps higher after the first minute.
In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4 steps - 2 steps = 4 steps.
Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is at 28 steps height.
Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at the top of the ladder.
Hence the answer is 15 minutes.

Question 3

A man throws a ball from certain height to the ground and it covers a distance of X meters on the ground. Each time it's bounce covers 2/3 of previous distance. The ball stops bouncing after 4 bounces. Then the total distance it covers on the ground is:

a)191X/43 meters b)250X/21 meters c)172X/75 meters d)211X/81 meters

Answer : d)211X/81 meters

Solution :

This problem is slightly different from the 1st one.
Here, we have to find the distance which covered by the ball on the ground.
Given that, it covers X meters by the through.
And it takes four bounce, each bounce covers 2/3 of previous distance.
Then the distance covered in 1st bounce = (2/3)X

The distance covered in 2nd bounce = (2/3) of (2/3)X = X(2/3)²
The distance covered in 3rd bounce = (2/3) of X(2/3)² = X(2/3)³
And the distance covered in 4th bounce = (2/3) of X(2/3)³ = X(2/3)⁴

Required distance = X + (2/3)X + X(2/3)² + X(2/3)³ + X(2/3)⁴
The above series is a G.P series of 5 terms with a = X and r = 2/3
Then sum = a(1 - rⁿ)/(1 - r)
= X[1 - (2/3)⁵]/(1 - 2/3)
= X[1 - 32/243]/(1/3)
= 3X(211/243) = 211X/81
Hence the required answer is 211X/81 meters.

TCS Measurement Problems

Solved Measurement Problems TCS Placement Tests

Dear Reader,

Below are three problems dealing with parameters of circles and squares.

Question 1

Let ABCD be a square. Alex wants to draw 5 circles of equal radius 'r' with their centres on BD such that the two extreme circles touch two sides AB, BC and AD, CD of the square respectively and each middle circle touches two circles on either side. Find the ratio of r to that of BD.

a) 2 : 11.828 b) 5 : 19.214 c) 1 : 10.828 d) 1 : 12.515

Answer : c) 1 : 10.828

Solution :

As per the statements in question, Alex would have drawn a diagram as shown below.

Let the radius of each circle be r.
Let the side of the square ABCD be a.
Then Diagonal BD = AC = a x sqrt 2 (You will get this formula by applying Pythagoras theorem)
We have to find this diagonal length as follows:
Given, he had drawn 5 circles on diagonal BD.
After sketching 5 circles according to the question, we see that some gap between corner of square and extreme circles. Now, let us calculate that space:

Draw perpendicular lines from sides AB and BC to extreme circle. Now you get a small square of side r and of diagonal r x sqrt2.
In the below diagram, Here, Y is the centre of left extreme circle, BXYZ is a small square with side r and diagonal r sqrt2.
Similar observation can be made on the other extreme circle as well.

Summing up our observations in the main diagram, we get,

Hence the diagonal of big square = r sqrt2 + r + 6r + r + r sqrt2 = 8r + 2r sqrt2
Then the required ratio = r : 8r + 2r sqrt2 = r : r(8 + 2sqrt2)
We can remove r on both sides as it is common. Therefore, we get,
= 1 : 8+2 sqrt2
= 1 : 8+2(1.414) = 1:10.828
Hence the required ratio is 1 : 10.828.

Question 2

In an exam, students were asked to find the perimeter of a square which contains three circles such that their centres on the diagonal of the square, the middle circle touches two extreme circles on either side and two extreme circles touch two sides of the square and the radius of the three are equal. And at what times the radius equals the perimeter?

a) 19.312 b) 12.312 c) 21.312 d) 17.312

Answer : a) 19.312

Solution:

Let the radius of circle be r.
Let the side of the square be a.
Then Diagonal = a(sqrt2)
By applying a similar logic as that of problem 1, we will get a diagram as follows.

a(sqrt2) = diagonal = r sqrt2 + r + 2r + r + r sqrt2 = 4r + 2r sqrt2
a = (4r + 2r sqrt2) / sqrt2
a = 4r/sqrt2 + 2r sqrt2 / sqrt2
a = (2 x 2r) / sqrt2 + 2r

But we know, 2/sqrt2 = sqrt2. Therefore, above equation becomes,
a = 2r sqrt2 + 2r = 2r(sqrt2+1)
Then perimeter of the square = 4a = 4 x 2r(sqrt2+1) = 8r(sqrt2+1)
= 8r(1.414+1) = 8r(2.414) = r(19.312)
Hence 19.312 time radius equals the perimeter of the square.

Question 3

Find the length of diagonal BD of a square ABCD when 7 circles of radius 2 cm are located in ABCD such that their centres on BD, two extreme circles touch two sides of ABCD and each middle circle touches two circles on either side.

a)32.98cm b)29.65cm c)31.12cm d)28.56cm

Answer : b)29.65cm

Solution :

Given that, ABCD is a square where AB, BC, CD and AD are sides of square and BD, AC are diagonals of the square.
Applying similar logic to that of first and second questions, we will get the below diagram corresponding to this question :

The length of the diagonal BD = r sqrt2 + r + 10r + r + r sqrt2
= 2 rsqrt2 + 12r
Given that r = 2 cm
Then BD = 2(2)sqrt2 + 12(2) = 4 sqrt2 + 24 = 4(1.414) + 24 = 29.65 cm.
Hence the required answer is 29.65cm

Simplification Problems HCL Aptitude Tests

Simplification Problems To Prepare For HCL Aptitude Tests

Dear Reader,

Below are three fraction based simplification problems which you can expect in HCL and all other companies aptitude tests.

Question 1

A man spent 5/16 of his age plus two as student, 1/40 plus 1 as a husband, 1/4 as a good politician and 3/40 as a father. And the remaining 6 years as a good grand father. Then the living days of the man is:

a) 72 b) 89 c) 80 d) 69

Answer : c)80

Solution :

Lets say his age is X.

So years spent as student = 5X / 16 + 2,
as Husband = X/4 + 1,
as Politician = X/4,
and as Father = 3X/40.

Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + X/4 + 3X/40}
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40}
6 = X - [13X / 16 + 3 + 3X / 40]
6 = X - [71X / 80 + 3]
9X/80 = 9
X = 80.
Hence the age of the man is 80 years.

Question 2

In a village, every weekend, 1/4 th of men and 1/6 th of women participate in the social activity. If the total number of participants is 75, out of them 25 are men then the total number of men and women in the village is:

a) 100 b) 200 c) 300 d) 400

Answer : d)400

Solution :

Number of men who participated in social activities = 25
Therefore, 1/4 th of the men = 25
Or, total number of men in the village = 100

Number of women who participated in social activities = 75 - 25 = 50
Therefore, 1/6 th of the women = 50
Or, total number of women in the village = 300
Hence the required answer is 300 + 100 = 400.

Question 3

A father has divided his properties in such a way that one-half of total properties goes to A, two-third of the remaining shared equally to B, C and D and the rest to E, F, G and H such they equally gets Rs.30,000 by their sharing property. Then the amount will D get is:

a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000

Answer : a) Rs.80,000

Solution :

A's share = 1/2
Then remaining = 1 - 1/2 = 1/2 ...(A)

2/3 of this remaining 1/2 (as in eq A) goes to B,C and D.
Total share of B, C and D = 2/3 of 1/2 = 1/3

Therefore, now remaining property = Total share of E,F,G and H
= value of (A) - total share of B,C and D
= 1/2 - 1/3 = 1/6

Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 1/24
Given 1/24 = Rs.30,000
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000
Total share of B, C and D = 1/3 x 7,20,000
Then their individual share amount = (1/3 x 7,20,000) / 3 = Rs.80,000.
Hence D's share is Rs.80000.

Java The Complete Reference Seventh Edition

Java The Complete Reference - 7th Edition
Herbert Schildt | 2006 | 1024 Pages | ISBN: 0072263857 | PDF | 6.67 MB

Download :Java Complt Refe7thEd (6.67 MB)

Description: In this comprehensive resource, top-selling programming author Herbert Schildt shows you everything you need to develop, compile, debug, and run Java programs. This expert guide has been updated for Java Platform Standard Edition 6 (Java SE 6) and offers complete coverage of the Java language, its syntax, keywords, and fundamental programming principles. 

You'll also find information on Java's key API libraries, learn to create applets and servlets, and use JavaBeans. Herb has even included expanded coverage of Swing--the toolkit that defines the look and feel of the modern Java GUI. Essential for every Java programmer, this lasting resource features the clear, crisp, uncompromising style that has made Herb the choice of millions of programmers worldwide. 

Download Link : Java Complete Reference7 Edition(6.67 MB)

Note : Share this post to your Friends & Make them also Get benefit. Give your            valuable Comments Below... Thank you !.

how to change label names in blogger

Change the name of a label

Let's say you have a number of posts with the label Hawaii but you've decided that you'd rather call it Vacation instead. You can't edit the name of a label directly, but there's a simple workaround to accomplish your goal:

1. Go to the Posts tab for your blog.
2. Click on the "All labels" drop-down on top-right and select "Hawaii." This will filter the posts, so you'll only see those with the "Hawaii" label.

3. Click the check box at the top of the list of posts to select all the posts.

4. Then, click the Label icon and select "Hawaii" to remove the label "Hawaii" from all the posts that are selected.

5. While the posts are still selected, click the Label icon again and select "Add new label." Then, type in "Vacation."

6. You're done! Now, all the posts that used to have the label "Hawaii" have the label "Vacation" instead.

IBPS Model Question Papers

IBPS-Dailythanthi part - 1 (Clerk Exam )

IBPS Clerk Exam Model Question Paper & Study Materials || IBPS Model Questions And Study Materials Free Download
* Answers in the bold Letters.





















Note : Please share this post  to know your friends. Thank you.

IBPS Clerk Exam Model Question Paper

IBPS-Dailythanthi part - 2 (Clerk Exam )

IBPS Clerk Exam Model Question Paper & Study Materials || IBPS Model Questions And Study Materials Free Download
* Answers in the bold Letters.





















Note : Please share this post  to know your friends. Thank you.

VAO Model Question Paper 2011

Tamilnadu tnpsc study materials . This study materials based on the VAO exams. Also you can study all other group exams. It is very helpful for tnpsc exam prepare students.

 VAO Model Question Paper 2011.

Download Link(12 MB)

Download Link (12 MB)

Please share this post  to know your friends. Thank you friends.

Tnpsc group-4 model question papers

Dinakaran news paper prepared TNPSC Group-IV 2013 study materials.

Download Link (6.12 MB)

This study materials are very useful to tnpsc group-IV VAO exams, Also it helps to other TNPSC group exams.

Download Link(6.12 MB)

Please share this post  to know your friends. Thank you friends.

Tnpsc , TET Exam Tamil Study Materials free Download

TNPSC, TED exam Tamil study materials free download, this study materials are prepared from,
 6th standard to 12th standard, samacheerkalvi 
text books.

This study materials contains 264 pages.

Download Pdf (3.33 Mb)

Download Pdf (3.33 Mb)

Please share this post  to know your friends. Thank you friends.

Infosys Aptitude Questions

3 Solved Sibling Problems To Prepare For Infosys Aptitude Tests

Dear Reader,

Below are three problems based on the concept of number of brothers and sisters, you have to find the number of siblings according to given statements.

Question 1 :

If Mr.X, Mr.Y and Mr.Z are three siblings and following three statements are true
1) Mr.X has one older brother and three younger sisters.
2) Mr.Y has two older brothers and two younger sisters
3) Mr.Z has three older brothers and one younger sister.

Then the least possible number of siblings in their family is:

a) 5 b) 6 d) 7 d) 8

Answer : b)6

Solution :

First, we need to think about X, Y and Z as who is the oldest among the three and who is the youngest among the three.
X has one older brother while Y has two. Therefore, X is older than Y. Similarly Y is older than Z. Therefore, among X,Y and Z
X is the oldest, Y is next and Z is the youngest.
X has an older brother and Z has a younger sister.

Try making a chart laying out all the three conditions given in question, and mark the gender of each known sibling.
Then, it is clear that X is male and Y is female.

 M M F M F F
 1st 2nd 3rd 4th 5th 6th
  X Y  Z

Based on the above chart, there should be at least 6 siblings.

Question 2:

If A, B and C are triblet. A,C are males and B is female,
1) A & C say " We have 2 more brothers than the number of sisters"
2) B says " I have 2 more brothers than twice the sisters "

If the above two statements are true then the number of siblings in their family is:

a) 6 b) 12 c) 8 d) 9

Answer : b) 12

Solution :

Let g be the number of girls in the family.
Let b be the number of boys in the family.

Consider statement 1, A and C would have b-2 brothers (excluding them). This b-2 is equal to two more than the number of sisters they have.
Therefore, b-2 = g+2
Or, b-g = 4 -----(1)

Now, from statement 2, B would have g-1 sisters (excluding herself) and b brothers. She has 2 more brothers than twice the number of sisters.
Therefore, Twice the number of sisters = 2(g-1) which is equal to 2 + number of brothers.
Then we have, 2(g-1) + 2 = b
2g - 2 + 2 = b
b-2g = 0 -------(2)
Solving (1) and (2), we have g = 4 and b = 8.
Hence the number of siblings = 4+8 = 12.

Question 3 :

Aravind and Bavya are friends.
1)Aravind's sister has thrice as many sisters of Bavya as brothers.
2)Aravind has 1 more brother than than that of Bavya
3)Twice Bavya's brother's sisters equals the number of boys in Aravind's family.

If the above all 3 statements are true then the number of brothers what Aravind has more than Bavya is:

a) 1 b) 2 c) 3 d) none of these

Answer : a) 1

Solution :

Let the number of boys in Aravind's family be a
Let the number of girls in Aravind's family be b
Let the number of boys in Bavya's family be c
Let the number of girls in Bavya's family be d.

consider statement 1, "Aravind's sister has thrice as many sisters of Bavya as brothers" :

Number of Bavya's sisters = d-1 (excluding her)
Thrice the Bavya's sisters = 3(d-1)
Aravind's sister has 'a' number of brothers.
Therefore, 3(d-1) = a
d-1 = a/3....(1)

Consider statement2,"Aravind has 1 more brother than that of Bavya" :

Number of Aravind's brothers = a-1
Number of Bavya's brother = c
Since, Aravind has 1 more brother than Bavya then a-1 = c+1
a-c = 2...(2)

Consider statement3,"Twice Bavya's brother's sisters equals the number of boys in Aravind's family"

The number of sisters of Bavya's brother = d
Number of boys in Aravind's family = a
Therefore 2d = a
Or d = a/2
Then, d-1 = a/2 - 1 ...(3)

Now, let us solve equations we obtained in previous steps :

From (1) and (3),
a/3 = a/2 - 1
a = 6
Put a value in (2), we have c = 4
Put a value in (3), we have d = 3

Therefore Aravind has a-1 = 6-1 = 5 brothers
And Bavya has c = 4 brothers.
Hence, Aravind has 1 more brother than Bavya.