PLACEMENT INFO
Tuesday, 12 May 2015
Monday, 11 May 2015
Siemens Recruitment For Freshers – May 2015
Company Profile : Siemens and researchers from the University of Genoa have built a power grid using a combination of conventional and renewable energy sources as well as energy storage devices. The smart grid, used at the university’s campus in Savona, has already reduced significantly the annual electricity costs.
Company Name : Siemens
Job Description :
Designation : .Net Developer
Qualification : BE/B.Tech
Location : Gurgaon
Experience : Freshers
Salary : As per the Industry
Skills :
1. BE/BTech in Computer Science or equivalent
1. BE/BTech in Computer Science or equivalent
2. Experience in any one or more programming language, e.g. VBscript, Powershell, C/ C++, Java
3. Knowledge of configuring and monitoring Windows components using a programming interface, e.g. Windows API, WMI, Powershell
4. Worked in Six Sigma and/or CMMi (at least level 3) process organization
5. Good communication skills
Wednesday, 1 April 2015
L&T JOB OPENINGS FOR FRESHERS/EXPERIENCED/ GRADUATES ON MARCH TO APRIL 2015: HUGE VACANCIES, APPLY NOW
Are you searching about L&T company Job openings? Want to apply for it? L&T has a huge opening for fresher and experienced candidates. Can you apply for it? now check below.
Company name: L&T
Company Profile: Larsen & Toubro Limited, also known as L&T, is an Indian multinational conglomerate head quartered in Mumbai, India. It was founded by Danish engineers taking refuge in India, as well as an Indian financing partner Company asks for 18,000 resumes for this recruitment
Position name: Engineer
Job Category: B.E, B.tech, Engineering Job
Required Skills: Good communication, subject knowledge, language skills and analytic skills
Required expected experience: Fresher to 12 years experienced candidates
Salary/ Pay scale: High in Industry
Work Place :Across India
Required Eligibility: any Graduates can also apply for this job
Interview mode: Written Test, face to face interview
Company Contact Address: Corporate Office - L&T IDPL
L&T Infrastructure Development Projects Limited (L&T IDPL)
TCTC Building, First Floor, Mount Poonamallee Road,
Manapakkam, Chennai - 600089,
Tamil Nadu, India.
Tel: (0) 44 2252 6666 Fax: (0) 44 2252 8724
Email: contactus@lntidpl.com
Selection mode: Interview
Apply mode: Online
Position name: Engineer
Job Category: B.E, B.tech, Engineering Job
Required Skills: Good communication, subject knowledge, language skills and analytic skills
Required expected experience: Fresher to 12 years experienced candidates
Salary/ Pay scale: High in Industry
Work Place :Across India
Required Eligibility: any Graduates can also apply for this job
Interview mode: Written Test, face to face interview
Company Contact Address: Corporate Office - L&T IDPL
L&T Infrastructure Development Projects Limited (L&T IDPL)
TCTC Building, First Floor, Mount Poonamallee Road,
Manapakkam, Chennai - 600089,
Tamil Nadu, India.
Tel: (0) 44 2252 6666 Fax: (0) 44 2252 8724
Email: contactus@lntidpl.com
Selection mode: Interview
Apply mode: Online
TCS Hiring for Freshers ( 2015 , 2014 , 2013 , 2012 ,2011 Batches ) Any Graduates – Apply Now
Eligibility : ME/M.Tech, BSc, BCA, BE/B.Tech, MSc, MCA
Location : Anywhere in India
Job Category : IT/Software, BSc/BCA/BCM
Job Type : Full Time
Hiring Process : Written-test, Face to Face Interview
Eligibility :
- BE/ B.Tech / ME/ M.Tech in any of the following disciplines :
- Computer Science
- Information Technology
- Electronics and Electrical Engineering
- Electronics and Communication
- Electronics and Telecommunication
- Mechanical Engineering
- Electronics/Electrical and Instrumentation Engineering
- Electrical Engineering
- Instrumentation Engineering
- Network Engineering
- MCA with BSc / BCA / BCom / BA (with Math / Statistics Background)
- M.Sc in Computer Science / InformationTechnology.
Marks Criteria
- Aggregate of 60% and above in class X, XII, Graduation and above
- http://fresherstime.in/tcs-hiring-for-freshers-2015-2014-2013-2012-2011-batches-any-graduates-apply-now/
Friday, 20 March 2015
Information regarding the android training by startup village
We spoke the other day about the training program that is to be conducted by the Startup Village. I was supposed to mail you sooner but got delayed, sorry for a bit of delay.
The Twenty day android training course will be conducted in the month of July.
--
The Twenty day android training course will be conducted in the month of July.
The course is a 20 days training in android. The student after the training shall get the opportunity to get into one of the startup companies incubated at Startup Village, as an intern.
The course fee is Rs 9,400/- and the student shall book their seats in advance by paying Rs 1,000/- in advance. The total number of seats are limited to 40, for the efficiency and quality of the program.
Even though there are no promises for helping the students get recruited, the skill level and the performance of the students might help them to get placed. The recruitment if any; solely depends only on the performance of the attendee and the company's interest to include them in their team.
The course fee is Rs 9,400/- and the student shall book their seats in advance by paying Rs 1,000/- in advance. The total number of seats are limited to 40, for the efficiency and quality of the program.
Even though there are no promises for helping the students get recruited, the skill level and the performance of the students might help them to get placed. The recruitment if any; solely depends only on the performance of the attendee and the company's interest to include them in their team.
For more details contact
Shrigovind S
+91-9447727039
Shrigovind S
+91-9447727039
With Regards,
Shrigovind.S

Managing Director, App incubator Innovation Labs
Innovation Ninja, Altersense .
______________________________________
Altersense
Click Here to visit our official website.

*Give Ideas to Life and Life to Ideas*
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WARNING: Computer viruses can be transmitted via email. The recipient should check this email and any attachments for the presence of viruses. The company accepts no liability for any damage caused by any virus transmitted by this email.Wednesday, 18 February 2015
HCL Technologies Off-Campus Recruitment Drive
Thursday, 12 February 2015
APTITUDE QUESTIONS & ANSWERS
1. Ray writes a two digit number.
He sees that the number exceeds 4 times the sum of its digits by 3.
If the number is increased by 18, the result is the same as the number
formed by reversing the digits. Find the number.
a) 35
b) 42
c) 49
d) 57
Solution: Let the two digit number be xy.
4(x + y) +3 = 10x + y .......(1)
10x + y + 18 = 10 y + x ....(2)
Solving 1st equation we get 2x - y = 1 .....(3)
Solving 2nd equation we get y - x = 2 .....(4)
Solving 3 and 4, we get x = 3 and y = 5
2. a, b, c are non negitive integers such that 28a+30b+31c = 365. a + b + c = ?
a) Greater than 14
b) less than or equal to 11
c) 13
d) 12
In a calender,
Number of months having 28 days = 1
Number of months having 30 days = 4
Number of months having 31 days = 7
28 x 1 + 30 x 4 + 31 x 7 = 365
Here, a = 1, b = 4, c = 7.
a+b+c = 12
3. George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete?
a) 11.30 am
b) 12 noon
c) 12.30 pm
d) 1 pm
Let the total work = 120 units.
As George completes this entire work in 8 hours, his capacity is 15 units /hour
Similarly, the capacity of paul is 12 units / hour
the capacity of Hari is 10 units / hour
All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74
Remaining work = 120 - 74 = 46
Now this work is to be done by paul and hari. 46 / (12 + 10) = 2 hours (approx)
So work gets completed at 1 pm
4. If x^y denotes x raised to the power y, Find last two digits of (1141^3843) + (1961^4181)
a) 02
b) 82
c) 42
d) 22
Remember 1 raised to any power will give 1 as unit digit.
To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power.
b) 42
c) 49
d) 57
Solution: Let the two digit number be xy.
4(x + y) +3 = 10x + y .......(1)
10x + y + 18 = 10 y + x ....(2)
Solving 1st equation we get 2x - y = 1 .....(3)
Solving 2nd equation we get y - x = 2 .....(4)
Solving 3 and 4, we get x = 3 and y = 5
2. a, b, c are non negitive integers such that 28a+30b+31c = 365. a + b + c = ?
a) Greater than 14
b) less than or equal to 11
c) 13
d) 12
In a calender,
Number of months having 28 days = 1
Number of months having 30 days = 4
Number of months having 31 days = 7
28 x 1 + 30 x 4 + 31 x 7 = 365
Here, a = 1, b = 4, c = 7.
a+b+c = 12
3. George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete?
a) 11.30 am
b) 12 noon
c) 12.30 pm
d) 1 pm
Let the total work = 120 units.
As George completes this entire work in 8 hours, his capacity is 15 units /hour
Similarly, the capacity of paul is 12 units / hour
the capacity of Hari is 10 units / hour
All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74
Remaining work = 120 - 74 = 46
Now this work is to be done by paul and hari. 46 / (12 + 10) = 2 hours (approx)
So work gets completed at 1 pm
4. If x^y denotes x raised to the power y, Find last two digits of (1141^3843) + (1961^4181)
a) 02
b) 82
c) 42
d) 22
Remember 1 raised to any power will give 1 as unit digit.
To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power.
So the Last two digits of the given expression =
21 + 61 = 82
5. J can dig a well in 16 days. P can
dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in
How many days?
a) 32
b) 48
c) 96
d) 24
Assume the total work = 48 units.
Capacity fo J = 48 / 16 = 3 units / day
Capacity of P = 48 / 24 = 2 units / day
Capacity of J, P, H = 48 / 8 = 6 units / day
From the above capacity of H = 6 - 2 - 3 = 1
So H takes 48 / 1 days = 48 days to dig the well
6. If a lemon and apple together costs
Rs.12, tomato and a lemon cost Rs.4 and an apple costs Rs.8 more than a
lemon. What is the cost of lemon?
L + A = 12 ...(1)
T + L = 4 .....(2)
L + 8 = A
Taking 1 and 3, we get A = 10 and L = 2
7. 3 mangoes and 4 apples costs Rs.85. 5
apples and 6 peaches costs 122. 6 mangoes and 2 peaches costs Rs.144.
What is the combined price of 1 apple, 1 peach, and 1 mango.
a) 37
b) 39
c) 35
d) 36
Note: It is 114 not 144.
3m + 4a = 85 ..(1)
5a + 6p = 122 ..(2)
6m + 2p = 114 ..(3)
(1) x 2 => 6m + 8a = 170
(3) => 6m + 2p = 114
Solving we get 8a - 2p = 56 ..(4)
(2) => 5a + 6p = 122
3 x (4) = 24a - 6p = 168
Solving we get a = 10, p = 12, m = 15
So a + p + m = 37
8. An organisation has 3 committees,
only 2 persons are members of all 3 committee but every pair of committee has 3
members in common. what is the least possible number of members on any one
committee?
a) 4
b) 5
c) 6
d) 1
Total 4 members minimum required to serve only on
one committee.
9. There are 5 sweets - Jammun, kaju,
Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to
Friday. A person eats one sweet a day, based on the following
constraints.
(i) Ladu not eaten on monday
(ii) If Jamun is eaten on Monday, Ladu
should be eaten on friday.
(iii) Peda is eaten the day following
the day of eating Jilebi
(iv) If Ladu eaten on tuesday, kaju
should be eaten on monday
based on above, peda can be eaten on any
day except
a) tuesday
b) monday
c) wednesday
d) friday
From the (iii) clue, peda must be eaten after
jilebi. so Peda should not be eaten on monday.
10. If YWVSQ is 25 - 23 - 21 - 19 - 17,
Then MKIGF
a) 13 - 11 - 8 - 7 - 6
b) 1 - 2-3-5-7
c) 9 - 8 - 7 - 6 - 5
d) 7 - 8 - 5 - 3
MKIGF = 13 - 11 - 9 - 7 - 6
Note: this is a dummy question. Dont answer these
questions
11. Addition of 641 + 852 + 973 = 2456
is incorrect. What is the largest digit that can be changed to make the
addition correct?
a) 5
b) 6
c) 4
d) 7
641
852
963
------
2466
largest among tens place is 7, so 7 should be
replaced by 6 to get 2456
12. Value of a scooter depriciates in
such a way that its value at the end of each year is 3/4th of its value at the
beginning of the same year. If the initial value of scooter is 40,000, what is
the value of the scooter at the end of 3 years.
a) 23125
b) 19000
c) 13435
d) 16875
value of the scooter at the end of the year = 40000×(34)3 = 16875
13. At the end of 1994, R was half as
old as his grandmother. The sum of the years in which they were born is
3844. How old R was at the end of 1999
a) 48
b) 55
c) 49
d) 53
In 1994, Assume the ages of GM and R = 2k, k
then their birth years are 1994 - 2k, 1994 -
k.
But given that sum of these years is 3844.
So 1994 - 2k + 1994 - k = 3844
K = 48
In 1999, the age of R is 48 + 5 = 53
14. When numbers are written in base b,
we have 12 x 25 = 333, the value of b is?
a) 8
b) 6
c) None
d) 7
Let the base = b
So, (b+2)(2b+5) = (b+2)(2b+5)=3b2+3b+3
2b2+9b+10=3b2+3b+3
b2−6b−7=0
Solving we get b = 7 or -1
So b = 7
15. How many polynomials of degree
>=1 satisfy f(x2)=[f(x)]2=f(f(x)
a) more than 2
b) 2
c) 0
d) 1
Let f(x) = x2
f(x2)=[x2]2=x4
(f(x))2=[x2]2=x4
f(f(x))=f(x2)=[x2]2=x4
Only 1
16. Figure shows an equilateral triangle
of side of length 5 which is divided into several unit triangles. A valid
path is a path from the triangle in the top row to the middle triangle in the
bottom row such that the adjacent triangles in our path share a common edge and
the path never travels up (from a lower row to a higher row) or revisits a
triangle. An example is given below. How many such valid paths are
there?
a) 120
b) 16
c) 23
d) 24
Sol:
Number of valid paths = (n-1) ! = (5-1)! = 24
17. In the question, A^B means, A raised
to power B. If x*y^2*z < 0, then which one of the following statements must
be true?
(i) xz < 0 (ii) z < 0 (iii)
xyz < 0
a) (i) and (iii)
b) (iii) only
c) None
d) (i) only
As y^2 is always positive, x*y^2*z <
0 is possible only when xz < 0. Option d is correct.
18. The marked price of a coat was 40%
less than the suggested retail price. Eesha purchased the coat for half
the marked price at the fiftieth anniversary sale. What percentage less
than the suggested retail price did Eesha pay?
a) 60
b) 20
c) 70
d) 30
Let the retail price is Rs.100. then market price
is (100-40) % of 100 = 60. Eesha purchased the coat for half of this
price. ie., 30 only. which is 70 less than the retail price. So Option C is
correct.
19. In a city 100% votes
are registered, in which 60% vote for congress and 40% vote for BJP.
There is a person A, who gets 75% of congress votes and 8% of BJP votes.
How many votes got by A?
Assume total votes are 100. So A got
75% of 60 = 45
8% of 40 = 3.2
A total of 48.2 %
20. Mean of 3 numbers is 10 more than the least of the numbers and 15 less than greatest of the 3. If the median of 3 numbers is 5, Find the sum of the 3 numbers?
Ans: Median is when the given numbers are arranged in ascending order, the middle one. Let the numbers are x, 5, y where x is the least and y is greatest.
Given that x+5+y3=x+10
and x+5+y3=y−15
Solving we get x = 0 and y = 25.
So sum of the numbers = 0 + 5 + 25 = 30
21. A and B start from house at 10am. They travel fro their house on the MG road at 20kmph and 40 kmph. there is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm.
Distnace between House and T junction = 20 x 2 = 40.
ie., B reached T at 11 am. B continued to right after 11 am and travelled upto 2. So distance covered by him = 3 x 40 = 120
A reached T at 12 noon and travelled upto 2 So distanced travelled by him = 2 x 20 = 40
So total distance between them = 120 + 40 = 160 km
Assume total votes are 100. So A got
75% of 60 = 45
8% of 40 = 3.2
A total of 48.2 %
20. Mean of 3 numbers is 10 more than the least of the numbers and 15 less than greatest of the 3. If the median of 3 numbers is 5, Find the sum of the 3 numbers?
Ans: Median is when the given numbers are arranged in ascending order, the middle one. Let the numbers are x, 5, y where x is the least and y is greatest.
Given that x+5+y3=x+10
and x+5+y3=y−15
Solving we get x = 0 and y = 25.
So sum of the numbers = 0 + 5 + 25 = 30
21. A and B start from house at 10am. They travel fro their house on the MG road at 20kmph and 40 kmph. there is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm.
Distnace between House and T junction = 20 x 2 = 40.
ie., B reached T at 11 am. B continued to right after 11 am and travelled upto 2. So distance covered by him = 3 x 40 = 120
A reached T at 12 noon and travelled upto 2 So distanced travelled by him = 2 x 20 = 40
So total distance between them = 120 + 40 = 160 km
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