Goforexam.in Guide for Placements, Interview, Aptitude Test...

The HSC chemistry paper, which was slated for February 27, will now be held on March 26.  The Biology paper has been postponed from March 4 to 17 Mumbai, Maharashtra: The HSC Board exam time table was issued quite long time before, in the month of October, 2012. After a certain time of discussions based on the demands from HSC science students and teachers across the State, the Maharashtra Board has announced the change in the Data sheet for Class 12 board examinations, which are scheduled to start from 21st of February, 2013 onwards.  The dates of Biology and Chemistry papers has been changed after teachers and students had pointed out that more time is required to study these two subjects. After raising this argument, students across Maharashtra wrote to the Education Minister based on which a debate was held at Mantralaya on Thursday and the new time table was announced.  Many officials from the state board and the education department had taken part in the debate. La

Question 1 Four concentric circles are drawn with the radii at interval of X units. If the radius of inner circle is X units then the ratio of the area between the 4th and 3rd circles to the area between the 2nd and 1st circles is: a)1:2 b)2:1 c)4:3 d)7:3 Answer : d)7:3 Solution : Let X be the radius of inner(1st) circle. Given that the radii of the circles are at the intervals of X units. Then X + X = 2X, 2X + X = 3X, 3X + X = 4X are the radii of successive circles respectively. Now, the area of first circle = pi(X) 2 = pi x X 2 Area of second circle = pi(2X) 2 = 4 x pi x X 2 Area of third circle = pi(3X) 2 = 9 x pi x X 2 Area of fourth circle = pi(4X) 2 = 16 x pi x X 2 Then, the area between 1st and 2nd circles = (4 x pi x X 2 ) - (pi x X 2 ) = 3piX 2 br /> And the area between 3rd and 4th circles = (16 x pi x X 2 ) - (9 x pi x X 2 ) = 7piX 2 Therefore the required ratio = 7piX 2 / 3piX 2 = 7/3 Hence the answer is 7:3 Question 2 88 and 66 are

Location : Bengaluru/Bangalore Category : IT Experience : 0 - 1 Year Qualification in Brief : B.Tech/B.E./Diploma - Computers, Electronics/Telecommunication Description in Brief : Candidates can apply for .Net Developer Openings at Metrico Soft Solutions. Requirements include good knowledge in .Net with C# ASP.Net, C# windows, C++, OOPS and databases like MS access,MS SQL Server 2003/2008, POSTGRESQL, SDLC, knowledge on Embedded Systems software, etc... link :http://jobsearch.naukri.com/job-listings-Software-Engineer-Developer-Metrico-Soft-Solutions-Pvt-Ltd--Bengaluru-Bangalore-0-to-1-090113004504?xz=10_0_5&xo=&xp=42&xid=135779625856103900&qjt=&qp=Software+Developer&id=&f=-090113004504

The State Education Board Maharashtra has declared the 10th class board exam time table 2013 on the starting of the academic year itself. Maharashtra SSC board exams are going to start from 2nd of march and will be end on 21st of march as general category. PDF file of time table Students who are appearing for Maharashtra Board 10th Examination 2013 are hereby informed to check 10th class Maharashtra board final exams time table . Students can see the SSC Board exams time table with date, day, paper name and timing. Secondary School Certificate (SSC) Examination March 2013 Time Table Paper Date Day Time Hindi/Marathi(First Language) 02-03-2013 Saturday 11.00 am to 2.00pm Hindi(second language) 05-03-2013 Tuesday 11.00 am to 2.00 pm English 07-03-2013 Thursday 11.00 am to 2.00 p Sanskrit/urdu 09-03-2013 Saturday 11.00 am to 2.00 pm Science 12-03-2013 Tuesday 11.00 am to 1.30 pm Algebra 14-03-2013 Thursday 11.00 am to 1.30 pm Geometry 16-03-201

Question 1 Four concentric circles are drawn with the radii at interval of X units. If the radius of inner circle is X units then the ratio of the area between the 4th and 3rd circles to the area between the 2nd and 1st circles is: a)1:2 b)2:1 c)4:3 d)7:3 Answer : d)7:3 Solution : Let X be the radius of inner(1st) circle. Given that the radii of the circles are at the intervals of X units. Then X + X = 2X, 2X + X = 3X, 3X + X = 4X are the radii of successive circles respectively. Now, the area of first circle = pi(X) 2 = pi x X 2 Area of second circle = pi(2X) 2 = 4 x pi x X 2 Area of third circle = pi(3X) 2 = 9 x pi x X 2 Area of fourth circle = pi(4X) 2 = 16 x pi x X 2 Then, the area between 1st and 2nd circles = (4 x pi x X 2 ) - (pi x X 2 ) = 3piX 2 br /> And the area between 3rd and 4th circles = (16 x pi x X 2 ) - (9 x pi x X 2 ) = 7piX 2 Therefore the required ratio = 7piX 2 / 3piX 2 = 7/3 Hence the answer is 7:3 Question 2 88 and 66 are t