Skip to main content

Posts

Showing posts from January, 2013

Cisco Sample Problems On Circles

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 X2
Area of second circle = pi(2X)2 = 4 x pi x X2
Area of third circle = pi(3X)2 = 9 x pi x X2
Area of fourth circle = pi(4X)2 = 16 x pi x X2
Then, the area between 1st and 2nd circles = (4 x pi x X2) - (pi x X2) = 3piX2br />
And the area between 3rd and 4th circles = (16 x pi x X2) - (9 x pi x X2) = 7piX2
Therefore the required ratio = 7piX2 / 3piX2 = 7/3
Hence the answer is 7:3

Question 2
88 and 66 are the sum and difference of the perimeter of two c…

Bangalore .Net Developer Opening

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

Maharashtra Board SSC Exam 2013 TimeTable | 10th Exam Schedule

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 TablePaperDateDayTimeHindi/Marathi(First Language)02-03-2013Saturday11.00 am to 2.00pmHindi(second language)05-03-2013Tuesday11.00 am to 2.00 pmEnglish07-03-2013Thursday11.00 am to 2.00 pSanskrit/urdu09-03-2013Saturday11.00 am to 2.00 pmScience12-03-2013Tuesday11.00 am to 1.30 pmAlgebra14-03-2013Thursday11.00 am to 1.30 pmGeometry16-03-2013Saturday11.00 am to 1.30 pmHistory & civics19-03-2013Tuesday11.00 am…

problems which are based on concentric circles.

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 X2
Area of second circle = pi(2X)2 = 4 x pi x X2
Area of third circle = pi(3X)2 = 9 x pi x X2
Area of fourth circle = pi(4X)2 = 16 x pi x X2
Then, the area between 1st and 2nd circles = (4 x pi x X2) - (pi x X2) = 3piX2br />
And the area between 3rd and 4th circles = (16 x pi x X2) - (9 x pi x X2) = 7piX2
Therefore the required ratio = 7piX2 / 3piX2 = 7/3
Hence the answer is 7:3

Question 2
88 and 66 are the sum and difference of the perimeter of two co…