## Maharashtra HSC March 2013 Time Table Changed

• 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. Later, the final decision was made that these exams shall be held on Sundays as they did not want to disturb other examinations.

On confirming the demands Krishnaroa Patil, in-charge secretary of State board assured that there had been several representations from teachers and students saying the syllabus were vast and the earlier time for preparation was not enough.

Mumbai Junior College Teacher's Association said "The state upgrade the syllabus to match up to the CBSE curriculum but didn't check if teachers could finish the course or not. The board has also done away with external examiners for practicals. The board must make changes or will boycott".

However, the teachers will be taking out a march to the Mumbai region board office on the coming Monday.

## 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
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

Question 2
88 and 66 are the sum and difference of the perimeter of two concentric circles respectively. Then the difference between the area of the circles is:
a)462 b)154 c)none of these d)cannot be determined
Solution :
Let R be the radius of the outer circle
And r be the radius of the inner circle.
Then their perimeters are 2 x pi x R and 2 x pi x r.
Now, the sum of their perimeters is given by
2 x pi x R + 2 x pi x r = 2pi(R + r) = 88
(R + r) = 44/pi ....eqn1
Similarly their difference is given by
2pi(R - r) = 66 ....eqn2
(R - r) = 33/pi
Now, we have to find the difference of their area
i.e., pi(R2) - pi(r2)
pi(R2 - r2) = pi(R + r)(R - r)
From eqn1 & 2, we have pi(R+r)(R-r) = pi x 44/pi x 33/pi
44 x 33 / pi
44 x 33 x 7/22 = 66 x 7 = 462

Question 3
Three concentric circles are given. The radius of inner circle is r, that of middle one is 2r and that of outer one is 3r. Given that the area of ring formed by 1st and 2nd is A and the area of ring formed by 2nd and 3rd is B. Then the relation between A and B is:
a) A = B B) B = 2A c) A < B d) A > B
Answer : c )A < B
Solution :
Given that A is the area of ring formed by 1st and 2nd circles.
i.e., to get the area of the ring, we have to subtract the area of the 1st from 2nd.
Similarly, the subtraction of 3rd and 2nd gives B.
Let r, 2r, 3r be the radius of the concentric circles.
Then, their area will be pi(r2), pi(2r)2 = 4pi(r2), pi(3r)2 = 9pi(r2).
Therefore, A = 4pi(r2) - pi(r2) = 3pi(r2)
and B = 9pi(r2) - 4pi(r2) = 5pi(r2)
From above, we have A < B.
Hence, the answer is option c.

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## 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.

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 16-03-2013 Saturday 11.00 am to 1.30 pm History & civics 19-03-2013 Tuesday 11.00 am to 1.00 pm Geography 21-03-2013 Thursday 11.00 am to 1.00 pm

Best of Luck!!!

## 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
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

Question 2
88 and 66 are the sum and difference of the perimeter of two concentric circles respectively. Then the difference between the area of the circles is:
a)462 b)154 c)none of these d)cannot be determined
Solution :
Let R be the radius of the outer circle
And r be the radius of the inner circle.
Then their perimeters are 2 x pi x R and 2 x pi x r.
Now, the sum of their perimeters is given by
2 x pi x R + 2 x pi x r = 2pi(R + r) = 88
(R + r) = 44/pi ....eqn1
Similarly their difference is given by
2pi(R - r) = 66 ....eqn2
(R - r) = 33/pi
Now, we have to find the difference of their area
i.e., pi(R2) - pi(r2)
pi(R2 - r2) = pi(R + r)(R - r)
From eqn1 & 2, we have pi(R+r)(R-r) = pi x 44/pi x 33/pi
44 x 33 / pi
44 x 33 x 7/22 = 66 x 7 = 462

Question 3
Three concentric circles are given. The radius of inner circle is r, that of middle one is 2r and that of outer one is 3r. Given that the area of ring formed by 1st and 2nd is A and the area of ring formed by 2nd and 3rd is B. Then the relation between A and B is:
a) A = B B) B = 2A c) A < B d) A > B
Answer : c )A < B
Solution :
Given that A is the area of ring formed by 1st and 2nd circles.
i.e., to get the area of the ring, we have to subtract the area of the 1st from 2nd.
Similarly, the subtraction of 3rd and 2nd gives B.
Let r, 2r, 3r be the radius of the concentric circles.
Then, their area will be pi(r2), pi(2r)2 = 4pi(r2), pi(3r)2 = 9pi(r2).
Therefore, A = 4pi(r2) - pi(r2) = 3pi(r2)
and B = 9pi(r2) - 4pi(r2) = 5pi(r2)
From above, we have A < B.