Answers to to Your Questions  Page 3 
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Question 26. solve for x:(x3)(x+9)(x7)(x+5)=208 by kanika khanna.
Email kanikakhanna............@hotmail.com answered on 22.1.07 at 8:43 p.m. Answer 26. These type of questions do not form part of the current class x syllabus. But still we are giving the solution. Step 1 rearrange (x+9)(x7)(x3)(x+5) = 208 Step 2 (x^2 + 2x  63)(x^2 + 2x  15) = 208 Step 3 Put x^2 + 2x = y Step 4 (y  63)(y  15) = 208 Step 5 y^2  78y + 945 = 208 Step 6 y^2  78y + 737 = 0 Step 6 (y  7)(y  71) = 0 Step 7 y = 7, y = 71 Step 8 putting value of y we get x^2 + 2x = 7 or x^2 + 2x = 71 now we solve for x 
We are back after Republic day break and will start answering questions from today. Posted on Jan 30,2007 at 9:04 I.S.T.
Question 27. find the weight of a lead pipe 3.5m iong if the external diameter of the pipe is 2.4m and the thickness is 2mm and 1 cubiic cm of lead weighs 11g by chaitanya kaul class x, Email add chaitanya................@gmail.com, answered on Jan 31, 2007 at 7:32 I.S.T. Answer 27. Height = 3.5m = 3500 mm, outer radius (R) = 1.2 m = 1200 mm, Inner radius(r) = 1200  2 = 1198 mm Volume = outer volume  inner volume = Pie h (R^2  r^2) = (22/7) 3500 (1200^2  1198^2) = 22*500(1200  1198)(1200 + 1198) using identity = 22*500*2*2398 = 22*1000*2398 = 52756000 mm^3 = 52756 cm^3 Density = 11g/cm^3 Mass = Volume * density = 52756*11 = 580316 grams or 580.316 k.g. 
Question 28. An isosceles triangle ABC is inscribed in a circle. If AB = AC = 13cm and BC = 10cm,find the radius of the circle. by Francis Thomas class x/1x, Email add francis..............@yahoo.co.in answered on Feb 1, 2007 at 11:08 p.m. I.S.T.
Answer 28. Draw AD perpendicular to BC intersecting BC at D. AD bisects BC , So BD = ½ BC = ½(10) = 5cm (since altitude to base in an isosceles triangle is also median) So AD becomes perpendicular bisector of BC and hence passes through centre of circle let it be O (since perpendicular bisector of a chord passes through centre of circle.) In right triangle ADB, AB^2 = AD^2 + BD^2 (pythagoras theorem) 13^2 = AD^2 + 5^2 169 = AD^2 + 25 AD^2 = 169  25 AD = 12 cm. Let radius OA = x cm. OD = (12  x) cm. OB = OA = x cm.(radii of same circle) In right triangle BDO, OB^2 = OD^2 + BD^2 x^2 = (12  x)^2 + 5^2 x^2 = 144 + x^2  24x + 25 24x = 169 x = 169/24 Radius = 169/24 cm. 
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Question 31. A round balloon of radius r subtends an angle 'b' at the eye of the observer while the angle of elevation of its centre is 'c'. Prove that the height of the centre of the balloon is (r sin b cosec c/2) by Arshad Email add ares.ar...........@gmail.com answered on Feb 2, 2007 at 9:09 I.S.T.
(Figures cannot be drawn on our webpages) b is angle subtended subtended on eye of observer, so angle APB = b, where in figure PA Answer31. and PB act as tangents also join OA, OB and OP, O being the centre of circle.Triangle OAP is congruent to triangle OBP so angle APO = angle BPO = b/2. angle OPC = c, because it is angle of elevation. sin c = h/OP and cosec b/2 = OP/r where h is height of centre of balloon sin c ×cosec b/2 = h/OP × OP/r = h/r r sin c ×cosec b/2 = h 
Question 32. Prove that if Cos A + Sin A = root 2×Sin A , Then Sin A  Cos A = Root2×sin A
i think this question is wrong, please tell me if i am right. by Piyush Pattanayak piyush..........@gmail.com, answered on Feb 3, 2007 at 10:09 p.m. I.S.T. Answer 32. The correct question is Prove that if Cos A + Sin A = root 2 Sin A , Then Sin A  Cos A = Root2 cos A Solution Cos A = root 2 Sin A  Sin A Cos A = (root 2  1 )Sin A Sin A = cos A/(root 2  1 ) and then rationalising the denominator we get sin A  cos A = root2 cosA 
Question 33. How to Factorise : 4 x^4 + y^4 (4 times x to the power of 4 plus y to the power of 4) by Saurabh , Email saurabh_bis........@yahoo.co.in, answered on Feb 8, 2007 at 8:52 p.m. I.S.T.
Answer 33. 4 x^4 + y^4 can be solved by method of completing the squares = [(2x)^2]^2 + (y^2)^2 = [(2x)^2]^2 + (y^2)^2 + 4x^2 y^2  4x^2 y^2 (adding and subtracting 2ab) = [2x^2 + y^2]^2  (2x^2y^2)^2 [ because a^2 + b^2 + 2ab = (a + b)^2] = [2x^2 + y^2  2x^2y^2][2x^2 + y^2 + 2x^2y^2] [ a^2  b^2 = (a  b)(a + b)] 
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Question 34. (tanA+secA1) /
(tanAsecA+1) = (1 + sinA) / cosA by preeti bhanwala, Email
em_pre010.........@yahoo.co.in, answered on 10 Feb 2007 at 1:22 p.m.
Answer 34. L.H.S. = [tanA+secA1] / [tanAsecA+1] = [tanA+secA  (sec^2 A  tan^2 A)] / [tanA  secA + 1] = [tanA+secA  (secA  tanA)(secA + tanA) / [tanA  secA + 1] [ a^2  b^2 = (a  b)(a + b)] = (tanA+secA) (tanA  secA + 1) / [tanA  secA + 1] = tanA+secA = (1 + sinA) / cosA = R.H.S. 
Question 35. Can is substitute the value
for tan 45* in a question which specifies that "we should not use
trigonometric table" ? by T.Santhosh Kumar santhoshney.......@gmail.com,
answered on Feb 11, 2007 at 1:20 p.m.
Answer 35. Yes, we put the value of tan 45* in these questions. Replace tan 45* by 1, its value. 
Question 36. For Income Tax, in which case
the surcharge will be calculated  when the gross income exceeds
Rs.10,00,000 or when the taxable income exceeds Rs.10,00,00 ?? And then
after that on what will be the edu cess calculated  edu cess will be
2% of tax (excluding surcharge) or 2% of total tax (including surcharge)
?? Plzz help. by Pinaakee Bhardwaj, Email pinaak.....@yahoo.co.in,
answered on Feb 12, 2007 at 11 p.m. I.S.T.
Answer 36. See table below which clears your doubt, if you still have any problem fill in the form on our website once again. 
Click to edit table header    
Taxable Income

Rate of Tax

Surcharge

Educational cess

Upto Rs.1,35,000

Nil

Nil

Nil

Rs.1,35,001 to
Rs.1,50,000 
10% of
Total Income
exceeding Rs.1,35,000 
Nil

2% of income tax

Rs. 1,50,001
to
Rs.2,50,000 
Rs.1500 + 20% of total
income exceeding Rs.1,50,000 
Nil

2% of income tax

Rs.2,50,001
to
Rs.10,00,00 
Rs.21,500 + 30% of total
income exceeding Rs.250000 
Nil

2% of income tax

Above Rs.10,00,000

Rs.246500 + 30%
of total
Income exceedingRs.10,00,000 
10% of income tax

2% of income tax and surcharge

Click to edit table header 