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Here you get answers of your CBSE mathematics 1X and X questions by our experienced teachers. To ask a question click here |

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Question 38. Sir, is there any changes in the income tax table for 2007 class 10 mathematics examination ? by suresh babu, Email sureshbabuko......@yahoo.co.in, answered on Feb 14, 2007 at 6:31 p.m.
Answer 38. There is no change. |

Question 39. simplify ___x___ + __y___ + ___z___
(x-y)(x-z) (y-z) ( y - x) ( z - x) ( z - y) by khushboo sharma Email khush_s_........@yahoo.co.in, answered on Feb 15, 2007 at 8:24 I.S.T. Answer 39. ___x___ - __y___ + ___z___ (x-y) (x-z) (y-z) ( - y + x) ( - z + x) (- z + y) [by taking -1 common from (y-x) (z-x) (z-y)] = [x(y - z) - y(x - z) + z(x - y)] / (x - y)(y - z)(x - z) = 0 / (x - y)(y - z)(x - z) = 0 |

Question 40. S is a point in the interior of a triangle ABC.Prove that AB+AC > SB+SC by Akshay Shreedhar class 1X, Email akshayshree.........@gmail.com, answered on Feb 19, 2007 at 7:45 p.m. Answer 40. Construction - Produce BS to intersect AC at T Proof - In triangle ABT, AB + AT > BT ........(i) In triangle STC, ST + TC > SC ........(ii) Adding (i) and (ii) AB + AT + ST + TC > BT + SC AB + AT + ST + TC > SB + ST + SC (BT = BS + ST from figure) AB + AT + TC > SB + SC ( cancelling ST) AB + AC > SB + SC (AT + TC = AC from figure) |

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Question 41. A peacock is sitting on a tree nine metres high. A snake at a distance of twenty Seven metres from pillar is coming to a pole at the base of the pillar. Seeing the Snake the peacock pounces upon it. If their speeds are equal find the distance from the hole at which the snake caught. by kartarlal class X, E mail kartar.l....@rediff.com, answered on Feb 20, 2007 at 8:32 p.m. Answer 41. Let AB represent tree, C is position of snake. It is moving along CB. Let D be the position where snake is caught by the peacock. Let CD = x metres. Then AD = x metres where AD is distance covered by the peacock. BD = (27 - x) metres Now in right triangle ABD, AB^2 + BD^2 = AD^2 9^2 + (27 - x)^2 = x^2 x^2 - (27 - x)^2 = 81 (x - 27 + x)(x + 27 - x) = 81 (2x - 27)27 = 81 (2x - 27) = 3 2x = 30 x = 15 Required distance = 27 - 15 = 12 metres |

Question 42. A triangle ABC is right angled at B. AB=a BC=b and AC=c.The perpendicular from B on AC is of length 'p'.Prove that 1/p^2=1/a^2 + 1/b^2 by Mustafa, E mail mustafa.k......@gmail.com, answered at Feb 21, 2007 at 5:41 p.m. Answer 42. ar(triangle ABC) = ar(right triangle ABC) 1/2 (cp) = 1/2(ab) [area of triangle = 1/2 base×height] cp = ab c = ab/p .........(i) In right triangle ABC, c^2 = b^2 + a^2 (ab)^2/p^2 = b^2 + a^2 using(i) 1/p^2=1/a^2 + 1/b^2 dividing both sides (ab)^2 |

Question 43. sir, can you please let me know, on which website can i see the new cbse syllabus for 2007-2008, for x standard. i have gone through all the possible sites but dont see the new syllabus anywhere. kindly help. by nayana anil, Email manjula_a.........@yahoo.com, answered on Feb 22, 2007 at 8:09 p.m. Answer 43. You can have it on our site at http://www.cbsemath.com/boarddownloads.htm which is given as cbse syllabus 2006 - 2008 |

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Question 44. ABC and ADC are two right triangles with common hypotenuse CD. Prove that angle CAD = angle CBD. by sandeep tiwari, Email sibu_789632.......@rediff.com , answered on Feb 23, 2007 at 6:19 p.m. Answer 44. There can be 2 cases (i) when triangles are on opposite sides of CD (ii)when triangles are on same sides of CD case i In quadrilateral ABCD, angle B + angle D = 90 + 90 = 180. Therefore quadrilateral ABCD is cyclic. So points A, B, C and D are concyclic. angle CAD = angle CBD (angles in same segment) case ii angle B and angle D are equal and on same side of AC. Therefore points A, B, C and D are concyclic. angle CAD = angle CBD (angles in same segment) |

Question 45. sir, I want to publish some sample papers in maths for class X cbse students in my name. Please tell me how I can upload my files or how can i do above, by Santanu Kumar Mishra, Email santanuchemist....@gmail.com, answered on Feb 25, 2007 at 11:57 a.m. Answer 45. You can send them to [email protected] as attachments, we will publish them on cbsemathspapers.com with your name. You can send papers from class V1 to X11. |

Question 46. Dear teacher, i'm a class 10 cbse student in the uae.i'm not at all clear about the lesson 'income tax'. Pls send an explanation to the lesson on solving those sums. by Lionel Nishant D'souza, Email [email protected] , answered on Feb 25, 2007 at 12:13 p.m. Answer 46. Step1 Find gross annual income. If monthly income is given multiply by 12 to get annual income. Step2 Find total donation taking 100% exemption for N.D.F., P.M.R.F. etc and 50% for C.T. etc example Rs 10000 to N.D.F. and Rs 10000 to C.T. gives total Rs 15000 Step3 Find total savings yearly Step4 Taxable Income = gross annual income - (donations + savings) [maximum saving = Rs 1,00,000] Step5 Find Income tax using the given table step6 Find surcharge if taxable income exceeds Rs 10,00,000. Surcharge = 10% of I.T. step7 Education cess = 2% of I.T. step8 If T.D.S is given subtract it from tax inclusive of edu cess. If you still have problem Contact us again. |

Question 47. I made a formula for doing instalment question for finding the principal corresponding to one month by applying Arithmetic progression
P = nb - n( n-1 ) x/2, where n is the no of instalments ,b is the balance and x the amount of equal monthly instalment.MY query is that whether in board exam evaluation whether any marks cut if I use this formula, by Deepa, Email renjinisom......@yahoo.com, answered on Feb 26, 2007 at 6:54 p.m. I.S.T. Answer 47. This formula or the other form n/2[2×balance - (n - 1)Instalment] is used by many teachers for finding the Principal. Our teachers have been doing evaluation of CBSE math papers from many years and full credit is given for this method. |