Answers to to Your Questions - Page 1 |

What do you get on this web page?
Here you get answers of your CBSE mathematics 1X and X questions by our experienced teachers. To ask a question click here |

See Answers to Your questions |

Question 1. What are Linear equations in two variables? by Vikrant class X, 16 Dec 2006
Answer 1. A Linear equation in two variables is an equation with two variables (for eg. x and y) each having exponent 1. (eg. x + 2y = 4). Definition: An equation of the form ax + by + c = 0 where a, b, c are real numbers and at least one of a, b is non zero. |

Question 2. When I plot x - y = 10 and 2x - 3y = - 20 on graph I get almost parallel lines but on solution I get an answer. How to get an answer on graph ? by simran X 18 Dec 2006
Answer 2. For these types of equations the scale has to be adjusted to a bigger value for example 1cm. = 10 units, using this scale you will see intersecting lies. |

Question 3. How to solve 1/x+a+b = 1/x +1/a +1/b ? by Ajay class x 19 Dec 2006
Answer 3. 1/x+a+b - 1/x = 1/b + 1/b x - (x + a + b)/x(x + a + b) = (a + b)/ab - (a + b)/x(x + a + b) = (a + b)/ab - 1/x(x + a + b) = 1/ab x^2 + ax + bx = - ab x^2 + ax + bx + ab = 0 x (x + a) + b (x + a) = 0 (x + a)(x + b) = 0 x = -a, -b |

See Answers to Your questions |

Question 4. Find p(x) and q(x) which are quadratic polynomials if their LCM is x^3 - 6x^2 + 11x - 6 and HCF is x - 2. by Rishabh, class x 20 Dec 2006
Answer 4. p(x) = x^2 - 5x + 6, q(x) = x^2 - 3x + 2 or q(x) = x^2 - 5x + 6, p(x) = x^2 - 3x + 2 |

Question 5. How many questions are expected from Instalments? by harjot singh class x 21 Dec 2006
Answer 5. We get 2 questions one from S.I. and the other from C.I. which carry 2 and 3 marks. |

Question 6. Solve 31x + 23y = 39, 23x + 31y = 15 by manik X, 22 Dec 2006
Answer 6. Here we find the coefficients of x and y have been interchanged in the two equations, in such cases we once add the equations and once subtract them adding the equations we get 54x + 54y = 54 which reduces to x + y = 1 ......i subtracting the equations we get 8x - 8y = 24 which reduces to x - y = 3 ......ii further adding i and ii we get 2x = 4 which gives x = 2 and y = - 1 |

See Answers to Your questions |

Question 7. I am very weak in mathematics I failed in the Dec exams. Which topics should I do to at least get passing marks? sahil class x 24.12.06
Answer 7. Dear sahil don not worry, you can easily pass in CBSE board exams 2007. For details of easy topics see our page ' pre exam tips' click here |

Question 8. Solve for x and y, 6x + 3y = 7xy, 3x + 9y = 11xy (given x,y not equal to 0) by Meena, x 25.12.06
Answer 8. 6x + 3y = 7xy ......i 3x + 9y = 11xy ....ii (i)*1 - (ii)*2 gives us 15y = 15xy which reduces to x = 1 substituting x = 1 in equation i we get y = 3/2 Detailed solution will be sent to you by E mail within 1 - 2 days |

Question 9. The dimensions of a rectangular box are in the ratio 2:3:4 and the difference between the cost of covering it with sheet of paper at the rate of Rs.4 and Rs. 4.50 per square m is Rs.416. Find the dimensions of the box. by Prabhat Aditya class x, answered on 2.12.2007 at 9 a.m. I.S.T. (+5.5 G.M.T.)
Answer 9. Let the sides be 2x, 3x and 4x in metres. S.A. = 2(lb+bh+lh) = 52x square metres 52x square(4.5 - 4)= 416 x ^2 = 16 x = 4 length = 2*4 = 8m, breadth = 3*4 = 12m and height = 4*4 = 16m |

Question 10. Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to six right angles. by Amitanshu Vishal cbse maths x answered on jan 3 2007 at 8:00 p.m. I.S.T.
Answer 10. Let ABCD be a cylclic quadrilateral. angles AEB, BFC,CGD and DHA are in respective segments of AB, BC, CD and DA. Join diagonal AC. angle H + angle ACD = 180 ........(i) (opp. angles of a cyclic quad.) angle E + angle ACB = 180 .........(ii) angle F + angle BAC = 180 ........(iii) angle G + angle CAD = 180 ........(iv) adding (i), (ii), (iii) and (iv) (angle ACD + angle ACB) + (angle BAC + angle CAD) + angle E + angle F angle G + angle H = 720 (angle BCD + angle BAD) + angle E + angle F angle G + angle H = 720 180 + angle E + angle F angle G + angle H = 720 angle E + angle F angle G + angle H = 720 - 180 angle E + angle F angle G + angle H = 540 (sum of opposite angles of a cyclic quadrilateral) angle E + angle F angle G + angle H = 6 * 90 |