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Answers to to Your Questions - Page 1
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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 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
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 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
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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
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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
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