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Answers to Your Questions - Page 17
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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 160.  What  is the possible date for  CBSE results 2008 class xii ?by Harshita. Answered on May13,  2008 at 6:42
Answer 160.   Taking into account the result dates of 2006 and 2007 the results xii expected around 23 May 2008
 Question 161. Please explain why 7x11x13+13 and 7x6x5x4x3x2x1+5 are composite numbers? I tried to explain in the following way .
7x11x13+13
=13{7x11+1}
=13x78
=13x2x3x13
and
7x6x5x4x3x2x1+5
=5{7x6x4x3x2x1+1)
=5x1009
now i don't know how to conclude. Please help me, by Sidharth. Answered on May 18, 2008 at 12:41 p.m.
Answer 161. A composite number has more than two factors or in other words two factors other than 1.
Now in question 1 you got 13x78 in second last step. Since it has two factors other than one, it is a composite no.
Similarly for second.
Question 162. Two poles 'a' and 'b' are 'p' metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by - ab/a+b metres by Namrata. Answered on May 23, 2008 at 5:29 p.m.
Answer 162. Let first pole 'a' be represented by AB and 'b' by CD and let they intersect at E, EF is perpendicular to BD. Now BD = 'p'. Let BF = x and FD = p - x. Also let EF = 'c'
Triangle BFE ~ BDC by AA .
Therefore c/b = x/p     ......(i)
Similarly c/a = (p-x)/p ......(ii)
Adding i and ii
c/b  +  c/a  = x/p + (p-x)/p
or   c(a+b)/ab = (x + p - x)/p                   taking LCM on both sides
or   c(a+b)/ab = p/p
or   c(a+b)/ab = 1
or   (a+b)/ab = 1/c
or   ab/(a+b) = c
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Question 163. There is a circular path around a sports field.sonia takes 18min to drive one around of the field,while ravi takes 12min for the same.suppose they both start at the same point at the same time,and go in the same direction.after how minutes will they meet again at the starting point? by Uthra from Chennai. Answered on May 24, 2008 at 5:37 p.m.
Answer 163. Required time = LCM of 18 and 12
                                                       36 minutes.
Question 164.  In triangle ABC, D and E are points on AB and AC such that DE || BC. if AD = 4x-3, AE = 8x-7, BD = 3x-1 and CE = 5x-3, find the value of x by Namrata. Answered on June 9, 2008 at 10:53 p.m.
Answer 164. In triangle ABC, D and E are points on AB and AC such that DE || BC
                         Therefore AD/DB = AE/EC
                                             (4x-3)/(3x-1) = (8x-7)/(5x-3)
                                  or       20x^2 - 12x - 15x + 9 = 24x^2 - 8x - 21x + 7
                                  or       4x^2 - 2x - 2 = 0
           (÷2)                or       2x^2 - x - 1 = 0
                                  or       (x - 1)(2x + 1) = 0
                                  or       x = 1, x = -1/2 (rejected)
                          Therefore x = 1
Question 165. 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 Brijesh kumar. Answered on June 16, 2008 at 7:17 p.m.
Answer 165. This question has already been answered. See question 41 on page 3 of question answers. Click Here
Question 166.   The sum of two number is 18 and the sum of their reciprocals is 1/4.Find the number, by Ujjawal. Answered on June 20, 2008 at 11:35PM
Answer 166   Let the numbers be x and 18 - x
                        according to problem
                        1/x  +  1/(18 - x)  = 1/4
                        x^2 - 18 x + 72 = 0
                        (x - 12)  (x - 6)  = 0
                         x = 12 or x = 6
                         If x = 12 the first number = 12 and second number = 6
                         If x = 6 the first number = 6 and second number = 12
                         therefore the numbers are 6 and 12
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Question 167.  Rationalise the denominator 4/(2+root3+root7) by Karan Batra. Answered on June 27, 2008 at 5:48 p.m.
Answer 167. 4/(2+root3+root7)   ×    (2+root3 - root7) /(2+root3 -  root7) 
                   = 4(2 + root3 -  root7)/ [(2+root3)^2 -  (root7)^2]
                       = 4(2 + root3 -  root7)/ [4 + 3 + 4root3 - 7]
                  =  4(2 + root3 -  root7)/  4root3
                       =  (2 + root3 -  root7)/  root3
                       =  (2 + root3 -  root7)/ root3   ×   root3 / root3
                       =   root3(2 + root3 -  root7)/ 3
Question 168.  Solve sec^2 10 - cos^2 80 by sidharth. Answered on July 6, 2008 at 12:01 a.m.
Answer 168.     sec^2 10 - cos^2 80
                        = cos^2 (90 - 10) - cos^2 80
                        = cos^2 80 - cos^2 80
                        = 0
Question 169.  Form the pair of linear equations for the following problem & find their solution by substitution method.
A fraction becomes 9/11, if 2 is added to both the  numerator & the denominator. If 3 is added to both the numerator & the denominator it becomes 5/6 . Find  the fraction, by Shakti Singh Delhi. Answered on July 15, 2008 at 10:40 p.m.
Answer 169. Let numerator = x and denominator = y
                        fraction = x/y
                        according to condition 1
                         (x + 2)/(y + 2)     =      9/11
                         11x + 22 = 9y + 18
                         11x - 9y = - 4     ......................(i)
                         according to condition 2
                         (x + 3)/(y + 3)     =      5/6
                         6x + 18 = 5y + 15
                         6x = 5y - 3
                         x   = (5y - 3)/6   ..................... (ii)
                         Substituting value of x from ii in i
                         11[(5y - 3)/6 ] - 9y = -4
                         (55y - 33)/6 - 9y = - 4
                         55y - 33 - 54y = - 24
                         y = 9
                          Substituting value of y in ii
                          x = (5×9 - 3)/6
                          x =  42/6
                          x = 7
                          Therefore required fraction = 7/9
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