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Answers to Your Questions - Page 18
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Question 170.  The sum of digits of a two digit number is 13.If the number is subtracted from one obtained by interchanging the digits,the result is 45 .what is the number by Robin Prasad, Varanasi. Answered on July 18, 2008 at 9:44 p.m.
Answer 170.   Let digit at units place = x
                                 digit at tens place = y
                                 Number = 10y + x
                                 Number formed by interchanging the digits = 10x + y
   condition 1            10x + y - (10y + x) = 45
                                 10x + y - 10y - x = 45
                                  9x - 9y = 45
                                  x - y = 5             ...............(i)
    condition 2           x + y = 13         ................(ii)
                                   (i) + (ii)
                                  2x = 18
                                   x = 9
                                   Substituting in (i)
                                   9 - y = 5
                                    y = 4
                                    Given No. = 49

 Question 171. The sum of a number and its +ve square root is 6/25. Find the number, by Manan Bhat. Answered on 18 August 2008 at 8:54 p.m.
Answer 171. Let the no = y
                       according to problem
                       y + square root y = 6/25
                       square root y = x
                       x^2 + x - 6/25 = 0
                       25 x^2 +25 x - 6 = 0
                       (5x + 6)(5x - 1) = 0
                       x = -6/5,  x = 1/5
                       Putting back the value of x in terms of y
                       square root y = -6/5 , square root y = 1/5
                       Squaring both sides
                       y = 36/25,    x = 1/25
Question 172. Factorize x^8-11x^4y^4-80y^8 by Vaxy Zoya. Answered on Sep 11, 2008 at 9:00 p.m.
Answer 172.    x^8 - 11x^4y^4 - 80y^8
                          = x^8 - 16x^4y^4 + 5x^4y^4 - 80y^8
                          = x^4 (x^4 - 16y^4) + 5y^4 (x^4 - 16y^4)
                          = (x^4 - 16y^4)(x^4 + 5y^4)
                          = [(x^2)^2 - (4y^4)^2 ](x^4 + 5y^4)
                          = (x^2 - y^2)(x^2 + y^2)(x^4 + 5y^4)
                          = (x - y)(x + y)(x^2 + y^2)(x^4 + 5y^4)
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Question 173. The age of father is equal to a sum of the ages of his 5 children . After 15 years sum of ages of the  children will  be twice the age of the father. Find the age of the father, by Manan. Answered on September 14, 2008.
Answer 173.   Let present age of father = x years
                         Present age of 5 children together = x years
                         fathers age after 15 years = ( x + 15 ) years
                         age of 5 children after 15 years = ( x + 15×5) years
                         according to question
                          x + 15×5 = 2 (x + 15)
                          x + 75 = 2 x + 30
                          x = 45
                          Therefore fathers age = 45 years 
Question 174. If sinA = 4/5, Find the value of 4tanA - 5cosA/secA + 4cotA by ashi singhai from chhindwara. Answered on September 23, 2008 at 11:18 a.m. IST.
Answer 174. Sin A = 4/5 . Let opposite side = 4 k and hypotenuse = 5 k.
                                              Using Pythagoras theorem adjacent side = 3 k
                                              4tanA - 5cosA/secA + 4cotA
                                           = 4×4/3 - 5(3/5 ÷ 5/3) + 4(3/4)
                                           = 16/3 - 5(9/25) + 3
                                           = 16/3 - 9/5 + 3
                                           = (80 - 27 + 45)/15
                                           = 98/15
Question 175. A person on tour has Rs. 360 for his daily expenses. If he exceeds his tour programme by 4 days, he must cut down his daily expense by Rs.3 per day. Find the number of days of his tour programme. Answered on Oct 9, 2008 at 4:05 p.m.
Answer 175.   Let number of days be x
                         according to question
                         360/x  -  360/(x +4)  =  3
      (÷3)          120/x  -  120/(x +4)  =  1
                         120(x + 4 - x)/(x^2 + 4x) = 0
                          x^2 + 4x - 480 = 0
                          (x + 24)(x - 20) = 0
                          x = - 24, x = 20
                          Original number of days = 20

Question 176. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of AP by Monila Jaipur. Answered on Oct 19, 2007 at 7:15 p.m. I.S.T
Answer 176.  T4 + T8 = 24
                         a + 3d + a + 7d = 24
               or       2a + 10d  = 24
               or       a + 5d  = 12        .................(i)
                          T6 + T10 = 44
                          a + 5d + a + 9d = 44
             or       2a + 14d  = 44
                or       a + 7d  = 22      .................(ii)
                Subtracting (i) from (ii)
                           2d = 10
                or        d = 5
                Substituting in (i)
                            a + 5× 5  = 12
                            a =  - 13
                 The first three terms are - 13, - 8, - 3

                                   See Answers to Your questions
Question 177. In a parallelogram ABCD angle A is 3x - 2 and angle C is 2x + 23.Find angle A and B, by Ketav Manik. Answered on November 8, 2008 at 9:11 p.m. I.S.T.
Answer 177. Download Answer.
Question 178. If the heigt of the tower is 100 metre and angle of the elevation is 30 degree. Find the distance of the tower from the point of observation? by Rahul Kumar, St. Josephs school, Muzaffarpur. Answered on 24 Dec 2008
Answer 178. Let AB represent tower and C point of observation. angle ABC = 90.
tan C = AB/BC
tan 30 = 100/BC
1/sqareroot3 = 100/BC
BC = 100(squareroot3)
BC = 173 m.
Question 179. Solve for x, 1/(x + a + b) = 1/a + 1/b + 1/x, by Simran Gill, Answered on 8 May 2009 at 5:00 p.m.
 Answer 1791/(x + a + b) = 1/a + 1/b + 1/x
                                        or    1/(x                        + a + b) - 1/x = 1/a + 1/b
                                        or    ( x - x - a - b) / x(x + a + b) = (a + b)/ab
                                        or     - (a + b) / x(x + a + b) = (a + b)/ab
                                        or     - 1/ x(x + a + b) = 1/ab
                                        or      x^2 + ax + bx + ab = 0
                                        or       x( x+ a ) + b ( x + a) = 0
                                        or       (x + a )( x + b ) = 0
                                                   x = - a , x = - b

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