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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
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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
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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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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

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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
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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                  
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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
                                          
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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

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