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Question 196. Solve for x and y: 10x+4y =3040 and 4x + 10y =765.50 by Aaradhna Tyagi, M.P.S, Ghaziabad. Answered on 05 Jan 2009 at 6:06 p.m.
Answer 196. 10x + 4y = 3040 ......... (i)
                     4x + 10y = 765.5 .......(ii)
                      (i) × 5 - (ii) × 2
                      50x + 20y = 15200
                      8x + 20y = 1531
                       -      -           -
______________________________________
                        42x = 13669
                or       x = 13669/42
                Substituting in (ii)
                      27338/21 + 10y = 765.5
                or   10y = - 11262.5/21
                         y = - 1126.25/21

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Question 197. In a frequency distribution, the mid value of a class is 10 and the width of the class is 8. The lower limit of the class is : (a) 6 (b) 7     (c) 8    (d) 12. By Rajendra. Answered on 7 March 2010 at 12:48pm.
Answer 197. Class Size = 8. Therefore Lower Limit = Mid Value - 8/2
                                                                        = 10 - 4
                                                                        = 6

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