manish8109 Posted March 10, 2006 Share Posted March 10, 2006 An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than its predesessor.If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and the actual path would be more than 6 mm but less than 30 mm.Find the time for which the ant moved(in sec). a) 5 sec b) 4 sec c) 6 sec d) 11 sec Quote Link to comment Share on other sites More sharing options...
bungholio Posted March 10, 2006 Share Posted March 10, 2006 where is this question from? whats the fifth choice? Quote Link to comment Share on other sites More sharing options...
manish8109 Posted March 11, 2006 Author Share Posted March 11, 2006 there is no fifth choice i got this question from one of my friend . i will ask him abt the source and let u know. Quote Link to comment Share on other sites More sharing options...
manish8109 Posted March 11, 2006 Author Share Posted March 11, 2006 The first pattern of movement is 3+7+11+15+19+23 The second Pattern of movement is 1+9+17+25+33+41 it is evident that the difference in the net movement b/w the second pattern and the first pattern is more than 6 mm and less than 30mm for a movement of 4 seconds. as, in 4 seconds firstpattern= 3+7+11+15=36mm second pattern= 1+9+17+25=52mm difference =52-36=16mm which lies b/w 6 & 30 Ans is 4sec(B) Quote Link to comment Share on other sites More sharing options...
bungholio Posted March 11, 2006 Share Posted March 11, 2006 NICE thanks! Quote Link to comment Share on other sites More sharing options...
luxsandy Posted March 15, 2006 Share Posted March 15, 2006 AP with n = time taken in seconds 6 Solving, 12 Only n = 4 satisfies the above inequality Quote Link to comment Share on other sites More sharing options...
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