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mingshum
Senior Member
USA
29 Posts 
Posted  02/25/2009 : 20:28:01

The average (arithmetic mean) weight of five students is 150.4 pounds. If no student weighs less than 130 pounds and if no two students' weights are within 5 pounds of each other, what is the most, in pounds that any one of the students can weigh?
Answer is 202
I know that the total weight is 752 and that's it! Please help. 


the_hill1962
Advanced Member
USA
1470 Posts 
Posted  02/26/2009 : 08:14:49

Here is an overview of how to work with averages in nonroutine problems: http://www.mathgoodies.com/lessons/vol8/advanced_mean.html Basically in your problem you want four out of the five to be as small as possible (because you want one of the five to be as large (maximum) as possible. Your problem spells out what the smallest can be. It is 130. Your problem also spells out that no two are within 5 pounds of each other. How much bigger is the "next one" than 130 (remember you want to keep them as small as possible). He/she can't be 131 (that would be "within 5 pounts). Go exactly 5 pounds at 135. You see, 130 and 135 are NOT "within" 5 pounds of each other (you see, "within" would not include exactly 5 pounds, "within" would include up to 134.9999999999999... pounds). Now, continuing, the next would be 140 pounds and 145 pounds. You have the average of 150.4 so add them up (denoting "M" to be the variable for the largest person: (M+145+140+135+130)/5 = 150.4 Solving for M gives 202



mingshum
Senior Member
USA
29 Posts 
Posted  02/26/2009 : 11:23:28

Thank you to "the hill1962" for pointing out what "winthin 5 pounds" means! I was adding 6 pounds. 



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