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A group of friends Tom, Tina, Dick, Diana, Harry, and Harriet go out to a fair three hundred meters from the McDonalds which is five KMs away. They see a weighing machine and decide to have some fun. However the girls refuse to step on the weighing machine. So Tom, Dick and Harry, weigh themselves in a particular order. First Tom, Dick, and Harry weigh themselves individually and then Tom and Dick, Dick and Harry, Tom and Harry and then Tom, Dick and Harry together respectively. The recorded weight for the last measure is 161 kgs. The average of all the 7 measures is:
O 92.00
O 115.00
O 207.00
O 53.67
Read Solution (Total 2)
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- let tom,dick and harry be a,b and c..
now there is totally 7 rounds of weight measure..
First Tom, Dick, and Harry weigh themselves individually and then Tom and Dick, Dick and Harry, Tom and Harry and then Tom, Dick and Harry together respectively.
sum of total 7 rounds written as,
a+b+c+(a+b)+(b+c)+(c+a)+(a+b+c) = 4(a+b+c)
4*161=644
average of 7 weighing is 644/7=92kg - 12 years agoHelpfull: Yes(13) No(1)
- sum of all 7 weighings is equal to 4 times the total weight of Tom, Dick and Harry together.
total weight of Tom, Dick and Harry together = 161 ( given as last weight)
so average of 7 weighings = 4*161/7= 92 kg - 12 years agoHelpfull: Yes(5) No(0)
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