Question : Five members of a team are weighted consecutively, and their average weight is calculated after each member is weighted. If the average weight increases by one kilogram each time, how much heavier is the last player than the first one?
Option 1: 4 kg
Option 2: 20 kg
Option 3: 8 kg
Option 4: 5 kg
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Correct Answer: 8 kg
Solution :
Let the average weight of 1st member be $x$ kg.
Given: The average weight increases by one kilogram each time.
Average weight of first two members $= (x+1)$ kg
Total weight of 1st two members $= 2(x+1) = 2x+2$
Weight of 2nd member $= 2x+2-x = (x+2)$
The average weight of second and third members $= (x+3)$ kg
Total weight of second and third members $= 2(x+3) = 2x+6$
Weight of 3rd member $= 2x+6-x-2 = (x+4)$
The average weight of third and fourth members $= (x+5)$ kg
Total weight of third and fourth members $= 2(x+5) = 2x+10$
Weight of 4th member $= 2x+10-x-4 = (x+6)$
Average weight of fourth and fifth members $= (x+7)$ kg
Total weight of fourth and fifth members $= 2(x+7) = 2x+14$
Weight of 5th member $= 2x+14-x-6 = (x+8)$
$\therefore$ The last member is 8 kg heavier than the first member.
Hence, the correct answer is 8 kg.
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