SOLUTION: The AVERAGE weight of the top 5 fish caught in the tournament was 13. The second fish weighed 13.5 lbs. the third fish weighed 12.7 lbs, the fourth fish weighed 12.1 and the 5th fi

Algebra ->  Average -> SOLUTION: The AVERAGE weight of the top 5 fish caught in the tournament was 13. The second fish weighed 13.5 lbs. the third fish weighed 12.7 lbs, the fourth fish weighed 12.1 and the 5th fi      Log On


   

Question 1047278: The AVERAGE weight of the top 5 fish caught in the tournament was 13. The second fish weighed 13.5 lbs. the third fish weighed 12.7 lbs, the fourth fish weighed 12.1 and the 5th fish weighed 11.8. What was the weight of the heaviest fish?
PLEASE EXPLAIN THE STEPS. My teacher does not explain how to arrive at the answer or how to figure out a an average with one of the numbers missing.
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
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The AVERAGE weight of the top 5 fish caught in the tournament was 13. The second fish weighed 13.5 lbs.
the third fish weighed 12.7 lbs, the fourth fish weighed 12.1 and the 5th fish weighed 11.8.
What was the weight of the heaviest fish?
PLEASE EXPLAIN THE STEPS. My teacher does not explain how to arrive at the answer or how to figure out
an average with one of the numbers missing.
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Let w1, w2, w3, w4 and w5 be the weights of the 5 fish . . . 
(We assume that these values are ordered in a descending order: w1 is greatest . . . ).

You are given that 

%28w1+%2B+w2+%2B+w3+%2B+w4+%2B+w5%29%2F5 = 13    ( "The AVERAGE weight of the top 5 fish caught in the tournament was 13." )

Hence, 

w1 + w2 + w3 + w4 + w5 = 13*5 = 65.

Agree ? 

You are also given all the values w2, w3, w4 and w5.

From this point, you can easily find the unknown w1.

Please complete the assignment on your own.