Question

5 members of a team are weighed consecutively and their average weight calculated after each member is weighed. if the average weight increase by one kg each time ,how much haviers is last player than first player..?​

Answer 1

Answer:

The last player is four more than twice the first player

Step-by-step explanation:

Let the weight of five players be a, b, , d,e respectively

By the given data , the average increases by 1 as each member is getting added

a+1=(a+b)÷2

a+2=b

(a+b)÷2 +1 =(a+b+c)÷3

(2a+2)÷2+1= (2a+2+c)÷3

(a+2)×3= 2a+ 2+c

4a=2+c

c=4a-2

(a+b+c)÷3+1=(a+b+c+d)÷4

(a+a+2+4a-2)÷3 +1= (a+a+2+4a-2+d)÷4

2a+1= (6a+d)÷4

8a+4= 6a+d

2a+4=d

(a+b+c+d)÷4+1=(a+b+c+d+e)÷5

(a+a+2+4a-2+2a+4)÷4+ 1=(a+a+2+4a-2+2a+4+e)÷5

2a+2= (8a+6+e)÷5

10a+10=8a+6+e

2a+4=e

The last player is four more than twice the first player