Question 1002068: Find four consecutive integers such that the sum of the first three is 54 more than the fourth.
Found 3 solutions by chipchocolate, Timnewman, MathTherapy:
Answer by chipchocolate(1)   (Show Source):
You can put this solution on YOUR website! Let x = the first number
(x+1) = the second
(x+2) = the third
and (x+3) = the fourth
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x+(x+1)+(x+2)=54+(x+3)
3x + 3 = 57 + x
2x = 54
x=27
••••••••••••••••••••••••
The numbers are:
27,28,29, and 30
Answer by Timnewman(323)  (Show Source):
You can put this solution on YOUR website! Hi dear,
let the interger be x,x+1,x+2,andx+3,.
since,the sum of the first three is 54 more than the fourth,
Then,
x+x+1+x+2=x+3+54
3x+3=x+57
2x=57-3
2x=54
x=27
The consecutive intergers are:27,28,29,and 30
OR
-27,-28,-29 and -30
HOPE THIS HELPS?
Answer by MathTherapy(10551)  (Show Source):
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