Question 362380: John and Marilyn are married and have two kids, Tom and Traci. If tom were one year younger, Traci would be half his age. Marilyn is four times Tom's age, and John is five years younger than five times Tom's age. All of them together have celebrated 68 birthdays. How old is each?
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! John and Marilyn are married and have two kids, Tom and Traci. If tom were one year younger, Traci would be half his age. Marilyn is four times Tom's age, and John is five years younger than five times Tom's age. All of them together have celebrated 68 birthdays. How old is each?
..
tom now x
one year younger = x-1
..
traci = 1/2 *(x-1)
..
marilyn = 4x
..
John 5x-5
...
x+1/2(x-1)+4x+5x-5=68
(2x+x-1+8x+10x-10)/2 = 68
21x-11=2*68
21x=136+11
21x=147
x=7 tom ( plug values of x to get other ages)
traci = 3
marilyn = 28
john = 30
...
7+3+28+30=68
...
m.ananth@hotmail.ca
|
|
|
