Question 94817: I have a worded problem that is giving me trouble and I would really appreciate if someone could assist me in making sense of it all. It goes like this.
Four brothers are born in successive intervals of three years. The eldest is twice the age of the youngest. What is the age of their father, who is eighteen years older than the eldest son?
I have given this a go, but all I am able to come up with is this
B1= Brother 1 =9
B2= Brother 2 =12
B3=Brother 3 =15
B4=Brother 4 =18
F=Father =36
B1+B2+3+B3+6B4+9
4B=+18
F=2B4+9
My problem is even though I think I can get an answer to this problem in my head. I am unable to sort it into an algebraic equation, which is what I desperately need. Any help would be greatly appreciated. Thank you so much.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Four brothers are born in successive intervals of three years. The eldest is twice the age of the youngest. What is the age of their father, who is eighteen years older than the eldest son?
:
Let x = age of the youngest
Then the remaining brothers' ages would be: x+3, x+6, x+9
:
It says,"The eldest is twice the age of the youngest."
(x+9) = 2x; you can do this in your head, but continuing in algebra
9 = 2x - x
x = 9 age of the youngest
:
Oldest, obviously is x + 9 = 18, + 18 = 36 is father
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