SOLUTION: A woman has a daughter and a son. The son is three years older than the daughter. In one year the woman will be six times as old as the daughter is now, and in ten years she will b
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Question 210024: A woman has a daughter and a son. The son is three years older than the daughter. In one year the woman will be six times as old as the daughter is now, and in ten years she will be fourteen years older than the combined ages(in ten years) of her children. What is the woman's present age?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A woman has a daughter and a son. The son is three years older than the daughter.
In one year the woman will be six times as old as the daughter is now, and in
ten years she will be fourteen years older than the combined ages(in ten years)
of her children. What is the woman's present age?
:
Let w = woman's present age
Let d = daughter's present age
Let s = son's present age
:
"The son is three years older than the daughter."
s = d + 3
:
"In one year the woman will be six times as old as the daughter is now,"
w + 1 = 6d
w = 6d - 1
:
"in ten years she will be fourteen years older than the combined ages
(in ten years) of her children."
w + 10 = (d+10) + (s+10) + 14
:
w + 10 = d + s + 20 + 14
:
w + 10 = d + s + 34
:
w = d + s + 34 - 10
:
w = d + s + 24
;
Replace w with (6d-1) and s with (d+3)
6d - 1 = d + (d+3) + 24
6d = 2d + 3 + 24 + 1
6d - 2d = 28
4d = 28
d =
d = 7 yr daughters present age
:
What is the woman's present age?\
:
we know w = 6d -1
w = 6(7) - 1
w = 42 - 1
w = 41 is woman's present age
;
;
Check:
s = d + 3
s = 7 + 3
s = 10 yrs is son's present age
:
Check solution in statement:
"in ten years she will be fourteen years older than the combined ages
(in ten years) of her children."
41 + 10 = (7+10) + (10+10) + 14
51 = 17 + 20 + 14
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