Question 1161067: A woman is 5 years older than her husband and 10 times as old as their daughter. In 14 years, the sum of their ages will be 100. How old are the woman, husband, and daughter now?
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A woman is 5 years older than her husband
w = h + 5
or
h = w - 5
and 10 times as old as their daughter.
w = 10d
or divide by 10
d = .1w
In 14 years, the sum of their ages will be 100.
(w+14) + (h+14) + (d+14) = 100
w + h + d + 42 = 100
w + h + d = 100 - 42
w + h + d = 58
replace h with (w-5), replace d with .1w
w + (w-5) + .1w = 58
2.1w = 58 + 5
2.1w = 63
w = 63/2.1
w = 30 yrs is the woman's age
then
h = 30 - 5
h = 25 yr is the husband's age
and
d = .1(30)
d = 3 yrs is the daughter's
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
A woman is 5 years older than her husband and 10 times as old as their daughter. In 14 years, the sum of their ages will be 100. How old are the woman, husband, and daughter now?
Let the daughter's age be D
Then woman's age is: 10D, and husband's is: 10D - 5
We then get: D + 10D + 10D - 5 + 3(14) = 100
21D + 37 = 100
21D = 63
Daughter, or 
You think you can now find the 2 other people's ages?
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