Question 1151468: A 40-year-old man has three daughters, ages 6, 3 and 1. In how many years will the combined ages of his daughters equal 80% of his age?
Found 2 solutions by jim_thompson5910, MathTherapy:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
The question asks "In how many years will the combined ages of his daughters equal 80% of his age?". Since we don't know right now, let's just call that x.
x = number of years that go by
Currently his daughters, who we'll call A, B, C for shorthand, are ages 6, 3 and 1.
A = 6
B = 3
C = 1
Those are the current ages or present day ages.
If we add on x years, then
A' = 6+x
B' = 3+x
C' = 1+x
represent the future ages of each daughter
Their combined ages in that time would be
A' + B' + C' = (6+x)+(3+x)+(1+x)
A' + B' + C' = 3x+10
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The father is currently 40 years old. After x years go by, he is going to be 40+x years old.
80% of this is (80/100)*(40+x) = 0.80*(40+x) = 32+0.8x
Set 32+0.8x and 3x+10 equal to each other, then solve for x
3x+10 = 32+0.8x
3x-0.8x = 32-10
2.2x = 22
x = 22/2.2
x = 10
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If x = 10 years go by, then the daughters' ages will update to
A' = 6+x = 6+10 = 16
B' = 3+x = 3+10 = 13
C' = 1+x = 1+10 = 11
Giving a combined sum of A'+B'+C' = 16+13+11 = 40
Let p = 40.
At the same time, the father goes from age 40 to age 50 after 10 years.
Let q = 50.
Note how p/q = 40/50 = 0.80 = 80%
Showing that the combined ages, 10 years into the future, of the daughters represents 80% of the father's future age during the same timespan.
Answer: 10 years
Answer by MathTherapy(10553)  (Show Source):
You can put this solution on YOUR website!
A 40-year-old man has three daughters, ages 6, 3 and 1. In how many years will the combined ages of his daughters equal 80% of his age?
Let number of years be y
Then at that time, father will be 40 + y, and with his 3 daughters' ages now summing to 10 (6 + 3 + 1), the daughters will be 10 + 3y, at that time
We then get: 10 + 3y = .8(40 + y)
10 + 3y = 32 + .8y
3y - .8y = 32 - 10
2.2y = 22
Number of years, or 
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