Question 1151468
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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: <font color=red size=4>10 years</font>
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