Question 936158: THE SUM OF AGES OF HUSBAND AND HIS WIFE IS 4 TIMES THE SUM OF AGES OF THEIR CHILDREN.4 YEARS AGO THE RATIO OF THEIR AGES TO THE AGES OF THEIR CHILDREN WAS 18:1.
2 YEARS HENCE THE RATIO WILL BE 3:1 . hOW MANY CHILDREN DO THEY HAVE?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! let p = present sum of the parents ages
let c = present sum of the children's ages
let n = number of children
:
THE SUM OF AGES OF HUSBAND AND HIS WIFE IS 4 TIMES THE SUM OF AGES OF THEIR CHILDREN.
p = 4c
4 YEARS AGO THE RATIO OF THEIR AGES TO THE AGES OF THEIR CHILDREN WAS 18:1
That will reduce the sum of the parents ages by 8, the children's age by 4n
p - 8 = 18(c-4n)
p - 8 = 18c - 72n
replace p with 4c
4c - 8 = 18c - 72n
subtract 18c from both sides and rearrange to:
-14c + 72n = 8
:
2 YEARS HENCE THE RATIO WILL BE 3:1.
that will increase the parents age sum by 4, the children's age by 2n
p + 4 = 3(c + 2n)
p + 4 = 3c + 6n
replace p with 4c
4c + 4 = 3c + 6n
subtract 3c from both sides and rearrange to:
c = 6n - 4
:
In the first equation replace c with (6n-4)
-14(6n-4) + 72n = 8
-84n + 56 + 72n = 8
-12n = 8 - 56
-12n = -48
n = -48/-12
n = +4 children
:
Check this out, find c
c = 6(4) - 4
c = 24 - 4
c = 20 is sum of the children's age
then
4(20) = 80 is the sum of parents
Check this in the 1st equation; p - 8 = 18(c-4n)
80 - 8 = 18(20 - 4(4))
72 = 18(20-16)
72 = 18(4)
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