SOLUTION: the sum of ages of husband and his wife is four times the sum of ages of their children.four years ago,the ratio of sum of their ages to the sum of ages of their children was 18:1.
Question 537055: the sum of ages of husband and his wife is four times the sum of ages of their children.four years ago,the ratio of sum of their ages to the sum of ages of their children was 18:1.two 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 h = husbands present age
Let w = wife's present age,
Let c = sum of children' age at this time
:
"the sum of ages of husband and his wife is four times the sum of ages of their children."
h + w = 4c
:
let n = number of children
four years ago, the ratio of sum of their ages to the sum of ages of their children was 18:1. = 18; (You subtract 4 for each person)
h+w-8 = 18(c-4n)
h+w-8 = 18c - 72n
h+w = 18c - 72n + 8
Replace h+w with 4c
4c = 18c - 72n + 8
4c - 18c = -72n + 8
-14c = -72n + 8
14c = 72n - 8; multiplied by -1
:
two years hence the ratio will be 3:1. = 3
h + w + 4 = 3(c+2n)
h + w = 3c + 6n - 4
Replace h+w with 4c
4c = 3c + 6n - 4
4c - 3c = 6n - 4
c = 6n - 4
:
14c = 72n - 8
replace c with (6n-4)
14(6n-4) = 72n - 8
84n - 56 = 72n - 8
84n - 72n = -8 + 56
12n = 48
n = 48/12
n = 4 children
:
:
Check this: find c,
c = 6(4) -4
c = 20; the sum of the children's ages
then
4(20) = 80 the sum of the parents ages
:
Use these values in the last statement:
two years hence the ratio will be 3:1. = = 3
