Question

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 the 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 1

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.
$\therefore$ $p=4c$ ....(1)
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
$\therefore$ $p-8=18(c-4n)$
$\therefore$ $p-8=18c-72n$ .....(2)
replace p with 4c (From (1))
$\therefore$ $4c-8=18c-72n$
subtract 18c from both sides and rearrange to:
$\therefore$ $-14c+72n=8$ .......(3)
2 YEARS HENCE THE RATIO WILL BE 3:1.
that will increase the parents age sum by 4, the children's age by 2n
$\therefore$ $p+4=3(c+2n)$
$\therefore$ $p+4=3c+6n$ .....(4)
replace p with 4c (From (1))
$\therefore$ $4c+4=3c+6n$
subtract 3c from both sides and rearrange to:
$c=6n-4$ .....(5)
In the (3) replace c with $(6n-4)$
$\therefore$ $-14(6n-4)+72n=8$
$\therefore$ $-84n+56+72n=8$
$\therefore$ $-12n=8-56$
$\therefore$ $-12n=-48$
$\therefore$ $n=-48/-12$
$\therefore$ $n=4$
Number if children's$=4$