The sum of ages of husband and his wife is four times the sum of ages of their children.four year ago, the ratio of sum of their ages to the sum of sum of ages of their children was 18:1.two year hence the ration will be3:1 how many children do they have?

1
by vpl

2015-10-16T13:42:31+05:30

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Husband's age = X  years.                  four years ago:  X - 4 yrs
wife's age = Y yrs                              four yrs ago = Y - 4 yrs

Let them have  N children.

Let sum of ages of their children = Z  yrs
this age four years ago = Z - 4 * N  yrs
this age after 2 yrs from now :  Z + 2 * N  years

Given
X + Y = 4 * Z          ---- (1)

Four years ago:
[ (X-4) + (Y-4) ] :  (Z - 4* N)  =  18  :  1
(X + Y - 8) = 18 (Z - 4 * N)
X + Y  = 18 Z - 72 N  + 8        --- (2)

from (1) and (2):
14 Z + 8 =  72 * N
N = (7 Z + 4)/ 36      --- (3)

As  N is an integer,   7 Z + 4 = 36 or 72 or 108 or 144  or 180 ...
Then  Z =  20     for  7 Z + 4 = 144.
Then N = 4

So  X + Y = 80   from (1)

Two years from now:
sum of ages of parents = X+2+Y+ 2  = 84  yrs
sum of ages of  children =  20 + 2 * 4 =  28 years

Their ratio  =  84: 28 = 3 : 1

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they have four children... the sum of the childrens ages = 20 yrs.
sum of parents ages = 80 yrs