Question

The average age of husband wife who were married 8 years ago was 26 years then including two children who were twins the present age of the family is 18 yrs after how many years of their marriage were the twins born?

Answer 1

Let $h$= present age of husband
Let $w$= present age of wife
Let $c$= present age of each child

$8$ years ago, the average of husband and wife is $26$
$\dfrac{h-8+w-8}{2}=26\implies h+w=68$

present average age of family is $18$ :
$\dfrac{h+w+2c}{4}=18\implies h+w=72-2c$

Since the left hand sides are equal in above equations, the right hand sides must also be equal :
$68=72-2c\implies c=2$

That means the twins were born after $8-2=\boxed{6}$ years of marriage.

Answer 2

Let the age of husband at the time of marriage = H yrs
let the age of wide at the time of marriage = W yrs.
average = (H+W)/2 = 26 yrs    =>  H+W = 52 yrs

 Present age of wife = W + 8 yrs
 present age of husband = H + 8 yrs
 present age of each of the twin  = T yrs
  Average of the family =  [ H + 8 + W + 8 + T + T ] / 4  = 18  given
 
               H + W + 2 T + 16 = 72  yrs
               52 + 2 T + 16 = 72  yrs.
                 =>  T  =  2 yrs
As the age of the children is 2 yrs each,  they were born  8 - 2 = 6 yrs after the marriage.

==================
another way:

Average age of husband and wife  NOW = 26 + 8 = 34 yrs.
So the Sum of their ages = 68 yrs

Total age of family = number of persons * average = 4 * 18 = 72 yrs.
So sum of the ages of twins = 72 - 2 * 34 = 4 yrs.
so age of 1 child = 2 yrs.
so children born after 8 - 6 = 2 yrs after marriage