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2015-04-24T19:01:48+05:30

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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.
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2015-04-25T05:24:40+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.

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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

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