A couple is taking their two daughters to a local fair. They are celebrating a birthday of one of the girls. It also marks the fathers’ favorite day of the year because it reminds him how lucky he is to not be barred from such festivities. Use the clues below to determine the ages each family member; which daughter is adopted; which daughter is the birthday girl; what is the difference in age of the two girls.

Bonus question: Why is it the fathers favorite day?

1. Dad is six times older than his youngest daughter.
2. The adopted daughters' age in days is 100 times her adopted mother's age in years.
3. In three years the dad will be exactly three times the difference in age between him and his younger wife.
4. The age of the two parents together is one less than ten times the eldest daughter's age.
5. The combined age of all four family members in whole years is seventy-one.

Note: Ages are to be taken in years unless otherwise stated and assume there is no leap year.

Dad = D, Mom = M , Youngest = Y , Eldest = E

1. D = 6Y

2.   (100)M = Adopted x 365

3. D+3= 3(D-M)

4.  D+M = (10)E - 1

5. M+D+Y+E = 71

Dad is 36, mom is 23, Eldest daughter is 6, youngest daughter is 6

The youngest has to be the birthday girl, else they wouldn't both be 6, The difference is age of the girls appears to be around 3.5 months, the adopted girl is the older child.

Posted by whome? on 2005-01-23 18:16:46