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.

We have...

1. D=6Y

2. A(days)=100M

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

4. D+W = 10(O)-1

5. D+W+O+Y=71

From 1,3,4,5 we have 4 equations with 4 unknowns. Solve them and you have D=36, W=23,Y=6, O=6.

Substitute 2, and you get the oldest adopted daughter is 6 years and 110 days since today is the youngest daughters birthday. The older daughter is older by 110 days.

As for Dad's favorite day... the only thing I can come up with is that maybe he wasn't married to his wife when they had their youngest daughter, for the mom (if that is the biological mother) would only be 17. 17 is too young in most states?? Maybe this was their wedding date too, but I'm grasping at straws here!!

Posted by Michael on 2004-11-27 04:46:26