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.

(In reply to Answer + Solution/Explanation by nikki)

 

You can get the parents ages as follows:

10 x daughter -1  = parents ages.

 

But 10D cannot be 70 because that doen't leave enough years for the others within the 71 total. and cannot be 50, because that would make the older D 5 and the younger 17, which is impossible.

 

So 10D must be 60 and parents ages total 59. That makes older daughter 6, but then there are only 6 years left, so younger daughter must be six.

 

Dad is 6 x D  = 36, so mom is 59-36 = 23.

 

This chekcs out because 36-23 = 13 and in three years dad will be 39 = 13x3

 

Posted by andrew on 2004-11-14 05:28:19