 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  The Fair (Posted on 2004-11-08) 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.

 See The Solution Submitted by Jeremy Trenary Rating: 3.6667 (6 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Answer + Solution/Explanation | Comment 3 of 19 | First, I will call their ages D (dad), M (mom), Y (youngest), E (eldest), A (adopted), B (birthday girl).

Ok, so since we are supposed to ignore leap years, I can say every year has 365 days in it. Let’s look clue #2. If it is the adopted daughter’s birthday, then her age in days is exactly 365*A. Then 100M = 365A. So 20M = 73A. In order for A to be an integer (since it is exactly her birthday), M must be a multiple of 73. But the sum of all their ages is 71 (clue #5). So that can’t be it. Therefore I don’t think that it is the adopted daughter’s birthday.

Anyways,
D = 6Y. Y = D/6
D+3 = 3(D-M). M = (2D-3)/3. M = 2D/3–1
D+M = 10E–1. E = (D+M+1)/10. E = (D + (2D/3–1)+ 1)/10. E = D/6

D+M+Y+E = 71 = D + 2D/3–1 + D/6 + D/6 = 2D – 1.
D = 36.

M = 23
Y = 6
E = 6

Ok, so both girls are 6, just one is exactly 6 (because it’s her birthday) and one is 6 and some days. So the elder one must be the 6 and some days. So we have it is the younger daughter’s birthday, she is the natural born daughter, and the elder daughter is the adopted daughter.

Now for their age difference. So the elder/adopted daughter's age in days is E = 6*365 + X. And remember, 100M = 6*365 + X = 100*23. So 2300 = 2190 + X. X = 110.

So their age difference is 110 days.

I hope I understood the problem correctly!

PS: I forgot about the Bonus question… I think Ruth is right =)

 Posted by nikki on 2004-11-08 13:40:01 Please log in:

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