Toughest Puzzel Ever
A man was dividing his property among his children. This was what
the man told to his first son “Take as many number of cattle as you
can care for and your wife may take one-ninth of the remaining
number.” To his second son he said “Take one more than what the first
son took and your wife will have one ninth of the cattle remaining after
you have taken.” This applied to the remaining sons too i.e. each son
would take one more than the next oldest brother and their wives
would have one ninth of the remaining after their husbands have got
their share.
After the cattle were divided, the man proceeded to divide the gold
bars he had among the children. Each gold bar was valued at 3.5 times
each cow's value. The gold bars were divided such that each couple had
equally valued inheritance. What are the number of cows, gold bars
and the number of sons the man had?
the man told to his first son “Take as many number of cattle as you
can care for and your wife may take one-ninth of the remaining
number.” To his second son he said “Take one more than what the first
son took and your wife will have one ninth of the cattle remaining after
you have taken.” This applied to the remaining sons too i.e. each son
would take one more than the next oldest brother and their wives
would have one ninth of the remaining after their husbands have got
their share.
After the cattle were divided, the man proceeded to divide the gold
bars he had among the children. Each gold bar was valued at 3.5 times
each cow's value. The gold bars were divided such that each couple had
equally valued inheritance. What are the number of cows, gold bars
and the number of sons the man had?
Replies
-
harshajd"After the cattle were divided, the man proceeded to divide the gold
bars he had among the children. Each gold bar was valued at 3.5 times
each cow's value. The gold bars were divided such that each couple had
equally valued inheritance"
I will not worry about the number of Gold bars as, if each couple should get equal valued inheritance then I assume that the number of cattles are also shared equally.
The no of cattles given to wives will be in multiples of 9 (as it's 1/9th part of remaining, and result should be a whole number)
With trial and error we get,
the number of cows = 64
and No of sons = 8
elder son takes 1 cow n his wife 7 (63/9) - leaves 56 cows
second one takes one more than his elder, so 2 n his wife 6 (54/9) - leaves 48 cows
third son takes 3 cows, n his wife 5(45/9) - leaves 40 cows
..
..
Seventh son takes 7 cows and his wife 1(9/9) - leaves 8 cown
Last (i.e eight) son takes 8 cows and his wife none -
zaveribut this is an age-old question
You are reading an archived discussion.
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