distinct lengths ??

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distinct lengths ??

by gmatpup » Thu Nov 10, 2011 11:51 am
Kim finds a one meter tree branch and marks it off in thirds and fifths. She then breaks the branch along all the markings and removes one piece of every distinct length. What fraction of the original branch remains?


Why are the distinct lengths 1, 2, and 3? :o

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by gmatclubmember » Thu Nov 10, 2011 11:56 am
The marking will be at :
1/5,1/3,2/5,3/5,2/3, and 4/5.
The lengths between any 2 successie markings would be
1/5,2/15,1/15,1/5,1/15, and 2/15
Out of these first 3 are unique.So upto 2/5 the lenghts would be unique. remaining is 3/5. Is that the answer?
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by neelgandham » Thu Nov 10, 2011 1:34 pm
Let us assume that the length of the branch is 15 units (Why, an assumption ? Because the question asks about 'Fraction of Original Length' which is a relative measure, so an assumption makes solving easy).

If such branch is broken as per the conditions in the question, it will result in

2 pieces of 1 Unit length,
2 pieces of 2 Units length, and
3 pieces of 3 Units length.

If one piece of every distinct length is removed from the branch, then it will result in

1 pieces of 1 Unit length,
1 pieces of 2 Units length, and
2 pieces of 3 Units length.

Length of the remains = (1*1) + (1*2) + (2*3) = 6+2+1 = 9
Original length of the branch = 15
Fraction of the original branch which remains = 9/15 = 3/5.

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