tree branches

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smodak: Senior | Next Rank: 100 Posts; Posts: 69; Joined: Fri Mar 25, 2011 11:08 am; Thanked: 1 times

by smodak » Fri Jun 24, 2011 11:29 am

Kim finds a 1-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?

A) 2/5
B) 7/15
C) 1/2
D) 8/15
E) 3/5

OA: E

Please explain your answer.

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Source: Beat The GMAT — Problem Solving | Fri Jun 24, 2011 11:29 am

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Fri Jun 24, 2011 12:03 pm

you need to produce ascending array for all the markings:
1/3, 2/3, 1
1/5, 2/5, 3/5, 4/5, 1
________________________
multiply everything by 5 and 3 (alowing LCM 15):
5,10,15
3,6,9,12,15 ---> 3,5,6,9,10,12,15 (remember we have common denominator of 15)
_________________________
now we get pieces of distinct sizes (distance between markings)
{3,2,1,3,1,2,3) and we get all repeats in 3,2 and 1, So we subtract 6 and get (15-6)/15 or 3/5

smodak wrote:Kim finds a 1-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?

A) 2/5
B) 7/15
C) 1/2
D) 8/15
E) 3/5

OA: E

Please explain your answer.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jun 24, 2011 3:49 pm

smodak wrote:Kim finds a 1-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?

A) 2/5
B) 7/15
C) 1/2
D) 8/15
E) 3/5

OA: E

Please explain your answer.

Since the question asks for a fraction, the length of the branch can be any value.
Let branch = 15.
Divided into thirds and fifths, we get:

Distinct lengths are 1, 2 and 3.
Total - distinct lengths = 15-1-2-3 = 9.
Remaining length/Total = 9/15 = 3/5.

The correct answer is E.

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