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Kim finds a 1-meter tree branch and marks it off in...

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by swerve » Wed Mar 14, 2018 1:23 pm
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

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
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Source: — Problem Solving |

by Jay@ManhattanReview » Thu Mar 15, 2018 9:23 pm
swerve 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

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
Pl. see the image below to get this better. The 1-meter tree branch is shown as a piece of 15 parts as the LCM of 1/3 and 1/5 is 1/15.

Image

The part of the branch after its 2/5th part is unwanted as the further cuts do not provide distinct lengths.

The fraction of the original branch remains = 1 - 2/5 = 3/5.

The correct answer: E

Hope this helps!

-Jay
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by GMATGuruNY » Fri Mar 16, 2018 12:57 am
swerve 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
Ignore the given length of 1-meter. The problem can be solved using any length.
Let length = 15 meters.
Dividing 15 into 3rds will yield markings at 5 and 10.
Dividing 15 into 5ths will yield markings at 3, 6, 9, and 12.

Listing the markings in order:
0......3....5..6......9..10....12......15

There are only 3 distinct lengths: 1 meter, 2 meters, and 3 meters.
Subtracting these 3 distinct lengths from 15, we get:
Length remaining = 15-1-2-3 = 9.
Remaining length/Total length = 9/15 = 3/5.
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