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xcise_science: Master | Next Rank: 500 Posts; Posts: 106; Joined: Tue Oct 09, 2007 12:25 pm; Thanked: 1 times

by xcise_science » Tue Nov 13, 2007 6:35 pm

I did this problem somewhere else, and I thought I got it right, but I guess not.
It looks pretty easy, but I got it wrong, can someone help?

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Source: Beat The GMAT — Data Sufficiency | Tue Nov 13, 2007 6:35 pm

gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member; Location: India; Thanked: 51 times

by gabriel » Wed Nov 14, 2007 11:03 pm

There seems to be something wrong with the image ... anyway it would be a lot easier if you type out the questions, will be a help to you in AWA too(you know.. by improving your typing speed)

..

Regards

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xcise_science: Master | Next Rank: 500 Posts; Posts: 106; Joined: Tue Oct 09, 2007 12:25 pm; Thanked: 1 times

by xcise_science » Thu Nov 15, 2007 7:08 am

Oops, sorry for that.
here it is.

0------!-|----!----|-!-----!-|----!----|-!----1

On the number line above, the segment from 0 to 1 has been divided into fifths, as indicated by the large tick marks, and also into sevenths as indicated by the small tick marks. What is the least possible difference between any two of the tick marks?

a) 1/70
b) 1/35
c) 2/35
d) 1/12
e) 1/7

correct answer is B

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swle24: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Oct 11, 2007 2:43 pm

by swle24 » Thu Nov 15, 2007 8:50 am

1/35 is the answer. 1/5=7/35 , 2/5=14/35, 3/5=21/35, 4/5=28/35, and then convert 1/7=5/35 , 2/7=10/35 , 3/7=15/35 , 4/7=20/25 , 5/7=25/35, 6/7=30/35 so the least difference is 15/35-14/35=1/35

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xcise_science: Master | Next Rank: 500 Posts; Posts: 106; Joined: Tue Oct 09, 2007 12:25 pm; Thanked: 1 times

by xcise_science » Thu Nov 15, 2007 9:12 am

Thanks, but I’m not clear on why you solved it this way, is this the only way?
Or maybe I didn't understand the question?

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swle24: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Oct 11, 2007 2:43 pm

by swle24 » Thu Nov 15, 2007 9:18 am

i dont know if this is the only way....

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samirpandeyit62: Master | Next Rank: 500 Posts; Posts: 460; Joined: Sun Mar 25, 2007 7:42 am; Thanked: 27 times

by samirpandeyit62 » Thu Nov 15, 2007 11:00 pm

LCM of 5 & 7 is 35

so let us assume that 1 corresponnds to 35

so scale is

0 7 14 21 28 35

0 5 10 15 20 25 30 35

smallest diff = 1/35 (14,15 & 21,20)

this would suffice as we are increaseing all values in the saeme proportion.

Regards
Samir

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