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Master | Next Rank: 500 Posts
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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: — Data Sufficiency |
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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) :) ..

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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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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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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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by swle24 » Thu Nov 15, 2007 9:18 am
i dont know if this is the only way....
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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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