Question on a number line

[dheeraj5519]

Hi,
I faced this question on GMAT prep test. I am rephrasing the question to avoid drawing a number line.

The question was a number line has evenly spaced marks with each mark at 1/5 of the length (so imagine marks at 1/5, 2/5, 3/5..). There are another set of marks on the same number line, which are also evenly spaced, but they are 1/7th of the length (imagine marks at 1/7, 2/7, 3/7 .. so on).

Now the question was what is the minimum distance between any 2 marks.

My first though on it was that its an inequality question
as in |x - 1/5| < 1/7, x being that mark whose distance from nearest 1/5th mark is less 1/7th .. however i could not convince myself that it would represent the minimum distance between the 2 points.

Alternatively i could have taken average of 1/5 and 1/7 and created an inequality like that but i am not sure what would be conceptually correct solution.

So can anyone please explain another method to solve this or the theory behind the inequality.

[LazyNK]

Hey Dhiraj,
First, I will acknowledge that I have worked out the solution, which I am demonstrating below, only after reading your post, so goes without saying that it would not be possible to derive such a relationship on real gmat in only 2 minutes !! . It would probably be much simpler to arrange all the ticks on the number line in ascending order, and identify the least or maximum distance between any two consecutive ticks.
Nevertheless, if it helps you to get a better feel, here is a solution which converts it to an "inequality like problem.

I couldn't convert it to a complete inequality like problem which we are normally used to- there are other general number theory concepts which I had to use.

1.) For this case, we apply ticks at 1/7 and 1/5, and last ticks would align at 1unit corresponding to 7th tick of 1/7 and 5th tick of 1/5. We need to note that 1/7 < 1/5 ( For any other numbers, we need to identify this inequality as a first step).
2.) The smallest distance can be for two cases - case-1 -> a tick for 1/7 is to the right of tick for 1/5 and case-2 -> a tick for 1/7 is to the left of tick for 1/5.
3.) For each 1/7 tick, the absolute number corresponding to that tick is 1/7*n=n/7, where n=1,2,3 etc.
4.)Thus for case 1, the least distance between any two ticks would corresponding to that value of n for which (n/7)/(1/5) leaves least remainder. i.e. 5n/7 leaves least remainder, We can see for n=1,2,3,4,5 and 6, that 5n/7 leaves a remainder of 5,3,1,6,4 and 2 respectively. Thus, the least remainder, which is 1, is corresponding to n=3 i.e the third tick of 1/7 will be the nearest tick to the right of any 1/5 tick, amongst all ticks, and the spacing will be 1/7 * 1/5=1/35
5.) For case 2, i.e. for the one for which we are trying to find that 1/7 tick which is nearest to a 1/5 tick to its right, the tick which will correspond to the least distance will be that 1/7 tick for which (n/7)/(1/5) leaves the maximum remainder. This corresponds to a remainder of 6 corresonding to n/4. And the spacing will be 1/5-6/7*1/5=1/35 again.

This can be used for such problems, in case you can memorize this result, but otherwise, arranging the ticks in ascending order (after converting the fractions to common denominator) is a more practical method.
-NK

[RonPurewal]

this is one of those problems whose purpose is to trap people who try to take theoretical/"textbook" approaches to absolutely everything. as you can see, any such approach to this problem is going to be extremely complicated.

this problem can be solved more practically by simply making a list of where all the tick marks are located. in order to compare them, convert all the denominators to the common denominator of 35.

the fifths are at
1/5 = 7/35
2/5 = 14/35
3/5 = 21/35
4/5 = 28/35

the sevenths are at
1/7 = 5/35
2/7 = 10/35
3/7 = 15/35
4/7 = 20/35
5/7 = 25/35
6/7 = 30/35

now, just locate the pair that is closest together. in this problem, that would be one of two pairs: 14/35 and 15/35, or 20/35 and 21/35.
so, the distance is 1/35.