MANHATTAN GMAT OFFICIAL SOLUTION:

When comparing fractional pieces of a whole, we must find a common denominator. In this case, the 1-mile stretch is divided in thirds, fifths, and eighths. The smallest common denominator of 3, 5, and 8 is 120. If the 1-mile highway is divided into 120 equal increments, where will the red, white, and blue marks fall?

Red (thirds): 40, 80 (out of 120 increments)
White (fifths): 24, 48, 72, 96 (out of 120 increments)
Blue (eighths): 15, 30, 45, 60, 75, 90, 105 (out of 120 increments)

The smallest distance between two marks is 75 – 72 = 3 or 48 – 45 = 3. This equates to 3/120, or 1/40 miles.

The correct answer is D.
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since iam not so good with decimals took the lcm of 8,5,3 which is 120 so if the length of the road is 120 the markings for 1/8 are
15,30,45,60,75,90,105,120
for 1/5 are
24, 48, 72,96 and 120.
for 1/3 are
40, 80,120
so the min difference is between 48 - 45=3
sinc3e the distance is to be found in 1 mile so the ans is 3/120 =1/40 hence D

Although I couldn't see any given Figures however the solution is as mentioned below:

1/8 = 0.125 i.e. Markings will be at {.125, .25, .375, .5, .625, .75, .875}
1/5 = 0.2 i.e. Markings will be at {.2, .4, .6, .8}
1/3 = 0.33 i.e. Markings will be at {.33, .66)

Let's say we have 1 mile stretch as mentioned below

----------------(.125)-----------(.2)-----(.25)------(.33)-------(.375)----(.4)------------(.5)-----------(.6)----(.625)-----(.66)--------(.75)---(.8)-----------(.875)----------------


The Set of all values become

{.125, .2, .25, .33, .375, .4, .5, .6, .625, .66, .75, .8, .875}

Hence the smallest possible difference between any two consecutive numbers will be the smallest piece cut and i.e. 0.4- 0.375 = .025 = 25/1000 = 1/40

Answer: Option
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Added the image. Thank you.
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red line at each third of a mile = \(\frac{1}{3} , \frac{2}{3} , \frac{3}{3}.\) white line at each fifth of a mile = \(\frac{1}{5} , \frac{2}{5} , \frac{3}{5} , \frac{4}{5} , \frac{5}{5}.\) blue line at each eighth of a mile = \(\frac{1}{8} , \frac{2}{8} , \frac{3}{8} , \frac{4}{8} , \frac{5}{8} ,\frac{6}{8} ,\frac{7}{8} ,\frac{8}{8}.\)

LCM of all three Denominators = 3 * 5 * 8 = 120 (I prefer the LCM approach too ,makes it much simpler to look at these numbers, + that's how answers are written too)

red line at each third of a mile = \(\frac{40}{120} , \frac{80}{120} , \frac{120}{120}.\)

white line at each fifth of a mile = \(\frac{24}{120} , \frac{48}{120} , \frac{72}{120} , \frac{96}{120} , \frac{120}{120}.\)

blue line at each eighth of a mile = \(\frac{15}{120} , \frac{30}{120} , \frac{45}{120} , \frac{60}{120} , \frac{75}{120} ,\frac{90}{120} ,\frac{105}{120} ,\frac{120}{120}.\)

look at the options that are closest to each other (\(\frac{48}{120}\), \(\frac{45}{120}\) ) and (\(\frac{72}{120}\) ,\(\frac{75}{120}\) )
closest with difference of = \(\frac{3}{120}\) = \(\frac{1}{3}\)

Ans: D

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Not sure if this works...but this is how I approached this question.

We have paint in 1/3, 1/5, and 1/8 increments.
The smallest distance between two lines has to have a denominator that is at most the denominator of two of these increments (e.g. 5*3=15, 8*3 = 24, 8*5 = 40). I say at most because we may be able to reduce the fraction.

A clearly can't be the answer, and using the logic above B and C cannot be the answer as well. Now I pick multiples of (1/8) and multiples of (1/5) to see if (1/40) works. If I have a denominator of 40, then for (1/5) I'm looking at intervals of 8, 16, 24, 32, 40. And for 1/8 I'm looking for intervals of 5, 10, 15, 20, 25, 30, 35, 40. As you can see this works for 16/40 (which is 2/5) and 15/40 (which is 3/8). (D)

(A) 0
(B) 1/120
(C) 1/60
(D) 1/40
(E) 1/30

Posted from my mobile device

Hi - why is it not 1/120? So if I look at thirds, fifths, eights then LCM = 120
so the lines will be at 3, 6, 9, 12, ... , 120. + 5, 10, 15, 20, 25....,120 + 8, 16..., 120. then the smallest distance is (6-5) = 1/120. The below link solution has taken this approach. Can you please explain what am I missing?
https://gmatclub.com/forum/kim-finds-a- ... 40038.html