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Bunuel
A one meter long wire is to be marked for cutting at intervals of fifths and thirds. How long, in meters, is the largest segment of the wire if the wire is cut at the marked intervals?

A. 1/15
B. 2/15
C. 1/5
D. 4/15
E. 9/15

So our 5ths are marked at 1/5, 2/5, 3/5 and 4/5.

Our 3rds are marked at 1/3 and 2/3.

If we use a common denominator of 15 our 5ths are marked at:

3/15, 6/15, 9/15, 12/15

And our 3rds at:

5/15, 10/15

So the largest segment 3/15 = 1/5 (occurred between the 1/5 mark and one end of the wire,between the 4/5 mark and the other end of the wire, or between the 2/5 mark and 3/5 mark).

Answer: C
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amanvermagmat
Bunuel
A one meter long wire is to be marked for cutting at intervals of fifths and thirds. How long, in meters, is the largest segment of the wire if the wire is cut at the marked intervals?

A. 1/15
B. 2/15
C. 1/5
D. 4/15
E. 9/15

So when we mark at intervals of thirds, the marks are at the positions 1/3 and 2/3 from the starting point.
And when we mark at intervals of fifths, the marks are at the positions 1/5, 2/5, 3/5, 4/5 from the starting point.

To make the comparison easier, we can write 1/3 and 2/3 as 5/15 and 10/15 respectively.
Similarly, we can write 1/5, 2/5, 3/5, 4/5 as 3/15, 6/15, 9/15 and 12/15 respectively.

So, now if we look at all the above 6 marks one by one from the starting point, the marks are at the following positions:
3/15 from start, 5/15 from start, 6/15, 9/15, 10/15 and 12/15 respectively. Cutting at exactly these 6 marks, we will get 7 pieces of wire which will be of following lengths (from the start):

3/15, 2/15, 1/15, 3/15, 1/15, 2/15 and 3/15 respectively. The largest length as we can see is that of 3/15 or 1/5. Hence C answer


I was not able to solve this.. Bunuel's soln seems most elegant and easy to understand for me! What i still can't wrap my head around is that why this question seems so hard!! is it just me? :sad:
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Bunuel
A one meter long wire is to be marked for cutting at intervals of fifths and thirds. How long, in meters, is the largest segment of the wire if the wire is cut at the marked intervals?

A. 1/15
B. 2/15
C. 1/5
D. 4/15
E. 9/15

So when we mark at intervals of thirds, the marks are at the positions 1/3 and 2/3 from the starting point.
And when we mark at intervals of fifths, the marks are at the positions 1/5, 2/5, 3/5, 4/5 from the starting point.

To make the comparison easier, we can write 1/3 and 2/3 as 5/15 and 10/15 respectively.
Similarly, we can write 1/5, 2/5, 3/5, 4/5 as 3/15, 6/15, 9/15 and 12/15 respectively.

So, now if we look at all the above 6 marks one by one from the starting point, the marks are at the following positions:
3/15 from start, 5/15 from start, 6/15, 9/15, 10/15 and 12/15 respectively. Cutting at exactly these 6 marks, we will get 7 pieces of wire which will be of following lengths (from the start):

3/15, 2/15, 1/15, 3/15, 1/15, 2/15 and 3/15 respectively. The largest length as we can see is that of 3/15 or 1/5. Hence C answer


I was not able to solve this.. Bunuel's soln seems most elegant and easy to understand for me! What i still can't wrap my head around is that why this question seems so hard!! is it just me? :sad:


Hi Krish

Have you looked at this solution:

https://gmatclub.com/forum/a-one-meter- ... l#p2001263

The above solution makes it easy I guess.
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I was not able to solve this.. Bunuel's soln seems most elegant and easy to understand for me! What i still can't wrap my head around is that why this question seems so hard!! is it just me? :sad:[/quote]


Hi Krish

Have you looked at this solution:

https://gmatclub.com/forum/a-one-meter- ... l#p2001263

The above solution makes it easy I guess.[/quote]

Yes, it does. thank you. quants solution always looks super easy.. i just couldn't visualize the question :cry: and was taken aback by the wording.
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Visualizing the scenario makes the problem easier.

Imagine a 15-meter long wire. It is marked at 5 meters interval (1/3 of 15)

But the wire is also marked at 3 meters interval (1/5 of 15)

If you draw it properly, you will see the largest segment is 3 meters (1/5 of the wire's length).

Option: C
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