gmatclub clever wire problem

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gmatclub clever wire problem

by arora007 » Tue Jul 06, 2010 9:19 am
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?

a) 1/6
b) 1/5
c) 3/10
d) 1/3
e) 2/5

[spoiler] OA is E, the clever thing is... a wire has 2 ends... and cutting within 1 meter from EITHER ends results in the reqired[/spoiler]
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by kvcpk » Tue Jul 06, 2010 10:02 am
good question!!

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by raunakrajan » Tue Jul 06, 2010 10:26 am
kvcpk wrote:good question!!
correct me if i am wrong please!

the only way of making a square would be if the dimensions are perfect squares?
so in that case if the longer end is 4 mts it would suffice ?
how do you get 2/5?

on another note, dimensions could be anyting right? cz root2 * root2 could also be the dimensions?

how do we go about gettng as 2/5

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by jeremy8 » Tue Jul 06, 2010 10:43 am
raunakrajan wrote:
kvcpk wrote:good question!!
correct me if i am wrong please!

the only way of making a square would be if the dimensions are perfect squares?
so in that case if the longer end is 4 mts it would suffice ?
how do you get 2/5?

on another note, dimensions could be anyting right? cz root2 * root2 could also be the dimensions?

how do we go about gettng as 2/5
Visualize a rope in your head and assume it's 5 mts. You can cut on both sides! That's the trick. So you have a 1/5 chance twice! For example, you can cut within that one meter range on the right side of the rope, OR on the left side of the rope.

I can't believe I got this one right after getting everything wrong today.

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by raunakrajan » Tue Jul 06, 2010 11:26 am
jeremy8 wrote:
raunakrajan wrote:
kvcpk wrote:good question!!
correct me if i am wrong please!

the only way of making a square would be if the dimensions are perfect squares?
so in that case if the longer end is 4 mts it would suffice ?
how do you get 2/5?

on another note, dimensions could be anyting right? cz root2 * root2 could also be the dimensions?

how do we go about gettng as 2/5
Visualize a rope in your head and assume it's 5 mts. You can cut on both sides! That's the trick. So you have a 1/5 chance twice! For example, you can cut within that one meter range on the right side of the rope, OR on the left side of the rope.

I can't believe I got this one right after getting everything wrong today.
ok agreed! how bout the same rope 5 meters long, u can cut it from either of the ends, but u can also cut it from the diameter so it becomes a thinner rope of the same 5 meter length! can we take that possibility?

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by kvcpk » Tue Jul 06, 2010 11:46 am
raunakrajan wrote: ok agreed! how bout the same rope 5 meters long, u can cut it from either of the ends, but u can also cut it from the diameter so it becomes a thinner rope of the same 5 meter length! can we take that possibility?
That will be out of scope for this problem.. We are not given the diameter/radius of the wire. So it might be very minute/ very large.

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by immaculatesahai » Fri Dec 16, 2011 1:55 am
strange problem. good for practice but doubt that we will see anything like this on exam.

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by ankush123251 » Fri Dec 16, 2011 2:45 am
Correct me if i am wrong here but where in the question it is given that the pieces cut must be integers.
Thus,we can have any length cut between length>4 and length<5 and have the prerequisite covered.please explain.

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by GmatMathPro » Fri Dec 16, 2011 3:13 am
ankush123251 wrote:Correct me if i am wrong here but where in the question it is given that the pieces cut must be integers.
Thus,we can have any length cut between length>4 and length<5 and have the prerequisite covered.please explain.
The pieces cut do not have to have integer lengths. Picture placing the 5 meter wire next to a number line with one end at zero and one end at five. If the cut is made anywhere on the interval from 0 to 1 or 4 to 5, the condition is satisfied. The sum of the lengths of these intervals is 2/5 of the length of the total interval from 0 to 5. Thus, there is a 2/5 chance the condition is met.
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