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Re M1417
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16 Sep 2014, 00:53
Official Solution: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. \(\frac{1}{6}\) B. \(\frac{1}{5}\) C. \(\frac{3}{10}\) D. \(\frac{1}{3}\) E. \(\frac{2}{5}\) In order the area of a square to be more than 1, its side must be more than 1, or the perimeter must be more than 4. Which means that the longer piece must be more than 4. Look at the diagram below: If the wire will be cut anywhere at the bolded region, then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is \(\frac{2}{5}\) (2 red pieces out of 5). Answer: E
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Re: M1417
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28 Oct 2014, 03:36
Bunuel wrote: Official Solution: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. \(\frac{1}{6}\) B. \(\frac{1}{5}\) C. \(\frac{3}{10}\) D. \(\frac{1}{3}\) E. \(\frac{2}{5}\) In order the area of a square to be more than 1, its side must be more than 1, or the perimeter must be more than 4. Which means that the longer piece must be more than 4. Look at the diagram below: If the wire will be cut anywhere at the bolded region, then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is \(\frac{2}{5}\) (2 red pieces out of 5). Answer: E Bunnel, I am confused in your explanation as , I got to the logic that in order for area of square to be greater than 1, its perimeter must be greater than 4. Now, Point of confusion is , That there are n ( no. of points ) in between point 45 of rope, How you came up with the idea of region ? Also in case of there are more similar question & you have link, pls share ... Regards LS
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Re: M1417
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28 Oct 2014, 04:59
lastshot wrote: Bunuel wrote: Official Solution: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. \(\frac{1}{6}\) B. \(\frac{1}{5}\) C. \(\frac{3}{10}\) D. \(\frac{1}{3}\) E. \(\frac{2}{5}\) In order the area of a square to be more than 1, its side must be more than 1, or the perimeter must be more than 4. Which means that the longer piece must be more than 4. Look at the diagram below: If the wire will be cut anywhere at the bolded region, then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is \(\frac{2}{5}\) (2 red pieces out of 5). Answer: E Bunnel, I am confused in your explanation as , I got to the logic that in order for area of square to be greater than 1, its perimeter must be greater than 4. Now, Point of confusion is , That there are n ( no. of points ) in between point 45 of rope, How you came up with the idea of region ? Also in case of there are more similar question & you have link, pls share ... Regards LS Don't understand what you mean by the red part above. There are infinite number of points from 4 and 5... Consider a number line: {total} is line segment of 5 units (from 0 to 5) and {favorable} is a line segment of 2 units (from 0 to 1 and 4 to 5), thus P = {favorable}/{total} = 2/5. Check Probability and Geometry Questions to practice from our Special Questions Directory. Hope it helps.
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Re: M1417
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21 Jun 2015, 04:02
the question says area>1 hence the cut must be beyond the red dots shown and there can infinite potints satistfying that condition eg. 4.27 ,4.2899 etc. so how can u say that the facourable cases are 2/5



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Re: M1417
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10 Aug 2015, 04:08
same doubt as others. All we know is the longer piece must be greater than 4.
Now I do not understand, how are you calculating probability. there can be infinite points between 4 to 5 where I can cut the rod such that longer piece can be 4.1, 4.2 , 4.22 , etc etc,
UPDATE: understood what you meant from the link above. Although, it was had to imagine this question in terms of probability.



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Hi is my reasoning correct below? 1. we are given that the wire was cut predeterminedly onto a longer and shorter pieces => the cut point should be further than 2.5 m 2. given that a square of an area more than 1 can be formed if its sides are longer than 1 and hence perimeter is longer by 4 => such a square could be formed in all those incidents when the cut point is located further than 4 3. therefore in total we have a number of outcomes for square area on the length from 2.5 m to 5 m depending on the location of the cutpoint = 2.5 m in total 4. and only 1 m out of this 2.5 m will give us the desired result hence we can form a fraction: \(\frac{1 m}{2.5 m}\) = \(\frac{2}{5}\)
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Re: M1417
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07 Nov 2015, 08:17
Bunuel,
I don't understand how cutting the wire like this x is different from cutting like this x (x=cut) since the wire is symmetrical (I can't tell where is the beginning and where is the end of the wire), so I choose 1/5 when I did this question in my CAT. Can you shed some light on this issue?
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Re: M1417
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10 Nov 2015, 00:41
Maybe this can simplify the question:
Main concept: Probability = (desired outcomes)/(possible outcomes)
Possible outcomes for the longer out of the 2 wires = 52.5 = 2.5 Desired outcomes for the longer wire to be greater than 4 = 54=1
Therefore the probability will be = 1/2.5 = 2/5.



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Re: M1417
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11 Jan 2016, 15:10
I thought of it like this:
In order for the area to be more than one, the long part of the string has to be longer than 4.
If the cut was at 0 (missed) and the string is 5 meters long that is possibility: 1 If the cut was anywhere less than 1 and the long string is anywhere more than 4 that is possibility: 2
2/5
Also, I'd like to add that this question is missing something. 1 what? Square meter I'm assuming.



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Re: M1417
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21 Jul 2016, 05:31
BunuelIs the method below correct ? shasadou wrote: Hi
is my reasoning correct below?
1. we are given that the wire was cut predeterminedly onto a longer and shorter pieces => the cut point should be further than 2.5 m
2. given that a square of an area more than 1 can be formed if its sides are longer than 1 and hence perimeter is longer by 4 => such a square could be formed in all those incidents when the cut point is located further than 4
3. therefore in total we have a number of outcomes for square area on the length from 2.5 m to 5 m depending on the location of the cutpoint = 2.5 m in total
4. and only 1 m out of this 2.5 m will give us the desired result hence we can form a fraction:
\(\frac{1 m}{2.5 m}\) = \(\frac{2}{5}\)
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Re M1417
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16 Oct 2016, 12:00
I think this is a highquality question and I agree with explanation. nice question.. its tricky



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Re: M1417
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21 Jan 2017, 20:08
My take is:
The not favorable condition is when the longer side of the wire is having length any where between 1 to 4...now if we consider there is 10 (or may be 100..any way it will b cancelled) decimal points in between two integers then the total is 5*10=50 (here we are also considering if the length of shorter side is less than 1) outcomes...and not favorable outcomes are 3*10=30....so probability of desired outcome is (1(30/50)) =2/5..
please correct my understanding if i am wrong..



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Re: M1417
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21 Jan 2017, 23:24
I get that probability of cutting a line so that the area is >4 is 1/5. and I get that we have 2/5 instead of 1/5 because the line can be cut from one end to the other in two ways.
But why should we consider two ways of cutting a line? the question is simply asking the probability of cutting a line to make the area >4.
Should we always consider two ways of cutting the line in questions like this?



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Re: M1417
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23 Feb 2017, 21:45
Hie Bunuel,
I see a flaw in the reasoning of the answer. Probability = fav outcome / total outcome if fav. outcome in this case = 2 ( 1 from each side), then total outcome = 10 (i.e. 5 from each side)
correct me if I am wrong



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Re: M1417
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23 Mar 2017, 12:07
Bunuel wrote: Official Solution: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. \(\frac{1}{6}\) B. \(\frac{1}{5}\) C. \(\frac{3}{10}\) D. \(\frac{1}{3}\) E. \(\frac{2}{5}\) In order the area of a square to be more than 1, its side must be more than 1, or the perimeter must be more than 4. Which means that the longer piece must be more than 4. Look at the diagram below: If the wire will be cut anywhere at the bolded region, then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is \(\frac{2}{5}\) (2 red pieces out of 5). Answer: E As there is no order specified in form of marking of ends of the wire then why are we considering 2 cases where wire is cut from left or from right



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Re: M1417
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24 Mar 2017, 04:46
smanujahrc wrote: Bunuel wrote: Official Solution: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. \(\frac{1}{6}\) B. \(\frac{1}{5}\) C. \(\frac{3}{10}\) D. \(\frac{1}{3}\) E. \(\frac{2}{5}\) In order the area of a square to be more than 1, its side must be more than 1, or the perimeter must be more than 4. Which means that the longer piece must be more than 4. Look at the diagram below: If the wire will be cut anywhere at the bolded region, then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is \(\frac{2}{5}\) (2 red pieces out of 5). Answer: E As there is no order specified in form of marking of ends of the wire then why are we considering 2 cases where wire is cut from left or from right Order has nothing to do eith it. Again: If the wire will be cut anywhere at the red region, then the rest of the wire (longer piece) will be more than 4 meter long.
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Re: M1417
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14 Nov 2017, 23:37
Excellent question!
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