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Five identical pieces of wire are soldered together endtoend to form
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Updated on: 24 Apr 2015, 03:40
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64% (02:55) correct 36% (01:42) wrong based on 216 sessions
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Five identical pieces of wire are soldered together endtoend to form one longer wire, with the pieces overlapping by 4cm at each joint. If the wire thus made is exactly 1 meter long, how long, in centimeters, is each of the identical pieces? (A) 21,2 (B) 22 (C) 23,2 (D) 24 (E) 25,4 The answer key says 23,3 cm. (C) According to the answer key there are four overlaps with 4cm so you have to calculate (100+16)/5.
Now, I am wondering whether there are not more overlaps. I would think there are 8 overlaps. As each overlapping section is overlapping with another section, so sections that are overlapped on 2 ends have to be added 2 times.
______ _______ ________ _______ _________
So I would think the true answer must be (100+32)/5 = 26,4 cm. Which unfortunately is not an answer choice. What do you think? Am I misunderstanding the problem?
Thanks for your help in advance.
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Originally posted by ailujkh on 24 Apr 2015, 03:36.
Last edited by Bunuel on 24 Apr 2015, 03:40, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Five identical pieces of wire are soldered together endtoend to form
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24 Apr 2015, 10:45
ailujkh wrote: Five identical pieces of wire are soldered together endtoend to form one longer wire, with the pieces overlapping by 4cm at each joint. If the wire thus made is exactly 1 meter long, how long, in centimeters, is each of the identical pieces? (A) 21,2 (B) 22 (C) 23,2 (D) 24 (E) 25,4 The answer key says 23,3 cm. (C) According to the answer key there are four overlaps with 4cm so you have to calculate (100+16)/5.
Now, I am wondering whether there are not more overlaps. I would think there are 8 overlaps. As each overlapping section is overlapping with another section, so sections that are overlapped on 2 ends have to be added 2 times.
______ _______ ________ _______ _________
So I would think the true answer must be (100+32)/5 = 26,4 cm. Which unfortunately is not an answer choice. What do you think? Am I misunderstanding the problem?
Thanks for your help in advance.
_______________________________________________ Feel free to share Kudos if this was helpful to you. I agree with OA, there are 4 overlapping. For first two pieces of wire, for e.g., the second wire overlaps by 4 cms. If you move the second wire from top of the first one and place it right at the very end of it, you'll gain just those 4 cms in length(and not 8 cms.). Same for 5 wires.



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Re: Five identical pieces of wire are soldered together endtoend to form
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24 Apr 2015, 16:03
Hi ailujkh, This is a rarer conceptual issue on the GMAT (you likely will NOT see it on Test Day), but there's a 'visual component' to the question that you can use to help you figure out how the math "works." From your notes, it looks like you started to visualize what the wires would look like, which is good. Before getting into the specifics of this prompt though, it would probably help to start with a simpler example.... What would it look like if you took two 10meter wires and overlapped them for 4 meters? Try drawing it. What would the total length be? Now what would happen if you added a third 10meter wire and overlapped it in the same way? Using this 'pattern' in logic, try applying it to the given prompt. GMAT assassins aren't born, they're made, Rich
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Re: Five identical pieces of wire are soldered together endtoend to form
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12 Jun 2015, 14:10
EMPOWERgmatRichC wrote: Hi ailujkh,
This is a rarer conceptual issue on the GMAT (you likely will NOT see it on Test Day), but there's a 'visual component' to the question that you can use to help you figure out how the math "works."
From your notes, it looks like you started to visualize what the wires would look like, which is good. Before getting into the specifics of this prompt though, it would probably help to start with a simpler example....
What would it look like if you took two 10meter wires and overlapped them for 4 meters? Try drawing it. What would the total length be? Now what would happen if you added a third 10meter wire and overlapped it in the same way? Using this 'pattern' in logic, try applying it to the given prompt.
GMAT assassins aren't born, they're made, Rich Hi Rich! I tried to solve this question using "Test the answers" strategy. I made a picture and found out that we have four overlaps here. So i looked for an answer that is equal to 116 when multiplied by 4. I started with D but 24 times 5 is 120. So i tested C and got the answer. Cool!=)). Looks like there a shift in my brain. I don't always try to solve, i try to test the answers and this shift is important for me. Your course helps me!
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Re: Five identical pieces of wire are soldered together endtoend to form
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12 Jun 2015, 17:25
Hi Konstantin1983, You are correct. TESTing THE ANSWERS is a great way to approach this question. Now that you're using it more often, you're going to see more potential opportunities to use this tactic in your homework, your practice CATs and on Test Day itself. GMAT assassins aren't born, they're made, Rich
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Re: Five identical pieces of wire are soldered together endtoend to form
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30 Jun 2015, 22:50
w1  (4)  w2  (4)  w3  (4)  w4  (4)  w5 so, 100+16 = 116/5 = 23.2 cm each.
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Re: Five identical pieces of wire are soldered together endtoend to form
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24 Oct 2016, 08:47
5(x)4(4) = 100 (convert 1m to cm)
x = 23.2



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Re: Five identical pieces of wire are soldered together endtoend to form
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24 Oct 2016, 12:41
ailujkh wrote: Five identical pieces of wire are soldered together endtoend to form one longer wire, with the pieces overlapping by 4cm at each joint. If the wire thus made is exactly 1 meter long, how long, in centimeters, is each of the identical pieces?
(A) 21,2 (B) 22 (C) 23,2 (D) 24 (E) 25,4 The wire will have 4 contact points soldered and since each overlapping joint is 4cm, the total length of the joint is 16 ( ie, 4*4 ) Length of the wire is 100 cm (Excluding the joint ) So, Total Length of the wire including the joints is 116 Now, Total length of each piece is \(\frac{116}{5}\) \(= 23.20 cm\) Hence, Answer will be (C)
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Five identical pieces of wire are soldered together endtoend to form
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15 Jul 2017, 01:37
Hello! Though the solutions given above makes sense  Can someone help me to understand what's wrong with my reasoning? I too am getting an answer that's not part of the answer choice. Let the length of each identical rod be A. With 4cms overlap on each of the rod, it will look something like the attached picture. From the Picture we get  100 = (A4) + (A8) + (A8) + (A8) + (A4) => 100 = 5A  32 => 5A = 132 => A = 132/5 = 26.4
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Re: Five identical pieces of wire are soldered together endtoend to form
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16 Jul 2017, 09:06
Hi susheelh,
You are subtracting 4cm twice.. from upper wires and from the lower wires. Now since you are deducting 4 cm from both sides from both upper wires, you don't have to deduct it from bottom wires because those bottom wires will compensate for the 4cm reduction from both ends of upper wires.
So the length of your wires would be A, A8, A, A8, A and their sum would be 100



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Re: Five identical pieces of wire are soldered together endtoend to form
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16 Jul 2017, 10:28
Kushal3692 wrote: Hi susheelh,
You are subtracting 4cm twice.. from upper wires and from the lower wires. Now since you are deducting 4 cm from both sides from both upper wires, you don't have to deduct it from bottom wires because those bottom wires will compensate for the 4cm reduction from both ends of upper wires.
So the length of your wires would be A, A8, A, A8, A and their sum would be 100 I came up with 26.4cm as well. The wires on the end only overlap on one end, while the other three overlap on both ends: 2(x4) + 3(x8)=100 If the wires overlap at each joint, there would need to be 4cm subtracted from both connecting wires. Maybe this is just a poorly worded question? Are you saying that they overlap by 2cm (2cm from each wire) then?



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Five identical pieces of wire are soldered together endtoend to form
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16 Jul 2017, 22:34
Thank you! Kushal3692 for taking time and responding. However, I still don't seem to actually "get" the question stem. What I understand from this part of the question stem  " with the pieces overlapping by 4cm at each joint" is each joint must be 4cms. Meaning 1st and 5th Rod has 4cms joint, 2nd, 3rd, 4th rod has 4+4 = 8 cms joint. In which case the answer has to be 26.4. I agree with mcdo on this. I think its a poorly worded question. if the above highlighted portion of the stem reads  " with the pieces overlapping by 4cm at each rod" or something on those lines then each joint must be 2cms. Meaning 1st and 5th Rod has 2cms joint, 2nd, 3rd, 4th rod has 2+2 = 4 cms joint. In which case the answer has to be 23.2. The trap I guess is the wording of the stem and how we interpret it. Thanks again to both of you for the response! Kudos to both of your for the response Kushal3692 wrote: Hi susheelh,
You are subtracting 4cm twice.. from upper wires and from the lower wires. Now since you are deducting 4 cm from both sides from both upper wires, you don't have to deduct it from bottom wires because those bottom wires will compensate for the 4cm reduction from both ends of upper wires.
So the length of your wires would be A, A8, A, A8, A and their sum would be 100
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Five identical pieces of wire are soldered together endtoend to form
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07 Sep 2017, 16:18
susheelh wrote: Hello!
Though the solutions given above makes sense  Can someone help me to understand what's wrong with my reasoning? I too am getting an answer that's not part of the answer choice.
Let the length of each identical rod be A. With 4cms overlap on each of the rod, it will look something like the attached picture.
From the Picture we get 
100 = (A4) + (A8) + (A8) + (A8) + (A4)
=> 100 = 5A  32
=> 5A = 132
=> A = 132/5 = 26.4 I'm surprised no one has responded yet to this... I think the error in your reasoning comes from the fact that you are only adding together the parts of the wires that are NOT overlapping. You are forgetting to add the four 'joints' (each measuring 4 cm) back into your calculations. Right now your equation only accounts for the lengths of each wire that is NOT a part of the joint. So really, the equation is part right: 100 = (parts of wire not in joints) + (length of joints)(4 joints) 100 = (5A32) + (4)(4) 100 = 5A  16 116 = 5A A = 23.2 Hope that helps!




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