Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 04 Apr 2015
Posts: 2
WE: Accounting (Internet and New Media)

Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
Updated on: 24 Apr 2015, 02:40
Question Stats:
63% (02:13) correct 37% (02:28) wrong based on 212 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by ailujkh on 24 Apr 2015, 02:36.
Last edited by Bunuel on 24 Apr 2015, 02:40, edited 1 time in total.
Renamed the topic and edited the question.



Manager
Joined: 03 Sep 2014
Posts: 73
Concentration: Marketing, Healthcare
Schools: Kellogg 1YR '17, Booth '16, McCombs '18, Tepper '18, INSEAD Jan '17, ISB '17, NUS '18, IIMA , IIMB, IIMC , IIML '15

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
24 Apr 2015, 09:45
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.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16149
Location: United States (CA)

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
24 Apr 2015, 15:03
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
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Senior Manager
Joined: 02 Dec 2014
Posts: 349
Location: Russian Federation
Concentration: General Management, Economics
WE: Sales (Telecommunications)

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
12 Jun 2015, 13:10
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!
_________________
"Are you gangsters?"  "No we are Russians!"



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16149
Location: United States (CA)

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
12 Jun 2015, 16:25
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
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Senior Manager
Joined: 28 Jun 2015
Posts: 274
Concentration: Finance
GPA: 3.5

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
30 Jun 2015, 21:50
w1  (4)  w2  (4)  w3  (4)  w4  (4)  w5 so, 100+16 = 116/5 = 23.2 cm each.
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 341

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
24 Oct 2016, 07:47
5(x)4(4) = 100 (convert 1m to cm)
x = 23.2



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4848
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
24 Oct 2016, 11:41
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)
_________________



Manager
Joined: 12 Jun 2016
Posts: 202
Location: India
Concentration: Technology, Leadership
WE: Sales (Telecommunications)

Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
15 Jul 2017, 00:37
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
Attachments
MGMAT_Page 126.png [ 24.37 KiB  Viewed 2381 times ]
_________________



Intern
Joined: 19 Jun 2017
Posts: 2

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
16 Jul 2017, 08:06
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



Current Student
Joined: 09 Jun 2017
Posts: 27
Location: United States (CA)
GMAT 1: 730 Q49 V41
GPA: 3.43
WE: Analyst (Real Estate)

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
16 Jul 2017, 09:28
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?



Manager
Joined: 12 Jun 2016
Posts: 202
Location: India
Concentration: Technology, Leadership
WE: Sales (Telecommunications)

Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
16 Jul 2017, 21:34
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
_________________



Intern
Joined: 29 Aug 2017
Posts: 1

Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
07 Sep 2017, 15:18
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!



Director
Joined: 04 Dec 2015
Posts: 721
Location: India
Concentration: Technology, Strategy
Schools: HEC Sept19 intake, ISB '19, Rotman '21, NUS '21, IIMA , IIMB, NTU '20, Bocconi '21, XLRI, Trinity MBA '20, Smurfit "21
WE: Information Technology (Consulting)

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
09 Dec 2018, 08:33
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 Let the length of wire be \(= x\) Total length of wires after joining \(= 1\) meter \(= 100\) cms Given the wires overlap at each joints by \(4\) cms. There are \(5\) wires hence there would be total of \(4\) joints overlap. \((x4) + (x4) + (x4) + (x4) + x = 100\) \(5x 16 = 100\) \(5x = 100 + 16 = 116\) \(x = 23.2\) Answer C



VP
Joined: 11 Feb 2015
Posts: 1034

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
14 Dec 2018, 09:07
If each joint is 4 cms then the total overlap is 4 joints * 4 cms = 16 cms. Total of the pieces is 100 cms + 16 cms = 116 cms. Length of each piece of wire is 116/5 = 23.2 cms. Ans C
_________________
________________ Manish "Only I can change my life. No one can do it for me"



Intern
Joined: 13 Dec 2018
Posts: 43
Location: India
GPA: 3.94

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
15 Dec 2018, 04:38
There would be 4 joints each using 4 cm of wire. Do, total wire wasted = 4 x 4 = 16 cm Hence, the length of 5 wires = 100 + 16 cm = 116 cm Length of each piece = 116/5 = 23.2 cm



NonHuman User
Joined: 09 Sep 2013
Posts: 14138

Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
Show Tags
25 Jan 2020, 06:14
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Five identical pieces of wire are soldered together endtoend to form
[#permalink]
25 Jan 2020, 06:14






