Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

24 Apr 2015, 03:36

20

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

64% (03:08) correct
37% (01:43) wrong based on 200 sessions

HideShow timer Statistics

Five identical pieces of wire are soldered together end-to-end 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?

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.

Re: Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

24 Apr 2015, 10:45

1

This post received KUDOS

ailujkh wrote:

Five identical pieces of wire are soldered together end-to-end 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?

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.

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 10-meter 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 10-meter wire and overlapped it in the same way? Using this 'pattern' in logic, try applying it to the given prompt.

Re: Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

12 Jun 2015, 14:10

1

This post received KUDOS

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 10-meter 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 10-meter 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!
_________________

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.

Re: Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

08 Aug 2016, 02:36

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 end-to-end to form [#permalink]

Show Tags

24 Oct 2016, 12:41

ailujkh wrote:

Five identical pieces of wire are soldered together end-to-end 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) _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

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 = (A-4) + (A-8) + (A-8) + (A-8) + (A-4)

=> 100 = 5A - 32

=> 5A = 132

=> A = 132/5 = 26.4

Attachments

MGMAT_Page 126.png [ 24.37 KiB | Viewed 556 times ]

Re: Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

16 Jul 2017, 09:06

1

This post received KUDOS

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, A-8, A, A-8, A and their sum would be 100

Re: Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

16 Jul 2017, 10:28

1

This post received KUDOS

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, A-8, A, A-8, 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(x-4) + 3(x-8)=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?

Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

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, A-8, A, A-8, A and their sum would be 100

Five identical pieces of wire are soldered together end-to-end to form [#permalink]

Show Tags

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 = (A-4) + (A-8) + (A-8) + (A-8) + (A-4)

=> 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 = (5A-32) + (4)(4) 100 = 5A - 16 116 = 5A A = 23.2