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:

A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

21 Dec 2010, 08:20

6

This post received KUDOS

23

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

63% (02:04) correct
37% (01:26) wrong based on 614 sessions

HideShow timer Statistics

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

A. \(\pi*r^2\) B. \(\pi*r^2 + 10\) C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\) D. \(\pi*r^2 + (40 - 2\pi*r)^2\) E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

A. \(\pi*r^2\) B. \(\pi*r^2 + 10\) C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\) D. \(\pi*r^2 + (40 - 2\pi*r)^2\) E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)

The area of a circle will be - \(\pi{r^2}\) and \(2\pi{r}\) meters of wire will be used; There will be \(40-2\pi{r}\) meters of wire left for a square. Side of this square will be \(\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}\), hence the area of the square will be \((10-\frac{\pi{r}}{2})^2\).

The total area will be - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Bunuel...how did u get side of the square as (40-2pi*r)/4 ?

\(40-2\pi{r}\) meters of wire are left for the square means that \(40-2\pi{r}\) is the perimeter of the square so the side of it will be \(\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}\). _________________

A thin piece of 40 m wire is cut in two. One piece is a circle with radius of r. Other is a square. No wire is left over. Which represents total area of circle and square in terms of r?

From a 40m rope, two equal parts of 20 represent, respectively, the circumference of Circle C and the perimeter of Square D.

1. Express the perimeter of D in terms of the circumference of C.

Perimeter + Circumference = 40

Perimeter = 40 - C

2. Substitute accepted identities for perimeter and circumference.

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius R, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and square regions in terms of R.

4) or 5) is a given, I just dont see how you make the calculation necessary.

Very minimal calculations are required if you assume a convenient value for R e.g. R = 7/2 since \(\pi = 22/7\) Circumference in this case \(= 2\pi*R = 2*(22/7)*(7/2) = 22\) So length of leftover wire = 18 Side of square = 18/4 Area of square = \((9/2)^2\)

Now put R = 7/2 in the second half of the options you want to check. i.e. say you want to check option (E) \((10 - 1/2\pi R)^2 = [10 - (1/2)*(22/7)*(7/2)]^2 = (9/2)^2\)

Option (E) is correct since the area of the square is (9/2)^2 when R = 7/2 (as shown above) _________________

Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20 perimeter of square = 4*side So length of side = 20/4 = 5.

How do I proceed further?

The length of each piece = 20;

Circumference of the circle is \(2\pi{r}=20\) --> \(r=\frac{10}{\pi}\) --> \(area=\pi{r^2}=\frac{100}{\pi}\); Perimeter of the square \(4*side=20\) --> \(side=5\) --> \(area=5^2=25\);

The total area is \(\frac{100}{\pi}+25\). Now, you should substitute the value of \(r=\frac{10}{\pi}\) in the answer choices and see which one gives \(\frac{100}{\pi}+25\). Answer choice E works.

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

01 Dec 2013, 21:52

Let the perimeter of circle be x---->2pi*r=x----(1) Let the perimeter of square be 40-x---->4S=40-x Substitute value of x from(1) ----->S=(40/4)-(2pi*r/4) ----->S=10-pi*r/2 ----->\(S^2=(10-pi*r/2)^2\) Therefore total area is \(pi*r^2+(10-pi*r/2)^2\)

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

11 Dec 2013, 20:31

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

Originally, I said that the perimeter of both objects was S (for the square) and 40-S for the circle. Why couldn't I solve the problem using S and 40-S? What is the thought process for this question or others that require similar reasoning?

Anyways, we are told that the wire is cut to form the shape (i.e. the perimeter of the square and the circle) The perimeter of the square is L*4 and the perimeter of the circle is pi*d. The perimeter for the square is 40-pi*d where pi*d represents the perimeter. To find the area, we can get the length of one of the sides of the square so we must find a leg length. (40-pi*d)/4 = 10-[(pi*d)/4]. The area of the circle is simply pi*r^2

A. \pi*r^2 B. \pi*r^2 + 10 C. \pi*r^2 + \frac{1}{4}*\pi^2*r^2 D. \pi*r^2 + (40 - 2\pi*r)^2 E. \pi*r^2 + (10 - \frac{1}{2}\pi*r)^2

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

12 Dec 2013, 02:46

Expert's post

WholeLottaLove wrote:

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

Originally, I said that the perimeter of both objects was S (for the square) and 40-S for the circle. Why couldn't I solve the problem using S and 40-S? What is the thought process for this question or others that require similar reasoning?

Anyways, we are told that the wire is cut to form the shape (i.e. the perimeter of the square and the circle) The perimeter of the square is L*4 and the perimeter of the circle is pi*d. The perimeter for the square is 40-pi*d where pi*d represents the perimeter. To find the area, we can get the length of one of the sides of the square so we must find a leg length. (40-pi*d)/4 = 10-[(pi*d)/4]. The area of the circle is simply pi*r^2

A. \pi*r^2 B. \pi*r^2 + 10 C. \pi*r^2 + \frac{1}{4}*\pi^2*r^2 D. \pi*r^2 + (40 - 2\pi*r)^2 E. \pi*r^2 + (10 - \frac{1}{2}\pi*r)^2

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

A. \(\pi*r^2\) B. \(\pi*r^2 + 10\) C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\) D. \(\pi*r^2 + (40 - 2\pi*r)^2\) E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)

The area of a circle will be - \(\pi{r^2}\) and \(2\pi{r}\) meters of wire will be used; There will be \(40-2\pi{r}\) meters of wire left for a square. Side of this square will be \(\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}\), hence the area of the square will be \((10-\frac{\pi{r}}{2})^2\).

The total area will be - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.

Bunuel- the perimeter of the circle and the square are the same, so: 2(pi)r=4a (a being one side of the square). a= (pi)r/2 area of square in terms of r: a^2= (pi)^2*r^2/4 wont this mean that option c is also correct? what am I missing?

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

A. \(\pi*r^2\) B. \(\pi*r^2 + 10\) C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\) D. \(\pi*r^2 + (40 - 2\pi*r)^2\) E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)

The area of a circle will be - \(\pi{r^2}\) and \(2\pi{r}\) meters of wire will be used; There will be \(40-2\pi{r}\) meters of wire left for a square. Side of this square will be \(\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}\), hence the area of the square will be \((10-\frac{\pi{r}}{2})^2\).

The total area will be - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.

Bunuel- the perimeter of the circle and the square are the same, so: 2(pi)r=4a (a being one side of the square). a= (pi)r/2 area of square in terms of r: a^2= (pi)^2*r^2/4 wont this mean that option c is also correct? what am I missing?

Hello.

The question only says: A thin piece of wire 40 meters long is cut into two pieces, NOT into same length pieces. Thus, your assumption 2(pi)r=4a is wrong.

Hope it's clear. _________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

01 Jun 2014, 23:44

A little confusion here: As the total length is 2πr + 4x = 40 => πr + 2x = 20

I assumed length 10 as the circumference of the circle and 10 as the perimeter of the square.

=> πr=10 and x=5 and computed the total area, which is πr²+x² => 100/π + 25 ----------(1)

I then substituted this values in the answer choice to find out the matches with (1), choices C and E seem to match. Can someone please tell me what am I doing wrong.

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

02 Jun 2014, 01:17

Expert's post

mamboe wrote:

A little confusion here: As the total length is 2πr + 4x = 40 => πr + 2x = 20

I assumed length 10 as the circumference of the circle and 10 as the perimeter of the square.

=> πr=10 and x=5 and computed the total area, which is πr²+x² => 100/π + 25 ----------(1)

I then substituted this values in the answer choice to find out the matches with (1), choices C and E seem to match. Can someone please tell me what am I doing wrong.

Thanks, a

Why did you assume that? We are NOT told that the piece is cut into the same length pieces. We are just told that the piece is cut into two pieces. _________________

Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20 perimeter of square = 4*side So length of side = 20/4 = 5.

How do I proceed further?

The length of each piece = 20;

Circumference of the circle is \(2\pi{r}=20\) --> \(r=\frac{10}{\pi}\) --> \(area=\pi{r^2}=\frac{100}{\pi}\); Perimeter of the square \(4*side=20\) --> \(side=5\) --> \(area=5^2=25\);

The total area is \(\frac{100}{\pi}+25\). Now, you should substitute the value of \(r=\frac{10}{\pi}\) in the answer choices and see which one gives \(\frac{100}{\pi}+25\). Answer choice E works.

Hope it helps.

Hi Bunuel,

Then in this case even answer choice C works. Right?

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

27 Oct 2014, 14:02

maggie27 wrote:

Bunuel wrote:

ENAFEX wrote:

Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20 perimeter of square = 4*side So length of side = 20/4 = 5.

How do I proceed further?

The length of each piece = 20;

Circumference of the circle is \(2\pi{r}=20\) --> \(r=\frac{10}{\pi}\) --> \(area=\pi{r^2}=\frac{100}{\pi}\); Perimeter of the square \(4*side=20\) --> \(side=5\) --> \(area=5^2=25\);

The total area is \(\frac{100}{\pi}+25\). Now, you should substitute the value of \(r=\frac{10}{\pi}\) in the answer choices and see which one gives \(\frac{100}{\pi}+25\). Answer choice E works.

Hope it helps.

Hi Bunuel,

Then in this case even answer choice C works. Right?

Can someone answer this ? Choice C & E both work with this method....how to reject one?

Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

Show Tags

10 Nov 2014, 12:21

Bunuel : Was unnecessarily complicating the question. Spent 5 min on it and still ended up guessing it wrong.. Might be the pressure of the practice test made me miss the whole point of the question. Definitely the best example to say Practice and FOCUS is the key on GMAT..

gmatclubot

Re: A thin piece of wire 40 meters long is cut into two pieces.
[#permalink]
10 Nov 2014, 12:21

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...