GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 30 Mar 2020, 08:53

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

A square wooden plaque has a square brass inlay in the

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 15 Jan 2010, 03:00
23
236
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

38% (02:31) correct 62% (02:49) wrong based on 1966 sessions

HideShow timer Statistics

The Official Guide For GMAT® Quantitative Review, 2ND Edition

A square wooden plaque has a square brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches, of the wooden strip?

I. 1
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II , and III

Problem Solving
Question: 175
Category: Geometry Area
Page: 85
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

_________________
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 15 Jan 2010, 05:14
47
41
hrish88 wrote:
so doest it mean any width <= 39/4 is possible.


this is the 2nd last problem in OG.so i thought it would be difficult. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) is possible. \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?\).

Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.

Hope it's clear.
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 15 Jan 2010, 03:26
23
1
9
hrish88 wrote:
A square wooden plaque has a square brass inlay in the center ,leaving a wooden strip of uniform width around the brass square.if the ratio of the brass area to the wooden area is 25 to 39,which of the following could be the width ,in inches ,of the wooden strip.

I. 1
II. 3
III. 4


A.I only
B.II only
C.III only
D.I and III only
e.I,II and III


Why would ANY width of the strip be impossible?

Answer: E.
_________________
Most Helpful Community Reply
Manager
Manager
User avatar
Joined: 13 Dec 2009
Posts: 173
Reviews Badge
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 24 Mar 2010, 05:10
6
2
hrish88 wrote:
A square wooden plaque has a square brass inlay in the center ,leaving a wooden strip of uniform width around the brass square.if the ratio of the brass area to the wooden area is 25 to 39,which of the following could be the width ,in inches ,of the wooden strip.

I. 1
II. 3
III. 4


A.I only
B.II only
C.III only
D.I and III only
e.I,II and III


Area of brass square/ area of wooden strip = 25 /39
lets say length of the wooden plaque= l and length of the square brass = x
then
x^2 / (l^2 - x^2) = 25/39
=>39x^2 = 25l^2 - 25x^2
=>64x^2 = 25l^2
=>8x = 5l
Width of wooden strip should be l-x
=>x = 5l/8
so l -x = l = 5l/8 = 3l/8

Now 3l/8 could be any value depending on the value of l
so answer is E.
General Discussion
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 15 Jan 2010, 09:44
6
atish wrote:
let the width of the wooden part = w
let the width of the brass part = b
given brass area/wooden area = 25/39
area of brass part = b^2
area of wooden part = (b+2w)^2 - b^2
Simplifying
64b^2=25(b+2w)^2
8b=5b+10w
b=10w/3

b has to be an integer or a terminating decimal. We can't have a width of 10/3 in real life (note the question doesn't ask for approximate width.) hence w has to be a multiple of 3.
Answer is B.


Are you saying that in real life everything has integer or terminating decimal length? Why cannot we have repeated decimal or even irrational number as width of something?

Take the square with side 1, diagonal would be \(\sqrt{2}\), it's not an integer or terminating decimal.

Also it's possible to divide the line segment into three equal parts, Google it and you find that it's quite easy.
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 01 Apr 2012, 08:16
6
3
Manager
Manager
User avatar
Joined: 25 Aug 2009
Posts: 102
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 15 Jan 2010, 10:25
2
Quote:
Are you saying that in real life everything has the integer or terminating decimal length? Why can not we have repeated decimal or even irrational number as width of something?

Take the square with side 1, diagonal would be \(\sqrt{2}\), it's not an integer or terminating decimal.

Also it's possible to divide the line segment into three equal parts, google it and you find that it's quite easy.


That takes the question to a whole new dimension, I do understand what you are saying though. If the width of something is 10/3 that means you can never (accurately) measure it. The width of the brass square can never be measured practically, it can be only measured mathematically. If such a square was to be made, the creator would have to take a square of 10/10 dimension, divide it into 9 exactly equal parts and use one of them, he/she could never make just the square as it would be impossible to measure 10/3 inches. I do get the concept, but don't like the fact that a question can be based on it.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 08 Aug 2010, 00:32
2
1
mainhoon wrote:
Bunuel wrote:
hrish88 wrote:
so doest it mean any width <= 39/4 is possible.


this is the 2nd last problem in OG.so i thought it would be difficult. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) is possible. \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?\).

Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.

Hope it's clear.


To generalize then, since the answer does not seem to depend on the fact that the ration is 25/39, can it be said that regardless of what the ratio is, the width of strip can be ANYTHING?


Yes, width can have any positive value: the larger the width is the larger the whole square would be.
_________________
Senior Manager
Senior Manager
User avatar
Joined: 13 Aug 2012
Posts: 392
Concentration: Marketing, Finance
GPA: 3.23
GMAT ToolKit User
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 01 Dec 2012, 20:18
2
2
This tells us that there is a particular ratio for the width of the wooden frame to the width of the brass inlay.

This means there is a f(x) = \((\frac{A}{B})(x)\). Any x is possible and will yield a corresponding f(x).
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 11 Sep 2015
Posts: 4549
Location: Canada
GMAT 1: 770 Q49 V46
A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post Updated on: 16 Apr 2018, 11:34
2
Top Contributor
1
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A square wooden plaque has a square brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches, of the wooden strip?

I. 1
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II , and III


Let's say we want to create a plaque with a square brass inlay in the center, and we want the brass to wood ratio to be 25:39

Let's begin a square wooden board with ANY dimensions.
Image


Now place a square brass inlay in the middle of the wooden board, and keep adjusting the size of the brass inlay until we have a brass to wood ratio that is 25:39
Image

At this point, if we shrink or expand the plaque . . .
Image
. . . the brass to wood ratio will remain at 25:39

So, as you can see, this plaque can be ANY size, which means the width of the wooden strip can have ANY measurement.

Answer:

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Image

Originally posted by GMATPrepNow on 15 Sep 2017, 11:19.
Last edited by GMATPrepNow on 16 Apr 2018, 11:34, edited 1 time in total.
Manager
Manager
User avatar
Joined: 25 Aug 2009
Posts: 102
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 15 Jan 2010, 08:49
1
let the width of the wooden part = w
let the width of the brass part = b
given brass area/wooden area = 25/39
area of brass part = b^2
area of wooden part = (b+2w)^2 - b^2
Simplifying
64b^2=25(b+2w)^2
8b=5b+10w
b=10w/3

b has to be an integer or a terminating decimal. We can't have a width of 10/3 in real life (note the question doesn't ask for approximate width.) hence w has to be a multiple of 3.
Answer is B.
Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 573
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 07 Aug 2010, 16:32
1
Bunuel wrote:
hrish88 wrote:
so doest it mean any width <= 39/4 is possible.


this is the 2nd last problem in OG.so i thought it would be difficult. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) is possible. \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?\).

Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.

Hope it's clear.


To generalize then, since the answer does not seem to depend on the fact that the ration is 25/39, can it be said that regardless of what the ratio is, the width of strip can be ANYTHING?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 01 Apr 2012, 07:43
1
mymbadreamz wrote:
I didn't understand the statement "We are asked which value of (y-x)/2 is possible" . Can someone explain?


Question asks about the possible width, in inches, of the wooden strip.

Let the the side of small square be \(x\) and the big square \(y\), then the width of the wooden strip would be \(\frac{y-x}{2}\), which means that we are asked to determine the possible values of this exact expression.

Hope it's clear.
_________________
Verbal Forum Moderator
User avatar
B
Joined: 10 Oct 2012
Posts: 569
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 01 Jun 2013, 00:31
1
hrish88 wrote:
A square wooden plaque has a square brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches, of the wooden strip.

I. 1
II. 3
III. 4

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III


The only restriction on the value of b(side of the brass square) is that b<w. From the given problem, we know that\(\frac{b^2}{(w^2-b^2)}\)= 25/39--> 39/25 = \((\frac{w}{b})^2-1\)-->
\(\frac{w}{b}\)= \(\sqrt{64/25}\)--> b = \(\frac{5}{8}*w\). Thus, any value of w, would still give a value where b<w.
E.
_________________
Manager
Manager
avatar
B
Joined: 09 Aug 2016
Posts: 61
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 27 Oct 2016, 09:01
1
Bunuel wrote:
mymbadreamz wrote:
Why is it not y-x? Why do we calculate (y-x)/2?


Consider the diagram below:
Attachment:
Wooden strip.png
As you can see the width of the wooden strip (the width of grey strip) is \(\frac{y-x}{2}\).



OMG !!! It took me ages to understand of what you said.... Is not too obvious at all.

For those that they are on the same boat with me it boils down to y-x / 2 BECAUSE...

Say d is the length of the of the wooden strip.... where Bunuel has the DOUBLE arrows. Then you know that Y = x + d (the left) + d (the right) so you end up having:

y = x + 2d hence d = (y - x) / 2
Manhattan Prep Instructor
User avatar
G
Joined: 04 Dec 2015
Posts: 929
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 16 Dec 2016, 11:20
1
hanyhamdani wrote:
Hi, I did it another way, but in my solution none of the given value satisfies.

Take wooden (bigger area) =\(a^2\)
let x be the width.
so brass area = \((a-2x)^2\)
so,

\((25/39)=( (a-2x)^2 )/ (a^2)\)


1. x=1
\((a-2) = 5\)
so, a=7
and \(a^2 = 49\)

giving ratio : \(25/49\) which is not as given one.

trying all these, give different ratios.

what did i do wrong?


Just because \((25/39)=( (a-2x)^2 )/ (a^2)\), doesn't necessarily mean that \(25=(a-2x)^2\).

That's a general truth about ratios. If x/y = 12/19, x isn't necessarily 12. For instance, x could be 24 and y could be 38.
_________________
Image

Chelsey Cooley | Manhattan Prep | Seattle and Online

My latest GMAT blog posts | Suggestions for blog articles are always welcome!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62351
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 27 Dec 2017, 21:06
1
Erjan_S wrote:
Bunuel wrote:
hrish88 wrote:
so doest it mean any width <= 39/4 is possible.


this is the 2nd last problem in OG.so i thought it would be difficult. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) is possible. \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?\).

Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.

Hope it's clear.


I have difficulties this this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)

How do you get it?


\(\frac{x^2}{y^2-x^2}=\frac{25}{39}\);

Cross multiply: \(39x^2=25y^2-25x^2\);

Re-arrange: \(64x^2=25y^2\);

\(\frac{x^2}{y^2}=\frac{25}{64}\).

Hope it's clear.
_________________
VP
VP
User avatar
D
Joined: 09 Mar 2016
Posts: 1220
Re: A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 05 May 2018, 04:34
1
Erjan_S wrote:
Bunuel wrote:
hrish88 wrote:
so doest it mean any width <= 39/4 is possible.


this is the 2nd last problem in OG.so i thought it would be difficult. :-D


No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) is possible. \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?\).

Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.

Hope it's clear.


I have difficulties this this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)

How do you get it?


hello generis :) i am honored to tag you and ask you three questions :)

Q # 1: i reviewed whole thread and all explanations but cant understand why do we divide by 2 here \(\frac{y-x}{2}\) ?

Q # 2: why do we subtract \(x^2\)from \(y^2\) here in the denominator \([m]\frac{x^2}{y^2-x^2}\) ? we know that ratio is \(\frac{25}{39}\), why to subtract ? :?

Q # 3: my approach to to tho problem was simple i rounded 39 to 40 so i simply got ratio \(\frac{25}{40} = \frac{5}{8}\) do you think its correct approach ? :? :)

Many thanks and have an awesome weekend ! :)

P.S. :cool: by the way in your signature isnt there a SC issue ? :)

it says "In the depths of winter, I finally learned
that within me there lay an invincible summer
."

but shouldnt we say so "In the depths of winter, I finally learned
that within me there is a thick layer of an invincible hot summer mood
" :)
Senior SC Moderator
avatar
V
Joined: 22 May 2016
Posts: 3667
A square wooden plaque has a square brass inlay in the  [#permalink]

Show Tags

New post 05 May 2018, 13:55
1
1
dave13 wrote:
Erjan_S wrote:
Bunuel wrote:
No I mean ANY width is possible.

Let the the side of small square be \(x\) and the big square \(y\).

Given: \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\) --> \(\frac{x}{y}=\frac{5}{8}\).

We are asked which value of \(\frac{y-x}{2}\) is possible. \(\frac{y-\frac{5}{8}y}{2}=\frac{3}{16}y=?\).

Well, expression \(\frac{3}{16}y\) can take ANY value depending on \(y\): 1, 3, 4, 444, 67556, 0,9, ... ANY. Basically we are given the ratios of the sides (5/8), half of their difference can be any value we choose, there won't be any "impossible" values at all.

Hope it's clear.
I have difficulties this part \(\frac{x^2}{y^2-x^2}=\frac{25}{39}\) --> \(\frac{x^2}{y^2}=\frac{25}{64}\)
How do you get it?
hello generis :) i am honored to tag you and ask you three questions :)
Q # 1: SEE BELOW Q # 2: SEE BELOW Q # 3: SEE BELOW

Many thanks and have an awesome weekend ! :)

P.S. :cool: by the way in your signature isnt there a SC issue ? :)

it says "In the depths of winter, I finally learned
that within me there lay an invincible summer
."

but shouldnt we say so "In the depths of winter, I finally learned
that within me there is a thick layer of an invincible hot summer mood
"
:)

:lol: :lol: :lol: OMG you are irrepressible.

Hi dave13 -
Your second question is hard to answer.
I cannot tell whether you are following the algebra.
Others have had a hard time. Maybe this write-up will help.

I am assigning capital letter variables in order to write as abbreviated an
answer as possible.
Let P = the whole PLAQUE (which is a "big square"). Area of \(P = y^2\)
Let B = the square BRASS inlay (which is a "little square). Area of \(B = x^2\)
Let W = the WOODEN part (which is a uniform strip, like a "frame"). Area of \(W = y^2-x^2\)
Quote:
Q # 1: i reviewed whole thread and all explanations but cant understand why do we divide by 2 here \(\frac{y-x}{2}\)?

Short answer: because there are TWO border widths between \(y\) and \(x\). TAKE A LOOK:
Attachment:
woodenborder.png
woodenborder.png [ 8.24 KiB | Viewed 1299 times ]

We are solving for the width of the border.
Does (side y) - (side x) = width of border?
No. \((y - x)\) = TWO times the width of the border

Why? On EACH side of \(x\), there is one border width, \(w\):
\(y = w + x + w\)
\(y - x = 2w\)
\(w = \frac{y-x}{2}\)


Try drawing the diagram. Let's say \(x = 2, y = 4\), and width of wooden frame, W = 1 (write values in)
Does \((y - x)\) = width of wooden border? \((y-x) = 2. W = 1\). Not correct.
\((y - x) = 2\) * width of the wooden border.
Width of border therefore equals \(\frac{y-x}{2}=\frac{4-2}{2}=1\)
Quote:
Q # 2: why do we subtract \(x^2\)from \(y^2\) here in the denominator \(\frac{x^2}{y^2-x^2}\) ? we know that ratio is \(\frac{25}{39}\), why to subtract ? :?

Short answer: We subtract because
1) \(y^2 - x^2\) equals the area of the wooden part, which
2) is the bottom part of the ratio given by the prompt
(I think you may be slightly confused about area of "the wooden part." It's the frame, the thin strip.)

We need this ratio: \(\frac{Area_{B}}{Area_{W}}\), in which \(Area_{W} = y^2-x^2\),
to get another ratio: side length of B to side length of P.

\(\frac{InnerSquareArea}{WoodenFrameArea} = \frac{25}{39}\)

1) inner square area of P = \(x^2\)
2) wooden frame area = \((y^2 - x^2)\)
Think of the area of the wooden frame as a "shaded region."
You would find the shaded region's area by subtracting B's area \((x^2)\) from P's area \((y^2)\).

We go from \(\frac{area of B}{area of W}\) to \(\frac{Side of B(=x)}{Side of P(=y)}\) to border width= some y expression

Border width depends on length of \(y\). That demonstration has steps

1) Use the given ratio of \(\frac{Barea}{Warea}\) to find ratio between the areas of the squares
The given ratio is NOT between the areas of the two squares (that's what we need to find)

GIVEN ratio is \(\frac{B}{W} = \frac{25}{39}\)

Area of W = (Area of P - Area of B) (think of W as a shaded region)
Area of P = \(y^2\)
Area of B = \(x^2\)
Area of W = \(y^2 - x^2\)
Substitute that RHS for W in the original ratio.

\(\frac{B}{W}=\frac{x^2}{y^2-x^2}=\frac{25}{39}\)
To find the ratio of areas of squares (P and B), we need \(x^2\) over \(y^2\)
Eliminate \(x^2\) in the denominator. Cross multiply:

\(\frac{x^2}{y^2-x^2}=\frac{25}{39}\)
\(39x^2 = 25y^2 - 25x^2\)
\(64x^2=25y^2\)

DIVIDE both sides by \(y^2\) and by \(64\)
\(\frac{x^2}{y^2}=\frac{25}{64}\)

(2) Use areas of squares to find ratio of side lengths of squares
Find ratio of side x to side y by taking the square root of \(\frac{x^2}{y^2}=\frac{25}{64}\):
\(\frac{x}{y}=\frac{5}{8}\)

(3) Express width of the border in terms of \(y\) only
At #2 we still have a ratio. Still no counting number.
Width of the border depends on \(y.\)
But \(y\) can be any length. NO limit on values for y = no limit on border width
Bunuel 's next steps simply detail THAT there are no limits on the longer side.
He shows that every answer depends on \(y\)

Defining border width in terms of \(y\)
Recall that border width = \(\frac{y-x}{2}\)
Define everything in terms of \(y\). Use the side ratio to do so
\(\frac{x}{y}=\frac{5}{8}\)
Multiply both sides by \(y\)
\(x = \frac{5}{8}y\)
Substitute that RHS for \(x\)

Border width calculation: \(\frac{y - x}{2} =\frac{y -\frac{5}{8}y}{2}\)

\(\frac{y-\frac{5}{8}y}{2}=(\frac{\frac{8}{8}y-\frac{5}{8}y}{2})=(\frac{\frac{3}{8}y}{2})\)

\(\frac{\frac{3}{8}y}{2}=(\frac{3}{8}y*\frac{1}{2})=\frac{3}{16}y\)

Border width equals \(\frac{3}{16}\) of \(y\)

That's nice, but how long can \(y\) be? \(y\)'s length can be ANY real number.

The width of the border, dependent on \(y\)'s length,
can be ANY value that is \(\frac{3}{16}\) of an infinite set of possibilities.
Quote:
Q # 3: my approach to to tho problem was simple i rounded 39 to 40 so i simply got ratio \(\frac{25}{40} = \frac{5}{8}\) do you think its correct approach ? :? :)

Excellent idea to try to find shortcuts. :thumbup:
But . . . wrong ratio. You got lucky.
To find side length, given area of squares, you must take a square root. AND you must be solving for the right parts.
\(\frac{25}{39}\) is the ratio of \(\frac{AreaB}{AreaW}\)
\(\frac{5}{8}\) is ratio of \(\frac{SideB}{SideP}\)

Your progression seems to be: \(\frac{25}{39}= \frac{AreaB}{AreaP}=\frac{SideB}{SideW}\)
The ratio you derived is not correct: \(P\neq{W}\)
Tip: be clear about the parts of the ratios you set up. WHICH side to WHICH side, e.g.

I hope that helps. :-)

(whew!)
You get kudos for making me laugh about the SC issues in my signature. I am a reading fool (fanatic). Elegant prose uses metaphor. SC questions are hard but . . . most do not display elegant prose. BTW, the author was an astonishing human being. But did you REALLY go after my pick for translation of “Au milieu de l'hiver, j'apprenais enfin qu'il y avait en moi un été invincible”? Thou art brave. :cool:
_________________
Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date.

Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has. -- Margaret Mead
Intern
Intern
avatar
Joined: 18 Jul 2009
Posts: 26
Re: hard problem OG Quant 2nd edition  [#permalink]

Show Tags

New post 15 Jan 2010, 04:03
so doest it mean any width <= 39/4 is possible.


this is the 2nd last problem in OG.so i thought it would be difficult. :-D
GMAT Club Bot
Re: hard problem OG Quant 2nd edition   [#permalink] 15 Jan 2010, 04:03

Go to page    1   2   3    Next  [ 53 posts ] 

Display posts from previous: Sort by

A square wooden plaque has a square brass inlay in the

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne