Last visit was: 11 Oct 2024, 07:39 It is currently 11 Oct 2024, 07:39
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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Joined: 12 Apr 2011
Status:D-Day is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Posts: 117
Own Kudos [?]: 1981 [142]
Given Kudos: 52
Location: Kuwait
Concentration: finance and international business
Schools:Columbia university
 Q18  V17 GMAT 2: 320  Q18  V19 GMAT 3: 620  Q42  V33
GPA: 3.48
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 96065
Own Kudos [?]: 667200 [66]
Given Kudos: 87603
Send PM
Board of Directors
Joined: 11 Jun 2011
Status:QA & VA Forum Moderator
Posts: 6035
Own Kudos [?]: 4864 [7]
Given Kudos: 463
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Send PM
General Discussion
User avatar
Joined: 12 Apr 2011
Status:D-Day is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Posts: 117
Own Kudos [?]: 1981 [0]
Given Kudos: 52
Location: Kuwait
Concentration: finance and international business
Schools:Columbia university
 Q18  V17 GMAT 2: 320  Q18  V19 GMAT 3: 620  Q42  V33
GPA: 3.48
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
GOT IT, 2ND APPROACH IS EASIER...THAAAAAANKS B.
Math Expert
Joined: 02 Sep 2009
Posts: 96065
Own Kudos [?]: 667200 [6]
Given Kudos: 87603
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
1
Kudos
5
Bookmarks
Expert Reply
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Geometry:
Triangles
Polygons
Circles
Coordinate geometry
3-D Geometries

All DS Geometry to practice: search.php?search_id=tag&tag_id=32
All PS Geometry to practice: search.php?search_id=tag&tag_id=53
avatar
Joined: 15 Jun 2013
Posts: 5
Own Kudos [?]: 8 [0]
Given Kudos: 6
Concentration: Strategy, Finance
GMAT Date: 02-13-2014
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
manalq8
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S
A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9


help needed please

We know Perimeter of a square (Ps) = 4*side
Perimeter of a rectangle (Pr) = 2(length+breath)

Let us assume 40 to be the perimeter of the square (since we know each side of a square is equal and the perimeter is divisible by 4, also take in to account the length and breadth of the rectangle is in the ration 2k:3k = 5k; we can assume such a number)

Therefore,
Ps = Pr = 40
Area of the square = 100 sq. units
We know 2(length+breadth) = 40
i.e. length + breadth = 20 (or 5k = 20 given that l:b (or b:l) = 2:3)
Therefore length = 8, breath = 12

Area of the rectangle = 8*12 = 96 sq. units

Question asked = Area of the rectangle : Area of the square = 96:100 ==> 24:25

Note : The explanation might be bigger, but it takes less than 15 seconds to solve this problem if you assume numbers and try the problem.
Joined: 09 Mar 2016
Posts: 1134
Own Kudos [?]: 1042 [0]
Given Kudos: 3851
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
Bunuel
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

Let the side of square be \(s\) and the side of rectangle \(a\) and \(b\)

Given: \(\frac{a}{b}=\frac{2}{3}\) --> \(b=\frac{3a}{2}\). Also: \(P=4s=2(a+b)\) --> \(2s=a+b=\frac{5a}{2}\) --> \(s=\frac{5a}{4}\).
Question: \(\frac{ab}{s^2}=?\)

\(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

Answer: B.


OR: you can pick numbers: let the sides of rectangle be 4 and 6 (ratio 2:3) then the perimeter of the rectangle will be 2(4+6)=20, thus the side of the square will be 20/4=5. Next, the area of the rectangle will be 4*6=24 and the area of the square will be 5^2=25, so the ratio of the areas will be 24/25.

Answer: B.


OR: if we take the side of rectangle to be 2x and 3x (for some positive multiple x), then the perimeter of the rectangle will be 2(2x+3x)=10x, thus the side of the square will be 10x/4=5x/2. Next, the area of the rectangle will be 2x*3x=6x^2 and the area of the square will be (5x/2)^2=25x^2/4, so the ratio of the areas will be 24/25.

Answer: B.

Similar question: https://gmatclub.com/forum/if-the-perim ... 96132.html


hi pushpitkc, can you explain how after this \(s=\frac{5a}{4}\). we got this --->

\(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

thanks :)
Joined: 26 Feb 2016
Posts: 2859
Own Kudos [?]: 5459 [1]
Given Kudos: 47
Location: India
GPA: 3.12
Send PM
If the perimeter of square region S and the perimeter of rec [#permalink]
dave13
Bunuel
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

Let the side of square be \(s\) and the side of rectangle \(a\) and \(b\)

Given: \(\frac{a}{b}=\frac{2}{3}\) --> \(b=\frac{3a}{2}\). Also: \(P=4s=2(a+b)\) --> \(2s=a+b=\frac{5a}{2}\) --> \(s=\frac{5a}{4}\).
Question: \(\frac{ab}{s^2}=?\)

\(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

Answer: B.


OR: you can pick numbers: let the sides of rectangle be 4 and 6 (ratio 2:3) then the perimeter of the rectangle will be 2(4+6)=20, thus the side of the square will be 20/4=5. Next, the area of the rectangle will be 4*6=24 and the area of the square will be 5^2=25, so the ratio of the areas will be 24/25.

Answer: B.


OR: if we take the side of rectangle to be 2x and 3x (for some positive multiple x), then the perimeter of the rectangle will be 2(2x+3x)=10x, thus the side of the square will be 10x/4=5x/2. Next, the area of the rectangle will be 2x*3x=6x^2 and the area of the square will be (5x/2)^2=25x^2/4, so the ratio of the areas will be 24/25.

Answer: B.

Similar question: https://gmatclub.com/forum/if-the-perim ... 96132.html


hi pushpitkc, can you explain how after this \(s=\frac{5a}{4}\). we got this --->

\(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

thanks :)

Hi dave13

Rectangle: Length - a , Breadth - b | Square: size - s.

We know that \(b=\frac{3a}{2}\) and \(s=\frac{5a}{4}\)

Ratio of area = \(\frac{ab}{s^2}\) = \(\frac{\frac{3a^2}{2}}{\frac{25a^2}{16}}\) = \({\frac{3a^2}{2}}*{\frac{16}{25a^2}} = \frac{24}{25}\)

Hope this clears your confusion!
Joined: 09 Mar 2016
Posts: 1134
Own Kudos [?]: 1042 [0]
Given Kudos: 3851
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
pushpitkc
dave13
Bunuel
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

Let the side of square be \(s\) and the side of rectangle \(a\) and \(b\)

Given: \(\frac{a}{b}=\frac{2}{3}\) --> \(b=\frac{3a}{2}\). Also: \(P=4s=2(a+b)\) --> \(2s=a+b=\frac{5a}{2}\) --> \(s=\frac{5a}{4}\).
Question: \(\frac{ab}{s^2}=?\)

\(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

Answer: B.


OR: you can pick numbers: let the sides of rectangle be 4 and 6 (ratio 2:3) then the perimeter of the rectangle will be 2(4+6)=20, thus the side of the square will be 20/4=5. Next, the area of the rectangle will be 4*6=24 and the area of the square will be 5^2=25, so the ratio of the areas will be 24/25.

Answer: B.


OR: if we take the side of rectangle to be 2x and 3x (for some positive multiple x), then the perimeter of the rectangle will be 2(2x+3x)=10x, thus the side of the square will be 10x/4=5x/2. Next, the area of the rectangle will be 2x*3x=6x^2 and the area of the square will be (5x/2)^2=25x^2/4, so the ratio of the areas will be 24/25.

Answer: B.

Similar question: https://gmatclub.com/forum/if-the-perim ... 96132.html


hi pushpitkc, can you explain how after this \(s=\frac{5a}{4}\). we got this --->

\(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

thanks :)

Hi dave13

Rectangle: Length - a , Breadth - b | Square: size - s.

We know that \(b=\frac{3a}{2}\) and \(s=\frac{5a}{4}\)

Ratio of area = \(\frac{ab}{s^2}\) = \(\frac{\frac{3a^2}{2}}{\frac{25a^2}{16}}\) = \({\frac{3a^2}{2}}*{\frac{16}{25a^2}} = \frac{24}{25}\)

Hope this clears your confusion!


Hi pushpitkc,

appreciate your explanation just one question :-) just one question we need to find ratio of area of rectangle to the area of square

but we know only breadth of rectangle \(b=\frac{3a}{2}\) i mean it doesnt represent the area of rectangle whereas area of square is aexprresed correctly \(s=\frac{5a}{4}\)

my question is: why do we divide only breadth of rectangle by area of square AND NOT area of rectangle by area of square :?
Joined: 26 Feb 2016
Posts: 2859
Own Kudos [?]: 5459 [1]
Given Kudos: 47
Location: India
GPA: 3.12
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
dave13
pushpitkc
dave13

Hi dave13

Rectangle: Length - a , Breadth - b | Square: size - s.

We know that \(b=\frac{3a}{2}\) and \(s=\frac{5a}{4}\)

Ratio of area = \(\frac{ab}{s^2}\) = \(\frac{\frac{3a^2}{2}}{\frac{25a^2}{16}}\) = \({\frac{3a^2}{2}}*{\frac{16}{25a^2}} = \frac{24}{25}\)

Hope this clears your confusion!


Hi pushpitkc,

appreciate your explanation just one question :-) just one question we need to find ratio of area of rectangle to the area of square

but we know only breadth of rectangle \(b=\frac{3a}{2}\) i mean it doesnt represent the area of rectangle whereas area of square is aexprresed correctly \(s=\frac{5a}{4}\)

my question is: why do we divide only breadth of rectangle by area of square AND NOT area of rectangle by area of square :?

Hi dave13

The area of the rectangle is length*breadth = a*b. We know that the breadth \(b=\frac{3a}{2}\).

The area of the rectangle will be \(a*b = a*\frac{3a}{2} = \frac{3a^2}{2}\)

Hope this helps you!
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3032
Own Kudos [?]: 6979 [4]
Given Kudos: 1646
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
3
Kudos
1
Bookmarks
Expert Reply
manalq8
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

We are given that the perimeters of square region S and rectangular region R are equal and that the sides of R are in the ratio 2 : 3. Let’s label the sides of our figures:

Width of rectangle R = 2x

Length of rectangle R = 3x

Side of square S = s

The perimeter of rectangular region R is 2(2x) + 2(3x) = 4x + 6x = 10x.

The perimeter of square region S is 4s.

Since the two perimeters are equal we can create the following equation:

4s = 10x

2s = 5x

s = (5/2)x

Lastly, we need to determine the areas of both rectangle R and square S.

Area of rectangle R = length * width

A = (3x)(2x) = 6x^2

Since s = (5/2)x, we can use (5/2)x for the side of S.

Area of square S = side^2

A = [(5/2)x]^2

A = (25x^2)/4

We must determine the ratio of the area of region R to the area of region S.

Area of R/Area of S

6x^2/[(25x^2)/4]

6/(25/4)

24/25

Alternate Solution:

We know the sides of the rectangle have a ratio of 2:3; thus we can express the sides of this rectangle as 2x and 3x for some number x. The perimeter of the rectangle, in terms of x, is then 3x + 2x + 3x + 2x = 10x. This is also the perimeter of the square, so taking x = 2 will give us easy numbers to work with.

When x = 2, the sides of the rectangle are 4 and 6; thus the area of the rectangle is 4 x 6 = 24.

Also, when x = 2, the perimeter of the square is 10x = 20; thus a side of the square will be 5. The area of the square will then be 5 x 5 = 25.

So, the ratio of the area of the rectangle to the area of the square is 24:25.

Answer: B
GMAT Club Legend
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 8083
Own Kudos [?]: 4380 [1]
Given Kudos: 243
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy)
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
1
Kudos
manalq8
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

given
4s=2*(2+3)
s=5/2
so area of R = 6
area of S = 25/4
ratio R/S ; 24/25
IMO B
Joined: 08 Feb 2020
Posts: 7
Own Kudos [?]: 31 [2]
Given Kudos: 49
Send PM
If the perimeter of square region S and the perimeter of rec [#permalink]
2
Bookmarks
I like to approach such questions by picking smart numbers.
Given that the sides of the rectangular are in the ratio of 2:3 we know that the perimeter of that rectangular would be (2x2)+(2x3)=10
Because the perimeter of the square should be the same, the sides have to be 4x2,5=10
I do want to work with integers so I don't like 2,5.
Let's just double the sides of the rectangular (2x4)+(2x6)=20
Now, each side of the square has to be 5
Area of rectangular R -> 4x6=24
Area of square S -> 5x5=25
Ratio 24:25
Hence, B
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6797
Own Kudos [?]: 31611 [1]
Given Kudos: 799
Location: Canada
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
1
Kudos
Expert Reply
Top Contributor
manalq8
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

Let's PLUG IN some values that meet the given conditions.

The sides of R are in the ratio 2:3
So, let the two sides have lengths 2 and 3.
This means the area of Region R = (2)(3) = 6
This means the ENTIRE perimeter of Region R is 2 + 2 + 3 + 3 = 10


The perimeters of square region S and rectangular region R are equal.
This means the perimeter of square region S is also 10
Since all 4 sides in a square are of equal length, each side must have length 2.5
So, the area of Region S = (2.5)(2.5) = 6.25

What is the ratio of the area of region R to the are of region S ?
We get: 6 : 6.25
Check the answer choices .... no matches. So, we need to take 6 : 6.25 and find an equivalent ratio.
If we multiply both parts by 4 we get: 24 : 25
So, the correct answer is B

Cheers,
Brent
Tutor
Joined: 30 Oct 2012
Status:London UK GMAT Consultant / Tutor
Posts: 75
Own Kudos [?]: 158 [0]
Given Kudos: 3
Send PM
If the perimeter of square region S and the perimeter of rec [#permalink]
Expert Reply
Hi GMATters,

Here is my video solution to this question:

NB: You can always plug numbers in here if you want to as Jeff suggested above. It's really up to you, but I personally lean into Algebra because thinking about the correct numbers seems like just about as much effort as setting up the variables and letting it rip. This is my way; do what works best for you.

Enjoy!

Originally posted by PGTLrowanhand on 02 Jul 2022, 08:27.
Last edited by PGTLrowanhand on 02 Jul 2022, 11:38, edited 1 time in total.
Tutor
Joined: 11 May 2022
Posts: 1080
Own Kudos [?]: 772 [0]
Given Kudos: 81
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
Expert Reply
manalq8
If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S

A. 25:16
B. 24:25
C. 5:6
D. 4:5
E. 4:9

I'll never understand why would solve this question any way aside from just picking numbers that make sense.

We could make the rectangle 2x3, but then the sides of the square would be 2.5. That doesn't sound all that awesome. How about we make the rectangle 4x6 and the square 5x5.
Area of the rectangle is 24. Area of the square is 25.
AreaR:AreaS = 24:25

Answer choice B.


ThatDudeKnowsPluggingIn
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 35194
Own Kudos [?]: 891 [0]
Given Kudos: 0
Send PM
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
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.
GMAT Club Bot
Re: If the perimeter of square region S and the perimeter of rec [#permalink]
Moderator:
Math Expert
96065 posts