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If the perimeter of square region S and the perimeter of rec

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If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 03 Feb 2012, 08:21
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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

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If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 03 Feb 2012, 08:29
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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:

The sides of R are in the ratio 2:3 --> \(\frac{a}{b}=\frac{2}{3}\) --> \(b=\frac{3a}{2}\).
The perimeter of square region S and the perimeter of rectangular region R are equal --> \(P=4s=2(a+b)\) --> \(2s=a+b=\frac{5a}{2}\) --> \(s=\frac{5a}{4}\).

Question: \(\frac{area \ of \ R}{area \ of \ S}=\frac{ab}{s^2}=?\)

Substitute b = 3a/2 and s = 5a/4 into the question: \(\frac{ab}{s^2}=\frac{3a^2}{2}*\frac{16}{25a^2}=\frac{24}{25}\).

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
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 03 Feb 2012, 10:17
GOT IT, 2ND APPROACH IS EASIER...THAAAAAANKS B.
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 20 Jun 2013, 06:39
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 24 Sep 2013, 01:59
manalq8 wrote:
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.
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 06 Oct 2014, 09:30
Bunuel has given a pretty good explanation. But still,, here is how i approached it :
For rectangle of length a and Breadth b: given is a/b = 2/3 ie a =2x/5 and b=3x/5(where x is some multiplication factor of the ratio). So, Perimeter of R = 2(a+b) = 2(2x/5 + 3x/5) = 2x
Perimeter of S = 4Side = 2x Therefore Side = x/2.
Area R (a*b):Area S side^2 = 6x^2/25 / x^2/4 = 24:25 .

:) Hope i could help :)
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 02 Nov 2016, 08:57
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manalq8 wrote:
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


In figure R -

Length = 3 ; Breadth = 2

Area of R = 6 and Perimeter of R = 2 ( 3 + 2 ) => 10

In figure S -

Sides = S (Say)

So, 4s = 10

Or, s = 5/2

Area of S = 25/4

Hence area of R to the area of S = 6 : 25/4 => 24 : 25

Hence, Correct answer will be (B) 24: 25
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 12 Jun 2018, 08:00
Bunuel wrote:
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 :)
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If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 12 Jun 2018, 09:48
1
dave13 wrote:
Bunuel wrote:
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!
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 12 Jun 2018, 14:08
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
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 :?
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 12 Jun 2018, 23:01
1
dave13 wrote:
pushpitkc wrote:
dave13 wrote:

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!
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 14 Jun 2018, 10:33
2
manalq8 wrote:
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
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 03 Jun 2019, 07:19
manalq8 wrote:
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
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Re: If the perimeter of square region S and the perimeter of rec  [#permalink]

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New post 25 Oct 2019, 13:10
Hi,

Is there a way to back solve this question? I have been learning this technique, but dont always know how to apply it.

If you use the "c" strategy and somehow plug in 5/6... how can we solve for this and what information do we need?

Thanks
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Re: If the perimeter of square region S and the perimeter of rec   [#permalink] 25 Oct 2019, 13:10
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