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# If a rectangular region has perimeter P inches and area A sq

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If a rectangular region has perimeter P inches and area A sq [#permalink]

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21 Sep 2013, 06:03
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If a rectangular region has perimeter P inches and area A square inches, is the region square?

(1) P = 4/3*A
(2) P = 4√A
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Sep 2013, 07:58, edited 2 times in total.
Renamed the topic and edited the question.
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Re: PS- Gmat Prep exam pack 1- If a rectangular region has [#permalink]

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21 Sep 2013, 06:29
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olifurlong wrote:
If a rectangular region has perimeter P inches and area A square inches, is the region square?
1) P=4/3A
2)P=4√A

Lets try finding the relation between area and perimeter of a square.
We know the area of a square with length x (A) = x^2
The perimeter of the square (P) = 4x

Since P = 4x --> x = P/4

So, A = (P/4)^2
A =(P^2)/16
Therefore, P^2 = 16A
and P = 4*sqrtA

Therefore, B is suff to answer this question
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Re: PS- Gmat Prep exam pack 1- If a rectangular region has [#permalink]

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21 Sep 2013, 06:34
I think 2) is pretty obvious, it satisfies length = breadth condition.
For 1) if you substitute length = P/2 - breadth,
it can be written as
Now substitute in Area formula A = length * breadth
the equation looks like this:

This clearly is not a form of (a-b)^2.
So the above equation has 2 distinct values of breadth. Hence it is not suff.
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Re: PS- Gmat Prep exam pack 1- If a rectangular region has [#permalink]

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21 Sep 2013, 06:40
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olifurlong wrote:
If a rectangular region has perimeter P inches and area A square inches, is the region square?
1) P=4/3A
2)P=4√A

for a rectangle Perimeter P=2(l+b)
Area A=l*b

From 1) 2(l+b)=4/3*(l*b)
6(l+b)=4(l*b)
3(l+b)=2(l*b)
this equation satisfies for l=3 and b=3, dimensions of a square and at l=6,b=2 dimensions of a rectangle..hence not sufficient

From 2) 2(l+b)=4√(l*b)
(l+b)=2√(l*b) squaring both sides
(l+b)^2=4l*b
(l-b)^2=0
l=b,which is a square, Sufficient
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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24 Mar 2014, 00:30
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olifurlong wrote:
If a rectangular region has perimeter $$P$$ inches and area $$A$$ square inches, is the region square?

(1) $$P = \frac{4}{3}*A$$
(2) $$P = 4\sqrt{A}$$

Let the rectangular region have sides $$x$$ and $$y$$.
--> $$Perimeter = P = 2(x+y)$$
--> $$Area = A = xy$$

Question: Is $$x = y$$ ?

(1) $$P = \frac{4}{3}*A$$

$$2(x+y) = (\frac{4}{3})xy$$

$$\frac{x+y}{xy} = \frac{4}{6} = \frac{2}{3}$$

$$\frac{1}{x} + \frac{1}{y} = \frac{4}{6} = \frac{2}{3}$$

So we have that the sum of two values equals $$\frac{2}{3}$$

If $$\frac{1}{x} = \frac{1}{y} = \frac{1}{3}$$, then the answer is YES

If $$\frac{1}{x} = \frac{1}{6}$$ and $$\frac{1}{y} =\frac{3}{6}$$, then the answer is NO

(2) $$P = 4\sqrt{A}$$

$$2(x + y) = 4\sqrt{xy}$$

$$x + y = 2\sqrt{xy}$$

$$x + y - 2\sqrt{xy} = 0$$

$$(\sqrt{x} - \sqrt{y})^2 = 0$$

$$\sqrt{x} - \sqrt{y} = 0$$

$$\sqrt{x} = \sqrt{y}$$

$$x = y$$

--> SUFFICIENT
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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30 Mar 2014, 08:50
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Let me show you what I tried to do with this one

Is the figure a square?

Well if its a square then perimeter is 4x and area is x^2, pretty obvious up to here.

Now, statement 1

P=4/3A

So if the figure is in fact a square then it must follow that 4x=4/3 (x^2)

Is 12x = 4x^2?
Is 4x (x-3) = 0?
Is x = 3?

Well we can't solve this so insufficient

Statement 2 says that P= 4 (sqrt) A

So is 4x = 4 (sqrt (x^2))?

Well is 4x = 4 abs (x)?

If x is the side of the square then x>0, so 4x = 4x, so yes this is in fact a square

Hope this helps
Cheers
J
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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30 Mar 2014, 09:33
jlgdr wrote:
Let me show you what I tried to do with this one

Is the figure a square?

Well if its a square then perimeter is 4x and area is x^2, pretty obvious up to here.

Now, statement 1

P=4/3A

So if the figure is in fact a square then it must follow that 4x=4/3 (x^2)

Is 12x = 4x^2?
Is 4x (x-3) = 0?
Is x = 3?

Well we can't solve this so insufficient

Statement 2 says that P= 4 (sqrt) A

So is 4x = 4 (sqrt (x^2))?

Well is 4x = 4 abs (x)?

If x is the side of the square then x>0, so 4x = 4x, so yes this is in fact a square

Hope this helps
Cheers
J

Dropping an "if and only if" (iff) assumption into a question and evaluating whether it holds is my favorite method.

If anyone wants some fun practice with that, try the following:

Prove that line y=mx+b is tangent to circle x^2+y^2=r^2 iff b^2=r^2(1+m^2).

This question is obviously beyond GMAT scope, but in terms of practicing for the exam, I thinks it's quite useful.

Hint: a quadratic equation in one variable has one solution iff the discriminant is zero.

Posted from my mobile device
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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03 Oct 2014, 05:40
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Reinfrank2011 wrote:
olifurlong wrote:
If a rectangular region has perimeter $$P$$ inches and area $$A$$ square inches, is the region square?

(1) $$P = \frac{4}{3}*A$$
(2) $$P = 4\sqrt{A}$$

Let the rectangular region have sides $$x$$ and $$y$$.
--> $$Perimeter = P = 2(x+y)$$
--> $$Area = A = xy$$

Question: Is $$x = y$$ ?

(1) $$P = \frac{4}{3}*A$$

$$2(x+y) = (\frac{4}{3})xy$$

$$\frac{x+y}{xy} = \frac{4}{6} = \frac{2}{3}$$

$$\frac{1}{x} + \frac{1}{y} = \frac{4}{6} = \frac{2}{3}$$

So we have that the sum of two values equals $$\frac{2}{3}$$

If $$\frac{1}{x} = \frac{1}{y} = \frac{1}{3}$$, then the answer is YES

If $$\frac{1}{x} = \frac{1}{6}$$ and $$\frac{1}{y} =\frac{3}{6}$$, then the answer is NO

(2) $$P = 4\sqrt{A}$$

$$2(x + y) = 4\sqrt{xy}$$

$$x + y = 2\sqrt{xy}$$

$$x + y - 2\sqrt{xy} = 0$$

$$(\sqrt{x} - \sqrt{y})^2 = 0$$

$$\sqrt{x} - \sqrt{y} = 0$$

$$\sqrt{x} = \sqrt{y}$$

$$x = y$$

--> SUFFICIENT

This is the clearest solution for this problem that I have seen! Well done, mate!
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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02 Nov 2014, 23:58
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My approach here :

From 1: 3(l+b)= 2lb

Put in
b= 10L (or anything that makes it a non square)

33L=20L^2
33=20L

There definitely is some value of L that will satisfy this..So a rectangle does exist..

b=l

6L = 2L^2
6=2L
L=3

So a square too exists...Done A is insuff

For B

the equation comes to: 2(L+B)=4\sqrt{LB}
Square and simplify
L^2 + B^2 - 2 LB = 0
which is (L-B)^2

you can also try the above approach here:
B=10 L
101L^2- 20L^2 = 0

Which aint true..So B is suff to say that it is indeed a square
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If a rectangular region has perimeter P inches and area A sq [#permalink]

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16 Dec 2014, 12:56
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olifurlong wrote:
If a rectangular region has perimeter P inches and area A square inches, is the region square?

(1) P = 4/3*A
(2) P = 4√A

I would solve this question by simply plugging in numbers.

1. Let's assume the rectangular is indeed a square.
2. Let's assume the side of the square is 4.

If the side of this square is 4:
The perimeter would be 16 and the area would be 16 as well.

(S1) P = 4/3√A
If the rectangular is a square, all values for A area should correspond with the right value for P.
In my assumed case: if the perimeter is 16 , the area should be 16 too.
In this case : 16 is not equal to 4/3*16 , so we can't say for sure that this rectangle indeed is a square.

(S2) P = 4√A
16=4*√16
We got a match!

If you want to be sure , try any other plugin for a square.
Lets say the side of the rectangle is 2.
For it to be a square the perimeter should be 8(2+2+2+2) and the area is 4(2*2)
8=4√4
So as you can see , S2 will hold true for any plugin.A
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If a rectangular region has perimeter P inches and area A sq [#permalink]

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30 Mar 2015, 11:01
kusena wrote:
olifurlong wrote:
If a rectangular region has perimeter P inches and area A square inches, is the region square?
1) P=4/3A
2)P=4√A

for a rectangle Perimeter P=2(l+b)
Area A=l*b

From 1) 2(l+b)=4/3*(l*b)
6(l+b)=4(l*b)
3(l+b)=2(l*b)
this equation satisfies for l=3 and b=3, dimensions of a square and at l=6,b=2 dimensions of a rectangle..hence not sufficient

From 2) 2(l+b)=4√(l*b)
(l+b)=2√(l*b) squaring both sides
(l+b)^2=4l*b
(l-b)^2=0
l=b,which is a square,Sufficient

This can be Rhombus as well right?Even Rhombus has all sides equal. So how can we confidently say that it is a square. We dont know whether the diagonals are equal

Would really appreciate if someone can clear my doubt

Thanks
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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30 Mar 2015, 17:36
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Expert's post
Hi buddyisraelgmat,

The prompt refers to a RECTANGULAR REGION, so the four corners must be 90 degrees each. Thus, we're dealing with either a rectangle of some kind or a square.

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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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30 Mar 2015, 21:12
Test scenarios

St.1 P=(4/3)A

take 2 as square side, we get P=8, A=4 - NO as square
take 3 as quare side, we get P=12, A=9 - YES as square

INSUFF

St.2 P=4√A

both 2 and 3 confirms this relation

SUFF

B
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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19 Oct 2015, 10:58
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olifurlong wrote:
If a rectangular region has perimeter P inches and area A square inches, is the region square?

(1) P = 4/3*A
(2) P = 4√A

I would also solve by plugging in values:

Let's create two rectangular regions which are squares:

Generally: P = 4s (where s is the side of a square) and $$A = s^2$$

First square with side 2:
P = 8 and A = 4

Second square with side 3:
P = 12 and A = 9

Statement 1: Gives you a yes on the square with side 3 because P = 4/3 * A but a no for the square with side 2.
Statement 2: $$P = 4*\sqrt{A}$$ - this is true for both squares we created. Fair enough, Answer B.
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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03 Feb 2016, 11:48
kusena wrote:
olifurlong wrote:
From 2) 2(l+b)=4√(l*b)
(l+b)=2√(l*b) squaring both sides
(l+b)^2=4l*b
(l-b)^2=0
l=b,which is a square, Sufficient

What is happening at: (l+b)^2=4l*b to entirely get rid of the right hand side and flip the sign on the left hand side?
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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03 Feb 2016, 12:34
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redfield wrote:
kusena wrote:
olifurlong wrote:
From 2) 2(l+b)=4√(l*b)
(l+b)=2√(l*b) squaring both sides
(l+b)^2=4l*b
(l-b)^2=0
l=b,which is a square, Sufficient

What is happening at: (l+b)^2=4l*b to entirely get rid of the right hand side and flip the sign on the left hand side?

Once you get,$$(l+b)^2=4l*b$$ ---> open the brackets by employing the relation $$(a+b)^2=a^2+b^2+2ab$$,

You get, $$l^2+b^2+2lb=4lb$$---> $$l^2+b^2-2lb =0$$ ---> recognize that this =$$(l-b)^2$$ as $$(l-b)^2 = l^2+b^2-2lb$$

Thus, $$(l-b)^2=0$$---> l=b. Hence, the given rectangle is a Square. Sufficient.

Hope this helps.
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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04 Sep 2016, 09:42
Can this problem be solved by picking numbers? Can someone show me how to? I tried to solve this by picking numbers and ended up getting into a rut. Thanks!
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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04 Sep 2016, 10:03
ruggerkaz wrote:
Can this problem be solved by picking numbers? Can someone show me how to? I tried to solve this by picking numbers and ended up getting into a rut. Thanks!

How can you take the real values of length and breath unless you know if they are equal or not? I don't think this question could be solved picking real numbers.
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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20 Sep 2016, 10:14
reto wrote:
olifurlong wrote:
If a rectangular region has perimeter P inches and area A square inches, is the region square?

(1) P = 4/3*A
(2) P = 4√A

I would also solve by plugging in values:

Let's create two rectangular regions which are squares:

Generally: P = 4s (where s is the side of a square) and $$A = s^2$$

First square with side 2:
P = 8 and A = 4

Second square with side 3:
P = 12 and A = 9

Statement 1: Gives you a yes on the square with side 3 because P = 4/3 * A but a no for the square with side 2.
Statement 2: $$P = 4*\sqrt{A}$$ - this is true for both squares we created. Fair enough, Answer B.

When you plug in numbers, you have to make sure that the numbers that you plug in lead you to a perimeter and an area that satisfy the condition mandated by the statement.

In the first statement, the values P = 8 and A = 4 do not satisfy the condition P = 4/3*A.

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If a rectangular region has perimeter P inches and area A sq [#permalink]

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15 Jun 2017, 10:35
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mh not quite satisfied with the explanations

So Ill give it a try

If a rectangular region has perimeter P inches and area A square inches, is the region square?

Perimeter of a Square = 4x
Area of a Square = x^2

(1) P = 4/3*A

4x = 4/3*x^2
3x=x^2
x=3

so in order to be a square x has to be 3, but we don't know --> insufficient

(2) P = 4√A

4x=4√x^2
x=x

--> sufficient

cheers
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If a rectangular region has perimeter P inches and area A sq   [#permalink] 15 Jun 2017, 10:35

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