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

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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]
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Aarav2706 wrote:
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

In statement 2; if all we have proved is x=y , then it could be a rhombus too right? Without knowing if the angles are 90 degrees, how can we be certain?

We are told that the region is a rectangle. So, angles are 90 degrees. A rhombus with 90 degrees angles is a square.
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If a rectangular region has perimeter P inches and area A sq [#permalink]
Hi avigutman - i know this was discussed on quantreasoning.com

While I do understand this problem - i am still un-easy with my steps.

Whenever I feel this, i try to create a 'variation' of the original problem, to get a deeper understanding of the concept

But i am struggling to create a version of this problem

Can you perhaps think of one ? i want to see if i can have truly understood the concept of this problem

If you can't think of a variation, thats totally fine, no worries

Thank you
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]
jabhatta2 wrote:

While I do understand this problem - i am still un-easy with my steps.

Whenever I feel this, i try to create a 'variation' of the original problem, to get a deeper understanding of the concept

But i am struggling to create a version of this problem

Can you perhaps think of one ?

jabhatta2 Can you articulate what your steps are? That should help you get a deeper understanding of the concept.
And if you mess it up, that'll be a good learning opportunity.
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If a rectangular region has perimeter P inches and area A sq [#permalink]
hI avigutman - here is what i did for S1

For S1 - attached is what I did for S1 . Found that the length of the region (represented by 'a' can be zero or can be three.

Now
-- a = zero. That too me doesnt make sense and thus "a"= zero can be dropped as a possible value of 'a'
-- Thus a = three only.

In Yes /No question - S1 should work for every value of "a". S1 only works when a = 3

Thus, Not sufficient.
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WhatsApp Image 2022-01-15 at 8.42.18 PM.jpeg [ 91.22 KiB | Viewed 10492 times ]

Originally posted by jabhatta2 on 15 Jan 2022, 18:33.
Last edited by jabhatta2 on 15 Jan 2022, 19:02, edited 2 times in total.
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If a rectangular region has perimeter P inches and area A sq [#permalink]
My question was on S2

avigutman - For S2, here is what I did

I am not sure what Zero = Zero represents (Specifically, how is zero=zero, different from a = zero, which is what i got in S1 and rejected)

I was not getting a value of "A" unlike in S1.

I think i was struggling because from S2 -- i was not getting values of "A"

Thus i wasnt sure what to make from S2 with regards to zero = zero.
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WhatsApp Image 2022-01-15 at 8.32.21 PM.jpeg [ 70.27 KiB | Viewed 10241 times ]

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If a rectangular region has perimeter P inches and area A sq [#permalink]
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jabhatta2

jabhatta2 wrote:

For S1: Found that the length of the region (represented by 'a' can be zero or can be three.

Are you using the question as if it were a statement (assuming the answer to the question is YES, and using that as if it were known?), and then testing what that means for S1?
If so, that's data validation, not data sufficiency.

jabhatta2 wrote:

In Yes /No question - S1 should work for every value of "a".

I'm not sure what you mean. The statements are always truthful. Again, seems like you're using the question to validate the statement.

jabhatta2 wrote:
S1 only works when a = 3

Thus, Not sufficient.

Why is that not sufficient?
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If a rectangular region has perimeter P inches and area A sq [#permalink]
avigutman wrote:
jabhatta2

Are you using the question as if it were a statement (assuming the answer to the question is YES, and using that as if it were known?), and then testing what that means for S1?
If so, that's data validation, not data sufficiency.

Hi avigutman - Yes, i think i was doing what you mentioned in the yellow

I am taking the question itself as 'TRUE' --> inserting it into the statements --> validating the statement instead [is the statement True or False]

• In the Case of validating S1 = i got True when length is 3 only. In case length was some other length, statement 1 would be FALSE.
• In case of the validating S2 -- I got S2 as True, irrespective of the value of the length of the square

Do you mind discusing my process (regarding what i did) in the next AMA (if you believe this mistake is perhaps usefull for the class to know about) - validating statements. Why is that not a legitimate strategy - uneven results / perhaps an analogy for a question WHERE it wont work ?

Below is a simpler example and the process i followed

Quote:

Is region ABCD a square ?
(S1) Volume = $$a^3$$

Here was the process i followed to proove sufficiency.
(Step 1) Write down what you know about a square - the volume is $$a^3$$ | area is $$a^2$$
(step 2) Start from statement 1.
(Step 3) Plug what you know about volume (regarding squares) INTO statement 1
(step 4) Insert Volume = $$a^3$$ in the LHS of the equation
(Step 5) a^3 = a^3
(Step 6) 0=0
(Step 7) Statement 2 is 'TRUE' always

I think i thought like this because i followed some version of this process in another quant question

Originally posted by jabhatta2 on 20 Jan 2022, 11:23.
Last edited by jabhatta2 on 20 Jan 2022, 12:21, edited 21 times in total.
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If a rectangular region has perimeter P inches and area A sq [#permalink]
Here is how i had solved S2 in the referenced question and i think i picked up this DS strategy because of this question

Quote:

If mv < pv <0, is v > 0 ?
(S2) m < 0

(Step 1) if question itself is true (i.e. if v > 0)
(Step 2) Insert that info into S2
(Step 3) m < 0 is true.
(Step 4) Then, mv must ALSO BE NEGATIVE (because v is positive)
(Step 5) that matches the given information [ mv < pv < 0]
(step 6) hence S2 is sufficinent

I think given i was allowed to do some version of 'going backwards' i thought i could do the same in the original question
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]
jabhatta2 wrote:
I think given i was allowed to do some version of 'going backwards' i thought i could do the same in the original question

Now that we've discussed in the AMA, would you like to try to explain here the errors in your approach and how you'd attack the problem differently now, jabhatta2?
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If a rectangular region has perimeter P inches and area A sq [#permalink]
Hi avigutman - last session on quantreasoning.com was very helpful. Here was my mistakes / insights from just this one problem

(a) my mistake was assuming the question as 'True'. Then plugging that information INTO S1 & S2 - That was a mistake. It should be the other way around

(b) The process should be instead
info from statement + free information == Can you proove the question

Originally posted by jabhatta2 on 25 Jan 2022, 19:36.
Last edited by jabhatta2 on 25 Jan 2022, 19:53, edited 1 time in total.
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If a rectangular region has perimeter P inches and area A sq [#permalink]
What i found interesting were the following insights
Quote:
(1) perimter / area CAN NEVER be a constant ratio for any figure (square / rectangle) by definition

Reason -- area will have an extra 'variable'.

That is why S1 is insufficient. Perimeter / Area will be 4/3 in one case (When length = 3) . In other cases, Perimeter / Area WILL NEVER BE 4/3. It will be a different ratio by definition.

This is why S1 is logically not sufficient.

Quote:
(2) Perimeter^2 / Area on the other hand is always constant given the variables are NOW Cancelled (2d to 2d)

Originally posted by jabhatta2 on 25 Jan 2022, 19:48.
Last edited by jabhatta2 on 25 Jan 2022, 19:54, edited 1 time in total.
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If a rectangular region has perimeter P inches and area A sq [#permalink]
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In fact avigutman --- i can make up a new problem and figure out whether the statement is sufficient

Quote:
(s2 variation) diagonal of square : perimter of square == this will be sufficient as this ratio is always going to be constant given both are 1-d
(s2 variation) diagonal^2 : area is == this will be sufficient as this ratio is always going to be constant as its now (2d to 2d)
(s2 variation) circumference of circle = square root of area == this will be sufficient as this ratio is always going to be constant (1d to 1d)

It is this kind of amazing insights that make quant reasoning AMA's unparallel in terms of building logic for prospective students
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]
Aarav2706 wrote:
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

In statement 2; if all we have proved is x=y , then it could be a rhombus too right? Without knowing if the angles are 90 degrees, how can we be certain?

It is given that it's a rectangular region, so we know it has 90-degree angles
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]
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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

Let's assume the sides of the region as 2 and 2. If any of the equations satisfy we can consider that the rectangular region is a square
Perimeter = 2(2+2) = 8
Area =l*b = 2*2 = 4

1) P=4/3*A
8 = 4/3 * 4
24 is not equal to 16

2) P = 4\sqrt{A}
8 = 4\sqrt{4}
8 = 4*2
8=8
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]
Hi Bunuel, Can you share more of such question types as you have done for various other question types ?
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