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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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New post 16 Jul 2017, 11:47
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


The goal is to determine if the shape is in fact a square.

Statement 1) P = 4/3*(A)

If true, then P=4x and A = x^2

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

Only true if the side is length 3. We don't know if the side of the shape has length 3, so the statement is insufficient.

Statement 2) P = 4sqrt(A)

4x = 4sqrt(x^2)
4x=4x
x = x

This statement is true, so the statement is sufficient.
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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New post 17 Jul 2017, 06:47
(B) in IMO. First assume the rectangle is a square. Then l=b. P=4l and Area=l^2. If you consider statement 1, LHS is not equal to RHS.
If you consider statement 2, LHS=RHS. So B is sufficient.
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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New post 29 Jan 2018, 04:13
This question raises very important point on DS questions.

First of all it is a YES/NO question.

When you plug in numbers statement (2) becomes YES. And when you plug numbers in statement (2) the answer is almost always no. Theoretically both statements give you concrete YES or NO, therefore each alone must be sufficient.

But two statements must both either say YES or NO. If one statement yields YES but the other NO, then the statement that you know for 100% is sufficient is the correct answer. And there must be at least on set of numbers that would give YES answer for statement one. But you should not bother to find that set of numbers.
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Re: If a rectangular region has perimeter P inches and area A sq [#permalink]

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New post 19 May 2018, 15:41
igotthis 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


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


I did the question in a similar way. Could someone please explain why A is not sufficient?

For a square, we know:
A = x^2
P = 4x
A/P = x/4 => P = 4A/x

So shouldn't any equation in that form (i.e. Statement A: P = 4A/3) be sufficient? According to A, our side x would be 3 but I don't think the side would matter since we just need the equation to be in the form P = 4A/x.
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If a rectangular region has perimeter P inches and area A sq [#permalink]

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New post 19 May 2018, 23:41
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dabaobao wrote:
I did the question in a similar way. Could someone please explain why A is not sufficient?

For a square, we know:
A = x^2
P = 4x
A/P = x/4 => P = 4A/x

So shouldn't any equation in that form (i.e. Statement A: P = 4A/3) be sufficient? According to A, our side x would be 3 but I don't think the side would matter since we just need the equation to be in the form P = 4A/x.


Hey dabaobao ,

You did a very BIG mistake of not reading the question properly.

Question says "If a rectangular region" but you took it a square.

For a rectangle, Perimeter = 2(l+b) and area = l * b.

Now, try substituting the values in statement A, you won't be able to determine whether l = b for the region to be a square. Hence, A is insufficient.

Does that make sense?
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If a rectangular region has perimeter P inches and area A sq   [#permalink] 19 May 2018, 23:41

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