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# Does rectangle A have a greater perimeter than rectangle B?

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Manager
Joined: 21 Feb 2010
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Does rectangle A have a greater perimeter than rectangle B? [#permalink]

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28 Jul 2010, 05:45
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Does rectangle A have a greater perimeter than rectangle B?

(1) The length of a side of rectangle A is twice the length of a side of rectangle B
(2) The area of rectangle A is twice the area of rectangle B
[Reveal] Spoiler: OA

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Manager
Joined: 16 Apr 2010
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28 Jul 2010, 06:04
Hi,

Statement one provides information about one side only. This is not enough since one side of rect A can be greater than one side of rect B while the second side of rectangle A can be either less or greater than the second side of rectangle B. Remember Perimeter = 2(L+W).
Consider:
Rect A: 20*20
Rect B: 1*2
or
Rect A: 20*1
Rect B: 1*40

Statement 2 is not sufficient as well. Consider:
Rect A: 20*1, P=42
Rect B: 1*10, P=22
or
Rect A: 2*2, P=8
Rect B: 20:0.1, P=40.2

Taking both conditions, the answer will be sufficient.
You can also solve this problem by equations.

regards,
Jack

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28 Jul 2010, 06:09
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tt11234 wrote:
hello all,
here's the question...
does rectangle A have a greater perimeter than rectangle B?
1) the length of a side of rectangle A is twice the length of a side of rectangle B
2) the area of rectangle A is twice the area of rectangle B

Let the sides of rectangle A be $$s$$ and $$t$$ and the side of rectangle B $$m$$ and $$n$$.

Question: is $$2(s+t)>2(m+n)$$? --> or is $$s+t>m+n$$?

(1) $$s=2m$$, clearly insufficient as no info about the other side of rectangles.

(2) $$st=2mn$$, also insufficient as if $$s=t=2$$, $$m=1$$ and $$n=2$$ then the answer would be YES, but if $$s=t=2$$, $$m=\frac{1}{2}$$ and $$n=4$$ then the answer would be NO.

(1)+(2) $$s=2m$$ and $$st=2mn$$ --> substitute $$s$$: $$2m*t=2mn$$, so $$t=n$$. Thus as $$s=2m$$ and $$t=n$$: $$s+t=2m+n$$ which is obviously more than $$m+n$$. Sufficient.

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29 Jul 2010, 12:13
I always love when the OP uses a [Reveal] Spoiler: but posts the OA in the question anyway...
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31 Jul 2010, 22:53
Bunuel wrote:
tt11234 wrote:
hello all,
here's the question...
does rectangle A have a greater perimeter than rectangle B?
1) the length of a side of rectangle A is twice the length of a side of rectangle B
2) the area of rectangle A is twice the area of rectangle B

Let the sides of rectangle A be $$s$$ and $$t$$ and the side of rectangle B $$m$$ and $$n$$.

Question: is $$2(s+t)>2(m+n)$$? --> or is $$s+t>m+n$$?

(1) $$s=2m$$, clearly insufficient as no info about the other side of rectangles.

(2) $$st=2mn$$, also insufficient as if $$s=t=2$$, $$m=1$$ and $$n=2$$ then the answer would be YES, but if $$s=t=2$$, $$m=\frac{1}{2}$$ and $$n=4$$ then the answer would be NO.

(1)+(2) $$s=2m$$ and $$st=2mn$$ --> substitute $$s$$: $$2m*t=2mn$$, so $$t=n$$. Thus as $$s=2m$$ and $$t=n$$: $$s+t=2m+n$$ which is obviously more than $$m+n$$. Sufficient.

Nice explanation, Bunuel! Thanks a lot

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Re: Does rectangle A have a greater perimeter than rectangle B? [#permalink]

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27 Dec 2013, 16:25
tt11234 wrote:
Does rectangle A have a greater perimeter than rectangle B?

(1) The length of a side of rectangle A is twice the length of a side of rectangle B
(2) The area of rectangle A is twice the area of rectangle B

Nice question, let me chip in

Let's call L,W of rectangle A (A,B) and LW, of rectangle B (C,D)

So question is is 2(A+B) > 2(C+D) or if you will A+B>C+D?

(1) A = 2C Insuff

(2) AB > CD Insuff too

(1) + (2) One ends up with 4C + 2B > 2B + CD (1)

On the other hand 2CB>CD so then 2B>D (2)

Now rearranging (1)

Is 2C + 2B > CD?

Well 2B > CD from (2) So given that sides have to be positive then yes

So C is our best choice

Hope it helps

Kudos rain!

Cheers!
J

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Re: Does rectangle A have a greater perimeter than rectangle B? [#permalink]

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16 Apr 2015, 12:14
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Re: Does rectangle A have a greater perimeter than rectangle B?   [#permalink] 16 Apr 2015, 12:14
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