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Does rectangle A have a greater perimeter than rectangle B?

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Does rectangle A have a greater perimeter than rectangle B? [#permalink] New post 28 Jul 2010, 04:45
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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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Re: rectangle [#permalink] New post 28 Jul 2010, 05:04
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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Re: rectangle [#permalink] New post 28 Jul 2010, 05:09
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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

the answer is C, could someone please explain why C is the answer? thanks!


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.

Answer: C.
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Re: rectangle [#permalink] New post 29 Jul 2010, 11:13
I always love when the OP uses a [Reveal] Spoiler: but posts the OA in the question anyway...
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Re: rectangle [#permalink] New post 31 Jul 2010, 21:53
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

the answer is C, could someone please explain why C is the answer? thanks!


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.

Answer: C.


Nice explanation, Bunuel! Thanks a lot
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Re: Does rectangle A have a greater perimeter than rectangle B? [#permalink] New post 27 Dec 2013, 15:25
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 :)
Re: Does rectangle A have a greater perimeter than rectangle B?   [#permalink] 27 Dec 2013, 15:25
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