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Is the perimeter of rectangle R greater than 28?

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Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 16 Jul 2018, 07:34
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Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.

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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 16 Jul 2018, 10:47
Harshgmat wrote:
Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.




Let x and y be the length and breath of the rectangle respectively.

Q: is \(x+y>14?\)

1) \(x*y=50\)
Taking cases,
\(x=25,y=2\) . Yes.
Reducing x further so as to make x and y almost equal ie \(x=7,y=50/7\) Yes.
If you reduce x further, y will go up and the sum will stay greater than 14. Hence Sufficient

2) \(x^2+y^2=100\)
\(x=8,y=6\) Yes
\(x=5\sqrt{3},y=5\) No
Hence Not Sufficient

A
Are you sure about the OA? IMO A is the correct answer.
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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 16 Jul 2018, 11:30
2a+2b>28?
a+b>14?

(1) ab=50
a=50/b
50/b+b>14?
50+b^2>14b?
b(b-14)>-50?
Yes
Sufficient

(2) sqrt(a^2+b^2)=10
a^2+b^2=100
a=sqrt(100-b^2)
Is sqrt(100-b^2)+b>14?
b is (0, 10) not always true on interval
Not Sufficient

IMO Answer A
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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 17 Jul 2018, 09:07
The answer should be D , that is each statement alone is sufficient - because of the following:
1) Area of rectangle is 50
Given l*b =50 , what the question asks is find the minimum value of l+b ( the question actually is " is l+b> 14" ..this can be translated to find the minimum value of l+b)

To find the minimum value of l+b , we can put b= 50/l in the expression l+b
New question is : Minimize l+50/l
if we differentiate this by l we get the value of l =5*sqrt(2) ~ 7.07... Hence the minimum value of l+b is 14.14

1 is hence sufficient

2) Given sqrt( l^2+b^2)= 100 , minimize l+b
Again, if we minimize this expression the value comes out to be the same 5*sqrt(2)...so this too is sufficient

However, I really doubt this is gmat problem
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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 17 Jul 2018, 10:15
bhanu88 wrote:
The answer should be D , that is each statement alone is sufficient - because of the following:
1) Area of rectangle is 50
Given l*b =50 , what the question asks is find the minimum value of l+b ( the question actually is " is l+b> 14" ..this can be translated to find the minimum value of l+b)

To find the minimum value of l+b , we can put b= 50/l in the expression l+b
New question is : Minimize l+50/l
if we differentiate this by l we get the value of l =5*sqrt(2) ~ 7.07... Hence the minimum value of l+b is 14.14

1 is hence sufficient

2) Given sqrt( l^2+b^2)= 100 , minimize l+b
Again, if we minimize this expression the value comes out to be the same 5*sqrt(2)...so this too is sufficient

However, I really doubt this is gmat problem


Maxima of 2(x+y)= 2(x+ \sqrt{(100-x^2)}..... once this is analyzed..... maximum value of the expression occur at x=5\sqrt{2}...... but this is not always true.... so (A)
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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 17 Jul 2018, 10:21
durgamadhab wrote:
bhanu88 wrote:
The answer should be D , that is each statement alone is sufficient - because of the following:
1) Area of rectangle is 50
Given l*b =50 , what the question asks is find the minimum value of l+b ( the question actually is " is l+b> 14" ..this can be translated to find the minimum value of l+b)

To find the minimum value of l+b , we can put b= 50/l in the expression l+b
New question is : Minimize l+50/l
if we differentiate this by l we get the value of l =5*sqrt(2) ~ 7.07... Hence the minimum value of l+b is 14.14

1 is hence sufficient

2) Given sqrt( l^2+b^2)= 100 , minimize l+b
Again, if we minimize this expression the value comes out to be the same 5*sqrt(2)...so this too is sufficient

However, I really doubt this is gmat problem


Maxima of 2(x+y)= 2(x+ \sqrt{(100-x^2)}..... once this is analyzed..... maximum value of the expression occur at x=5\sqrt{2}...... but this is not always true.... so (A)



you are finding minima not maxima
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Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 17 Jul 2018, 10:26
Harshgmat wrote:
Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.



(1) For a given area..... perimeter is minimum at side=square root of area.....minimum perimeter>28.....so sufficient
(2) For a given diagonal .... perimeter is maximum at side=diagonal/1.44.......after calculation....maximum perimeter is 28.8.......so there can be other lower values. so not sufficient.

In case of statement (2)....as u increase one side from 7....the perimeter will go down actually....plug in values and see for yourself.

Lets say 3xsquare root of 11 and 1...their sum of squares is 100. but perimeter is less than 22.
For calculation of maxima and minima....same procedure is applied..... so be careful about what u got..... it may be maxima...it may be minima.
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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 25 Jul 2018, 14:07
chetan2u What are your thoughts on this one?
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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 25 Jul 2018, 14:19
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Quadrilateral with minimun perimeter for a given area is a square. So, if we can assume the rectangle to be a square and calculate it's perimeter, that would be the minimum possible perimeter.

St.1: Area 50. That means, each side is √50. Minimun Perimeter= 4*√50 is greater than 4*√49= 4*7= 28. Sufficient.

St.2: Diagonal 10. That means, each side is 10/√2. Minimun Perimeter= 40/√2= 20*√2= 20*1.414 which is greater than 20*1.4= 28. Sufficient.

Hence, Ans D

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Re: Is the perimeter of rectangle R greater than 28?  [#permalink]

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New post 25 Jul 2018, 14:40
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Thanks Shobhit7. This is the key knowledge I was missing:
Shobhit7 wrote:
Quadrilateral with minimun perimeter for a given area is a square. So, if we can assume the rectangle to be a square and calculate it's perimeter, that would be the minimum possible perimeter.
Posted from my mobile device


If we assume the rectangle to be a square with side x, question becomes:
Is 28<4x, i.e. is 7<x, or equivalently, is 49<x^2?

(1)
Area of rectangle (assume it's a square) is 50.
so x^2 = 50. Sufficient

(2)
Diagonal of rectangle (assume it's a square) is 10.
Area = (diagonal^2)/2 = (10^2)/2 = 50
Same info as in (1). Sufficient

Answer: D
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Re: Is the perimeter of rectangle R greater than 28? &nbs [#permalink] 25 Jul 2018, 14:40
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