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by sanju09 » Sat May 29, 2010 5:59 am
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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Source: — Data Sufficiency |
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by selango » Sat May 29, 2010 7:52 am
We need to find whether 2l+2b>28

-->l+b>14

From statement 1 ,we know l*b=50

50=2*5*5

the possible scenarios for l*b are,

2*25

10*5

5*10

25*2

Whatever value l and b are , l+b>14,sufficient

Diagonal^2=l^2+b^2

100=l2+b2

Only one scenario exists for this l=8,b=6 or l=6,b=8,from this we can find l+b so sufficient


Hence D is answer

what is OA
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by akhpad » Sat May 29, 2010 8:55 am
sanju09 wrote:Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.
Side = a and b; Perimeter = P, Area = A, Diagonal = D

Statement 1:
A = 50, a = 1, b = 50, P = 102 > 28
A = 50, a = 6, b = 50/6, P = 86/3 > 28
A = 50, a = 8, b = 50/8, P = 57/2 > 28
Suppose A = 49, a = b = 7, P = 28; So, if A=50, P>28

Sufficient

Statement 2:
D = 10, a=6, b=8, P = 28
D = 10, a=5, b=5*sqrt(3) = 5*1.732 = 8.66, P = 27.32 < 28
Suppose D*D = 98, a = b = 7, P = 28; So, if D*D = 100, P>28

Insufficient

Answer: A
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by gmatjedi » Sat May 29, 2010 9:08 am
to minimize perimeter, figure dimension must approach that of a square
note that a square of 7x7 will have a perimeter of 28, therefore, as the figure is a rectangle and per st 1 has an area of 50, then one dimension (along with its opposing face) is >7 which would make perimeter >28

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by clock60 » Sat May 29, 2010 1:21 pm
sanju09 wrote:Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.
finally come to answer A. but initial answer was wrong
is a+b>14, or is
a^2+2ab+b^2>196

(1) ab=50
is a^2+b^2+100>196
a^2+b^2>96
to make a^2+b^a least as possible with attention that ab=50
let us a=7 and b=7 (7*7=49 so restriction with area=50 doesn`t violate)
49+49=98, that is greater than 96
any other pairs of a,b will result in even greater square sum.
consider 6,8
36+64=100. other examples also possible
so 1 st sufficient

(2)here a^2+b^2=100
is 2ab+100>196
2ab>96
is ab>48 with a^2+b^2=100
7^2+x^2=100
x^2=100-49
x^2=51
x=51^1/2
51^1/2 is slightly more than 7.say 7,a
7,a*7>48

and perimeter is more that 14
1^2+x^2=100
x^2=99
x=99^1/2=3*(11)^1/2
1*3*11^1/2<48
so various answers are possible
2 st insuffiicent
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by kstv » Mon May 31, 2010 6:14 am
Is the perimeter of rectangle R greater than 28?
Let the sides be a & b so a+b > 14

(1) Area of rectangle R is 50.
ab = 50 = 5*5*2 , one side has to be 5 or multiple of 5 here it can be 10
if a = 5 , b = 10
Sufficient

(2) Diagonal of rectangle R is 10.
a²+b² = 100
Not sufficient as u need knowledge of Pythagoras triplets :wink: . I can think of 8²+6², 4²+7²
But in both of them a+b = < 14 , which is diff from solutions in 1) this is troubling me.
IMO A.
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by liferocks » Mon May 31, 2010 6:21 am
kstv wrote:Is the perimeter of rectangle R greater than 28?
Let the sides be a & b so a+b > 14

(1) Area of rectangle R is 50.
ab = 50 = 5*5*2 , one side has to be 5 or multiple of 5 here it can be 10
if a = 5 , b = 10
Sufficient

(2) Diagonal of rectangle R is 10.
a²+b² = 100
Not sufficient as u need knowledge of Pythagoras triplets :wink: . I can think of 8²+6², 4²+7²
But in both of them a+b = < 14 , which is diff from solutions in 1) this is troubling me.
IMO A.
You are considering only rational numbers for a and b..consider irrationals as well

From1
ab=50..since ab is constant to minimize a+b ,a must be equal to b i.e. a=b=5V2
so perimeter=4*5V2=20V2..V2=1.41 approx so 20V2=28.2 approx..sufficient

From 2
a^2+b^2 =100..a,b can be anything.As you have got they can be 8²+6², 4²+7² or (5V2)^2+(5V2)^2=100 as well
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by ssuarezo » Wed Jun 02, 2010 5:58 am
Hi:
Please, what is the OA? It seems we have many different logical answers.
Silvia
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by lavinia » Sun Jan 02, 2011 11:14 am
I believe that the answer is A.

Question: Is 2(L+W)>28? Is L+W>14?

(1) L*W=50
L= 50
W= 1 -> L+W=51 Yes

L= 25
W= 2 -> L+W= 27 Yes

L= 10
W=5 -> L+W= 15 Yes

This is sufficient

(2) Diagonal of rectangle R is 10 -> Diagonal^2= L^2+W^2

L=8
W= 6 -> 14>14?
L= 3root 11
W= 1 -> 3 root 11+1>14? No

Two different answers (equal or smaller): Insufficient
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by lavinia » Sun Jan 02, 2011 11:17 am
Hi Liferocks,

How is it possible for statement 2: 4²+7²=100? 16+49=100? Probably you had other numbers in your mind :)
liferocks wrote:
kstv wrote:Is the perimeter of rectangle R greater than 28?
Let the sides be a & b so a+b > 14

(1) Area of rectangle R is 50.
ab = 50 = 5*5*2 , one side has to be 5 or multiple of 5 here it can be 10
if a = 5 , b = 10
Sufficient

(2) Diagonal of rectangle R is 10.
a²+b² = 100
Not sufficient as u need knowledge of Pythagoras triplets :wink: . I can think of 8²+6², 4²+7²
But in both of them a+b = < 14 , which is diff from solutions in 1) this is troubling me.
IMO A.
You are considering only rational numbers for a and b..consider irrationals as well

From1
ab=50..since ab is constant to minimize a+b ,a must be equal to b i.e. a=b=5V2
so perimeter=4*5V2=20V2..V2=1.41 approx so 20V2=28.2 approx..sufficient

From 2
a^2+b^2 =100..a,b can be anything.As you have got they can be 8²+6², 4²+7² or (5V2)^2+(5V2)^2=100 as well
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by GMATGuruNY » Sun Jan 02, 2011 2:23 pm
sanju09 wrote:Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.
Statement 1:
LW = 50.
Let's assume that the perimeter = 28.
Then 2(L+W) = 28, L+W=14, and W=14-L.
Substituting into LW=50, we get:
L(14-L) = 50.
14L - L^2 -50 = 0.
L^2 - 14L + 50 = 0.
For any quadratic equation in the form of ax^2 + bx + c, the determinant is b^2 - 4ac. If the determinant is negative, the equation has no real solutions. In the equation above, a=1, b= -14 and c=50. Since the determinant of the equation is b^2-4ac = (-14)^2 - 4*1*50 = 196-200 = -4, the equation does not have a real solution.
This shows us that in order for the equation above to have a real solution, the perimeter has to be greater than 28.
Sufficient.

Statement 2:
L^2 + W^2 = 100.
If L=6 and W=8, then p = 2*(6+8) = 28. Is the perimeter greater than 28? No.
If L=7 and W=√51, then p = 2*(7 + √51). Recognizing that √51>7, is the perimeter greater than 28? Yes.
Since the answer can be both no and yes, insufficient.

The correct answer is A.
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by vishnuchaithanya » Mon May 30, 2011 1:22 am
I have a new approach

L*W = 50 implies 2*L*W = 100 ---(1)

L^2 + W^2 = 100 ---(2)

adding (1) and (2)
we get

L^2 + W^2 + 2LW = 100+100
i.e. (L+W)^2 = 200
(L+W) = +/- 10 * sqrt(2)
i.e. perimeter = 2(L+W) = 20* sqrt(2) (taking only positive quantity)
and this gives perimeter greater than 28
therefore answer should be C
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by SoCan » Mon May 30, 2011 7:50 am
While you can prove it mathematically, I think it's faster to think about statement 2 conceptually. Think about a rectangle with diagonals that are hinged at their midpoints. You can change the shape of the rectangle by "opening" and "closing the diagonals" like scissors. You can make the short side infinitely small and the longer side infinitely close to 10 - so the lower limit of the perimeter is 20.
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by MM_Ed » Fri Aug 19, 2011 5:00 pm
Is 2l + 2b > 28?
Or, is l + b > 14?

l*b = 50

l^2 + b^2 = 100
or l^2 + 2500/l^2 = 100
or l^4 - 100*l^2 + 2500 = 0
or l^2 = 50
or l = sqrt(50) so b = sqrt (50)

Perimeter = 4*sqrt(50) > 28
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by Himanshu Vyas » Sun Sep 25, 2011 6:39 am
The Correct Answer is A.

This is how it goes:
We have to find out if 2(L+B)>28 OR L+B>14
OPTION A

The Area is given as 50. For a rectangle the area will be LxB=50.
There are two options here when this rectangle is a "conventional rectangle" i.e where Length and Breadth are different
(i) 25x2 OR
(ii)10x5
For both of the above Perimeter which is 2(L+B)>28 CONDITION SATISFIED!!

But hang on, there is another scenario to be considered i.e
The other option for which we need to ascertain the above inequality is when the Rectangle is not the conventional rectangle but a "square" because even a square can be a rectangle.

Therefore, LxB =50 implies only one set of values that is L=B= sqrt50 which is nearly equal to 7.1. Here also L+B >14 CONDITION SATISFIED!!
Hence A is sufficient

Now, for Option B:
The diagonal is 10 units implies Sqrt(L^2+B^2)=10
=>L^2+B^2 =100
Some people might think that only "8 and 6" is the best answer which is equal to 14 and a definite NO,which would mean that for them even B is sufficient. However, we still need to consider the Square is a Rectangle Theory and check
The L=B Therefore, L^2x L^2 =100 (Putting B=L)
The only option here again is: L=B= Sqrt50, but for this the individual values of L&B will be slightly greater than 7, hence the Perimeter will be sightly greater than 28. Thus there is a contradiction. HENCE INSUFFICIENT!!!
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