Is the perimeter of rectangle R greater than 28?

Question

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.

Project PS Butler 
 Subscribe to get Daily Email -   Click Here  |  Subscribe via RSS -   RSS

Are You Up For the Challenge: 700 Level Questions

nick1816 · Answer

Length of rectangle = L
Breadth of rectangle = B

Is (L+B)> 14

Statement 1- L*B= 50

\frac{(L+B)}{2}  ≥  \sqrt{LB}

\frac{(L+B)}{2}  ≥  \sqrt{50}  > \sqrt{49}

(L+B)> 14

Sufficient

Statement 2-

L+B > \sqrt{(L^2+B^2)}

Sum of any 2 sides of triangle is greater than the 3rd side

L+B>10

L+B can be smaller than 14 or greater than 14.

Insufficient

GMATBusters · Answer

For constant Perimeter, Area of a rectangle is max, when it is a square 
 For Constant Area, Perimeter of a rectangle is min, when it is a square

(1) Area of rectangle R is 50.
Since, Area of rectangle is given, perimeter will be minimum when it is a square
if it is square, side =  \sqrt{Area}= \sqrt{50}= 5\sqrt{2} 
Perimeter = 4*Side = 4*5 \sqrt{2} = 20*1.414= 28.28
Hence the Perimeter must be greater than 28
SUFFICIENT
(2) Diagonal of rectangle R is 10.
Hence L^2+B^2 = 100
Now, L+B will be max when L = B
so, L = B = 5 \sqrt{2} 
it will be a square , perimeter =4*Side = 4*5 \sqrt{2} = 20*1.414= 28.28 : Greater than 28 
but if the sides L = 8, B = 6
Perimeter = 2(6+8)= 28: NOT Greater than 28
Hence, It is NOT SUFFICIENT

Answer = A

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.

Project PS Butler 
 Subscribe to get Daily Email -   Click Here  |  Subscribe via RSS -   RSS

Are You Up For the Challenge: 700 Level Questions

eakabuah · Answer

We are to determine whether the perimeter of the rectangle R is greater than 28.

Statement 1: Area of rectangle R is 50.
Sufficient.
This is because irrespective of the possible values of lengths and breadths of R, the perimeter is always greater than 28. For example length can be 50, and breadth can be 1. Perimeter =  2(51)=102 
length and breadth can be  \sqrt{50}  which means that length and breadth are greater than 7, then perimeter  > 2(14)>28 ,

Statement 2: the diagonal of R is 10
Insufficient.
Assuming that the rectangle is a square, then length=breadth=5\sqrt{2} Hence perimeter =2(10\sqrt{2})  =  14.14*2=28.28  Yes
Assuming that the diagonal is part of the Pythagorean triplets  6,8,10  then the length will be 8 while the breadth will be 6. perimeter  = 2(14)=28  No.

the answer is A.

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
Let l and b be the length and breadth of the rectangle
We need to find whether the perimeter of rectangle R is greater than 28 or not.
 • 2(l+b) > 28
• l + b >14
 o We need to find whether the sum of l and b is greater than 14 or not.  
 Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:  Area of rectangle R is 50
 • l*b = 50
 o For a given rectangle the perimeter is minimum when the length is equal to the breadth
   l^2 = 50 
  l > 7 
  l+l= l+ b > 14  
o The perimeter of rectangle is more than 28.  
Hence, statement 1 is sufficient, we can eliminate the answer options B, C, and E.
 Statement 2:  Diagonal of rectangle R is 10
 •  l^2 + b^2 = 100 
 o Case I: If l =b, then 
o  l^2 = 50 
 o  l > 7 
o  l + b > 14  
o Case II: If l =8 and b = 6,
 o Then,  8^2 + 6^2 = 100 
o  l + b = 14  
 
• We are getting different results 
Hence, statement 2 is not sufficient, the correct answer is  Option A.

Kinshook · Answer

Asked: Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.
lw = 50 
Since AM>GM
 l+w > 2\sqrt{lw} = 2\sqrt{50} = 10\sqrt{2} = 14.1 
Perimeter of rectangle = 2(l+w) > 28.2
SUFFICIENT

(2) Diagonal of rectangle R is 10.
l^2 + w^2 =100
(l + w)^2 - 2lw = 100
l + w > 10
Perimeter of rectangle = 2(l+w) > 20
NOT SUFFICIENT

IMO A

GMATinsight · Answer

Question: Is Perimeter 2*(l+b) > 28?

Statement 1 : Area of rectangle R is 50

Case 1: l = 5 and b = 10, perimeter = 30
Case 2:  FOr Minimum Perimeter   l = 5√2 and b = 5√2, perimeter = 2*10√2 = 28.2 i.e. Greater than 28

SUFFICIENT

Statement 2 : Diagonal of rectangle R is 10.

Case 1: l = 6 and b = 8, Perimeter = 28 i.e. answer to the question is NO
Case 2:  FOr Maximum Perimeter   l = 5√2 and b = 5√2, perimeter = 2*10√2 = 28.2 i.e. Greater than 28

NOT SUFFICIENT

Answer: Option A

sauravleo123 · Answer

Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.

Is a+b>14?

Statement 1:
ab=50
If a=7, b=7
ab=49
so definitely a+b has to be greater than 14
Sufficient

Statement 2: a^2 + b^2= 100
(a+b)^2- 2ab=100
a+b= (100+2ab)^1/2
So a+b>10
We cannot say if a+b>14
Insufficient
A

Posted from my mobile device

EgmatQuantExpert · Answer

Solution 
  To find  
 • If the Area of Rectangle R > 50

Approach and Working out   
The process skills are instrumental in solving questions on Data Sufficiency. This question can be solved using the process skill of ‘inference’. Let’s see how.
As per the question, knowledge of following concepts is needed.
 • Perimeter of rectangle R = 2(length + breadth) =  2(l+b) 
• Area of R =  length*breadth =lb .
• Diagonal of R =  \sqrt{(l^2+b^2)} 
 
We need to find: If  2(l+b) > 28

Application of ‘inference’:
 • If  2(l+b) >28 , therefore  l+b >14 .
 • Therefore problem reduces to finding whether the statements are sufficient to check if   l+b>14 .

Now as per Statement 1, 
 •  lb = 50 
•  l+b = (l+50/l) . We cannot say if it is greater than 14.
 • Therefore statement 1 is not sufficient.

Now as per Statement 2,
 •  \sqrt{( l^2 + b^2)} = 10 . 
•  l^2 + b^2 = 100 
• Using the conceptual knowledge that  (a+b)^2 =a^2 +b^2 +2ab , we can write it as  (l+b)^2 – 2lb =100 .
•  (l+b)^2 = 100 + 2lb .
•  (l+b) = \sqrt{(100+2lb)} . We still cannot say if it is greater than 14. Therefore statement 2 is not sufficient to answer the question.
 • But we can infer that Since  14^2 =196 , if  l+b>\sqrt{(196)} ,  l+b  will be  >14 .

When we combine both statements 1 and 2, 
 • Using  lb = 50  with statement 2 as solved above, 
•  (l+b) = \sqrt{(100 + 2lb) }  =   \sqrt{(100+100)} = \sqrt{(200)} > \sqrt{(196)}. 
 • We thus used the inference derived in statement 2 without having to solve for  \sqrt{(200)} .

Correct Answer: Option C

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Is the perimeter of rectangle R greater than 28?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

Is the perimeter of rectangle R greater than 28?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards