Is the perimeter of rectangle R greater than 28?

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

(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.




Are You Up For the Challenge: 700 Level Questions[/quote]
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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.

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.
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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
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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
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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.

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

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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
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