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by ArunangsuSahu » Fri Jan 06, 2012 9:14 pm
Approach:

Statement 1:
ab=50

The Inflection point a= 7, b=50/7..Here also (a+b) > 14

In all other cases, (a+b) is greater than 14...

SO SUFFICIENT

Statement 2:

c=10
Case I: a=6 ,b=8 implies a+b=14 so perimeter is 2*(a+b)=28
Case II : a=5*sqrt(3), b=5 a+b=5*(1+sqrt(3)) <14..so perimeter <28

Two different answer

INSUFFICIENT

So (A) is the answer
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by mjmehta81 » Tue Feb 14, 2012 4:53 am
Hi,

Can you please guide me why my below explanation is false..

(a+b)^2 = a^2+b^2+2ab
= 10^2 + 2x50
= 100+100
= 200
Therefore = (a+b) = 10sqrt(2)

so perimeter 2(a+b) = 20 x sqrt(2) > 28

So both Statement together help to find the solution.


GMATGuruNY wrote:
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 ronnie1985 » Fri Mar 30, 2012 6:49 am
Given: Rectangle

To Find: Perimeter?

S1: Area = 50
Not Sufficient

S2: Diagonal = 10
Not Sufficient

Combining, can solve for sides
Sufficient.

(C) is ans
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by sanju09 » Sat Mar 31, 2012 5:36 am
vishnuchaithanya wrote: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
Why do we need to take two statements together, if the question can be answered using statement (1) alone?
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by Lifetron » Thu Aug 02, 2012 8:34 pm
Good One !

A :)
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by Ganesh hatwar » Thu Aug 23, 2012 1:17 am
good one but got wrong B.

Did not add the sides for A and did not try with different values for B
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by rajeshsinghgmat » Sun Mar 10, 2013 4:55 am
A for Answer.

7*7=49
6*8=48
5*10=50

Let the perimeter be 2(x+y)=28

i.e. x+y=14

Let, x*(14-x)=50

i.e., x^2-14x+50=0

Here, x has no real solutions.

So, we can conclude that the sum of the sides x,y of the rectangle is greater than 14.
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by mariofelixpasku » Thu Mar 14, 2013 3:34 pm
GMATGuruNY wrote:
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.


Can you please explain why you can deduct that there is no solution if there is no solution for your equation? and why do you assume that the equation is = 28 ... when the original sentence states >28 ?
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by sanju09 » Fri Aug 23, 2013 12:04 am
mariofelixpasku wrote:
GMATGuruNY wrote:
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.


Can you please explain why you can deduct that there is no solution if there is no solution for your equation? and why do you assume that the equation is = 28 ... when the original sentence states >28 ?
Quadratic Formula is attached for help. The b^2 - 4ac part under the square root is the DISCRIMINANT, once D is negative, we get no real solution for the variable L as taken by Mitch. If we take perimeter to be 28 or less, we'll keep getting a negative D, hence no real solution. That makes us believe that the perimeter of this rectangle R must have to be greater than 28. Hence, statement 1 is sufficient by its own.

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by Java_85 » Wed Oct 02, 2013 10:30 am
IMO A.
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by NeilWatson » Mon Apr 07, 2014 3:20 pm
GMATGuruNY wrote:
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.
I understand your explanation for statement 2, but regarding statement 1 how do you know that based on all that calculation, the perimeter can't be less than 28 instead of greater?
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by GMATGuruNY » Mon Apr 07, 2014 3:47 pm
NeilWatson wrote:
GMATGuruNY wrote:
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.
Regarding statement 1, how do you know that based on all that calculation, the perimeter can't be less than 28 instead of greater?
When p=28, the value of the determinant is negative because the value of b² (196) is TOO SMALL.
If p<28, then the value of b² will become even SMALLER.

Case 2: p=20
Follow the values in red:
Since p = 2(L+W) = 20, L+W=10 and W=10-L.
Substituting W=10-L into LW=50, we get:
L(10-L) = 50.
10L - L² - 50 = 0.
L² - 10L + 50 = 0.
In this case, b² = 100, with the result that b² - 4ac = 100 - 4*1*50 = -100.
When p<28, the value of the determinant becomes MORE NEGATIVE.

Implication:
For the determinant to be NONNEGATIVE, p>28.
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by nikhilgmat31 » Mon Jul 27, 2015 10:59 pm
Answer is A.

This problem can be more difficult if it is given as 4 sided polygon rather than rectangle. than A& B are not sufficient. We need to go with C in that case.
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by Max@Math Revolution » Fri Jul 31, 2015 10:07 am
In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and equations ensures a solution.

In original condition, we have 2 variables because rectangle should know "length and width" so we need 2 equations. 1) and 2) are 2 equations. now it is sufficien so C is the answer.

Why C? If you know the logics of Variable approach, you can easily find the answer. You don't need to actually solve the problem.

www.mathrevolution.com

Our world's first Variable Approach (DS) and IVY Approach (PS) help students dramatically reduce their time spent per question and improve accuracy. You will have 10 min. to spare before the test ends.
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by GMATGuruNY » Fri Jul 31, 2015 10:55 am
Max@Math Revolution wrote:In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and equations ensures a solution.

In original condition, we have 2 variables because rectangle should know "length and width" so we need 2 equations. 1) and 2) are 2 equations. now it is sufficien so C is the answer.

Why C? If you know the logics of Variable approach, you can easily find the answer. You don't need to actually solve the problem.

www.mathrevolution.com

Our world's first Variable Approach (DS) and IVY Approach (PS) help students dramatically reduce their time spent per question and improve accuracy. You will have 10 min. to spare before the test ends.
The correct answer is not C but A.
Statement 1 on its own is sufficient to determine that the perimeter of rectangle R must be greater than 28.
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