In the original condition, there is a rectangle whose length and width we need to know, so there are two variables.
We need the number of equations to match the number of variables to solve for the variables, so we need 2 equations.
Conditions 1 and 2 each provide an equation each, so the answer is C.

Max@Math Revolution wrote:In the original condition, there is a rectangle whose length and width we need to know, so there are two variables.
We need the number of equations to match the number of variables to solve for the variables, so we need 2 equations.
Conditions 1 and 2 each provide an equation each, so the answer is C.

This line of reasoning does not apply to the posted problem.
The question stem does not ask us to solve for the length and width.
It asks only whether we have sufficient information to determine that the perimeter of the rectangle -- the value of 2(L+W) -- must be greater than 28.

From my post above:

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.

As you can see, Statement 1 provides sufficient information to determine that 2(L+W) > 28.
The correct answer is not C but A.

An alternate -- and perhaps easier -- way to prove that Statement 1 is SUFFICIENT.

RULE:
If Rectangle R has a perimeter of x units, then the greatest possible area will be yielded if R is a SQUARE with a side of length x/4.

Example: Let p = 40
If L=10 and W=10, then A = 10*10 = 100.
If L=11 and W=9, then A = 11*9 = 99.
If L=12 and W=8, then A = 12*8 = 96.
As the example above illustrates, the greatest possible area is yielded when L=W=10 and R is a SQUARE.

Question stem: Is the perimeter of Rectangle R greater than 28?
If p=28, then the greatest possible area will be yielded if R is a square with a side of 7:
7*7 = 49.
Implication.
To have an area greater than 49, R must have a perimeter GREATER THAN 28.

Statement 1: The area of rectangle R is 50.
Since the area is greater than 49, R must have a perimeter greater than 28.
SUFFICIENT.

Forget conventional ways of solving math questions. 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.



Is the perimeter of rectangle R greater than 28?

(1) Area of rectangle R is 50.

(2) Diagonal of rectangle R is 10.

In this problem, (1)=(2), and when the area of the rectangle is fixed, the perimeter of the square is minimum. In other words, the length and the width are identical, meaning a^2=50, a=5sqrt 2. The perimeter is 4a=20sqrt2=20*1.41>28, thus the conditions are sufficient.
Therefore the answer is D.

But since the original condition has 2 variables (width, length) we need 2 equations to match the number of variables, and since we have 1 for each conditions 1), 2), C is likely the answer and it often is the answer to such problems.


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Hello everyone, I opened an old thread.
but if anyone can help, I have a different approach.
we know that if you are given an Area , the square has the least perimeter.
so what I did I transferred it into square , if I got more than 28 then the rectangular will diff be greater than 28,
St.A:
Area=50 , lets make it a square then sides will equal 5*2^1/2 , then multiply by 4 we get 28.sth then rec is diff greater than 28, hence St A sufficient.

St.B:
the same, dia=10 , make it a square we have sides 5*2^1/2 , multiply by 4 then we will get 28.sth then rec is greater than 28.hence B suff

therefore answer is D

please help.