Check my two posts here:

https://www.beatthegmat.com/difference-b ... 68610.html

IMO C. i.e. both 1 and 2 are necessary to solve it.

What is the OA?

There are 2 variables in the isosceles triangle and 1 in a square, so we need 3 variables in order to solve the problem.
There are only 2 equations given from the conditions, so there is high chance (E) will be our answer.
Looking at the conditions together, AC and AB passes through the midpoint of SE and QR, so the area of ABC is always greater that that of square SQRE, so this answers the question 'yes' and is sufficient.
The answer is (C).

For cases where we need 3 more equation, such as original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

Similar problem:

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR.

(2) The area of triangle ABC is 75% of the area of square SQRE.

Since AB=AC and AS=AQ, both triangle ABC and square SQRE are SYMMETRICAL about height AM in the figures below.

Statement 1: The length of RE is 4 times the length of BR.
RE = 4BR implies FIGURE A below:

In right triangle ABM, since MB = 3x and AM = 4x, triangle ABM is a 3-4-5 triangle.
Thus, AB = 5x, implying that AC = 5x.
Result:
Perimeter of square SQRE = 4x+4x+4x+4x = 16x.
Perimeter of triangle ABC = 5x+x+2x+2x+x+5x = 16x.
Thus, the difference between the perimeters = 16x-16x = 0.
SUFFICIENT.

Statement 2: The area of triangle ABC is 75% of the area of square SQRE.
In FIGURE A above, the area of triangle ABC is 75% of the area of square SQRE:
(area of triangle ABC)/(area of square SQRE) = (1/2*6x*4x)/(4x*4x) = 12x²/16² = 3/4 = 75%.
Thus, FIGURE A satisfies statement 2.
To determine whether the FIGURE A is the ONLY case that will satisfy statement 2, consider FIGURE B:

In FIGURE B, RE = 6BR.
Here, (area of triangle ABC)/(area of square SQRE) = (1/2*8x*6x)/(6x*6x) = 24x²/36x² = 2/3 = 66 2/3%.
Doesn't work.
The implication is that, for the area of triangle ABC to be 75% of the area of square SQRE, it must be true that RE=4BR, as in statement 1.
Thus, since statement 1 is sufficient, statement 2 must also be SUFFICIENT.

The correct answer is D.

Alternate way to evaluate Statement 2:
Plug in a value for the side of the square and SOLVE for the dimensions of the triangle.

Square:
Let s = 8.
Area = s² = 8² = 64.
Perimeter = 4s = 4*8 = 32.

Triangle:
Since the area of the triangle is equal to 75% of the area of the square, we get:
Area = (3/4)64 = 48.
As in FIGURE A above, let AM = the height of triangle ABC.
Since AM = 8, we get:
(1/2) * b * 8 = 48.
b = 12.

Thus, MB = (1/2)b = (1/2)12 = 6.
Since MB=6 and AM=8, triangle ABM is a 6-8-10 triangle, implying that AB=10.
Thus:
Perimeter of the triangle = 10+12+10 = 32.

Result:
Perimeter of the square - perimeter of the triangle = 32-32 = 0.

GMATGuruNY wrote:

Max@Math Revolution wrote:There are 2 variables in the isosceles triangle and 1 in a square, so we need 3 variables in order to solve the problem.
There are only 2 equations given from the conditions, so there is high chance (E) will be our answer.
Looking at the conditions together, AC and AB passes through the midpoint of SE and QR, so the area of ABC is always greater that that of square SQRE, so this answers the question 'yes' and is sufficient.
The answer is (C).

The OA is not C but D.

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR.

(2) The area of triangle ABC is 75% of the area of square SQRE.

Hi Mitch,

I think the question u have solved is different from the question posted. The figures look similar though.

Thanks.

binit wrote: Hi Mitch,

I think the question u have solved is different from the question posted. The figures look similar though.

Thanks.

Good catch.
I've amended my post accordingly.