The most concrete piece of given information on this problem is that the area of a rectangle is . You know that Area = Length × Width, so you can say here that . You're also told how the length and width relate to each other. The longer side (which is the one they ask you to solve) is  inches longer than the shorter side, so you can call those  (for the longer) and  for the shorter. This then means that:

You can then expand the multiplication:

Which becomes a quadratic if you subtract  from each side:

And now you have a choice: you could solve this algebraically by factoring the quadratic, but of course  may not be the easiest number to quickly factor. Instead, you could test the answer choices to see which potential  multiplies with  to yield a product of .

If you start in the middle with , you'll see that  would be . Before you calculate, first check to see whether you will indeed get a units digit of  (otherwise why do the math?). You will, but when you do do that math you'll see that  is , which is too small since you need a larger number in .

So then assess the remaining larger answers. If the answer were E, the sides would be 20×16 which will not end in a , so that's out. If it were D, then your sides would be , which does end in a , and which does yield .

Had you wished to factor the quadratic, you would find that  factors to , again yielding  as the answer for the longer side.

Since the monitor in question is square, its diagonal creates a  isosceles right triangle. Recall that the sides of such a triangle are in the ratio . In this case, the hypotenuse has length , so  and . Thus, the sides of the square monitor have length , and the area of the square is .

We are told that the screen area is three times the frame area. Now, the temptation may be to divide  by , but in fact  represents the area of the total monitor – screen plus frame – and the ratio of screen to total area is  (we can compute the ratio to the total by adding up the component ratios  and ; screen:frame:total = ). So we can get the screen area by multiplying the monitor area, , by . The screen area is .

To get the diagonal length for the screen, recall once again that . So the screen has side length . And the diagonal again creates a  right triangle, so the hypotenuse is .

While the math looks a little ugly to start, it cleans up nicely (as usual on the GMAT). Since the longest distance between two points on a square is its diagonal, if that distance for the current plot of land is , then that means that the length of each side of that smaller square is . And since we know that the width of the new property is the same but the length is double, that makes the length of the new property . When combined, the width stays the same at  and the new length becomes , and then to find the area just multiply them together. The roots in the denominator will multiply out to just , so the fraction is

 for a total of .

The relationships between side lengths and area for two dimensional figures, and side lengths and volume for three dimensional figures can be confusing. It is very easy to think that if the lengths all double, the area and volume should follow suit, but if every dimension is doubled in a 2-dimensional shape (a square or rectangle, where you're talking about area) the area is multiplied by  (the square of the change for the sides) and in a 3-dimensional shape (a cube, sphere, or box, where you're talking about volume) the volume is multiplied by 8 (the cube of the change of the sides).

Suppose that the length of each side of the smaller cube is . That would mean that the volume would be  If you double that length across all dimensions, then in calculating the length, width, and depth you'd multiply: 

This simplifies to . And note the relationship between the larger and smaller cubes: , the volume of the larger cube, . So with the given problem, if the volume of the smaller cube is , then the volume of the larger cube is 8 times that, so the correct answer is .

This problem asks you to find the surface area for  sides of the box, since the top side will not have area. You should then determine the dimensions of each side that you'll be using.

For the left and right sides, the measurement will be  square inches, and since you'll have two of those sides you'll multiply by  to have  square inches of sides.

The front and back will measure  square inches, and since you'll have two of those sides you should multiply by  to have  square inches of front/back.

Then you'll need to account for the bottom, which measures  square inches.

So your total calculation is  square inches.

The surface area of a cylinder has three components: the area of the top, the area of the bottom, and the area of the "side."

The areas of the top and bottom are classic circles, so you'll use  to calculate. Here the radius is , so each circle will have a volume of . One important key here is to remember to multiply that by 2 to account for both circles. So combined, the top and bottom have an area of .

For the "side," it is important to think conceptually about what constitutes that area. If you were to unroll the circular nature of the cylinder, the side would form a rectangle. Quite clearly the height will be the same as the height of the cylinder, but what about the length? The length is the circumference of the circle, the distance along the top (or bottom) for the material to stretch exactly around the circle.

Circumference is , so here that's . Multiply that by the height of  and you have  as the area of the side. So your area is now , which sums to .

Importantly here, the greatest distance between two points in the cube (from one corner to the opposite corner) will equal the diameter of the sphere. Because the cube is perfectly inscribed within the circle, a line that travels through the center of the cube will travel through the center of the sphere, and if it touches two corners of the cube then it's touching the outside of the sphere, satisfying the definition of the diameter.

With that, your goal should be to use the volume of the cube to determine the diameter of the sphere. This can be done quickly if you know the rule for the greatest distance in a rectangular box: . Here since length, width, and height are all the same, , you have a quick calculation:

.

Since the circumference of a circle can be expressed as , your answer is simply .

The volume of a rectangular box is Length × Width × Height. Here you're given those three dimensions as , but then told that the volume of water is only of the total. So your calculation is , which comes out to .

An important lesson from this problem involves the relationship between length and volume. Since volume is three-dimensional and length is only one-dimensional, when you reduce the length of all sides in a 3-D shape, you have to account for that change along all three dimensions. In this case, the box is scaled down by a linear factor of , so its volume scales down by a factor of .

So the new box is  the volume of the old box, meaning that it decreased in volume by .  expressed as a percentage is .

Alternatively, you could avoid the abstraction by choosing your own numbers and playing out the scenario that way. Imagine a  cubical box (volume ) being scaled down to a  cubical box (volume ). You'd go from a volume of  to a volume of , losing  of the volume, again equating to an  reduction.

The volume of the box is Base * Height, where Height is the length of the box and Base is the area of the triangular face. Thus, . The area of the base, then, is . Now recall that the area of an equilateral triangle is .

So . and  and , which is the value of .