The most concrete piece of given information on this problem is that the area of a rectangle is \(\displaystyle 252\). You know that Area = Length × Width, so you can say here that \(\displaystyle LW=252\). 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 \(\displaystyle 4\) inches longer than the shorter side, so you can call those \(\displaystyle x\) (for the longer) and \(\displaystyle (x-4)\) for the shorter. This then means that:

\(\displaystyle x(x-4)=252\)

You can then expand the multiplication:

\(\displaystyle x^{2}-4x=252\)

Which becomes a quadratic if you subtract \(\displaystyle 252\) from each side:

\(\displaystyle x^{2}-4x-252=0\)

And now you have a choice: you could solve this algebraically by factoring the quadratic, but of course \(\displaystyle 252\) may not be the easiest number to quickly factor. Instead, you could test the answer choices to see which potential \(\displaystyle x\) multiplies with \(\displaystyle (x-4)\) to yield a product of \(\displaystyle 252\).

If you start in the middle with \(\displaystyle x=16\), you'll see that \(\displaystyle (x-4)\) would be \(\displaystyle 12\). Before you calculate, first check to see whether you will indeed get a units digit of \(\displaystyle 2\) (otherwise why do the math?). You will, but when you do do that math you'll see that \(\displaystyle 16*12\) is \(\displaystyle 192\), which is too small since you need a larger number in \(\displaystyle 252\).

So then assess the remaining larger answers. If the answer were E, the sides would be 20×16 which will not end in a \(\displaystyle 2\), so that's out. If it were D, then your sides would be \(\displaystyle 18*14\), which does end in a \(\displaystyle 2\), and which does yield \(\displaystyle 252\).

Had you wished to factor the quadratic, you would find that \(\displaystyle x^{2}-4x-252=0\) factors to \(\displaystyle (x-18)(x+14)=0\), again yielding \(\displaystyle 18\) as the answer for the longer side.

Since the monitor in question is square, its diagonal creates a \(\displaystyle 45^{\circ}-45^{\circ}-90^{\circ}\) isosceles right triangle. Recall that the sides of such a triangle are in the ratio \(\displaystyle x:x:x\sqrt{2}\). In this case, the hypotenuse has length \(\displaystyle 20\), so \(\displaystyle x\sqrt{2}=20\) and \(\displaystyle x=\frac{20}{\sqrt{2}}\). Thus, the sides of the square monitor have length \(\displaystyle \frac{20}{\sqrt{2}}\), and the area of the square is \(\displaystyle (\frac{20}{\sqrt{2}})^{2}=\frac{400}{2}=200\).

We are told that the screen area is three times the frame area. Now, the temptation may be to divide \(\displaystyle 200\) by \(\displaystyle 3\), but in fact \(\displaystyle 200\) represents the area of the total monitor – screen plus frame – and the ratio of screen to total area is \(\displaystyle 3:4\) (we can compute the ratio to the total by adding up the component ratios \(\displaystyle 1\) and \(\displaystyle 3\); screen:frame:total = \(\displaystyle 3:1:4\)). So we can get the screen area by multiplying the monitor area, \(\displaystyle 200\), by \(\displaystyle \frac{3}{4}\). The screen area is \(\displaystyle 150\).

To get the diagonal length for the screen, recall once again that \(\displaystyle area=side^{2}\). So the screen has side length \(\displaystyle \sqrt{(150)}\). And the diagonal again creates a \(\displaystyle 45^{\circ}-45^{\circ}-90^{\circ}\) right triangle, so the hypotenuse is \(\displaystyle x\sqrt{(2)}=\sqrt{(150)}*\sqrt{(2)}=\sqrt{(150*2)}=\sqrt{(300)}\).

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 \(\displaystyle 4\), then that means that the length of each side of that smaller square is \(\displaystyle \frac{4}{\sqrt{2}}\). 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 \(\displaystyle \frac{8}{\sqrt{2}}\). When combined, the width stays the same at \(\displaystyle \frac{4}{\sqrt{2}}\) and the new length becomes \(\displaystyle \frac{12}{\sqrt{2}}\), and then to find the area just multiply them together. The roots in the denominator will multiply out to just \(\displaystyle 2\), so the fraction is

\(\displaystyle \frac{48}{2}\) for a total of \(\displaystyle 24\).

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 \(\displaystyle 4\) (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 \(\displaystyle x\). That would mean that the volume would be \(\displaystyle x^{3}\) If you double that length across all dimensions, then in calculating the length, width, and depth you'd multiply: 

\(\displaystyle 2x*2x*2x\)

This simplifies to \(\displaystyle 8x^{3}\). And note the relationship between the larger and smaller cubes: \(\displaystyle 8x^{3}\), the volume of the larger cube, \(\displaystyle x^{3}*8\). So with the given problem, if the volume of the smaller cube is \(\displaystyle a\), then the volume of the larger cube is 8 times that, so the correct answer is \(\displaystyle 8a\).

This problem asks you to find the surface area for \(\displaystyle 5\) 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 \(\displaystyle 10*8=80\) square inches, and since you'll have two of those sides you'll multiply by \(\displaystyle 2\) to have \(\displaystyle 160\) square inches of sides.

The front and back will measure \(\displaystyle 10*16=160\) square inches, and since you'll have two of those sides you should multiply by \(\displaystyle 2\) to have \(\displaystyle 320\) square inches of front/back.

Then you'll need to account for the bottom, which measures \(\displaystyle 16*8=128\) square inches.

So your total calculation is \(\displaystyle 160+320+128=608\) 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 \(\displaystyle \pi (r^{2})\) to calculate. Here the radius is \(\displaystyle 3\), so each circle will have a volume of \(\displaystyle (3^{2})\pi =9\pi\). 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 \(\displaystyle 18\pi\).

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 \(\displaystyle 2\pi (r)\), so here that's \(\displaystyle 6\pi\). Multiply that by the height of \(\displaystyle 88\) and you have \(\displaystyle 48\pi\) as the area of the side. So your area is now \(\displaystyle 18\pi +48\pi\), which sums to \(\displaystyle 66\pi\).

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: \(\displaystyle \sqrt{l^{2}+w^{2}+h^{2}}\). Here since length, width, and height are all the same, \(\displaystyle 4\), you have a quick calculation:

\(\displaystyle {\sqrt{4^{2}+4^{2}+4^{2}}}=\sqrt{48}=4\sqrt{3}\).

Since the circumference of a circle can be expressed as \(\displaystyle \pi (d)\), your answer is simply \(\displaystyle 4\pi \sqrt{3}\).

The volume of a rectangular box is Length × Width × Height. Here you're given those three dimensions as \(\displaystyle 4*3*2\), but then told that the volume of water is only \(\displaystyle \frac{3}{4}\)of the total. So your calculation is \(\displaystyle 4*3*2*\frac{3}{4}\), which comes out to \(\displaystyle 18\).

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 \(\displaystyle \frac{1}{2}\), so its volume scales down by a factor of \(\displaystyle (\frac{1}{2})^{3}=\frac{1}{8}\).

So the new box is \(\displaystyle \frac{1}{8}\) the volume of the old box, meaning that it decreased in volume by \(\displaystyle \frac{7}{8}\). \(\displaystyle \frac{7}{8}\) expressed as a percentage is \(\displaystyle 87.5%\).

Alternatively, you could avoid the abstraction by choosing your own numbers and playing out the scenario that way. Imagine a \(\displaystyle 2*2*2\) cubical box (volume \(\displaystyle 8\)) being scaled down to a \(\displaystyle 1*1*1\) cubical box (volume \(\displaystyle 1\)). You'd go from a volume of \(\displaystyle 8\) to a volume of \(\displaystyle 1\), losing \(\displaystyle \frac{7}{8}\) of the volume, again equating to an \(\displaystyle 87.5%\) 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, \(\displaystyle 135=Base*5\sqrt{3}\). The area of the base, then, is \(\displaystyle \frac{135}{(5\sqrt{3})}=\frac{27}{\sqrt{3}}=9\sqrt{3}\). Now recall that the area of an equilateral triangle is \(\displaystyle \frac{side^{2}\sqrt{3}}{4}\).

So \(\displaystyle 9\sqrt{3}=\frac{side^{2}\sqrt{3}}{4}\). and \(\displaystyle side^{2}=36\) and \(\displaystyle side = 6\), which is the value of \(\displaystyle x\).