Exercise Answers
Discrete Quantitative Questions
- B Let e = the edge of the cube. Using the formulas for surface area and
volume, divide both sides by e^2 : ee^32 =^6 ee 22 → e = 6
e^3 = 6e^2
e = 6 - C To solve for surface area, use the formula surface area of a cube = 6e^2 , where
e = edge length. If the volume is 125, then
e^3 = 125
e = 5
Substitute e into the formula for surface area: 6(5)^2 = 150. - 13,824 cubic inches First, convert from feet to inches. 2 feet = 24 inches.
The volume of the cube, in cubic inches, is thus 24^3 = 13,824 cubic inches. - D Let e = the length of the cube’s edge. Thus the radius of the cylinder
also = e. Now create an equation: π(e^2 )h = e^3. Isolate h: h = eπ. - D Recall that the length of the diagonal of a cube = e√3, where e = length of
an edge of the cube. To solve for the edge of the cube, we’ll use the fact that the
surface area of the cube = 216.
Use the formula for surface area to solve for the edge:
6 e^2 = 216
e^2 = 36
e = 6
Now, substitute 6 for e into the formula for the diagonal of a cube to arrive
at 6√ 3. - C The volume of the tank is π(r^2 )h = π(5^2 )10 = 250π. To solve for time, use the
r × t = w formula, where the rate is 75 cubic feet/second, and the work is 250π:
r × t = w
75 t = 250π
T = 250π/75 ≈ 10.5 seconds
The closest answer is C. - 18 Plug in numbers. Let the edge of cube y = 4. The edge of Cube X is thus 2. In
this case, the volume of Cube Y is 4^3 = 64, and the volume of Cube X is 2^3 = 8.
Volume of X
Volume of Y^ =
8
64 =1
8.- D Plug in numbers. Let the original length = 2, the original width = 3, and
the original height = 4. The original volume is thus 2 × 3 × 4 = 24. The new
length will be 2 × 2 = 4. The new width will be 3 × 3 = 9. The new height will
be 4 × 4 = 16. The new volume is thus 4 × 9 × 16 = 576. Now use the percent
greater formula: Percent of – 100%: (^57624 × 100) – 100 = 2,300%.
CHAPTER 13 ■ GEOMETRY 42904-GRE-Test-2018_313-462.indd 429 12/05/17 12:06 pm