GED Math - How to Get the Right Answers on the 2024 Test (1) - Free Printable

Problem Analysis:

The problem asks us to calculate the volume of a cone-shaped paper drinking cup. The dimensions provided are:
- Diameter of the base: \( 2 \frac{3}{4} \) inches
- Height of the cone: 4 inches

We need to find the volume of the cone to the nearest tenth of a cubic inch.

Step-by-Step Solution:

#### Step 1: Recall the formula for the volume of a cone
The volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius of the base,
- \( h \) is the height of the cone.

#### Step 2: Determine the radius of the base
The diameter of the base is given as \( 2 \frac{3}{4} \) inches. To convert this mixed number to an improper fraction:
\[
2 \frac{3}{4} = 2 + \frac{3}{4} = \frac{8}{4} + \frac{3}{4} = \frac{11}{4}
\]
The radius \( r \) is half of the diameter:
\[
r = \frac{1}{2} \times \frac{11}{4} = \frac{11}{8} \text{ inches}
\]

#### Step 3: Substitute the values into the volume formula
We have:
- \( r = \frac{11}{8} \) inches,
- \( h = 4 \) inches.

Substitute these values into the volume formula:
\[
V = \frac{1}{3} \pi \left( \frac{11}{8} \right)^2 \cdot 4
\]

#### Step 4: Simplify the expression
First, calculate \( \left( \frac{11}{8} \right)^2 \):
\[
\left( \frac{11}{8} \right)^2 = \frac{11^2}{8^2} = \frac{121}{64}
\]

Now substitute this back into the volume formula:
\[
V = \frac{1}{3} \pi \cdot \frac{121}{64} \cdot 4
\]

Simplify the multiplication:
\[
V = \frac{1}{3} \pi \cdot \frac{121 \cdot 4}{64} = \frac{1}{3} \pi \cdot \frac{484}{64}
\]

Simplify \( \frac{484}{64} \):
\[
\frac{484}{64} = \frac{121}{16}
\]

So the expression becomes:
\[
V = \frac{1}{3} \pi \cdot \frac{121}{16}
\]

Multiply the fractions:
\[
V = \frac{121 \pi}{48}
\]

#### Step 5: Approximate the value
Use \( \pi \approx 3.14159 \):
\[
V \approx \frac{121 \cdot 3.14159}{48}
\]

Calculate the numerator:
\[
121 \cdot 3.14159 \approx 379.94039
\]

Now divide by 48:
\[
V \approx \frac{379.94039}{48} \approx 7.91542
\]

Round to the nearest tenth:
\[
V \approx 7.9
\]

Final Answer:

\[
\boxed{B}
\]