Jeff tests how the total volume occupied by a fluid contained in a graduated cylinder changes when round marbles of various size are added. He found that the total volume occupied by the fluid, V, in cubic centimeters, can be found using the equation below, where x equals the number of identical marbles Jeff added,one at a time, to the cylinder , and r is the radius of one of the marbles .
`V=24pi+x(4/3pir^3)`
If the volume of the graduated cylinder is `96 pi` cubic centimeters ,then , what is the maximum number of marbles with a radius of 3 centimeters that Jeff can add without the volume of the fluid exceeding that of the graduated cylinder?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will follow the given equation and constraints.

### Step 1: Write down the equation for total volume.
The total volume \( V \) occupied by the fluid and the marbles is given by the equation:
\[ V = 24\pi + x \left(\frac{4}{3}\pi r^3\right) \]
where \( x \) is the number of marbles and \( r \) is the radius of one marble.

### Step 2: Identify the maximum volume of the graduated cylinder.
The maximum volume of the graduated cylinder is given as:
\[ V = 96\pi \text{ cubic centimeters} \]

### Step 3: Set up the equation for maximum volume.
We want to find the maximum number of marbles \( x \) that can be added without exceeding the volume of the graduated cylinder. Therefore, we set up the equation:
\[ 96\pi = 24\pi + x \left(\frac{4}{3}\pi r^3\right) \]

### Step 4: Substitute the radius of the marbles.
The radius \( r \) of the marbles is given as 3 centimeters. We can calculate \( r^3 \):
\[ r^3 = 3^3 = 27 \]
Now substitute \( r \) into the equation:
\[ 96\pi = 24\pi + x \left(\frac{4}{3}\pi \cdot 27\right) \]

### Step 5: Simplify the equation.
First, simplify the term involving \( x \):
\[ \frac{4}{3} \cdot 27 = 36 \]
So the equation becomes:
\[ 96\pi = 24\pi + 36\pi x \]

### Step 6: Isolate \( x \).
Subtract \( 24\pi \) from both sides:
\[ 96\pi - 24\pi = 36\pi x \]
This simplifies to:
\[ 72\pi = 36\pi x \]

### Step 7: Divide both sides by \( 36\pi \).
\[ \frac{72\pi}{36\pi} = x \]
The \( \pi \) cancels out:
\[ \frac{72}{36} = x \]
This simplifies to:
\[ x = 2 \]

### Conclusion:
The maximum number of marbles that Jeff can add without exceeding the volume of the graduated cylinder is:
\[ \boxed{2} \]