Professor Malingowski, a chemist and teacher at to communite college, is organizing his graduated cylinders in the hopes of keeping his office tidy and setting a good example for his students. He has beakers with diameters, in inches, of `1/2,3/4,4/5,1` and `5/4`.
Professor Malingowski notices one additional cylinder lying on the ground, and can recall certain facts about it, but not its actual diameter. If he knows that the value of the additional graduated cylinder's diameter ,x, will not create any modes and will make the mean of the set equal to `5/6`, what is the value of the additional cylinder's diameter?

Answer

To find the diameter of the additional graduated cylinder (denoted as \( x \)), we need to follow these steps:

### Step 1: List the given diameters
The diameters of the existing graduated cylinders are:
- \( \frac{1}{2} \)
- \( \frac{3}{4} \)
- \( \frac{4}{5} \)
- \( 1 \)
- \( \frac{5}{4} \)

### Step 2: Calculate the sum of the existing diameters
We first convert all the fractions to a common denominator or calculate their decimal equivalents for easier addition:
- \( \frac{1}{2} = 0.5 \)
- \( \frac{3}{4} = 0.75 \)
- \( \frac{4}{5} = 0.8 \)
- \( 1 = 1.0 \)
- \( \frac{5}{4} = 1.25 \)

Now, we sum these values:
\[ 0.5 + 0.75 + 0.8 + 1.0 + 1.25 = 4.3 \]

### Step 3: Set up the equation for the mean
The mean of the set of diameters (including the additional cylinder) is given to be \( \frac{5}{6} \). Since there are now 6 cylinders (5 existing + 1 additional), we can set up the equation for the mean:
\[ \frac{4.3 + x}{6} = \frac{5}{6} \]

### Step 4: Eliminate the denominator
To eliminate the denominator, we multiply both sides of the equation by 6:
\[ 4.3 + x = 5 \]

### Step 5: Solve for \( x \)
Now, we isolate \( x \):
\[ x = 5 - 4.3 \]
\[ x = 0.7 \]

### Step 6: Convert to fraction (if needed)
The decimal \( 0.7 \) can be expressed as a fraction:
\[ x = \frac{7}{10} \]

### Final Answer
The diameter of the additional graduated cylinder is \( 0.7 \) inches or \( \frac{7}{10} \) inches.

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