If X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal...

1. Translate the problem requirements: Understand what \(0.1\mathrm{X}\) and \(0.02\mathrm{Y}\) represent in terms of their decimal place values, and identify what we need to maximize
2. Express the decimals in terms of X and Y: Convert the decimal representations into fractional form to make the relationship clear
3. Apply maximization logic: Use the principle that to maximize a fraction, we make the numerator as large as possible and denominator as small as possible
4. Calculate and compare to answer choices: Compute the maximum value and select the closest answer choice

Execution of Strategic Approach

1. Translate the problem requirements

Let's first understand what these decimal notations actually mean in plain English.

When we see \(0.1\mathrm{X}\), this means we have a decimal number where:

• The tenths place is 1
• The hundredths place is X (where X is some nonzero digit)

So \(0.1\mathrm{X}\) could be 0.11, 0.12, 0.13, ..., up to 0.19.

When we see \(0.02\mathrm{Y}\), this means we have a decimal number where:

• The tenths place is 0
• The hundredths place is 2
• The thousandths place is Y (where Y is some nonzero digit)

So \(0.02\mathrm{Y}\) could be 0.021, 0.022, 0.023, ..., up to 0.029.

We need to find the greatest possible value of the fraction \(0.1\mathrm{X} \div 0.02\mathrm{Y}\).

Process Skill: TRANSLATE - Converting the decimal notation into concrete understanding

2. Express the decimals in terms of X and Y

Now let's write these decimals in a way that makes the math easier to work with.

For \(0.1\mathrm{X}\):

• This is \(0.1 + 0.0\mathrm{X}\)
• Which equals \(0.1 + \frac{\mathrm{X}}{100}\)
• So \(0.1\mathrm{X} = 0.1 + \frac{\mathrm{X}}{100}\)

For \(0.02\mathrm{Y}\):

• This is \(0.02 + 0.00\mathrm{Y}\)
• Which equals \(0.02 + \frac{\mathrm{Y}}{1000}\)
• So \(0.02\mathrm{Y} = 0.02 + \frac{\mathrm{Y}}{1000}\)

Therefore, our fraction becomes:

\(\left(0.1 + \frac{\mathrm{X}}{100}\right) \div \left(0.02 + \frac{\mathrm{Y}}{1000}\right)\)

3. Apply maximization logic

To maximize a fraction, we use this simple principle: make the top (numerator) as large as possible and make the bottom (denominator) as small as possible.

For the numerator \(\left(0.1 + \frac{\mathrm{X}}{100}\right)\):

• Since X can be any nonzero digit from 1 to 9
• To maximize this, we choose \(\mathrm{X} = 9\)
• This gives us: \(0.1 + \frac{9}{100} = 0.1 + 0.09 = 0.19\)

For the denominator \(\left(0.02 + \frac{\mathrm{Y}}{1000}\right)\):

• Since Y can be any nonzero digit from 1 to 9
• To minimize this, we choose \(\mathrm{Y} = 1\)
• This gives us: \(0.02 + \frac{1}{1000} = 0.02 + 0.001 = 0.021\)

Process Skill: APPLY CONSTRAINTS - Using the fact that X and Y must be nonzero digits

4. Calculate and compare to answer choices

Now we calculate the maximum value:

Maximum value = \(0.19 \div 0.021\)

To make this division easier, let's convert to fractions:

\(0.19 = \frac{19}{100}\)

\(0.021 = \frac{21}{1000}\)

So we have: \(\frac{19}{100} \div \frac{21}{1000} = \frac{19}{100} \times \frac{1000}{21} = \frac{19000}{100 \times 21} = \frac{19000}{2100}\)

Simplifying: \(\frac{19000}{2100} = \frac{190}{21} \approx 9.05\)

Looking at our answer choices:

• A. 4
• B. 5
• C. 6
• D. 9
• E. 10

The value 9.05 is closest to 9.

Final Answer

The greatest possible value of \(\frac{0.1\mathrm{X}}{0.02\mathrm{Y}}\) is approximately 9.05, which is closest to answer choice D. 9.