If x , which of the following could be greater than 1? (60% text{ of } x 0.125 * 2x...

1. Translate the problem requirements: We need to determine which of three expressions could exceed 1, given that \(\mathrm{x} < \frac{8}{5}\) (which equals 1.6). The key insight is that since x is bounded above by 1.6, we need to check if any expression can exceed 1 within this constraint.
2. Establish the constraint boundary: Convert \(\frac{8}{5}\) to decimal form (1.6) and understand that x can be any value less than 1.6, including negative values.
3. Test each expression at the boundary: Evaluate each expression using x approaching 1.6 to find the maximum possible value for each expression.
4. Verify with strategic test values: Check expressions with both positive values near the boundary and negative values to ensure no expression can exceed 1.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have three mathematical expressions, and we want to know which ones could possibly be greater than 1, given that x is less than \(\frac{8}{5}\).

First, let's convert that constraint to a decimal to make it easier to work with: \(\frac{8}{5} = 1.6\). So we know that \(\mathrm{x} < 1.6\).

The three expressions we need to check are:

• I. 60 percent of x (which means \(0.6 \times \mathrm{x}\))
• II. 0.125 times 2x (which means \(0.125 \times 2\mathrm{x} = 0.25\mathrm{x}\))
• III. \(\frac{1}{\mathrm{x}}\) percent of 16x (which means \(\frac{\frac{1}{\mathrm{x}}}{100} \times 16\mathrm{x} = \frac{16}{100} = 0.16\))

Process Skill: TRANSLATE - Converting percentages and verbal descriptions into mathematical expressions

2. Establish the constraint boundary

Since \(\mathrm{x} < 1.6\), we need to think about what values x can take. It can be any number less than 1.6 - this includes positive numbers like 1.5, 1, 0.5, and also negative numbers like -1, -10, etc.

To find the maximum possible value for each expression, we should test what happens when x gets very close to 1.6 (but stays less than it). We should also consider what happens with negative values of x, since sometimes negative inputs can produce surprising results.

Process Skill: CONSIDER ALL CASES - Recognizing that x can be negative as well as positive

3. Test each expression at the boundary

Let's check each expression by using x approaching 1.6:

Expression I: 60% of x = \(0.6\mathrm{x}\)
When x approaches 1.6: \(0.6 \times 1.6 = 0.96\)
Since x must be less than 1.6, this expression must be less than 0.96, which is less than 1.

Expression II: \(0.125 \times 2\mathrm{x} = 0.25\mathrm{x}\)
When x approaches 1.6: \(0.25 \times 1.6 = 0.4\)
This is clearly less than 1.

Expression III: \(\left(\frac{1}{\mathrm{x}}\right)\) percent of \(16\mathrm{x}\)
Let's work this out step by step:

• "\(\frac{1}{\mathrm{x}}\) percent" means \(\frac{\frac{1}{\mathrm{x}}}{100}\)
• "of \(16\mathrm{x}\)" means we multiply by \(16\mathrm{x}\)
• So we get: \(\frac{\frac{1}{\mathrm{x}}}{100} \times 16\mathrm{x} = \frac{1}{\mathrm{x}} \times 16\mathrm{x} \div 100 = \frac{16}{100} = 0.16\)

Interestingly, this expression equals 0.16 regardless of what value x takes (as long as \(\mathrm{x} \neq 0\))!

4. Verify with strategic test values

Let's double-check our work with some specific values and consider negative cases:

Testing with \(\mathrm{x} = 1\):

• I: \(0.6 \times 1 = 0.6\) ✓ (less than 1)
• II: \(0.25 \times 1 = 0.25\) ✓ (less than 1)
• III: \(\frac{\frac{1}{1}}{100} \times 16 \times 1 = 0.16\) ✓ (less than 1)

Testing with \(\mathrm{x} = -2\):

• I: \(0.6 \times (-2) = -1.2\) (less than 1)
• II: \(0.25 \times (-2) = -0.5\) (less than 1)
• III: \(\frac{\frac{1}{(-2)}}{100} \times 16 \times (-2) = \frac{(-0.5)}{100} \times (-32) = 0.16\) (less than 1)

Notice that even with negative values, none of our expressions exceed 1.

Process Skill: APPLY CONSTRAINTS - Systematically checking that our constraint \(\mathrm{x} < 1.6\) is maintained throughout

4. Final Answer

After testing all three expressions:

• Expression I has a maximum value approaching 0.96 (less than 1)
• Expression II has a maximum value approaching 0.4 (less than 1)
• Expression III always equals 0.16 (less than 1)

Since none of the expressions can be greater than 1, the answer is A. None.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "\(\frac{1}{\mathrm{x}}\) percent" notation
Students often struggle with the compound percentage notation "\(\frac{1}{\mathrm{x}}\) percent of \(16\mathrm{x}\)" and may interpret it as simply "\(\frac{1}{\mathrm{x}}\) of \(16\mathrm{x}\)" (forgetting to divide by 100), or may incorrectly parse the order of operations. The correct interpretation is \(\frac{\frac{1}{\mathrm{x}}}{100} \times 16\mathrm{x}\), but students might calculate \(\frac{1}{\mathrm{x \text{ percent}}}\) or other variations.

2. Overlooking that x can be negative
The constraint \(\mathrm{x} < \frac{8}{5}\) allows for negative values of x, but students typically only consider positive values when testing expressions. This oversight could lead them to miss cases where negative values might produce different results, especially for expressions involving fractions like \(\frac{1}{\mathrm{x}}\).

3. Misunderstanding what "could be greater than 1" means
Students may think they need to find values where the expressions are always greater than 1, rather than understanding they need to determine if there exists any valid value of x that makes each expression greater than 1. This leads to testing only boundary values instead of exploring the full range of possibilities.

Errors while executing the approach

1. Arithmetic errors when simplifying Expression III
When simplifying \(\frac{\frac{1}{\mathrm{x}}}{100} \times 16\mathrm{x}\), students often make algebraic mistakes. They might incorrectly cancel terms or forget to divide by 100, leading them to get 16 instead of 0.16. The correct simplification requires careful attention to: \(\frac{1}{\mathrm{x}} \times \frac{1}{100} \times 16\mathrm{x} = \frac{16}{100} = 0.16\).

2. Incorrectly handling negative values in calculations
When testing negative values of x, students frequently make sign errors, especially when dealing with products of negative numbers. For example, with \(\mathrm{x} = -2\) in Expression III: \(\frac{\frac{1}{(-2)}}{100} \times 16 \times (-2)\), students might incorrectly handle the multiplication of \((-0.5) \times (-32)\).

Errors while selecting the answer

1. Confusing "could be greater than 1" with "is always greater than 1"
Even after correctly calculating that all expressions have maximum values less than 1, students might second-guess themselves and think that since some expressions get close to 1 (like Expression I approaching 0.96), this somehow counts as "could be greater than 1." They fail to recognize that "less than the constraint" means "definitely cannot exceed 1."

Alternate Solutions

Smart Numbers Approach

Step 1: Choose strategic test values for x

Since \(\mathrm{x} < \frac{8}{5} = 1.6\), let's test with specific values that will help us evaluate each expression clearly:

• \(\mathrm{x} = 1.5\) (close to the upper bound)
• \(\mathrm{x} = 1\) (a convenient positive value)
• \(\mathrm{x} = -2\) (a negative value to test different scenarios)

Step 2: Test Expression I with \(\mathrm{x} = 1.5\)

Expression I: 60% of x = \(0.6 \times 1.5 = 0.9\)

Since \(0.9 < 1\), this doesn't exceed 1.

Let's verify with the maximum possible value. As x approaches 1.6:

\(0.6 \times 1.6 = 0.96\), which is still less than 1.

Step 3: Test Expression II with \(\mathrm{x} = 1.5\)

Expression II: \(0.125 \times 2\mathrm{x} = 0.125 \times 2(1.5) = 0.125 \times 3 = 0.375\)

Since \(0.375 < 1\), this doesn't exceed 1.

At the boundary: \(0.125 \times 2(1.6) = 0.125 \times 3.2 = 0.4\), still less than 1.

Step 4: Test Expression III with \(\mathrm{x} = 1\)

Expression III: \(\left(\frac{1}{\mathrm{x}}\right)\%\) of \(16\mathrm{x} = \frac{1}{\mathrm{x}} \times \frac{1}{100} \times 16\mathrm{x} = \frac{16}{100} = 0.16\)

With \(\mathrm{x} = 1.5\): \(\left(\frac{1}{1.5}\right)\%\) of \(16(1.5) = \frac{2}{3} \times \frac{1}{100} \times 24 = 0.16\)

Notice that this expression simplifies to \(\frac{16}{100} = 0.16\) regardless of the positive value of x.

Step 5: Test with negative values

Let's check \(\mathrm{x} = -2\) for Expression III:

\(\left(\frac{1}{(-2)}\right)\%\) of \(16(-2) = \left(\frac{-1}{2}\right) \times \frac{1}{100} \times (-32) = 0.16\)

Even with negative values, Expression III equals 0.16.

Conclusion:

Using our strategic test values:

• Expression I: Maximum value approaches 0.96 (< 1)
• Expression II: Maximum value approaches 0.4 (< 1)
• Expression III: Always equals 0.16 (< 1)

Therefore, none of the expressions can be greater than 1.

Answer: A