A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on...

1. Translate the problem requirements: We have two measurement scales (R-scale and S-scale) that are related linearly, meaning they have a straight-line relationship. We know two coordinate pairs: \(\mathrm{R=6}\) corresponds to \(\mathrm{S=30}\), and \(\mathrm{R=24}\) corresponds to \(\mathrm{S=60}\). We need to find what R-value corresponds to \(\mathrm{S=100}\).
2. Identify the linear relationship pattern: Examine how the R-scale and S-scale values change together to understand the rate of change between the two scales.
3. Set up the linear equation: Use the two known points to establish the mathematical relationship between R and S scales.
4. Apply the relationship to find the target value: Use the established linear equation to determine what R-value corresponds to \(\mathrm{S=100}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have here. Think of this like converting between different units of measurement - like how Celsius and Fahrenheit are two different temperature scales that relate to each other in a predictable way.

We're told that the R-scale and S-scale are "related linearly" - this simply means that if we were to plot points on a graph, they would form a straight line. We have two specific examples:

• When \(\mathrm{R = 6}\), then \(\mathrm{S = 30}\)
• When \(\mathrm{R = 24}\), then \(\mathrm{S = 60}\)

Our goal is to find: What value on the R-scale gives us \(\mathrm{S = 100}\)?

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Identify the linear relationship pattern

Let's look for the pattern by examining how both scales change together. This is like figuring out the "exchange rate" between the two scales.

From our two data points:

• R changes from 6 to 24, which is an increase of 18
• S changes from 30 to 60, which is an increase of 30

So when R increases by 18, S increases by 30. This means the "rate of change" is \(\frac{30}{18} = \frac{5}{3}\).

In everyday terms: for every 3 units R increases, S increases by 5 units.

3. Set up the linear equation

Now let's express this relationship mathematically. Since we know the rate of change \(\left(\frac{5}{3}\right)\), we can write the relationship as:

\(\mathrm{S} = \frac{5}{3} \times \mathrm{R} + \text{some starting value}\)

To find that starting value, let's use one of our known points. Using \(\mathrm{R = 6}\) and \(\mathrm{S = 30}\):

\(30 = \frac{5}{3} \times 6 + \text{starting value}\)

\(30 = 10 + \text{starting value}\)

\(\text{starting value} = 20\)

So our complete relationship is: \(\mathrm{S} = \frac{5}{3} \times \mathrm{R} + 20\)

Let's verify with our second point \((\mathrm{R = 24, S = 60})\):

\(\mathrm{S} = \frac{5}{3} \times 24 + 20 = 40 + 20 = 60\) ✓

4. Apply the relationship to find the target value

Now we can find what R-value gives us \(\mathrm{S = 100}\). We substitute \(\mathrm{S = 100}\) into our equation:

\(100 = \frac{5}{3} \times \mathrm{R} + 20\)

Solving for R:

\(100 - 20 = \frac{5}{3} \times \mathrm{R}\)

\(80 = \frac{5}{3} \times \mathrm{R}\)

To solve this, multiply both sides by \(\frac{3}{5}\):

\(\mathrm{R} = 80 \times \frac{3}{5} = \frac{240}{5} = 48\)

Final Answer

The measurement on the R-scale that corresponds to 100 on the S-scale is 48.

This matches answer choice C: 48.

To double-check: \(\mathrm{S} = \frac{5}{3} \times 48 + 20 = 80 + 20 = 100\) ✓

Common Faltering Points

Errors while devising the approach

Students often confuse which variable is independent and which is dependent. They might set up the equation as \(\mathrm{R} = \text{(some slope)} \times \mathrm{S} + \text{constant}\) instead of \(\mathrm{S} = \text{(some slope)} \times \mathrm{R} + \text{constant}\). This happens because the question asks "What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?" which can mislead students into thinking R depends on S.

Seeing the pairs (6,30) and (24,60), students might assume a simple multiplicative relationship exists (like \(\mathrm{R} \times 5 = \mathrm{S}\)) without recognizing that linear relationships often include a constant term. This leads them to set up incorrect proportions that don't account for the y-intercept of the linear equation.

Errors while executing the approach

When finding the rate of change, students frequently make mistakes in the fraction arithmetic. For example, they might incorrectly calculate \(\frac{30}{18}\) as \(\frac{3}{2}\) instead of \(\frac{5}{3}\), or struggle with simplifying fractions properly.

When substituting known values to find the constant term, students often make algebraic mistakes. For instance, from \(30 = 10 + \text{constant}\), they might incorrectly conclude that \(\text{constant} = 10\) instead of \(\text{constant} = 20\), or get confused about which side to subtract from.

When solving \(80 = \frac{5}{3} \times \mathrm{R}\), students frequently struggle with multiplying by the reciprocal. They might incorrectly calculate \(80 \times \frac{3}{5}\) or make errors in the arithmetic, leading to wrong values like 60 or 36.

Errors while selecting the answer

No likely faltering points - the calculation directly yields an integer that matches one of the answer choices, making selection straightforward.