Question:C = A/(5r - 1)The given equation relates the positive numbers A, C, and r, where r gt 1/5. Which...

1. TRANSLATE the problem information

• Given equation: \(\mathrm{C = \frac{A}{5r - 1}}\)
• Need to solve for r in terms of A and C

2. INFER the solution strategy

• The variable r is trapped in the denominator of a fraction
• Strategy: Eliminate the denominator first by multiplying both sides by \(\mathrm{(5r - 1)}\)
• Then use standard algebraic techniques to isolate r

3. SIMPLIFY by eliminating the denominator

• Multiply both sides by \(\mathrm{(5r - 1)}\):

\(\mathrm{C \times (5r - 1) = A}\)

• This removes r from the denominator

4. SIMPLIFY to isolate the r-term

• Divide both sides by C:

\(\mathrm{5r - 1 = \frac{A}{C}}\)

• Add 1 to both sides:

\(\mathrm{5r = \frac{A}{C} + 1}\)

5. SIMPLIFY the fraction combination

• Express 1 as \(\mathrm{\frac{C}{C}}\) to get common denominator:

\(\mathrm{5r = \frac{A}{C} + \frac{C}{C} = \frac{A + C}{C}}\)

• Divide both sides by 5:

\(\mathrm{r = \frac{A + C}{5C}}\)

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students struggle with combining \(\mathrm{\frac{A}{C} + 1}\) into a single fraction.

Many students write \(\mathrm{5r = \frac{A}{C} + 1}\) and then incorrectly combine this as \(\mathrm{5r = \frac{A + 1}{C}}\), forgetting that 1 needs to be written as \(\mathrm{\frac{C}{C}}\) before adding. This leads to \(\mathrm{r = \frac{A + 1}{5C}}\), which would correspond to choice (D) if it had been an option, or causes confusion when this result doesn't match any given choice.

Second Most Common Error:

Poor INFER reasoning about strategy: Students try to isolate r without first eliminating the denominator.

Some students attempt to work directly with the fraction, trying to "flip" the equation or manipulate it without the systematic approach of multiplying both sides by \(\mathrm{(5r - 1)}\). This leads to getting stuck early in the problem and ultimately guessing among the choices.

The Bottom Line:

This problem tests systematic algebraic manipulation skills. Success requires recognizing that rational equations need the denominator eliminated first, then following through with careful fraction arithmetic to combine terms with common denominators.