A retailer sets the sale price S of a product using the formula \(\mathrm{S = L - \frac{d}{100}(L - R)}\),...

1. TRANSLATE the problem requirements

• Given: \(\mathrm{S = L - \frac{d}{100}(L - R)}\)
• Find: Express \(\mathrm{d}\) in terms of \(\mathrm{L}\), \(\mathrm{S}\), and \(\mathrm{R}\)
• This means we need to rearrange the formula to get \(\mathrm{d}\) by itself on one side

2. SIMPLIFY by rearranging to isolate the d term

• Start with: \(\mathrm{S = L - \frac{d}{100}(L - R)}\)
• Subtract \(\mathrm{L}\) from both sides: \(\mathrm{S - L = -\frac{d}{100}(L - R)}\)
• Multiply both sides by -1: \(\mathrm{L - S = \frac{d}{100}(L - R)}\)

3. SIMPLIFY by solving for d

• We now have: \(\mathrm{L - S = \frac{d}{100}(L - R)}\)
• To get \(\mathrm{d}\) by itself, multiply both sides by \(\mathrm{\frac{100}{(L - R)}}\):
• \(\mathrm{d = \frac{100(L - S)}{(L - R)}}\)

Answer: A

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make sign errors when rearranging the equation, particularly struggling with the negative sign in front of \(\mathrm{\frac{d}{100}(L - R)}\).

Instead of correctly getting \(\mathrm{L - S = \frac{d}{100}(L - R)}\), they might end up with \(\mathrm{S - L = \frac{d}{100}(L - R)}\), leading them to \(\mathrm{d = \frac{100(S - L)}{(L - R)}}\). This causes them to select Choice B.

Second Most Common Error:

Poor SIMPLIFY reasoning: Students correctly rearrange to get \(\mathrm{L - S = \frac{d}{100}(L - R)}\) but make an error when solving for \(\mathrm{d}\). They might flip the fraction incorrectly or confuse which terms go in the numerator versus denominator.

This leads to confusion about the final form and may cause them to select Choice C with \(\mathrm{(R - L)}\) in the denominator, or get stuck and guess.

The Bottom Line:

This problem tests systematic algebraic manipulation skills. Success requires careful attention to signs and methodical step-by-step rearrangement without rushing through the algebra.