A lab device displays a value D that depends linearly on the sample mass m. For every increase of 4.0...

1. TRANSLATE the problem information

• Given information:
  • Linear relationship between display value D and mass m
  • Rate: +7.0 units in D for every +4.0 g in m
  • When m = 0, D = 5.0 units (starting point)
  • Find: increase in D when m increases by 8.4 g

2. INFER the solution approach

• Since this is a linear relationship, I need the rate of change (slope)
• The question asks for change in D, not the final value of D
• The starting point (5.0 units) won't affect the change calculation

3. SIMPLIFY to find the rate of change

• Rate: \(\mathrm{Rate = \frac{change\:in\:D}{change\:in\:m} = \frac{7.0\:units}{4.0\:g} = 1.75\:units\:per\:gram}\)

4. SIMPLIFY to find the increase for 8.4 g

• Increase in D: \(\mathrm{Increase\:in\:D = rate \times mass\:change = 1.75 \times 8.4 = 14.7\:units}\) (use calculator)

Answer: C. 14.7

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students try to incorporate the y-intercept (5.0 units) into their change calculation, thinking they need to find the final value rather than just the change.

They might calculate: \(\mathrm{(1.75 \times 8.4) + 5.0 = 14.7 + 5.0 = 19.7\:units}\)

This may lead them to select Choice D (19.7)

Second Most Common Error:

Poor TRANSLATE reasoning: Students set up an incorrect proportion, perhaps confusing which values correspond to which variables.

They might calculate: \(\mathrm{\frac{8.4}{4.0} = \frac{x}{7.0}}\), giving \(\mathrm{x = 14.7 \times \frac{8.4}{4.0} \approx 30.87}\), or get confused about the setup entirely.

This leads to confusion and guessing among the remaining choices.

The Bottom Line:

The key insight is distinguishing between finding a change versus finding an absolute value in linear relationships. The y-intercept affects absolute values but never affects how much the function changes.