Question:Company A charges customers a $30 setup fee plus $15 per hour of service. Company B charges customers a $45...
GMAT Algebra : (Alg) Questions
Company A charges customers a \(\$30\) setup fee plus \(\$15\) per hour of service. Company B charges customers a \(\$45\) setup fee plus \(\$\mathrm{k}\) per hour of service, where \(\mathrm{k}\) is a positive constant. For what value of \(\mathrm{k}\) will the two companies never charge the same total amount for any positive number of hours of service?
Express your answer as an integer or fraction in simplest form.
1. TRANSLATE the problem information
- Given information:
- Company A: $30 setup fee + $15 per hour
- Company B: $45 setup fee + $k per hour
- Want: Value of k so companies never charge same amount for any positive hours
- What this tells us: We need to find k such that the two cost equations are never equal for \(\mathrm{h \gt 0}\)
2. TRANSLATE into mathematical expressions
- Let h = number of hours of service (\(\mathrm{h \gt 0}\))
- Company A's total cost: \(\mathrm{C_A = 30 + 15h}\)
- Company B's total cost: \(\mathrm{C_B = 45 + kh}\)
3. INFER what "never charge the same amount" means mathematically
- For the costs to never be equal: \(\mathrm{30 + 15h \neq 45 + kh}\) for all \(\mathrm{h \gt 0}\)
- This is equivalent to asking: when do these linear functions never intersect?
- Answer: When they are parallel lines (same slope, different y-intercepts)
4. INFER the slope requirement for parallel lines
- Company A's slope = 15 (coefficient of h)
- Company B's slope = k (coefficient of h)
- For parallel lines: \(\mathrm{k = 15}\)
5. APPLY CONSTRAINTS to verify the solution
- When \(\mathrm{k = 15}\): Company A costs \(\mathrm{30 + 15h}\), Company B costs \(\mathrm{45 + 15h}\)
- Same slope (15) but different y-intercepts (\(\mathrm{30 \neq 45}\)) ✓
- These lines are parallel and never intersect ✓
Answer: 15
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students try to solve \(\mathrm{30 + 15h = 45 + kh}\) for a specific intersection point instead of recognizing this is about parallel lines.
They might solve:
\(\mathrm{15h - kh = 15}\)
\(\mathrm{h(15-k) = 15}\)
\(\mathrm{h = \frac{15}{15-k}}\)
Then they pick a specific value like \(\mathrm{h = 1}\) to solve for k, getting:
\(\mathrm{1 = \frac{15}{15-k}}\)
leading to \(\mathrm{k = 0}\). This creates a fundamental misunderstanding of the problem's requirement.
This leads to confusion and incorrect reasoning about the problem's constraints.
The Bottom Line:
The key insight is translating "never charge the same amount" into the geometric concept of parallel lines, rather than trying to solve for a specific intersection point. This requires understanding that linear cost functions are represented as lines on a graph.