Choosing a Tutoring Payment Plan
A tutoring center compares payment models before signing a contract. Plan A: $40 enrollment fee plus $18 per session. Plan B: $95 enrollment fee plus $11 per session. Plan C: no enrollment fee, $12 per session, plus a one-time $60 materials kit. A staff member solves 40 + 18n = 95 + 11n, obtains 7n = 55, so n ≈ 7.9, and writes: "Plan A becomes the cheaper option after about 8 sessions." The center expects most students to attend between 6 and 10 sessions.
1. Identify the error in the staff member's conclusion and verify the correct crossover with a specific session count.
Read the worked answer
Check the algebra, then the interpretation. 40 + 18n = 95 + 11n gives 7n = 55, n ≈ 7.86 - that part is correct. The error is in the direction of the inequality. Plan A has the higher per-session rate ($18 vs. $11); its only advantage is the smaller upfront fee. So Plan A can only be cheaper at small n, not large n.
Verify at n = 10: A = 40 + 180 = $220; B = 95 + 110 = $205. B is cheaper. Verify at n = 5: A = 40 + 90 = $130; B = 95 + 55 = $150. A is cheaper.
Correct statement: Plan B becomes cheaper once n exceeds about 7.9 sessions, i.e., from 8 sessions on; Plan A is cheaper below that. The staff member applied the crossover but reversed which side of it belongs to which plan.
2. Include Plan C and determine which plan the center should offer students attending 6 to 10 sessions. Show values at both ends of the range.
Read the worked answer
Model C: cost = 60 + 12n.
| n | A: 40 + 18n | B: 95 + 11n | C: 60 + 12n |
|---|---|---|---|
| 6 | $148 | $161 | $132 |
| 8 | $184 | $183 | $156 |
| 10 | $220 | $205 | $180 |
Comparison: C beats A whenever 60 + 12n < 40 + 18n, i.e., 20 < 6n, so n > 3.33 (from 4 sessions on). C beats B whenever 60 + 12n < 95 + 11n, i.e., n < 35. Since the expected range is 6–10 sessions, both conditions hold throughout, so Plan C is cheapest at every point in the range - between $48 and $73 cheaper than the next-best plan at n = 8.
Recommendation: offer Plan C for students expected in the 6–10 session band, and note that the answer is range-specific: outside that band (for example, 1–3 sessions) Plan A is cheapest, and above 35 sessions Plan B overtakes C.
3. The center director says, "A student will attend 7.9 sessions, so we are right at the switch point." Respond, and explain how the non-whole-number solution should be used.
Read the worked answer
A student cannot attend 7.9 sessions. Sessions are counted in whole numbers, so the algebraic solution is a boundary marker, not a prediction. It tells you the switch happens between n = 7 and n = 8, and nothing more.
Test the two integers the boundary falls between: at n = 7, A = $166 and B = $172, so A is cheaper. At n = 8, A = $184 and B = $183, so B is cheaper - by exactly $1. That $1 margin is worth pointing out: near the crossover the plans are nearly equivalent, so a $1 difference should not by itself drive a multi-year contract decision. Fee waivers, cancellation terms, or a student who quits after three sessions would swamp it.
How to use the solution: round to the nearest whole session on each side and compare, then treat the crossover as a threshold to communicate to families ("Plan B pays off from your eighth session"), not as an expected value.