Question 111·200 Super-Hard SAT Math Questions·Algebra

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A debate workshop (Plan A) charges a one-time facility fee of $360 for the entire group. In addition, Plan A charges $110 per student for the first week and $70 per student for each additional week.

Another workshop (Plan B) charges a one-time facility fee of $200 for the entire group and $80 per student per week for every week.

For a group of 12 students, which choice describes all positive integer values of $w$ for which Plan A costs less than Plan B?

A break-even point is where two plans cost the same. Keep a fee paid once separate from charges that repeat, and count the first week separately if it has a different price. Write each total, then use a Desmos regression to find the break-even point. Compare the added cost of another week to decide which side is cheaper; the strict inequality and whole-week rule determine the answer.

Hints

Hint 1
Plan A charges one price for the first week and another for later weeks. If the workshop lasts $w$ weeks, it has $w-1$ additional weeks. How do those two charges combine for one student?
Hint 2
A one-time fee is paid once for the group, not once per student. Multiply each per-student charge by 12, but add each facility fee only once.
Hint 3
The break-even point is where the plans cost the same. After finding it, compare what each plan charges for one more week to decide which side makes Plan A cheaper. Then take the first whole week on that side.

Step-by-step

Find where the costs cross
Step 1
Write Plan A's cost
Let $w$ count weeks. Plan A charges $110 per student for the first week, then $70 per student for each of the other $w-1$ weeks. For 12 students, add the $360 fixed fee once: $A(w)=360+12(110+70(w-1))$ . The fee is for the whole group, so it does not get multiplied by 12.
Step 2
Write Plan B's cost
Plan B's weekly rate is $80 per student for every week, so 12 students cost $12(80w)$ for $w$ weeks. Add its $200 group fee once: $B(w)=200+12(80w)$ .
Step 3
Compare the two totals
The words costs less call for a strict inequality, not an equal sign. Put Plan A on the less-than side: $360+12(110+70(w-1))<200+12(80w)$ . A week when the costs are equal does not qualify.
Step 4
Find the boundary in Desmos
At the break-even point, the costs are equal. Type $360+12(110+70(w_1-1))\sim200+12(80w_1)$ in Desmos. The subscript makes $w_1$ the value to find, and $\sim$ tells Desmos to fit it. Under PARAMETERS, Desmos shows $w_1=5.333\ldots$ . That's the boundary, not a week when Plan A costs less.
Step 5
Choose the cheaper side
Each additional week adds $12(70)$ to Plan A but $12(80)$ to Plan B. Since $70<80$ , Plan A's cost rises more slowly for the same 12 students. So after the break-even point, Plan A costs less and keeps that advantage.
Step 6
Choose the positive integer weeks
The boundary $5.333\ldots$ lies between 5 and 6, and Plan A is cheaper above it. For whole-number answers, move from the boundary in the allowed direction instead of rounding it. The first qualifying week is 6, and every later week qualifies too. Plan A costs less for $w\ge6$ . Choice B.

Question 111·200 Super-Hard SAT Math Questions·Algebra

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