Question 111·200 Super-Hard SAT Math Questions·Algebra
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 for which Plan A costs less than Plan B?
When comparing two pricing plans, write each total cost as a linear function of (weeks). Then solve the inequality (or find the intersection point and compare which plan is lower). If is restricted to integers, convert the inequality solution into the correct integer range (often an off-by-one trap).
Hints
Model each plan with an expression
Write and for 12 students. Be careful with the “first week” versus “additional weeks” language for Plan A.
Compare the plans with an inequality
Set up and solve for .
Remember must be an integer
After solving the inequality, convert the result into the correct statement about positive integer values of .
Desmos Guide
Create a slider for weeks
In Desmos, type w=1 to create a slider, then restrict yourself to positive integers for testing.
Define the two cost functions
Enter
A = 360 + 12*(110 + 70*(w-1))B = 200 + 12*(80*w)
Find where Plan A becomes cheaper
Compute A-B (or graph both and look for when is below ). The switch happens just after , so for positive integers Plan A is cheaper for .
Step-by-step Explanation
Write Plan A’s cost for 12 students
For weeks, each student pays $110 for the first week and $70 for each of the remaining weeks.
So Plan A’s total cost is
Write Plan B’s cost for 12 students
Plan B charges $80 per student per week plus a $200 facility fee, so
Solve the inequality
First simplify each expression:
and
Now solve:
Use the fact that is a positive integer
Since must be a positive integer and , the solutions are
So the correct choice is .