A tutoring service offers two billing plans. - Plan A: A flat fee of \21 per hour. For any time beyond 120 minutes, additional time costs \0.65 per minute for the first 80 minutes, then \$0.30 per minute for each additional minute after 80 minutes. At how many minutes of tutoring will the two plans have the same total cost?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 43 | Free SAT Prep | Aniko

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Question 43·200 Super-Hard SAT Math Questions·Algebra

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A tutoring service offers two billing plans.

Plan A: A flat fee of $50 for up to the first 60 minutes. From 60 to 120 minutes, the cost increases at a rate of $21 per hour. For any time beyond 120 minutes, additional time costs $15 per hour.
Plan B: No flat fee. A rate of $0.65 per minute for the first 80 minutes, then $0.30 per minute for each additional minute after 80 minutes.

At how many minutes of tutoring will the two plans have the same total cost?

For break-even problems with changing rates, write each plan as a piecewise linear function in the same unit (minutes). Evaluate costs at the breakpoints to determine which interval contains the intersection, then solve by setting the appropriate linear expressions equal (and make sure each expression includes the correct accumulated cost from earlier intervals).

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Convert units first

Convert $21 per hour and $15 per hour into dollars per minute by dividing by 60.

Compute Plan A at 120 minutes

Find $A(120)$ by starting from the $50 flat fee and adding the cost of the next 60 minutes at the middle rate.

Use breakpoints to choose the right formulas

Compare $A(120)$ and $B(120)$ . Then compare the per-minute slopes after 120 minutes to decide whether the intersection happens for $t>120$ .

Solve a single linear equation in the final interval

For $t>120$ , use the $t>120$ formula for Plan A and the $t>80$ formula for Plan B, then set them equal and solve for $t$ .

Desmos Guide

Graph Plan A (piecewise)

Enter the piecewise function (with $x$ in minutes):

y=50{0<=x<=60}+(50+0.35(x-60)){60<x<=120}+(71+0.25(x-120)){x>120}

Graph Plan B (piecewise)

Enter:

y=0.65x{0<=x<=80}+(52+0.30(x-80)){x>80}

Find the intersection point

Use the intersection tool (or click the intersection) and read the $x$ -coordinate. That $x$ -value is the break-even time in minutes; match it to the answer choices.

Step-by-step Explanation

Convert hourly rates to per-minute rates

Convert the hourly rates in Plan A:

$21 per hour $=\frac{21}{60}=0.35$ dollars per minute
$15 per hour $=\frac{15}{60}=0.25$ dollars per minute

Write Plan A as a piecewise function

Let $t$ be the number of minutes.

If $t\le 60$ , then $A(t)=50$ .
If $60<t\le 120$ , then

A(t)=50+0.35(t-60).

In particular,

A(120)=50+0.35(60)=71.

If $t>120$ , then Plan A starts from 71 dollars at 120 minutes and adds $0.25$ dollars per minute:

A(t)=71+0.25(t-120)=0.25t+41 \quad (t>120).

Write Plan B as a piecewise function

If $t\le 80$ , then $B(t)=0.65t$ .
If $t>80$ , then Plan B charges 0.65 per minute for the first 80 minutes (which is $0.65\cdot 80=52$ dollars), then 0.30 per minute after that:

B(t)=52+0.30(t-80)=0.30t+28 \quad (t>80).

Identify the interval where the break-even occurs

Check costs at $t=120$ :

$A(120)=71$
Since $120>80$ , use $B(t)=0.30t+28$ : $B(120)=0.30(120)+28=64$

So at 120 minutes, Plan B is cheaper.

For $t>120$ , Plan A increases with slope $0.25$ while Plan B increases with slope $0.30$ , so Plan B will eventually catch up exactly once after 120 minutes.

Solve for the intersection (and interpret)

For $t>120$ , set the expressions equal:

\begin{aligned} 0.25t+41 &= 0.30t+28 \\ 41-28 &= 0.30t-0.25t \\ 13 &= 0.05t \\ t &= 260 \end{aligned}

Therefore, the two plans have the same total cost at 260 minutes.

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Step-by-step Explanation

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