Volunteer Cleanup: When Will Both Groups Finish Equal Rows?

Question

PROBLEM

Two groups of volunteers are cleaning up the football stadium after the Homecoming game. Volunteers from the Band Booster Club have already cleaned 8 rows of bleachers and will continue to clean at a rate of 1 row per minute. The leadership class has completed 2 rows and will continue working at 4 rows per minute. Once the two groups get to the point where they have cleaned the same number of rows, they will take a break. How long will that take? Write a system of equations, graph them, and find the solution.

Skills This Problem Builds

Setting up systems of linear equations from real-world scenarios
Recognizing when two quantities with different starting points and rates will be equal
Graphing linear systems and interpreting intersection points
Solving systems using substitution and elimination methods
Connecting algebraic solutions to their geometric representations

Picture This

Two groups of volunteers are cleaning up the football stadium after the Homecoming game. Volunteers from the Band...

The graph shows both groups' progress over time. The intersection at (2, 10) reveals when they'll have cleaned equal rows.

Solution: Method 1 — System Setup and Substitution

Step 1 — Define the variables and write equations

Let t = time in minutes from now (when both groups continue working)

Let B = total rows cleaned by Band Boosters

Let L = total rows cleaned by Leadership class

Since total rows = starting amount + (rate × time):

B = 8 + 1t
L = 2 + 4t

Step 2 — Set up the equal condition

At break time, both groups have cleaned the same number of rows, so B = L:

8 + 1t = 2 + 4t

Step 3 — Solve for t

Subtract 1t and 2 from both sides:

8 - 2 = 4t - 1t
6 = 3t
t = 2

Step 4 — Find the number of rows

Substitute t = 2 back into either equation:

B = 8 + 1(2) = 10 rows
L = 2 + 4(2) = 10 rows ✓

Solution: Method 2 — Gap-Closing Analysis

Step 1 — Find the initial gap

At the start, Band Boosters have cleaned 8 rows and Leadership has cleaned 2 rows.

Initial gap = 8 - 2 = 6 rows

Step 2 — Calculate the closing rate

Leadership works faster, so they close the gap at a rate of:

Closing rate = 4 - 1 = 3 rows per minute

Step 3 — Find when the gap closes

Time to close gap = Initial gap ÷ Closing rate:

Time = 6 ÷ 3 = 2 minutes

Step 4 — Calculate final row count

After 2 minutes, each group has cleaned:

Band Boosters: 8 + 1(2) = 10 rows
Leadership: 2 + 4(2) = 10 rows

Both groups will take a break after 2 minutes, when each has cleaned 10 rows.

Verification

Let's check our answer by substituting t = 2 back into both original equations:

Band Boosters: B = 8 + 1(2) = 8 + 2 = 10 rows ✓
Leadership: L = 2 + 4(2) = 2 + 8 = 10 rows ✓

Perfect! Both groups have cleaned exactly 10 rows after 2 minutes. We can also verify this makes physical sense: Leadership started 6 rows behind but works 3 times faster, so they need 6 ÷ 3 = 2 minutes to catch up.

Watch Out For These

✗ Mistake 1: Setting up equations as B = 8 + t and L = 2 + t

This assumes both groups work at the same rate. The problem clearly states different rates: 1 row/min vs 4 rows/min. The correct equations must include these different coefficients.

✗ Mistake 2: Forgetting the initial amounts

Some students write B = 1t and L = 4t, ignoring that both groups have already cleaned some rows before this timing begins. Always account for starting values in rate problems.

✗ Mistake 3: Misinterpreting the graph intersection

The intersection point (2, 10) means "after 2 minutes, both have 10 rows." Students sometimes read this as "2 rows after 10 minutes" by switching the coordinates.

The Pattern Behind This

This problem follows the classic "catch-up" pattern that appears throughout algebra and real-world scenarios. The general structure is:

Group A: Total = Start₁ + Rate₁ × Time
Group B: Total = Start₂ + Rate₂ × Time
Solution: When Total_A = Total_B

The intersection occurs at time t = (Start₁ - Start₂) ÷ (Rate₂ - Rate₁), provided Rate₂ > Rate₁ (the group that starts behind must work faster to catch up).

If the slower group starts ahead by too much, or if they work at the same rate, the lines will never intersect - there's no catch-up point.

This Calculation in the Real World

You'll encounter this same mathematical structure in numerous real scenarios:

Business: Two companies with different starting revenues growing at different rates - when will the smaller company overtake the larger?

Technology: Two servers processing different backlogs at different speeds - when will their queue sizes match?

Sports: A runner starting behind but moving faster than the leader - at what point in the race will they be tied?

What If?

1

Faster Leadership

Same starting positions (Band has 8 rows, Leadership has 2 rows), but Leadership works even faster at 6 rows per minute while Band Boosters still work at 1 row per minute. When will they have cleaned equal rows?

Step 1 — Set up equations

Band Boosters: B = 8 + 1t
Leadership: L = 2 + 6t

Step 2 — Set equal and solve

8 + 1t = 2 + 6t
8 - 2 = 6t - 1t
6 = 5t
t = 6/5 = 1.2 minutes

Step 3 — Find total rows

B = 8 + 1(1.2) = 9.2 rows
L = 2 + 6(1.2) = 9.2 rows

Step 4 — Verify

Both groups reach 9.2 rows after 1.2 minutes ✓

2

Different Starting Lead

Band Boosters start with 12 rows already cleaned and work at 1 row per minute. Leadership starts with 2 rows and works at 4 rows per minute. When will they be tied?

Step 1 — Write the equations

Band Boosters: B = 12 + 1t
Leadership: L = 2 + 4t

Step 2 — Solve the system

12 + 1t = 2 + 4t
12 - 2 = 4t - 1t
10 = 3t
t = 10/3 ≈ 3.33 minutes

Step 3 — Calculate rows cleaned

B = 12 + 1(10/3) = 12 + 10/3 = 46/3 ≈ 15.33 rows
L = 2 + 4(10/3) = 2 + 40/3 = 46/3 ≈ 15.33 rows

Step 4 — Final answer

They'll be tied after 3⅓ minutes at 15⅓ rows each

3

Three Teams

Band (8 rows, 1 row/min), Leadership (2 rows, 4 rows/min), and now Parent Volunteers join with 0 rows cleaned, working at 3 rows per minute. When will all three groups have cleaned the same number of rows?

Step 1 — Set up three equations

Band: B = 8 + 1t
Leadership: L = 2 + 4t
Parents: P = 0 + 3t

Step 2 — Set Band = Parents

8 + 1t = 0 + 3t
8 = 2t
t = 4 minutes

Step 3 — Check if Leadership equals them too

At t = 4: L = 2 + 4(4) = 18 rows
But Band and Parents each have: 8 + 1(4) = 12 rows

Step 4 — Conclusion

All three groups are never simultaneously tied. Leadership's head start and high rate mean they're always ahead when the other two meet.

4

Reverse Engineering

You observe both groups taking their break after exactly 3 minutes, at which point they've both cleaned 15 rows. If Leadership started with 3 rows and worked at 4 rows/min, what must have been Band Boosters' starting amount and rate?

Step 1 — Use Leadership's known data

Leadership: L = 3 + 4t
At t = 3: L = 3 + 4(3) = 15 rows ✓

Step 2 — Set up Band's equation

Band: B = s + rt where s = starting rows, r = rate
We know at t = 3, B = 15

Step 3 — Create constraint equation

s + r(3) = 15
s + 3r = 15

Step 4 — Find possible solutions

Multiple solutions exist: If r = 1, then s = 12
If r = 2, then s = 9
If r = 3, then s = 6
Any combination where starting rows + 3 × rate = 15

Frequently Asked Questions

Create two equations where each represents the total progress of one group. Use the form "total = starting amount + (rate × time)". In this problem, Band Boosters get B = 8 + 1t and Leadership gets L = 2 + 4t. The solution occurs when B = L.

The intersection point shows the exact moment when both equations have the same value. Here, when the lines cross at (2, 10), it means after 2 minutes both groups have cleaned exactly 10 rows - that's when they take their break.

These problems model real situations where one group has a head start but the other works faster. The mathematics reveals whether and when the faster group will catch up, which applies to everything from race problems to business scenarios.

NJ

Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-05-28