Find Paint Cans Needed for Wall Minus Whiteboard

Geometry & Area 7th-8th Grade
Problem
Ms. Baker wants to paint a wall in her classroom. The wall is 24 feet long and 15 feet tall. She has a whiteboard on this wall that is 12 feet long and 7 feet tall. If one can of paint covers 75 square feet, how many cans of paint does she need to paint the wall around her whiteboard?

What You Will Learn

  • How to calculate area when you need to exclude sections
  • When and why to round up in real-world problems
  • Converting area calculations into practical purchasing decisions
  • The difference between total coverage and usable coverage
  • How geometric visualization helps solve word problems

Let's Draw It

Here's what Ms. Baker's wall looks like with the whiteboard that won't be painted:

Ms. Baker wants to paint a wall in her classroom. The wall is 24 feet long and 15 feet tall. She has a whiteboard on...

Solution: Method 1 — The Subtraction Approach

The most intuitive way is to find the total wall area, then subtract the whiteboard area that won't be painted.

Step 1 — Calculate the total wall area

The wall is a rectangle, so its area is length times width:

Wall area = 24 ft × 15 ft = 360 ft²

Step 2 — Calculate the whiteboard area

The whiteboard is also rectangular:

Whiteboard area = 12 ft × 7 ft = 84 ft²

Step 3 — Find the area that needs paint

Subtract the whiteboard area from the total wall area:

Paintable area = 360 ft² - 84 ft² = 276 ft²

Step 4 — Calculate how many cans are needed

Divide the paintable area by the coverage per can:

Number of cans = 276 ft² ÷ 75 ft²/can = 3.68 cans

Step 5 — Round up to whole cans

Since Ms. Baker can't buy 0.68 of a can, she needs to round up to the next whole number:

3.68 cans → 4 cans needed

Solution: Method 2 — The Panel-by-Panel Approach

Another way is to think of the wall as separate rectangular panels around the whiteboard.

Step 1 — Identify the four panels

Looking at our diagram, we can divide the paintable area into four rectangles:

  • Top panel: above the whiteboard
  • Bottom panel: below the whiteboard
  • Left panel: to the left of the whiteboard
  • Right panel: to the right of the whiteboard

Step 2 — Calculate each panel area

If the whiteboard is centered on the wall (we'll assume this for simplicity):

Left panel = 6 ft × 15 ft = 90 ft² Right panel = 6 ft × 15 ft = 90 ft² Top panel = 12 ft × 4 ft = 48 ft² Bottom panel = 12 ft × 4 ft = 48 ft²

Step 3 — Add up all panels

Total paintable area = 90 + 90 + 48 + 48 = 276 ft²

Step 4 — Convert to cans and round up

276 ft² ÷ 75 ft²/can = 3.68 → 4 cans

Notice we get the same answer! This confirms our subtraction method was correct.

Ms. Baker needs 4 cans of paint.

Verification

Let's verify our answer makes sense:

Check the area calculation:

  • Wall area: 24 × 15 = 360 ft²
  • Whiteboard area: 12 × 7 = 84 ft²
  • Paintable area: 360 - 84 = 276 ft²

Check the paint calculation:

  • 3 cans cover: 3 × 75 = 225 ft² (not enough)
  • 4 cans cover: 4 × 75 = 300 ft² (enough for 276 ft²) ✓

The verification confirms that 4 cans is correct.

Does This Seem Reasonable?

Let's do a quick sanity check on our answer.

The whiteboard takes up 84 ÷ 360 = 23% of the wall. So we're painting about 77% of the wall.

If the whole wall needed painting, it would require 360 ÷ 75 = 4.8 cans, so 5 cans total. Since we're painting about 77% of it, we'd expect roughly 5 × 0.77 = 3.85 cans.

Our answer of 4 cans is right in that ballpark, so it passes the smell test!

Common Pitfalls

✗ Mistake 1: Forgetting to subtract the whiteboard
Some students calculate 360 ÷ 75 = 4.8 → 5 cans by painting the whole wall. But the whiteboard won't be painted, so this wastes money on an extra can.
✗ Mistake 2: Rounding down instead of up
Calculating 276 ÷ 75 = 3.68 and rounding to 3 cans. But 3 cans only cover 225 ft², leaving 51 ft² unpainted. Always round up for purchasing problems.
✗ Mistake 3: Adding the areas instead of subtracting
Computing 360 + 84 = 444 ft² because the problem mentions both areas. But we need wall area minus whiteboard area, not their sum.
✗ Mistake 4: Confusing perimeter with area
Calculating (24 + 15) × 2 = 78 because they're thinking about the wall's edges. But paint coverage is always about area (square feet), not perimeter.

The General Formula

This problem follows the standard "area with exclusion" pattern:

Paintable Area = Total Area - Excluded Area
Cans Needed = ⌈Paintable Area ÷ Coverage per Can⌉

The ceiling symbol ⌈⌉ means "round up to the next integer." This formula works for any paint problem where you need to avoid certain areas.

You'll see this pattern again with:

  • Painting rooms with windows and doors
  • Carpeting floors with built-in furniture
  • Fertilizing lawns with gardens or pools
  • Any "coverage minus obstacles" problem

Where This Shows Up in Real Life

Home renovation: Contractors constantly calculate paint, tile, or carpet needed while excluding fixtures, cabinets, and appliances. Getting this wrong means costly return trips to the store.

Commercial flooring: Office managers need to know how much carpet to order for a floor plan with cubicles, pillars, and permanent fixtures already in place.

Landscaping: Calculating mulch or grass seed needed for a yard minus the area occupied by trees, flower beds, patios, and walkways.

What If?

1
Different Coverage
Same wall (24 ft × 15 ft) and whiteboard (12 ft × 7 ft), but now one can of paint covers only 60 square feet. How many cans does Ms. Baker need?
Step 1 — Find paintable area

This is the same as before: 360 - 84 = 276 ft²

Step 2 — Calculate cans with new coverage

276 ÷ 60 = 4.6 cans

Step 3 — Round up

Since we can't buy 0.6 of a can, round up to 5 cans

Step 4 — Verify

5 cans cover 5 × 60 = 300 ft², which is more than the needed 276 ft² ✓

2
Add a Window
Same wall (24 ft × 15 ft) has both a whiteboard (12 ft × 7 ft) and a rectangular window (4 ft × 6 ft) that won't be painted. Paint covers 75 ft² per can. How many cans are needed?
Step 1 — Calculate wall area

24 × 15 = 360 ft²

Step 2 — Calculate excluded areas

Whiteboard: 12 × 7 = 84 ft²
Window: 4 × 6 = 24 ft²
Total excluded: 84 + 24 = 108 ft²

Step 3 — Find paintable area

360 - 108 = 252 ft²

Step 4 — Calculate cans needed

252 ÷ 75 = 3.36 → 4 cans (rounded up)

Step 5 — Verify

4 cans cover 4 × 75 = 300 ft², which exceeds 252 ft² ✓

3
Two Walls
Ms. Baker wants to paint two identical walls. Each wall is 24 ft × 15 ft with a 12 ft × 7 ft whiteboard. Paint covers 75 ft² per can. How many cans for both walls?
Step 1 — Find paintable area per wall

From our original problem: 276 ft² per wall

Step 2 — Calculate total paintable area

276 × 2 = 552 ft²

Step 3 — Calculate cans needed

552 ÷ 75 = 7.36 cans

Step 4 — Round up

8 cans are needed for both walls

Step 5 — Verify

8 cans cover 8 × 75 = 600 ft², which exceeds 552 ft² ✓

4
Reverse Problem
Ms. Baker has exactly 3 cans of paint (75 ft² per can). The wall is 24 ft × 15 ft. What's the largest square whiteboard she can install and still have enough paint for the rest of the wall?
Step 1 — Find maximum paintable area

3 cans × 75 ft²/can = 225 ft²

Step 2 — Calculate wall area

24 × 15 = 360 ft²

Step 3 — Find maximum whiteboard area

If paintable area = 225 ft², then:
Whiteboard area = 360 - 225 = 135 ft²

Step 4 — Find square side length

For a square: side² = 135
side = √135 ≈ 11.6 ft

Step 5 — Verify

11.6 ft × 11.6 ft = 134.6 ft² whiteboard
360 - 134.6 = 225.4 ft² to paint ≈ 3 cans ✓

Frequently Asked Questions

Calculate the total wall area, then subtract the area of any objects that won't be painted (like windows, whiteboards, or doors). In this problem: wall area is 24 × 15 = 360 ft², whiteboard area is 12 × 7 = 84 ft², so paintable area is 360 - 84 = 276 ft².
Always round up to the next whole can because you can't buy a fraction of a can. If your calculation gives 3.68 cans, you need to buy 4 cans. Here, 276 ÷ 75 = 3.68, so Ms. Baker needs 4 cans total.
Paint coverage refers to area (length × width), not perimeter. You need to know how many square feet the paint covers, then divide the total paintable area by the coverage per can. Don't confuse this with measuring around the edges.
DN

Dr. Neven Jurkovic

Mathematics educator with expertise in problem-solving and mathematical reasoning

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.

2026-07-12