One can of deck sealant covers 150 square feet of surface per coat. A rectangular deck has length L feet...
GMAT Algebra : (Alg) Questions
One can of deck sealant covers \(150\) square feet of surface per coat. A rectangular deck has length \(\mathrm{L}\) feet and width \(\mathrm{W}\) feet. The contractor plans to apply three coats uniformly. Which equation represents the number of cans \(\mathrm{C}\) needed to complete the job?
\(\frac{\mathrm{LW}}{150}\)
\(\frac{\mathrm{LW}}{75}\)
\(\frac{\mathrm{LW}}{50}\)
\(150\mathrm{LW}\)
\(450\mathrm{LW}\)
1. TRANSLATE the problem information
- Given information:
- Deck is rectangular with length L feet and width W feet
- One can covers 150 square feet per coat
- Three coats will be applied uniformly
- Need to find number of cans C
2. TRANSLATE the deck area
- Rectangular area formula gives us: \(\mathrm{Area = L \times W}\) square feet
3. INFER the total coverage needed
- Here's the key insight: Since three coats are applied, we need to cover the deck area three times
- Total surface area to be covered = \(\mathrm{3 \times (deck\ area) = 3 \times LW = 3LW}\) square feet
4. INFER the relationship for number of cans
- We need: Number of cans = Total coverage needed ÷ Coverage per can
- Therefore: \(\mathrm{C = \frac{3LW}{150}}\)
5. SIMPLIFY the expression
- \(\mathrm{C = \frac{3LW}{150}}\)
- Factor out 3 from numerator and denominator: \(\mathrm{C = \frac{3 \times LW}{3 \times 50} = \frac{LW}{50}}\)
Answer: (C) \(\mathrm{\frac{LW}{50}}\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE reasoning: Students correctly identify the deck area as \(\mathrm{LW}\) but miss that three coats means the total surface area to be covered is \(\mathrm{3LW}\), not just \(\mathrm{LW}\).
They set up: \(\mathrm{C = \frac{LW}{150}}\) instead of \(\mathrm{C = \frac{3LW}{150}}\)
This leads them to select Choice (A) \(\mathrm{\frac{LW}{150}}\)
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly set up \(\mathrm{C = \frac{3LW}{150}}\) but make arithmetic errors when simplifying the fraction.
They might incorrectly simplify to \(\mathrm{\frac{LW}{75}}\) by dividing only the denominator by 2, or get confused about how to reduce the fraction properly.
This may lead them to select Choice (B) \(\mathrm{\frac{LW}{75}}\)
The Bottom Line:
This problem tests whether students can properly translate "three coats" into mathematical terms (multiplying the base area by 3) and then perform the division setup correctly. The key insight is recognizing that multiple coats increase the total coverage requirement.
\(\frac{\mathrm{LW}}{150}\)
\(\frac{\mathrm{LW}}{75}\)
\(\frac{\mathrm{LW}}{50}\)
\(150\mathrm{LW}\)
\(450\mathrm{LW}\)