The function h is defined by \(\mathrm{h(x) = \frac{x}{3} + 2}\). For what value of x does \(\mathrm{h(x) = 7}\)?5152127
GMAT Algebra : (Alg) Questions
The function \(\mathrm{h}\) is defined by \(\mathrm{h(x) = \frac{x}{3} + 2}\). For what value of \(\mathrm{x}\) does \(\mathrm{h(x) = 7}\)?
- 5
- 15
- 21
- 27
1. TRANSLATE the problem information
- Given information:
- Function: \(\mathrm{h(x) = \frac{x}{3} + 2}\)
- We need to find x when \(\mathrm{h(x) = 7}\)
- What this tells us: We need to set the function expression equal to 7 and solve for x
2. TRANSLATE the question into an equation
- "For what value of x does h(x) = 7?" becomes:
\(\mathrm{\frac{x}{3} + 2 = 7}\) - This equation represents the core relationship we need to solve
3. SIMPLIFY through algebraic manipulation
- Subtract 2 from both sides to isolate the term with x:
\(\mathrm{\frac{x}{3} = 5}\) - Multiply both sides by 3 to solve for x:
\(\mathrm{x = 15}\)
4. Verify the solution
- Substitute x = 15 back into the original function:
\(\mathrm{h(15) = \frac{15}{3} + 2 = 5 + 2 = 7}\) ✓
Answer: B) 15
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students may misinterpret what "h(x) = 7" means, thinking they need to substitute 7 for x in the original function instead of setting the function equal to 7.
They might calculate \(\mathrm{h(7) = \frac{7}{3} + 2 = 2.33... + 2 = 4.33...}\), then get confused when this doesn't match any answer choice. This leads to confusion and guessing.
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly set up \(\mathrm{\frac{x}{3} + 2 = 7}\) but make arithmetic errors in the solving process. Common mistakes include:
- Subtracting 2 incorrectly: getting \(\mathrm{\frac{x}{3} = 9}\) instead of \(\mathrm{\frac{x}{3} = 5}\)
- Multiplying by 3 incorrectly: getting \(\mathrm{x = 27}\) instead of \(\mathrm{x = 15}\)
This may lead them to select Choice D) (27) if they made the subtraction error.
The Bottom Line:
This problem tests whether students understand the relationship between function notation and equation solving. The key insight is recognizing that "h(x) = 7" creates an equation where the function expression equals 7, not where x equals 7.