The function f is defined by \(\mathrm{f(x) = \frac{x+15}{5}}\), and \(\mathrm{f(a) = 10}\), where a is a constant. What is...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(x) = \frac{x+15}{5}}\)
  • \(\mathrm{f(a) = 10}\)
  • Need to find the value of \(\mathrm{a}\)

• What this tells us: We need to find the input value \(\mathrm{a}\) that makes the function output equal to 10.

2. TRANSLATE function notation into an equation

• Since \(\mathrm{f(a) = 10}\), substitute \(\mathrm{a}\) for \(\mathrm{x}\) in the function:
  \(\mathrm{f(a) = \frac{a+15}{5} = 10}\)

• This gives us the equation: \(\mathrm{\frac{a+15}{5} = 10}\)

3. SIMPLIFY by solving the linear equation

• Multiply both sides by 5 to eliminate the fraction:
  \(\mathrm{5 \times \frac{a+15}{5} = 5 \times 10}\)
  \(\mathrm{a + 15 = 50}\)

• Subtract 15 from both sides to isolate \(\mathrm{a}\):
  \(\mathrm{a + 15 - 15 = 50 - 15}\)
  \(\mathrm{a = 35}\)

Answer: C. 35

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Misunderstanding what \(\mathrm{f(a) = 10}\) means and thinking that \(\mathrm{a = 10}\) directly.

Students might confuse the input and output of the function, reasoning "if \(\mathrm{f(a) = 10}\), then \(\mathrm{a}\) must equal 10." This bypasses the need to actually use the function definition and substitute.

This may lead them to select Choice B (10).

Second Most Common Error:

Poor SIMPLIFY execution: Making sign errors during algebraic manipulation.

Students correctly set up \(\mathrm{\frac{a+15}{5} = 10}\) and multiply by 5 to get \(\mathrm{a + 15 = 50}\), but then add 15 to both sides instead of subtracting: \(\mathrm{a = 50 + 15 = 65}\).

This may lead them to select Choice D (65).

The Bottom Line:

This problem tests whether students truly understand function notation (that \(\mathrm{f(a)}\) means substitute \(\mathrm{a}\) into the function) and can perform reliable algebraic manipulation. The key insight is recognizing that finding \(\mathrm{a}\) when \(\mathrm{f(a) = 10}\) is essentially working backwards through the function.