Let a and b be positive numbers. If a/(sqrt(b)) = 5 and 3a/(sqrt(kb)) = 5 for some positive constant k,...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{a/\sqrt{b} = 5}\) (first equation)
  • \(\mathrm{3a/\sqrt{kb} = 5}\) (second equation)
  • a, b, and k are all positive
• We need to find the value of k

2. INFER the solution strategy

• Key insight: Both equations equal 5, so we can use substitution
• From the first equation, we can solve for 'a' in terms of 'b'
• Then substitute this expression into the second equation
• This will give us an equation with only k and b, where b will cancel out

3. SIMPLIFY to find the expression for 'a'

From \(\mathrm{a/\sqrt{b} = 5}\):

• Multiply both sides by \(\mathrm{\sqrt{b}}\): \(\mathrm{a = 5\sqrt{b}}\)

4. SIMPLIFY by substituting into the second equation

Substitute \(\mathrm{a = 5\sqrt{b}}\) into \(\mathrm{3a/\sqrt{kb} = 5}\):

• \(\mathrm{3(5\sqrt{b})/\sqrt{kb} = 5}\)
• \(\mathrm{15\sqrt{b}/\sqrt{kb} = 5}\)

5. SIMPLIFY using radical properties

Apply \(\mathrm{\sqrt{kb} = \sqrt{k} \cdot \sqrt{b}}\):

• \(\mathrm{15\sqrt{b}/(\sqrt{k} \cdot \sqrt{b}) = 5}\)

The \(\mathrm{\sqrt{b}}\) terms cancel (since b is positive, \(\mathrm{\sqrt{b} \neq 0}\)):

• \(\mathrm{15/\sqrt{k} = 5}\)

6. SIMPLIFY to solve for k

Multiply both sides by \(\mathrm{\sqrt{k}}\):

• \(\mathrm{15 = 5\sqrt{k}}\)

Divide both sides by 5:

• \(\mathrm{3 = \sqrt{k}}\)

Square both sides:

• \(\mathrm{k = 9}\)

Answer: C) 9

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students often struggle with the radical properties and cancellation steps. They might incorrectly handle \(\mathrm{\sqrt{kb}}\), thinking it equals \(\mathrm{\sqrt{k} + \sqrt{b}}\) instead of \(\mathrm{\sqrt{k} \cdot \sqrt{b}}\), or they may be hesitant to cancel the \(\mathrm{\sqrt{b}}\) terms.

This conceptual confusion with radical arithmetic can lead them to set up incorrect equations, potentially leading to wrong values like Choice A (1/9) or causing them to get stuck and guess.

Second Most Common Error:

Poor INFER reasoning: Some students attempt to solve the system by setting the two equations equal to each other without using substitution effectively. They might try \(\mathrm{a/\sqrt{b} = 3a/\sqrt{kb}}\) but then get confused about how to proceed, especially when trying to eliminate 'a' from both sides.

This leads to algebraic dead ends and may cause them to select Choice B (3) if they incorrectly conclude that \(\mathrm{\sqrt{k} = 3}\) without properly squaring.

The Bottom Line:

This problem tests your ability to work systematically with radical equations and use substitution strategically. The key breakthrough is recognizing that substitution will eliminate variables cleanly, and having confidence with radical properties to execute the cancellation steps correctly.