To find the value that, when multiplied by \(5\sqrt{27}\), results in a rational number, we first simplify \(5\sqrt{27}\).
We can write \(\sqrt{27}\) as: \[ \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}. \] Thus, \[ 5\sqrt{27} = 5 \cdot 3\sqrt{3} = 15\sqrt{3}. \]
Now we need to determine which of the given options, when multiplied by \(15\sqrt{3}\), results in a rational number. A product of two expressions will be rational if the factor involving the square root has a square root that can combine with \(\sqrt{3}\) to produce a whole number. In other words, we want to find \(x\) such that \(15\sqrt{3} \cdot x\) is rational.
Let’s analyze each option:
A. \( \sqrt{12} \): \[ 15\sqrt{3} \cdot \sqrt{12} = 15\sqrt{3} \cdot \sqrt{4 \cdot 3} = 15\sqrt{3} \cdot 2\sqrt{3} = 30 \cdot 3 = 90 \quad (\text{rational}) \]
B. \( \sqrt{9} \): \[ 15\sqrt{3} \cdot \sqrt{9} = 15\sqrt{3} \cdot 3 = 45\sqrt{3} \quad (\text{not rational}) \]
C. \( 4\sqrt{2} \): \[ 15\sqrt{3} \cdot 4\sqrt{2} = 60\sqrt{6} \quad (\text{not rational}) \]
D. \( \sqrt{15} \): \[ 15\sqrt{3} \cdot \sqrt{15} = 15\sqrt{45} = 15 \cdot 3\sqrt{5} = 45\sqrt{5} \quad (\text{not rational}) \]
The only option that produces a rational product is option A, \(\sqrt{12}\).
Therefore, the answer is: \[ \boxed{\sqrt{12}}. \]