The simplified value of \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is (a) 4 (b) 3 (c) 2 (d) 6

To simplify the given expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\), let’s start by simplifying each square root by expressing them in terms of their prime factors and then simplify:

\[
\sqrt{32} = \sqrt{2^5} = \sqrt{16 \cdot 2} = 4\sqrt{2}
\]
\[
\sqrt{48} = \sqrt{2^4 \cdot 3} = 4\sqrt{3}
\]
\[
\sqrt{8} = \sqrt{2^3} = \sqrt{4 \cdot 2} = 2\sqrt{2}
\]
\[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
\]

Substituting these simplified forms back into the original expression gives:

\[
\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}
\]

We can take a 2 out of both the numerator and the denominator to simplify further:

\[
= \frac{2(2\sqrt{2} + 2\sqrt{3})}{2(\sqrt{2} + \sqrt{3})}
\]

Upon simplification, we find:

\[
= \frac{2\sqrt{2} + 2\sqrt{3}}{\sqrt{2} + \sqrt{3}}
\]

Since the terms in the numerator are double those in the denominator, simplifying the expression incorrectly suggested the terms could cancel out directly. The correct simplification should consider factoring correctly and assessing the common terms. The step where the 2 is factored out both in the numerator and denominator simplifies the expression directly, leading to:

\[
= 2
\]

Thus, the simplified value of the expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is indeed \(2\).

The correct answer is (c) 2.