`KMnO_(4)underset(R.A.)overset(H^(+))toMn^(x)`
`KMnO_(4)underset(R.A.)overset(OH^-)toMn^(y)`
`x+y`

Answer

To solve the problem, we need to determine the oxidation states of manganese (Mn) in two different reactions involving potassium permanganate (KMnO₄) in acidic and basic media.

### Step-by-Step Solution:

1. **Identify the Reaction in Acidic Medium:**
- The reaction is given as:
\[ KMnO_4 \xrightarrow{R.A.} Mn^{(x)} \]
- In acidic medium, KMnO₄ is reduced to MnO₂.

2. **Determine the Oxidation State of Mn in MnO₂:**
- The formula for MnO₂ can be expressed as:
\[ x + 2(-2) = 0 \]
- Solving for \(x\):
\[ x - 4 = 0 \implies x = +4 \]

3. **Identify the Reaction in Basic Medium:**
- The reaction is given as:
\[ KMnO_4 \xrightarrow{OH^-} Mn^{(y)} \]
- In basic medium, KMnO₄ is reduced to MnO₄²⁻.

4. **Determine the Oxidation State of Mn in MnO₄²⁻:**
- The formula for MnO₄²⁻ can be expressed as:
\[ y + 4(-2) = -2 \]
- Solving for \(y\):
\[ y - 8 = -2 \implies y = +6 \]

5. **Calculate \(x + y\):**
- Now that we have both oxidation states:
\[ x = +4 \quad \text{and} \quad y = +6 \]
- Therefore,
\[ x + y = 4 + 6 = 10 \]

### Final Answer:
\[ x + y = 10 \]