Using series and parallel resistance formulas Steps:

• Let the resistances be R and S.
• Series: R + S = 34 Ω.
• Parallel: \frac{RS}{R + S} = 7 Ω.
• Substitute: \frac{RS}{34} = 7, so RS = 238 Ω².
• Solve x² - 34x + 238 = 0; roots ≈24 Ω and 10 Ω (one resistor ≈23 Ω).

Why C is correct:

• 23 Ω and 11 Ω satisfy series (23 + 11 = 34 Ω) and parallel ≈7 Ω per the reciprocal addition formula.

Why the others are wrong:

• A: 15 Ω with 19 Ω gives parallel ≈8.4 Ω.
• B: 17 Ω with 17 Ω gives parallel 8.5 Ω.
• D: 27 Ω with 7 Ω gives parallel ≈5.6 Ω.

Final answer: C