Two hoses, opened simultaneously, can fill a swimming pool in $10$ hours. If hose A working alone would take $26$ hours, how long would hose B working alone take to fill the same pool?

Let $t$ be hose B's solo time. The combined-rate equation: $\tfrac{1}{26} + \tfrac{1}{t} = \tfrac{1}{10}$. Isolate $\tfrac{1}{t}$: $\tfrac{1}{t} = \tfrac{1}{10} - \tfrac{1}{26}$. Common denominator $130$: $\tfrac{13}{130} - \tfrac{5}{130} = \tfrac{8}{130} = \tfrac{4}{65}$. So $t = \dfrac{65}{4} = 16.25$ hours $= 16$ hr $15$ min. Check: $\tfrac{1}{26} + \tfrac{4}{65}$. Convert to common denominator $130$: $\tfrac{5}{130} + \tfrac{8}{130} = \tfrac{13}{130} = \tfrac{1}{10}$ ✓. - Trap B ($16$ hours) rounds down — but the question requires the exact value. - Trap D ($36 = 10 + 26$) adds times, wrong model. - Trap C ($18$) gives a combined fill time of $1/26 + 1/18 = 11/117 \ne 1/10$. Sanity: hose B must work *faster* than hose A's $26$ hours but *slower* than the combined $10$ hours. $16.25$ sits squarely in that window.