A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. The speed of the boat in still water is (I.I.F.T., 2008)

### Concept & Linear Systems This classic boat problem presents two journey scenarios, forming a linear system of two equations. We solve for the upstream speed ($U$) and downstream speed ($D$) to find the still water speed. $$ \text{Still Water Speed} = \frac{D + U}{2} $$ ### Step-by-Step Solution * Let the upstream speed be $U$ and the downstream speed be $D$. * Scenario 1: $$ \frac{30}{U} + \frac{44}{D} = 10 \quad \text{--- (Equation 1)} $$ * Scenario 2: $$ \frac{40}{U} + \frac{55}{D} = 13 \quad \text{--- (Equation 2)} $$ * To eliminate $U$, multiply Equation 1 by $4$ and Equation 2 by $3$: $$ \frac{120}{U} + \frac{176}{D} = 40 \quad \text{--- (Equation 3)} $$ $$ \frac{120}{U} + \frac{165}{D} = 39 \quad \text{--- (Equation 4)} $$ * Subtract Equation 4 from Equation 3: $$ \frac{176}{D} - \frac{165}{D} = 40 - 39 $$ $$ \frac{11}{D} = 1 \implies D = 11 \text{ km/hr} $$ * Substitute $D = 11$ into Equation 1: $$ \frac{30}{U} + \frac{44}{11} = 10 $$ $$ \frac{30}{U} + 4 = 10 $$ $$ \frac{30}{U} = 6 \implies U = 5 \text{ km/hr} $$ * The speed of the boat in still water is: $$ \frac{D + U}{2} = \frac{11 + 5}{2} = \frac{16}{2} = 8 \text{ km/hr} $$ ### Exam Strategy & Shortcut Look for common factors in the downstream distances. $44$ and $55$ are both multiples of $11$. Guessing $D = 11$ is a highly educated shortcut. If $D=11$, then $\frac{44}{11} = 4$ hours. Leaving $10 - 4 = 6$ hours for $30$ km upstream, so $U = 5$. Test this on the second equation: $\frac{40}{5} + \frac{55}{11} = 8 + 5 = 13$. It works perfectly, bypassing all complex algebra. ### Common Pitfall Solving for the stream speed instead of the boat's speed in still water. The stream speed would be $\frac{11 - 5}{2} = 3$, which is prominently listed as an incorrect trap option (a). ### Final Answer Therefore, the correct answer is **8 km/hr**.