Question:
Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is:
Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is:
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In upstream–downstream problems:
\begin{itemize}
\item First find the stream speed using time differences and the given rower’s speed.
\item Then apply those speeds to other rowers, using the relation \(\text{time} = \frac{\text{distance}}{\text{speed}}\).
\item For round trips, total time is the sum of upstream and downstream times.
\end{itemize}
Updated On: Jul 7, 2026
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Correct Answer: 8
Approach Solution - 1
Approach: Sneha's upstream-vs-downstream time gap is the only clue to the current speed \(c\); find \(c\) first. Then Rita's 100-minute round trip (up and back to the same point) fixes the one-way distance, and the total distance is twice that.
Step 1 (find the current): Sneha rows at \(6\) km/h still water. Upstream \(=6-c\), downstream \(=6+c\). She takes \(48\) min \(=0.8\) h more upstream over \(14\) km: \[ \frac{14}{6-c}-\frac{14}{6+c}=0.8. \]
Step 2: Combine: \(\dfrac{14\cdot 2c}{36-c^2}=0.8\Rightarrow 28c=0.8(36-c^2)\Rightarrow c^2+35c-36=0\Rightarrow (c-1)(c+36)=0\). Since \(c>0\), \(c=1\) km/h.
Step 3 (Rita): Rita rows at \(5\) km/h still water, so upstream \(=5-1=4\) km/h and downstream \(=5+1=6\) km/h. Let the one-way distance be \(d\). Total time \(=100\) min \(=\tfrac53\) h: \[ \frac{d}{4}+\frac{d}{6}=\frac53. \]
Step 4: \(\dfrac{3d+2d}{12}=\dfrac{5d}{12}=\dfrac53\Rightarrow 5d=20\Rightarrow d=4\) km one way.
Step 5: She goes up \(4\) km and comes back \(4\) km, so total distance \(=2\times 4=8\) km.
Answer: 8 km.
Step 1 (find the current): Sneha rows at \(6\) km/h still water. Upstream \(=6-c\), downstream \(=6+c\). She takes \(48\) min \(=0.8\) h more upstream over \(14\) km: \[ \frac{14}{6-c}-\frac{14}{6+c}=0.8. \]
Step 2: Combine: \(\dfrac{14\cdot 2c}{36-c^2}=0.8\Rightarrow 28c=0.8(36-c^2)\Rightarrow c^2+35c-36=0\Rightarrow (c-1)(c+36)=0\). Since \(c>0\), \(c=1\) km/h.
Step 3 (Rita): Rita rows at \(5\) km/h still water, so upstream \(=5-1=4\) km/h and downstream \(=5+1=6\) km/h. Let the one-way distance be \(d\). Total time \(=100\) min \(=\tfrac53\) h: \[ \frac{d}{4}+\frac{d}{6}=\frac53. \]
Step 4: \(\dfrac{3d+2d}{12}=\dfrac{5d}{12}=\dfrac53\Rightarrow 5d=20\Rightarrow d=4\) km one way.
Step 5: She goes up \(4\) km and comes back \(4\) km, so total distance \(=2\times 4=8\) km.
Answer: 8 km.
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Approach Solution -2
Setting up the equations. Let the river flow at \(v\) km/h. Sneha's speeds are \(6-v\) upstream and \(6+v\) downstream. Rowing 14 km, the time difference is 48 minutes \(=\tfrac{4}{5}\) hour:
\[ \frac{14}{6-v}-\frac{14}{6+v}=\frac{4}{5}. \]
This simplifies to \(\frac{14\times 2v}{36-v^2}=\frac{4}{5}\), so \(140v=4(36-v^2)\Rightarrow v^2+35v-36=0\Rightarrow (v-1)(v+36)=0\). Since \(v>0\), \(v=1\) km/h.
Rita's still-water speed is 5 km/h, so her upstream and downstream speeds are \(5-1=4\) km/h and \(5+1=6\) km/h. If she covers a distance \(d\) upstream and returns the same \(d\) downstream in a total of 100 minutes \(=\tfrac{5}{3}\) hour:
\[ \frac{d}{4}+\frac{d}{6}=\frac{5}{3}\Rightarrow \frac{5d}{12}=\frac{5}{3}\Rightarrow d=4 \text{ km}. \]
Total distance covered \(=2d=8\) km.
Answer: 8 km.
\[ \frac{14}{6-v}-\frac{14}{6+v}=\frac{4}{5}. \]
This simplifies to \(\frac{14\times 2v}{36-v^2}=\frac{4}{5}\), so \(140v=4(36-v^2)\Rightarrow v^2+35v-36=0\Rightarrow (v-1)(v+36)=0\). Since \(v>0\), \(v=1\) km/h.
Rita's still-water speed is 5 km/h, so her upstream and downstream speeds are \(5-1=4\) km/h and \(5+1=6\) km/h. If she covers a distance \(d\) upstream and returns the same \(d\) downstream in a total of 100 minutes \(=\tfrac{5}{3}\) hour:
\[ \frac{d}{4}+\frac{d}{6}=\frac{5}{3}\Rightarrow \frac{5d}{12}=\frac{5}{3}\Rightarrow d=4 \text{ km}. \]
Total distance covered \(=2d=8\) km.
Answer: 8 km.
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