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:

Hint:

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}

Answer 1

Correct Answer: 8

To solve the problem, we need to analyze both Sneha's and Rita's rowing scenarios.

Sneha's Rowing:

Let the speed of the river's current be c km/h. Sneha's effective speed upstream is (6 - c) km/h and downstream is (6 + c) km/h.

The time taken by Sneha to row 14 km upstream is \( \frac{14}{6 - c} \) hours, and the time downstream is \( \frac{14}{6 + c} \) hours.

According to the problem, Sneha takes 48 minutes (or 0.8 hours) more to row upstream:
\( \frac{14}{6 - c} = \frac{14}{6 + c} + 0.8 \)

Solving the equation:

\( 14(6 + c) = 14(6 - c) + 0.8(6 - c)(6 + c) \)
\( 84 + 14c = 84 - 14c + 0.8(36 - c^2) \)
\( 28c = 28.8 - 0.8c^2 \)
\( 0.8c^2 + 28c - 28.8 = 0 \)

Dividing by 0.8, we get:
\( c^2 + 35c - 36 = 0 \)

Using the quadratic formula, \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( c = \frac{-35 \pm \sqrt{1225 + 144}}{2} \)
\( c = \frac{-35 \pm \sqrt{1369}}{2} \)
\( c = \frac{-35 \pm 37}{2} \)

The solutions are \( c = 1 \) and \( c = -36 \). Since speed can’t be negative, c = 1 km/h.

Rita's Rowing:

Rita's effective speed upstream is (5 - 1) = 4 km/h and downstream is (5 + 1) = 6 km/h.

Let d be the distance she rows upstream and downstream (total 14 km). As it’s a round trip, she takes two-way trips of the same segment.

The total time given is 100 minutes, which is \( \frac{100}{60} = \frac{5}{3} \) hours:
\( \frac{d}{4} + \frac{d}{6} = \frac{5}{3} \)

Finding a common denominator and simplifying:
\( \frac{3d + 2d}{12} = \frac{5}{3} \)
\( \frac{5d}{12} = \frac{5}{3} \)
\( 5d = \frac{5 \times 12}{3} \)
\( 5d = 20 \)
\( d = 4 \) km per trip.

So, the total distance covered by Rita in two trips (upstream and downstream) is \( 2 \times 4 = 8 \) km.

This distance falls within the specified range (8,8).

Conclusion: Rita covers a total distance of 8 km.

Answer 2

Step 1: Find the speed of the river. Sneha’s speed in still water: \[ u_s = 6 \text{ km/h}. \] Let the speed of the river be \(v\) km/h. Then: \[ \text{Upstream speed} = 6 - v,\quad \text{Downstream speed} = 6 + v. \] Distance in each direction is 14 km, and the difference in time between upstream and downstream is 48 minutes: \[ \frac{14}{6 - v} - \frac{14}{6 + v} = 48 \text{ minutes} = \frac{4}{5} \text{ hours}. \] Compute: \[ \frac{14}{6 - v} - \frac{14}{6 + v} = 14 \left( \frac{(6+v) - (6-v)}{(6 - v)(6 + v)} \right) = 14 \left( \frac{2v}{36 - v^2} \right) = \frac{28v}{36 - v^2}. \] So, \[ \frac{28v}{36 - v^2} = \frac{4}{5}. \] Divide both sides by 4: \[ \frac{7v}{36 - v^2} = \frac{1}{5}. \] Cross-multiply: \[ 5 \cdot 7v = 36 - v^2 \Rightarrow 35v = 36 - v^2 \Rightarrow v^2 + 35v - 36 = 0. \] Factor: \[ (v + 36)(v - 1) = 0 \Rightarrow v = 1 \text{ (taking positive, feasible speed)}. \] Thus, the river speed is \(1\) km/h.
Step 2: Use Rita’s speeds to find the distance. Rita’s speed in still water: \[ u_r = 5 \text{ km/h}. \] So: \[ \text{Upstream speed} = 5 - 1 = 4 \text{ km/h},\quad \text{Downstream speed} = 5 + 1 = 6 \text{ km/h}. \] Total time taken for her round trip: \[ 100 \text{ minutes} = \frac{100}{60} = \frac{5}{3} \text{ hours}. \] Let the one-way distance be \(D\) km. Then: \[ \text{Upstream time} = \frac{D}{4},\quad \text{Downstream time} = \frac{D}{6}. \] Given: \[ \frac{D}{4} + \frac{D}{6} = \frac{5}{3}. \] Take LCM \(12\): \[ \frac{3D}{12} + \frac{2D}{12} = \frac{5}{3} \Rightarrow \frac{5D}{12} = \frac{5}{3}. \] Divide both sides by 5: \[ \frac{D}{12} = \frac{1}{3} \Rightarrow D = \frac{12}{3} = 4 \text{ km}. \]
Step 3: Total distance covered. Rita rows from the starting point to the turning point and back: \[ \text{Total distance} = 2D = 2 \times 4 = 8 \text{ km}. \] Therefore, Rita will cover a total distance of \(8\) km.