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A motorboat covers a given distance in 6...

A motorboat covers a given distance in `6h` moving downstream on a river. It covers the same distance in `10h` moving upstream. The time it takes to cover the same distance in still water is

A

(a)9.5 h

B

(b)7.5 h

C

(c)6.5 h

D

(d)8 h

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The correct Answer is:
To solve the problem, we need to determine the time it takes for a motorboat to cover the same distance in still water, given the time taken to cover the distance downstream and upstream. ### Step-by-Step Solution: 1. **Define Variables:** - Let the speed of the motorboat in still water be \( V \) (in km/h). - Let the speed of the river current (stream) be \( x \) (in km/h). 2. **Determine Downstream and Upstream Speeds:** - When moving downstream, the effective speed of the boat is \( V + x \). - When moving upstream, the effective speed of the boat is \( V - x \). 3. **Set Up Distance Equations:** - The distance covered downstream in 6 hours can be expressed as: \[ \text{Distance} = \text{Speed} \times \text{Time} = (V + x) \times 6 \] - The distance covered upstream in 10 hours can be expressed as: \[ \text{Distance} = (V - x) \times 10 \] 4. **Equate the Distances:** - Since the distances are the same, we can set the two expressions equal to each other: \[ 6(V + x) = 10(V - x) \] 5. **Expand and Rearrange the Equation:** - Expanding both sides gives: \[ 6V + 6x = 10V - 10x \] - Rearranging the equation to isolate terms involving \( V \) and \( x \): \[ 6V + 6x + 10x = 10V \] \[ 6V + 16x = 10V \] - Simplifying further: \[ 4V = 16x \] \[ V = 4x \] 6. **Calculate the Distance:** - Substitute \( V = 4x \) back into the distance equation for downstream: \[ \text{Distance} = 6(V + x) = 6(4x + x) = 6(5x) = 30x \] 7. **Calculate Time in Still Water:** - The speed in still water is \( V = 4x \). - The time taken to cover the same distance in still water is given by: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{30x}{4x} = \frac{30}{4} = 7.5 \text{ hours} \] ### Conclusion: The time it takes to cover the same distance in still water is **7.5 hours**. ---
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