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Speeds of Boat A and B in still water are in the ratio of 3:2 Rate of current is 10 Km/hr. Both Boats started from Point P to point Q downstream at the same time. After Boat B reaching Point Q, in return journey, it is powered by engine due to which the speed of the boat in still water is increased by 70%, while retuned Boat A returned to Point Q as usual. Both the boats returned back to point P at the same time. Then what is the speed of Boat A?
  • a)
    20 Km/hr
  • b)
    30 Km/hr
  • c)
    40 Km/hr
  • d)
    50 Km/hr
  • e)
    Cannot be determined
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Speeds of Boat A and B in still water are in the ratio of 3:2 Rate of ...
Answer – 2. 30 Km/hr Explanation : S 1 /S 2 = 3/2
R = 10
then (1/3x+10)+ (1/3x-10) = (1/2x+10) + (1/3.4x-10) x = 10 Speed of boat A =3*10 = 30
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Most Upvoted Answer
Speeds of Boat A and B in still water are in the ratio of 3:2 Rate of ...
Given information:
- Ratio of speeds of boats A and B in still water = 3:2
- Rate of current = 10 km/hr
- Boat B increases speed by 70% on return journey

To find:
- Speed of Boat A

Approach:
1. Use the concept of relative speed to find the speed of each boat with respect to the water.
2. Use the given ratio of speeds of the two boats to set up equations for their speeds in still water.
3. Use the time taken by each boat to travel from P to Q and back to P to set up equations for the distances traveled.
4. Solve the equations to find the speed of Boat A.

Calculation:
Let the speeds of boats A and B in still water be 3x and 2x, respectively.
Speed of Boat A with respect to the water = Speed of Boat A in still water + Rate of current = 3x + 10
Speed of Boat B with respect to the water = Speed of Boat B in still water + Rate of current = 2x + 10

Let the distance from P to Q be d km.
Time taken by Boat A to travel from P to Q = d / (3x + 10)
Time taken by Boat B to travel from P to Q = d / (2x + 10)
Time taken by Boat B to travel from Q to P with increased speed = d / (2.7x + 10)

Since both boats return to P at the same time, we can set up the following equations:

Time taken by Boat A to travel from Q to P = Time taken by Boat B to travel from Q to P with increased speed
d / (3x - 10) = d / (2.7x + 10)

Solving for x, we get x = 10.
Therefore, the speed of Boat A in still water = 3x = 30 km/hr.

Answer: The speed of Boat A is 30 km/hr, option (b) is correct.
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Speeds of Boat A and B in still water are in the ratio of 3:2 Rate of current is 10 Km/hr. Both Boats started from Point P to point Q downstream at the same time. After Boat B reaching Point Q, in return journey, it is powered by engine due to which the speed of the boat in still water is increased by 70%, while retuned Boat A returned to Point Q as usual. Both the boats returned back to point P at the same time. Then what is the speed of Boat A?a)20 Km/hrb)30 Km/hrc)40 Km/hrd)50 Km/hre)Cannot be determinedCorrect answer is option 'B'. Can you explain this answer?
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