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A boat covers 25 km upstream and 44 km downstream in t hours. Also, it covers 15 km upstream and 22 km downstream in 5 hours. Find the speed of the boat in still water and that of the stream.​

Sagot :

9514 1404 393

Answer:

  boat speed: (7t -95)/(6t² -110t +500)

  stream speed: (37t -345)/(6t² -110t +500)

Step-by-step explanation:

The relation between time, speed, and distance is ...

  time = distance/speed

The boat speed (b) is added to the stream speed (s) going downstream. The stream speed is subtracted going upstream. The given relations tell us ...

  25/(b-s) +44/(b+s) = t

  15/(b-s) +22/(b+s) = 5

This gives us 2 equations in 3 unknowns, so we can only provide a solution in terms of one of the unknowns. Since we are asked for the speeds, they will be provided in terms of t.

Subtracting the first equation from 2 times the second gives ...

  2(15/(b-s) +22/(b+s)) -(25/(b-s) +44/(b+s)) = 2(5) -(t)

  5/(b-s) = 10 -t

  5/(10 -t) = b -s

Substituting for b-s in the second equation gives ...

  15/(5/(10-t)) +22/(b+s) = 5

  30 -3t +22/(b+s) = 5

  22/(b+s) = 3t -25

  22/(3t -25) = b+s

Adding the equations for the sum and difference of speeds, we get ...

  (b-s) +(b+s) = 5/(10-t) +22/(3t -25)

  2b = (5(3t -25) +22(10 -t))/((10-t)(3t-25)) = (95 -7t)/((10-t)(3t-25))

  b = (7t -95)/(6t² -110t +500)

Subtracting the equations for sum and difference of speeds, we get ...

  (b+s) -(b-s) = 22/(3t -25) -5/(10 -t)

  2s = (22(10-t) -5(3t -25))/((10-t)(3t-25))

 s = (37t -345)/(6t² -110t +500)

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The speed of the boat in still water is (7t -95)/(6t² -110t +500) km/h.

The speed of the stream is (37t -345)/(6t² -110t +500) km/h.

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Note that these solutions only make sense for values of t between 25/3 and 345/37, approximately 8.33 < t < 9.32. For t=9 hours, the boat speed is 8 km/h and the stream speed is 3 km/h.