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anonymous

  • one year ago

Two towns sealand and beachland are on the banks of a river . Amit takes 13 hours to row and back . Sachin , who rows at twice Amit's speed, covers same distance in 6 hours. Find the ratio of Amit's speed to that of flow of river.

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  1. IrishBoy123
    • one year ago
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    |dw:1443009359676:dw|

  2. IrishBoy123
    • one year ago
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    |dw:1443009923304:dw|

  3. phi
    • one year ago
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    I don't exactly understand Irish's hint. But I would his basic idea, but with distances. Let's say Amit rows at speed v for 13 hours. using rate * time = distance he rowed a distance of 13v (I guess we don't care about the units) I assume he is trying to go straight across to the other side, but the current would carry him downstream, so he rows at an angle, as shown in the picture.

  4. phi
    • one year ago
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    in 13 hours, the current moves him a distance of 13u (u is how fast the current goes) and we will call the distance straight across the river "w" now we can use pythagoras |dw:1443011236342:dw|

  5. phi
    • one year ago
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    Sachin, rows for 6 hours, but at a rate of 2v, using rate*time = distance 2v * 6 = 12 v he rows a distance of 12 v the current carries him a distance of 6u (6 hours at a speed of u) |dw:1443011437401:dw|

  6. phi
    • one year ago
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    we have two equations \[169 v^2 = 169u^2 + w^2\\ 144 v^2 = 36u^2 + w^2 \] we can subtract the two equations (this will get rid of the w^2) we get \[ 25v^2 = 133 u^2 \]

  7. phi
    • one year ago
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    Amit's speed is v and the river's speed is u we have \[ \frac{v^2}{u^2}= \frac{133}{25} \] take the square root of both sides \[ \frac{v}{u}= \frac{\sqrt{133}}{5} \] is the ratio we want

  8. IrishBoy123
    • one year ago
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    straight from the drawing \[(2w)^2 = (13v)^2 - (13u)^2 = (6\times 2v)^2 - 6u^2\] \[25v^2 = 133u^2\] one more line and you're sone

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