Latest Accenture Aptitude Question SOLUTION: 23. The horizontal distance of a kite from the boy flying it is 30 m and 50 m of cord is out from the roll. If the wind moves the kite horizontally at the rate of 5 km per hour di

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23. The horizontal distance of a kite from the boy flying it is 30 m and 50 m of cord is out from the roll. If the wind moves the kite horizontally at the rate of 5 km per hour directly away from the boy, how fast is the cord being released?

Asked In Accenture X22 (12 years ago)

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selventhiran irtt  (12 years ago)

If the horizontal distance is 30m and the string is 50m, the vertical distance is 40m (3,4,5 triangle).
Kite is moving away at 5000/3600 horizontally (in m/s)

Ratio of string extension to horizontal distance is 5 to 3
So 5000/3600 = 1.39 m/s
String released at
(5/3)x 1.39 = 2.32 m/s (Roughly 2 1/3 m/s)

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Payel Das  (7 years ago)

Initial position of kite:
horizontal: 30
vertical: 40
hypotenuse: 50

After 1hr, wind made the kite move 5km directly away from the boy, so, position of kite after 1 second:
horizontal: 30 + 1.39=31.39(because of wind, 5kmph = 1.39m/s)
vertical: 40
hypotenuse= Sqrt( 31.39*31.39+40*40) = 50.846

now in 1sec, cord has moved 0.846m => 3kmph
So, cord is released at the rate of 3kmph.
Hence 1st option

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Khilesh  (7 years ago)

Parallel way is always the shortest way.. Keeping that in mind its clear that the speed with which cord is being released is more then the horizontal speed.. For travel of every 3m horizontally, cord released is 5m..so solving the equation we get 8.33km/hr as an ans.. Ny ways M unable to see options here so nt sure..

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Khilesh  (7 years ago)

Parallel way is always the shortest way.. Keeping that in mind its clear that the speed with which cord is being released is more then the horizontal speed.. For travel of every 3m horizontally, cord released is 5m..so solving the equation we get 8.33km/hr as an ans.. Ny ways M unable to see options here so nt sure..

Is this solution Helpfull? Yes | No   

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