anonymous
  • anonymous
There are two taps in the bottom of a pool. One tap empties the pool with a speed 2 times more than that of the other tap. The two taps empty the pool in 20 hours. In how many hours the tap with slow speed alone empties the pool?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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imqwerty
  • imqwerty
let the faster tap empty the pool with a speed of 2L litres/hour the slower tap will empty it with L litres/hour total time to empty the pool=20hrs water taken out by faster tap=2L x 20=40L water taken out by slower tap=Lx 20=20L total volume of pool=40L+20L=60L so now we have total volume =60L and slower tap empties pool @L litres/hour so total time taken by small tap to empty pool=60/L=60hrs
anonymous
  • anonymous
no the answer is 40 i thought if the speed of slow tap is V, the speed of faster tap should be 3V cause it says two times more than
imqwerty
  • imqwerty
ok lets go reverse so it says that the small tap will take 40hrs let the speed of small tap be V so total volume=40V let the speed of fast tap be Y so 40V=20V+20Y 20V=20Y V=Y so this means both taps have same speed but this is not possible :D

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