There are t taps numbered 1,2 and so on till t, each of which can fill a cistern. the rate of filling of the nth ta is such that it is equal to twice that of all the taps from 1 to (n-1) put together. If the 18th tap can fill the cistern in 2 minutes, then find the time in which the 15th tap alone can fill the empty cistern.

1) 27
2) 34
3) 54
4) 72

• Let 1st tap can fill the cistern in T mins.
  2nd tap: T/2 mins
  3rd tap: half the time taken by 1st and 2nd tap combined. : (T/3)/2 = T/6
  4th tap: half the time taken by 1st, 2nd and 3r combined. : (T/9)/2 = T/18
  
  Similarly,
  5th tap : (T/27)/2 = T/54
  ..
  nth tap: T/(3^(n-2)*2)
  
  Now we have,
  18th tap: 2 mins
  => T/(3^(18-2)*2) = 2
  Hence, T = 2^2 * 3^16
  
  Now, for 15th tap,
  Time taken = T/(3^(n-2)*2) = (2^2 * 3^16) / (3^(15-2)*2) = 2 * 3^3 = 54
  Hence time taken by 15th tap is 54 minutes.
• 9 years agoHelpfull: Yes(2) No(2)