Two drainpipes working together can drain a pool in 12 hours. Working​ alone, the smaller pipe would take 18 hours longer than the larger pipe to drain the pool. How long would it take the smaller pipe alone to drain the​ pool?

2 Answers
PJ
Jul 21, 2018

36hr

Explanation:

Time of both working together #t_b=12#hr

Time of small pipe # t_s# = (time of large pipe) # + 18 #

Let time of large pipe #= x#

# t_s=x + 18 #

For pipes in parallel #1/T=1/t_1+1/t_2#

#1/12= 1/ (x +18)+ 1/x#

#x/12= x/ (x +18) + x/ x#

#(x*(x +18))/12= x/ 1 + (x +18)#

#x^2+18x = 12*x + 12*x +216#

#x^2-6x - 216= 0 #

#(x+12)(x-18)=0#

#x = -12 or 18#
but time can't be -ve so

#x=18#hr

so # t_s =18 + 18 = 36#hr

Harish Chandra Rajpoot
Jul 21, 2018

#36\ \text{hrs#

Explanation:

Let #x\ \text{hrs}# be the time taken by the smaller pipe alone to drain a pool of volume say #V# . Then the time taken by the larger pipe alone to drain the same pool of volume #V# will be #(x-18)\ text{hrs}#

Now, draining rate of smaller pipe #=V/x#

Similarly, draining rate of larger pipe #=V/{x-18}#

Given that both the pipes working together can drain the same pool of volume #V# in #12# hours hence we have

#\text{total volume drained by both pipes together in 12 hrs}#

#= \text{volume of pool}#

#\therefore 12(V/x+V/{x-18})=V#

#12({x-18+x}/{x(x-18)})=1#

#12(2x-18)=x(x-18)#

#x^2-42x+216=0#

#x^2-36x-6x+216=0#

#x(x-36)-6(x-36)=0#

#(x-36)(x-6)=0#

#x-36=0, x-6=0#

#x=36, 6#

But #x>18# hence

#x=36#

hence, the time taken by the smaller pipe alone to drain the pool is #36\ \text{hrs#

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