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=12hr

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=18hr

so t_s =18 + 18 = 36hr

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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