Question

Two pipes can fill a tank with water in 15 and 12 hours respectively and the third pipe can empty it in 4 hours. If the pipes are opened in order at 8, 9 and 11 A.M. respectively, the tank will be emptied at 

(a) 11 : 40 a.m. 

(b) 12 : 40 p.m. 

(c) 1 : 40 p.m. 

(d) 2 : 40 p.m

Answer 1

(d) 2 : 40 p.m.

Let the tank be emptied in x hrs after 8 A.M.

\(\therefore\) Tank filled by first pipe in x hours = \(\frac{X}{15}\)

Tank filled by second pipe in (x – 1) hours = \(\frac{(X-1)}{12}\)

Tank emptied by third pipe in (x – 3) hours = \(\frac{(X-3)}{4}\)

\(\therefore\) \(\frac{X}{15}\) + \(\frac{(X-1)}{12}\)- \(\frac{(X-3)}{4}\) = 0

\(\Rightarrow\) \(\frac{4X+5(X-1)-15(X-3)}{60}\) = 0

\(\Rightarrow\) 4x + 5x – 5 – 15x + 45 = 0 \(\Rightarrow\) – 6x + 40 = 0

\(\Rightarrow\) x = \(\frac{40}{6}\)= \(\frac{20}{3}\) = \(6\frac{2}{3}\) hrs. = 6 hrs. 40 min.

\(\therefore\) The tank will be emptied at 2:40 p.m.