Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to $$\frac{7}{8}$$ of full flow in pipe A and $$\frac{5}{6}$$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.
A. 8 min
B. 3 min
C. 5.6 min
D. 4.5 min
E. None of these
Answer: Option D
Solution (By Examveda Team)
Let the obstruction remain for x min. Hence, Part of cistern filled in X min + part of cistern filled in 3 min = full cistern $$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$$ $$ + $$ $$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$$ = 1 $$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$$ Thus, X = 4.5 min.This Question Belongs to Arithmetic Ability >> Time And Work
Ans will be 7.5 that makes last option i.e nota correct
Plz correct the equation 12X /96 +7/16 =1
(12X/96 + 7/6) =1
Then how to come X=4.5 min
Above eqn slove X value in negative