Question

There are two inlets pipes A and B .Pipe A alone can fill the tank in 6 hours efficiency of B is 25% more than that of pipe A while the pipe C alone can empty the full tank in 12 hours.If all the pipes opened together then in how many hours will be 83.33% of the tank be filled?. ​

Answer 1

Answer:

The time taken to fill the 83.33% of tank when all pipe are open is 2 hours 16 minutes .

Step-by-step explanation:

Given as :

Pipe A can fill tank in 6 hours

So, The time taken by pipe A to fill tank = $\dfrac{1}{6}$

pipe B is 25 % more efficient than pipe A

So, The time taken by pipe B to fill the tank = 6 - 25% of 6

i.e The time taken by pipe B to fill the tank = 6 - $\dfrac{3}{2}$

or, The time taken by pipe B to fill the tank = 4.5 hours

So, The time taken by pipe B to fill tank = $\dfrac{1}{4.5}$

Again

The time taken by pipe C to empty the tank = 12 hours

So, The time taken by pipe C to empty the tank = $\dfrac{1}{12}$

Now, if All the pipes opened together than the full will full in x hours

or, Time taken to fill the full tank = $\dfrac{1}{6}$ +  $\dfrac{1}{4.5}$ - $\dfrac{1}{12}$

Or, $\dfrac{1}{x}$ = $\dfrac{1}{6}$ +  $\dfrac{1}{4.5}$ - $\dfrac{1}{12}$

or,  $\dfrac{1}{x}$ = $\dfrac{1}{6}$ + $\dfrac{2}{9}$ - $\dfrac{1}{12}$

Or,   $\dfrac{1}{x}$ = $\dfrac{6+8-3}{36}$

Or,   $\dfrac{1}{x}$ = $\dfrac{11}{36}$

So, The time taken to fill the full  tank when all pipe are open = x = $\dfrac{36}{11}$ hours

Or, The time taken to fill the full tank when all pipe are open = 3 $\dfrac{3}{11}$ hours

Now,

Let The time taken to fill 83.33 % of time = x hours

So, x = $\frac{3.27\times 83.33}{100}$

∴    x = 2.27 hours

Hence, The time taken to fill the 83.33% of tank when all pipe are open = 2 hours 16 minutes . Answer