Question

Two pipes are used to fill the astern completely. The second pipe opened one hour after the
first pipe. Three hours after the first plpe has opened, there is still (9/20) of the cistern is to
be filled. When the cistern is completely filled, it was found that each pipe has filled half of
the astern. How many hours would it take each one to completely filled the cistem
individually?​

Answer 1

Given:

The 2nd pipe is opened 1 hr after the 1st pipe to fill the cistern

3 hours after the first pipe is opened, there is still (9/20) of the cistern to

be filled.

When the cistern is completely filled, it was found that each pipe has filled half of  the cistern

To find:

How many hours would it take each one to completely filled the cistern

individually?

Solution:

Let's assume,

A be the first pipe & B be the second pipe

"A" hrs → represent the time taken by pipe A to fill the cistern

"B" hrs → represent the time taken by pipe B to fill the cistern

So,

In 1 hr, the part of cistern filled by pipe A = $\frac{1}{A}$

In 1 hr, the part of cistern filled by pipe B = $\frac{1}{B}$

Let the cistern is completely filled in "x" hrs.

Since the pipe B is opened 1 hr after the pipe A so if, pipe A takes "x" hrs then pipe B takes "(x - 1)" hrs.

Also, it is given that when the cistern is completely filled, each of the pipe

has filled half of the cistern. So, we have

$\frac{x}{A} = \frac{(x-1)}{B} = \frac{1}{2}$

∴ A = 2x and B = 2(x - 1) ...... (i)

Now, after 3 hrs, the part of the cistern filled is given by the following equation,

$\frac{3}{A} \:+\:\frac{2}{B} = 1\: -\: \frac{9}{20}$

⇒ $\frac{3}{A} \:+\:\frac{2}{B} = \: \frac{20 - 9}{20}$

⇒ $\frac{3}{A} \:+\:\frac{2}{B} = \: \frac{11}{20}$

substituting from eq. (i)

⇒ $\frac{3}{2x} \:+\:\frac{2}{2(x - 1)} = \: \frac{11}{20}$

⇒ $\frac{3}{x} \:+\:\frac{2}{(x - 1)} = \: \frac{11}{10}$

⇒ $\frac{3(x - 1)+2x}{x(x - 1)}\: = \: \frac{11}{10}$

⇒ $10[3x - 3 + 2x]\: = \: 11[x^2 - x]$

⇒ $30x - 30 +20x\: = \: 11x^2 - 11x$

⇒ $11x^2 - 61x + 30 = 0$

⇒ $11x^2 -55x - 6x + 30 = 0$

⇒ $11x(x-5)-6(x-5) = 0$

⇒ $(x - 5)(11x -6) = 0$

⇒ $x = 5 \:or\: \frac{6}{11}$

Neglecting fractional value of x and substituting x = 5 in eq. (i), we get

A = 2x = 2 × 5 = 10 hrs

and

B = 2(x - 1) = 2(5 - 1) = 2 × 4 = 8 hrs

Thus, the first pipe would take 10 hrs and the second pipe would take 8 hrs, individually to fill the cistern.

---------------------------------------------------------------------------------------------

Also View:  

Two pipes a and b can fill an empty cistern in 18 and 27 hours respectively. Pipe c can drain the entire cistern in 45 hours when no other pipe is in operation. Initially, when the cistern was empty pipe a and pipe c were turned on. After a few hours pipe a was turned off and pipe b was turned on instantly. In all, it took 55 hours to fill the cistern. For how many hours was pipe b turned on?  

https://brainly.in/question/8502414

 

Pipes a and b can fill a tank in 5 and 6 hours respectively. Pipe c can empty it in 12 hours. The tank is half fill. All 3 pipes are in operation simultaneously. After how much time will the tank be full ?

https://brainly.in/question/5807736

6 pipes are required to fill a tank in 1 hour 20 minutes how long will it take if only 5 types of the same type are used.  

https://brainly.in/question/3150761  

Answer 2

Given

The second pipe opened one hour after than the first pipe. After three hours after the first pipe has opened there is still 9/20 of the cistern that is to be filled. When the cistern is completely filled it is found that that both pip has filled the cistern equally.

Find

How any hours would it take each one to completely filled the cistern individually.

Step by step explanation

Let assume that A represents the time taken by the pipe A and B represents the time taken by the pipe B

So in 1 hour the part of cistern filled by pipe A is 1/A

In 1 hour the part of cistern filled by pipe B is 1/B

Let the cistern is filled in x hours

Since the pipe B is opened after 1 hour of opening of pipe A so if pipe A takes x hours then pipe B takes x-1 hours.

It is given that when the cistern is completely filled each ogf the pipe has filled it equally. So we have  

x/A = (x-1)/B =1/2

A=2x   and B = 2(x-1) ....1

After 3 hours the part of the cistern filled is given by following equation

3/A +2/B = 1-9/20

3/A +2/B=(20-9)/20

3/A +2/B=11/20

10(3X-3+2X]=11[$x^{2}$-X]

30X-30+20X=11$x^{2}$-x

11$x^{2}$-55x-6x+30=0

11x(x-5)-6(x-5)=0

x=5 or 6/11

we ignore the fraction term and use x=5

Let take x=5

A=2x=2*5=10

B=2(x-1)=2*4=8

Thus the first pipe would take 10 hours while the other take 8 hour to fill the cistern completely itself.

#SPJ2