Question

It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled.How long would it take for each pipe to fill the pool separately?

Answer 1

Hey mate...

here is your answer

let the time taken by the first pipe be x hours and the time taken by the second pipe be y hours.

in 1 hour the first pipe can fill it = 1/x

in 1 hour the second pipe can fill it = 1/y

1/x +1/y = 1/12 --------------(1)

4/x = 9/9 =1/2---------------(2)

Consider 1/x be a and 1/y be b.

a + b = 1/12 -------------(3)

4a + 9b = 1/2 ----------------(4)

multiply 3 by equ (4) and then subtract equ(5) from equ(4).

4a + 4b =1/3 --------------(5)

we get, b = 1/30

y = 30.

sustituting the value of y in equ(3) we get

a + 1/30 = 1/12

x = 20.

Hence the first pipe would take 20hours and the second pipe would take 30 hours.

hope it helps you!!!!

please mark my answers as the brainliest answer

Answer 2

It takes 12 hrs to fill a swimming pool with 2 pipes .

Let the first pipe fill the pool in x hours .

Let the second pipe fill the pool in y hours .

∴ 1/x will be the part of the pool filled in 1 hr by the second pipe .

∴ 1/y will be the part of the pool filled in 1 hr by the second pipe .

1/x + 1/y = 1/12 .........( 1 )

Now if the first pipe is open for 4 hrs and the second pipe for 9 hrs , then the pool is half filled .

4/x + 9/y = 1/2 .......( 2 )

Multiply eq ( 1 ) by 4 :

4/x + 4/y = 4/12

⇒ 4/x + 4/y = 1/3 ........( 2 )

Subtract eq ( 2 ) from eq ( 1 ) :

⇒ 4/y - 9/y = 1/3 - 1/2

⇒ ( 4 - 9 )/y = ( 2 - 3 )/6

⇒ - 5/y = - 1/6

⇒ y = - 5 × - 6

⇒ y = 30

The second pipe fills it in 30 hrs .

1/x + 1/30 = 1/12

⇒ 1/x = 1/12 - 1/30

⇒ 1/x = ( 10 - 4 )/120

⇒ 1/x = 6/120

⇒ 1/x = 1/20

⇒ x = 20

The time taken by the first pipe is 20 hrs .