Question

A cistern can be filled by two taps in 20 min and 30 min respectively and be emptied by a third in 48 min. They are all turned on at once. When will the cistern be half-full? ​

Answer 1

Let the first pipe be closed after x min.

Part filled by two pipes in x minutes = x(120+130)=5x60=x12

Part filled by the second pipe in (18−x) = 18−x30

x12+18−x30=1

10x+72−4x=120

6x = 120 - 72 = 48

x = 8 minutes

Answer 2

Given:

Time taken to fill 3 taps = 20 min ,30 mins 48 mins

To Find:

When the cistern be half filled.

Explanation:

T₁ = 20

T₂ = 30

T₃ = 48

Now be the first pipe closed after x min.

Part filled by two pipes in x minutes will be

= x(20+30)

=60/5x

=12x

Part filled by the second pipe will be

(18−x) = 18−x30

12x+18−30x=1

10x+72−4x=120

6x = 120 - 72

6x= 48

x = 8 minutes

Answer is 8 minutes