Question

two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes

Answer 1

Fraction of cistern filled by P in 1 min =

$\frac{1}{24}$

Fraction of cistern filled by Q in 1 min =

$\frac{1}{32}$

Fraction of cistern filled by P and Q opened together =

$\frac{1}{24} + \frac{1}{32} = \frac{4 + 3}{96} = \frac{ 7 }{96}$

Let the number of minutes both pipes are opened = x

Therefore, number of minutes only Q is opened = 16 - x

So,

$\frac{7}{96} x + \frac{1}{32} (16 - x) = \frac{7x - 3x}{96} + \frac{16}{32} = 1$

$\frac{x}{24} = \frac{1}{2}$

$x = 12$

Therefore,

The first pipe must be turned off after 12 minutes.

Answer 2

Answer:

Step-by-step explanation:

Fraction of cistern P = 1/24

Fraction of cistern Q = 1/32

Lcm..... = 4+3/96 = 7/96

Let it be 7x/96

7x/96+1/32×(16-x) = 1

Therefore no. Of minute pipe opened = (16-x)

7x-3x/96+16/32= 1

4x/96+16/32 = 1

x/24+1/2= 1

x+12/24=1

x+12 = 24

x = 12

After 12 minute we turn off the first pipe