Question

Two inlet pipes a and b can fill an empty cistern in 22 and 33 hours respective.they are switched on together but pipe b had to be closed 7 hours before the cistern was full.how many hours in all did it take the two pipes to fill the cistern

Answer 1

Answer:

Total hours taken to fill the cistern is 15 hours

Step-by-step explanation:

Inlet pipe $a$ can fill empty cistern in 22 hours

Thus, pipe $a$ can empty parts of cistern 1 hour $=\frac{1}{22}$

Inlet pipe $b$ can fill empty cistern in 33 hours

Thus, pipe $b$ can empty parts of cistern 1 hour $=\frac{1}{33}$

Total parts of cistern is LCM of 22 and 33 $=66\hspace{0.1cm}parts$

Pipe $a$ can fill parts of cistern in 1 hour $=\frac{66}{22}=3\hspace{0.1cm}parts$

Pipe $b$ can fill parts of cistern in 1 hour $=\frac{66}{33}=2\hspace{0.1cm}parts$

Given that pipe $b$ is closed 7 hours before the cistern was full

This means, for last 7 hours pipe a only fills the cistern.

Parts of cistern filled by pipe $a$ in 7 hours $=7 \times 3 = 21\hspace{0.1cm}parts$

Parts of cistern remaining $=66 - 21 = 45\hspace{0.1cm}parts$

45 parts of cistern are filled by pipe $a$ and $b$ together

Pipe $(\,a + b)\,$ can fill parts of cistern in 1 hour $= 2 + 3 = 5\hspace{0.1cm}parts$

Thus, 5 parts of cistern filled by pipe $(\,a + b)\,$ in 1 hour

45 parts of cistern filled by pipe $(\,a + b)\,$ in $\frac{45}{5} = 9$ hours

Total hours taken to fill the cistern $=7 + 9 = 15\hspace{0.1cm}hours$