Question

Four pipes p,q, r and s can fill a cistern in 20,25, 40 and 50 hours respectively.The first pipe p was opened at 6:00 am, q at 8:00 am, r at 9:00 am and s at 10:00 am. When will the cistern be full?

Answer 1

Answer:

The cistern is full at 3:08 PM

Step-by-step explanation:

Given that four pipes p, q, r and s can fill cistern 20, 25, 40 and 50 hours respectively

Pipe p's 1 hour work $=\frac{1}{20}$

Pipe q's 1 hour work $=\frac{1}{25}$

Pipe r's 1 hour work $=\frac{1}{40}$

Pipe s's 1 hour work $=\frac{1}{50}$

Total work to fill a cistern is calculated as LCM of 20, 25, 40 and 50 $=200\hspace{0.1cm}parts$

Pipe p's can fill cistern in 1 hour $=\frac{200}{20} = 10\hspace{0.1cm}parts$

Pipe q's can fill cistern in 1 hour $=\frac{200}{25} = 8\hspace{0.1cm}parts$

Pipe r's can fill cistern in 1 hour $=\frac{200}{40} = 5\hspace{0.1cm}parts$

Pipe s's can fill cistern in 1 hour $=\frac{200}{50} = 4\hspace{0.1cm}parts$

Now, pipe p was opened at 6:00 am, q was opened at 8:00 am, r was opened at 9:00 am and s was opened at 10:00 am

After 10:00 am, all pipes are open to fill the cistern.

Now, calculate the parts of cistern filled before 10:00 am

Pipe p fills cistern for 4 hours $=4 \times 10 = 40\hspace{0.1cm}parts$

Pipe q fills cistern for 2 hours $=2 \times 8 = 16\hspace{0.1cm}parts$

Pipe r fills cistern for 1 hours $=1 \times 5 = 5\hspace{0.1cm}parts$

Parts of cistern filled at 10:00 am $=40 + 16 + 5 = 61\hspace{0.1cm}parts$

Parts of cistern not filled $=200 - 61 = 139\hspace{0.1cm}parts$

After 10:00 am, all four pipes are filling the cistern

(p + q + r + s)'s can fill cister in 1 hour $=10 + 8 + 5 + 4 = 27 \hspace{0.1cm}parts$

27 parts of cistern filled in 1 hour

139 parts of cistern filled in $\frac{139}{27} = 5\frac{4}{27}hours = 5 hours\hspace{0.1cm}and\hspace{0.1cm} (\,\frac{4}{27} \times 60 = 9 \hspace{0.1cm}mins\hspace{0.1cm}approximately)\,$

Thus, cistern is full at 3:08 PM