Pipes and cisterns-Hard test 2

Answer: Option B

Explanation

 Water filled by first pipe in 1 min $=\frac{1}{12}$

Water filled by second pipe in 1 min $=\frac{1}{15}$

Water filled by both the pipes in 5 min $=5\left(\frac{1}{12}+\frac{1}{15}\right)=\frac{3}{4}$

Water filled when all the three pipes work together in 1 min $=\left(\frac{1}{12}+\frac{1}{15}\right)-\frac{1}{6}=-\frac{1}{60}$

Here negative sign implies that $\frac{1}{60}$^th part is emptied in 1minute

 ∴ $\frac{3}{4}$^th part is emptied in $60\times \frac{3}{4}=45 \mathrm{minutes}$

Hence option B is correct. 

Answer: Option B

Explanation

As per the question,

The two filler piper can fill the cistern in $\frac{10\times 15}{10+15}=\frac{150}{25}=6 \mathrm{minutes}$

A leak empties the tank in 18 minutes

So, 

$\frac{18\times 6}{18-6}=\frac{108}{12}=9 \mathrm{minutes}$

Hence in 9 Minutes the waste pipe can empty the full cistern.

Hence option B is correct.

Answer: Option A

Explanation

As per the question,

In 5 minutes both the pipes together can fill,

$5\times \left(\frac{1}{20}+\frac{1}{25}\right)=\frac{5}{20}+\frac{5}{25}=\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$^th pat of tank

Part filled by first pipe,

$1-\frac{9}{20}=\frac{11}{20}$^th part

So, 

$\frac{18\times 6}{18-6}=\frac{108}{12}=9 \mathrm{minutes}$

Hence in 9 Minutes the waste pipe can empty the full cistern.

Hence option A is correct.

Answer: Option D

Explanation

Pipe A fills the tank normally in 2 hours
So, Pipe A can fill half of the tank in 1 hour
Let us consider, the leak takes x hours to empty the full tank when pipe A is shut.
So, the leak will empty $\frac{1}{\mathrm{x}}$of the tank in 1 hour.
The net amount of water that gets filled in the tank in 1 hour when pipe A and the leak are open$=\frac{1}{2}-\frac{1}{\mathrm{x}}$ of the tank---Eq 1

As per the question,
when the leak is open, time taken to fill the tank $=2\frac{1}{2}=\frac{5}{2} \mathrm{hours}$

Therefore, in 1 hour, $\frac{2}{5}$^th of the tank gets filled--Eq 2

Equating Eq1 and Eq2, we get

$$\begin{gathered} \Rightarrow \frac{1}{2}-\frac{1}{\mathrm{x}}=\frac{2}{5} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{1}{2}-\frac{2}{5} =\frac{1}{10} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{1}{10} \\ \Rightarrow \mathrm{x}=10 \mathrm{hours} \end{gathered}$$

Hence option D is correct. 

Answer: Option B

Explanation

Let the number of filling pipes be ‘x’

So, the number of draining pipes will be (8 – x)

As per the question,
Each of the filling pipes can fill the tank in 8 hours.
So, in 1 hour each of the filling pipes fill $\frac{1}{8}$^th of the tank

Hence, in 1 hour x filling pipes will fill $\frac{\mathrm{n}}{8}$^th of the tank

Similarly,
each of the waste pipes can drain the full tank in 6 hours.
So in 1 hour each of the waste pipes will drain $\frac{1}{6}$^th of the tank

Hence, in 1 hour (8 – x) waste pipes will drain $\frac{8-x}{6}$^th of the tank

If for 1 hour, all filling and draining pipes are opened $\frac{1}{6}$^th of the tank gets drained

(water filled in 1 hour – water drained 1 hour) = $\frac{1}{6}$^th of the tank

Therefore,

$\frac{x}{8}-\frac{8-x}{6}=-\frac{1}{6}$

(As the tank is getting drained we have negative sign)

Upon cross multiplication and solving the equations we get

$$\begin{gathered} \Rightarrow 14\mathrm{n}-64=-8 \\ \Rightarrow 14\mathrm{n}=64-8 \\ \Rightarrow 14\mathrm{n}=56 \\ \Rightarrow \mathrm{n}=4 \end{gathered}$$

 Hence option B is correct.