Pipes and cisterns-Hard test 1

Answer: Option D

Explanation

1 pipe can completely fill the tank in 6 hours 

therefore, $\frac{6 \mathrm{parts}}{6 \mathrm{hours}}=\frac{1 \mathrm{part}}{1 \mathrm{hour}}$

After first pipe fills 3 parts (half tank), 3 more pipes are added.

Now altogether four pipes are filling the remaining 3 parts (half tank)

3 parts of the tank can be filled in 3 hours (180 min) by each of the pipes

1 pipe -- 3 parts -- 180 min

4 pipes -- 3 parts -- X

$$\begin{gathered} =\frac{1\times 3\times 180}{4\times 3\times \left(\mathrm{x}\right)} \\ =\frac{180}{4\mathrm{x}} \\ \therefore \mathrm{x}=\frac{180}{4}=45 \min \end{gathered}$$

So it takes 45 minutes to fill the remaining tank.

Hence total time taken by all the four pipes to fill the tank is,

3 hours by first pipe + 45 min by all four pipes together

Total 3 hours 45 minutes.

Hence option D is correct.

Answer: Option D

Explanation

The capacity of the tank can be obtained by finding LCM of 6, 2 and 7 which is 42.

Total parts of the tank is 42

A, B, C together can fill the tank in 6 hours

$\frac{42 \mathrm{parts}}{6 \mathrm{hours}}=\frac{7 \mathrm{parts}}{1 \mathrm{hour}}$

A,B and C can fill 7 parts in 1 hour

From the question A, B and C together are filling only for 2 hours

$=\frac{14 \mathrm{parts}}{2 \mathrm{hour}}$

Together they complete 14 parts in 2 hours

And the remaining 28 part will be filled by A and B in 7 hours

$\frac{28 \mathrm{parts}}{7 \mathrm{hours}}=\frac{4 \mathrm{parts}}{1 \mathrm{hour}}$

A and B together can fill 4 parts in 1 hour

From this we could find C’s filling parts per hour
(A+B+C)-( A+B)=C

7-4= 3 parts

Total 42 parts of the tank will be filled by C alone

$\frac{42}{3}=14 \mathrm{hours}$

Hence option D is correct.

Answer: Option C

Explanation

According to the question,

Pipe A fills the tank in 2 hours (in 1 hour it fills $\frac{1}{2}$ part of the tank)

Pipe B fills the tank in 6 hours (in 1 hour it fills $\frac{1}{6}$ part of the tank)

As pipe A was opened at 10:00 am and it worked alone till 11:00 am. In 1 hour it will fill $\frac{1}{2}$ part of the tank.

the remaining $\frac{1}{2}$ part of the tank will be filled by both Pipe A and B. 

According to the formula, if pipe A can comletely fill a tank in x hours and pipe B can comletely fill a tank in y hours, then the time taken by both pipes together to fill the tank is $\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$ hours.

So, $\frac{2\times 6}{2+6}=\frac{12}{8}=\frac{3}{2} \mathrm{part}$ (both the pipes together will fill $\frac{3}{2}$ part of the tank)

So, Time taken by both pipes to fill $\frac{3}{2}$ part of the tank = 60 minutes
Time taken to fill $\frac{1}{2}$ part = $\frac{60\times 3}{2}\times \frac{1}{2}$ = 45 minutes

Therefore, the tank will be filled at 11:45 AM

Hence option C is correct.

Answer: Option B

Explanation

Let us assume the third pipe empties the cistern in x min.

Part of the cistern filled in 1 min when all the three pipes are opened simultaneously

$=\frac{1}{60}+\frac{1}{75}-\frac{1}{\mathrm{x}}$

According to the question,

$$\begin{gathered} \frac{1}{50}=\frac{1}{60}+\frac{1}{75}-\frac{1}{\mathrm{x}} \\ \Rightarrow \frac{1}{\mathrm{x}}=\frac{5+4-6}{300}=\frac{3}{300} \\ \Rightarrow \mathrm{x}=\frac{300}{3}=100 \mathrm{minutes} \end{gathered}$$

Hence option B is correct. 

Answer: Option D

Explanation

According to the question,

part of the tank filled by pump in 1 hour=$\frac{1}{2}$

Because of the leak, part of the tank emptied by pump in 1 hour=$\frac{1}{{\displaystyle 2\frac{1}{3}}}=\frac{1}{{\displaystyle \frac{7}{3}}}=\frac{3}{7}$

part of the tank emptied by the leak in 1 hour$=\frac{1}{2}-\frac{3}{7}=\frac{7-6}{14}=\frac{1}{14}$

The tank will be emptied in 14 hours.

Hence option D is correct.