Pipes and cisterns-Hard test 1

Answer: Option D

Explanation

According to the question,

part of the tank filled by pipe x in1 hour=$\frac{1}{\mathrm{x}}$

part of the tank emptied by pipe y in1 hour=$\frac{1}{\mathrm{y}}$

part of the tank filled in1 hour when both pipes x and y are opened=$=\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{y}}=\frac{\mathrm{y}-\mathrm{x}}{\mathrm{xy}}$

Hence tank will be filled in $\frac{\mathrm{xy}}{\mathrm{y}-\mathrm{x}} \mathrm{hours}$

Hence option D is correct.

Answer: Option B

Explanation

According to the above given formula, If pipe A  and B fills a tank in x and y hours respectively, but a third pipe C empties the full tank in z hours then the net portion filled in 1 hour when all of them are opened together is $\frac{\mathrm{xyz}}{\mathrm{yz}+\mathrm{xz}-\mathrm{xy}}$ hours.

According to the question x+y+z=10

So,

 $$\begin{gathered} 10=\frac{30\times 40\times \mathrm{z}}{(30\times 40)+(40\times \mathrm{z})+(\mathrm{z}\times 30)} \\ 10=\frac{1200\mathrm{z}}{1200+30\mathrm{z}+40\mathrm{z}} \\ 10=\frac{1200\mathrm{z}}{1200+70\mathrm{z}} \end{gathered}$$

Upon cross multiplication,

$$\begin{gathered} 10\left(1200+70\mathrm{z}\right)=1200\mathrm{z} \\ 12000+700\mathrm{z}=1200\mathrm{z} \\ 12000=1200\mathrm{z}-700\mathrm{z} \\ 12000=500\mathrm{z} \\ \mathrm{z}=\frac{12000}{500} \\ \mathrm{z}=24 \mathrm{minutes} \end{gathered}$$

Hence, the tank will be emptied in 24 minutes.

Hence option B is correct.

Answer: Option B

Explanation

According to the formula, If pipe A  and B fills a tank in x and y hours respectively, but a third pipe C empties the full tank in z hours then the net portion filled in 1 hour when all of them are opened together is $\frac{\mathrm{xyz}}{\mathrm{yz}+\mathrm{xz}-\mathrm{xy}}$ hours.

So,

 $$\begin{gathered} \frac{3\times 4\times 5}{(4\times 5)+(3\times 5)-(3\times 4)} \\ =\frac{60}{20+15-12}=\frac{60}{23} \mathrm{or} 2\frac{14}{23}\mathrm{hours} \end{gathered}$$

Hence, the tank will be filled in $2\frac{14}{23}$ hours when all the pipes are opened simultaneously.

Hence option B is correct.

Answer: Option D

Explanation

According to the formula, if pipe can comletely fill a tank in x hours, then the portion filled in 1 hour is $\frac{1}{x}$

So,

Time taken by A and B together is 6 h $\mathrm{A}+\mathrm{B}=\frac{1}{6}$---Eq 1

Time taken by B and C together is 6 h $\mathrm{B}+\mathrm{C}=\frac{1}{10}$---Eq2

Time taken by C and A together is 6 h $\mathrm{C}+\mathrm{A}=\frac{1}{7{\displaystyle \frac{1}{2}}}=\frac{2}{15}$---Eq3

Adding up all the equations,

$$\begin{gathered} \mathrm{A}+\mathrm{B}+\mathrm{B}+\mathrm{C}+\mathrm{C}+\mathrm{A}=\frac{1}{6}+\frac{1}{10}+\frac{2}{15} \\ 2\mathrm{A}+2\mathrm{B}+2\mathrm{C}=\frac{5+3+4}{30} \\ 2\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{12}{30} \\ \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{12}{30\times 2}=\frac{1}{5} \end{gathered}$$

Therefore, pipes A, B and C together can fill the tank in 5 h

∴ $\frac{1}{10}$^th part of the tank will be filled by the pipe in 1 hour.

Hence option D is correct.

Answer: Option B

Explanation

According to the formula, if pipe A can comletely fill a tank in x hours and pipe B can empty the same tank in y hours, then the time taken to fill the tank when both the pipes are opened simultaneously is $\frac{\mathrm{xy}}{\mathrm{y}-\mathrm{x}}$ hours.

so, $\frac{12\times 20}{20-12}=\frac{280}{8}=30 \min$

Hence, full tank will be filled in 30 min when both the pipes are opened simultaneously.

But the question is to calculate the time taken to fill half tank.

So, $\frac{30}{2}=15 \min$

Hence option B is correct.