Pipes and cisterns-Hard test 2

Answer: Option C

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=$=\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{y}}=\frac{\mathrm{x}-\mathrm{y}}{\mathrm{xy}}$

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

Hence option C is correct.

Answer: Option A

Explanation

In one minute, 29 taps fill $\frac{1}{60}$^th part of water tank

similarly, in one minute, 1 tap fills $\frac{1}{60\times 29}=\frac{1}{1740}$^th part of water tank

Let the number of taps used to fill the tank in 5 min be x.

So, $\mathrm{x} \times \frac{1}{1740}=\frac{1}{5}$

upon cross multiplication,

$\mathrm{x}=\frac{1740}{5}=348 \mathrm{taps}$

Therefore, 348 taps can fill the tank in 5 minutes

Hence option A is correct.

Answer: Option A

Explanation

In one minute, 29 taps fill $\frac{1}{60}$^th part of water tank

similarly, in one minute, 1 tap fills $\frac{1}{60\times 29}=\frac{1}{1740}$^th part of water tank

Let the number of taps used to fill the tank in 5 min be x.

So, $\mathrm{x} \times \frac{1}{1740}=\frac{1}{5}$

upon cross multiplication,

$\mathrm{x}=\frac{1740}{5}=348 \mathrm{taps}$

Therefore, 348 taps can fill the tank in 5 minutes

Hence option A 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, $\frac{120\times 80}{120-80}=\frac{9600}{40}=240 \min$

Hence, full tank will be filled in 240 min when both the pipes are opened simultaneously.

But the question is to calculate the time taken to fill $\frac{3}{4}$^th tank.

So, $240\times \frac{3}{4}=180 \min$

Hence option B is correct.

Answer: Option B

Explanation

6According 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,

 $$\begin{gathered} \frac{30\times 40\times 50}{(40\times 50)+(30\times 50)-(30\times 40)} \\ =\frac{60,000}{2000+1500-1200}=\frac{60,000}{2300} \mathrm{or} 26\frac{2}{3}\mathrm{hours} \end{gathered}$$

Hence, the tank will be filled in $26\frac{2}{3}$ hours when all the pipes are opened simultaneously.

Hence option B is correct.