Pipes and cisterns-Hard test 2

Answer: Option B

Explanation

Tap M fills 5 buckets in 40 minutes ⇒ $\frac{40}{5}=8 \mathrm{minutes}$

Tap N fills 7 buckets in 49 minutes ⇒ $\frac{49}{7}=7 \mathrm{minutes}$

Tap O fills 1 bucket in 30 minutes ⇒ $\frac{30}{1}=30 \mathrm{minutes}$

In 3 hours,

Number of buckets filled by tap M ⇒ $\frac{180}{8}=22.5 \mathrm{buckets}$ 

Number of buckets filled by tap N ⇒ $\frac{180}{7}=25.7 \mathrm{buckets}$ 

Number of buckets filled by tap O ⇒ $\frac{180}{30}=6 \mathrm{buckets}$ 

Total number of buckets filled ⇒  (22.5 + 25.7 + 6) = 54.2 buckets

Total capacity of the tank = Total number of buckets filled X Capacity of each bucket

 ⇒  54.2 x 6 = 325.2 L

Hence option B is correct.

Answer: Option C

Explanation

Since the leak can empty the tank in 8 hours,
In 1 hour, part of the tank emptied $=\frac{1}{8}$

After opening the tap
In 1 hour, part of the tank emptied $=\frac{1}{12}$

Let us consider the tap fills the tank in x hours.
So, In 1 hour, part of the tank filled by the tap$=\frac{1}{x}$

According to the question,

$\Rightarrow \frac{1}{x}-\frac{1}{8}=\frac{1}{12}$

upon cross multiplication and solving we get x = 24

Since the tap admits 6 liters of water per hour, it will admit $6\times 24=144$liters of water

Hence option C is correct.

Answer: Option B

Explanation

As per the question,

Capacity of three small pumps = Capacity of two large pumps

Also, if we want to express the capacity of three small pumps + one large pump in terms of large pump, we should add one large pump on both sides of the above equation

If we add one large pump on both sides of the above equation, we get

Three small pumps + one large pump = Three large pumps

Thus, if all the four pumps are open together, they would fill the tank in $\frac{1}{3}$^rd of the time large pump would have taken alone.

Hence option B is correct.

Answer: Option D

Explanation

Let the time taken by B alone to fill the tank be x hours

Time taken by A to fill the tank is $\mathrm{x}+\frac{\left(\mathrm{x}\right)\times 80}{100}=\frac{9\mathrm{x}}{5}\mathrm{hours}$

According to the formula, time taken by both the filling pipes to fill the tank is $\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$

$$\begin{gathered} \Rightarrow 45=\frac{\left(\mathrm{x}\right)\times {\displaystyle \frac{9\mathrm{x}}{5}}}{\left(x\right)+{\displaystyle \frac{9x}{5}}} \\ \Rightarrow 45+\frac{14\mathrm{x}}{5}=\frac{9{\mathrm{x}}^2}{5} \\ \Rightarrow \mathrm{x}=\frac{45\times 14}{9}=70 \mathrm{hours} \end{gathered}$$

Hence option D is correct.

Answer: Option D

Explanation

Let the time taken by B alone to fill the tank be x hours

Time taken by A to fill the tank is $\mathrm{x}+\frac{\left(\mathrm{x}\right)\times 80}{100}=\frac{9\mathrm{x}}{5}\mathrm{hours}$

According to the formula, time taken by both the filling pipes to fill the tank is $\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$

$$\begin{gathered} \Rightarrow 45=\frac{\left(\mathrm{x}\right)\times {\displaystyle \frac{9\mathrm{x}}{5}}}{\left(x\right)+{\displaystyle \frac{9x}{5}}} \\ \Rightarrow 45+\frac{14\mathrm{x}}{5}=\frac{9{\mathrm{x}}^2}{5} \\ \Rightarrow \mathrm{x}=\frac{45\times 14}{9}=70 \mathrm{hours} \end{gathered}$$

Hence option D is correct.