Arithmetic aptitude :: Pipes and Cisterns :: Data Sufficiency 1

• Pipes and Cisterns - Solved Examples
• Pipes and Cisterns - Data Sufficiency 1
• Pipes and Cisterns - Data Sufficiency 2
• Pipes and Cisterns - Important Formulas and Facts

1. Inlet:

An Inlet is a pipe which is connected with a tank or cistern or reservoir, which fills it.

2. Outlet:

 An Outlet is a pipe which is connected with a tank or cistern or reservoir, which empties it.

3. If an inlet can fill a tank in $x$ hours, then, part filled in $1$ hour $=\frac{1}{x}$.

4. If an outlet can empty a tank in $y$ hours, then: part emptied in $1$ hour $=\frac{1}{y}$.

5. If a pipe $A$ can fill a cistern in $x$ hours and another pipe $B$ can empty the full tank in $y$ hours (where $y>x$), then on opening both

the pipes, then, the net part filled in $1$ hour $=\left(\frac{1}{x}-\frac{1}{y}\right)$.

6. If an inlet can fill a reservoir in $x$ hours and another outlet can empty the full tank in $y$ hours (where $x>y$), then on opening both

inlet and outlet, then, the net part emptied in $1$ hour $=\left(\frac{1}{y}-\frac{1}{x}\right)$.

A tank is filtered with two inlet pipes $A$ and $B$. Both the pipes are kept open for $10$ minutes so that the tank is $\frac{2}{3}$ full and then pipe $A$ is closed. How much time will $B$ take to fill the remaining part of the tank ?

I. Pipe $A$ is thrice as fast as pipe $B$.

II. Pipe $B$ alone can fill the tank in $60$ minutes.

Answer:

Explanation:

I. Let $B$’s $1$ min. work $=\frac{1}{x}$. Then, $A$’s $1$ min. work $=\frac{3}{x}$.

$(A+B)$’s $1$ min. work $=\left(\frac{1}{x}+\frac{3}{x}\right)$ $=\frac{4}{x}$.

$(A+B)$’s $10$ min. work $=\left(\frac{4}{x}\times 10\right)$ $=\frac{40}{x}$.

$\therefore$ $\frac{40}{x}=\frac{2}{3}$

$\Leftrightarrow x=60$.

$\therefore$ $B$’s $1$ min. work $=\frac{1}{60}$.

$\frac{1}{60}$ part is filled by $B$ in $1$ min.

$\frac{1}{3}$ part is filled by $B$ in $\left(\frac{4}{x}\times 10\right)$ $=\frac{40}{x}$.

$\therefore$ $\frac{40}{x}=\frac{2}{3}$

$\Leftrightarrow x=60$.

$\therefore$ $B$’s $1$ min. work $=\frac{1}{60}$.

$\frac{1}{60}$ part is filled by $B$ in $1$ min.

$\frac{1}{3}$ part is filled by $B$ in $\left(60\times\frac{1}{3}\right)$ = 20 min.

II. $B$’s $1$ min. work $=\frac{1}{60}$.

$\frac{1}{60}$ part is filled by $B$ in $1$ min.

$\frac{1}{3}$ part is filled by $B$ in $\left(60\times\frac{1}{3}\right)$ = 20 min.

Hence, the correct answer is (C).

Two taps $A$ and $B$ , when opened together, can fill a tank in $6$ hours. How long will it take for the pipe $A$ alone to fill the tank ?

I. $B$ alone take $5$ hours more than $A$ to fill the tank.

II. The ratio of the time taken by $A$ to that taken by $B$ to fill the tank is $2:3$

Answer:

Explanation:

$(A+B)$’s $1$ hour filling work $=\frac{1}{6}$

I. Suppose $A$ takes $x$ hours to fill the tank.

Then, $B$ takes $(x+5)$ hours to fill the tank.

$\therefore$  ($A$’s $1$ hour work) + ($B$’s $1$ hour work) = $(A+B)$’s $1$ hour work

$\Leftrightarrow\frac{1}{x}+\frac{1}{(x+5)}$  $=\frac{1}{6}$

$\Leftrightarrow\frac{(x+5)+x}{x(x+5)}$ $=\frac{1}{6}$

$\Leftrightarrow x^{2}-5x$ $=12x+30$

$\Leftrightarrow x^{2}-7x-30=0$

$\Leftrightarrow x^{2}-10x+3x-30$ $=0$

$\Leftrightarrow x(x-10)+3(x-10)$ $=0$

$\Leftrightarrow (x-10)(x+3)$ $=0$

$\Leftrightarrow x=10$.

So, $A$ alone takes 10 hours to fill the tank.

II. Suppose $A$ takes $2x$ hours and $B$ takes $3x$ hours to fill the tank. Then,

$\frac{1}{2x}+\frac{1}{3x}=\frac{1}{6}$

$\Leftrightarrow\left(\frac{1}{2}+\frac{1}{3}\right).\frac{1}{x}$ $=\frac{1}{6}$

$\Leftrightarrow\frac{5}{6x}$ $=\frac{1}{6}$

$\Leftrightarrow x=5$.

So, $A$ alone takes $(2\times 5)=10$ hours to fill the tank.

Thus, each one of I and II gives the answer.

$\therefore$ the correct answer is (C).  

• Pipes and Cisterns - Solved Examples
• Pipes and Cisterns - Data Sufficiency 1
• Pipes and Cisterns - Data Sufficiency 2