10
Q:

# Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

 A) 10 B) 12 C) 14 D) 16

Explanation:

Part filled in 2 hours =${\color{Black}&space;\frac{2}{6}=\frac{1}{3}}$

Remaining part =${\color{Black}&space;\left&space;(&space;1-\frac{1}{3}&space;\right&space;)=\frac{2}{3}}$

${\color{Black}&space;\therefore&space;}$ (A + B)'s 7 hour's work =${\color{Black}&space;\frac{2}{3}}$

(A + B)'s 1 hour's work =${\color{Black}&space;\frac{2}{21}}$

${\color{Black}&space;}$${\color{Black}\therefore&space;}$C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=${\color{Black}&space;\left&space;(&space;\frac{1}{6}-\frac{2}{21}&space;\right&space;)=\frac{1}{14}}$

${\color{Black}&space;\therefore&space;}$C alone can fill the tank in 14 hours.

Q:

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 ${\color{Blue} m^{3}}$. The emptying capacity of the tank is 10 $\inline {\color{Blue} m^{3}}$ per minute heigher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

 A) 50 m^3/min B) 60 m^3/min C) 72 m^3/min D) None of these

Explanation:

Let the filling capacity of the pump be x $\inline {\color{Black} m^{3}}$/min.

Then, emptying capacity of the pump=(x+10) $\inline {\color{Black} m^{3}}$/min.

so,$\inline {\color{Black} \frac{2400}{x}-\frac{2400}{x+10}=8\; \; \Leftrightarrow x^{2}+10x-3000=0}$

$\inline {\color{Black} \Leftrightarrow \left ( x-50 \right )+\left ( x+60 \right )=0\; \; \Leftrightarrow x=50}$

11 2620
Q:

Three taps A,B and C can fill a tank in 12,15 and 20 hours respectively. If A is open all the time and B ,C are open for one hour each alternatively, the tank will be full in:

 A) 6 hrs B) 20/3 hrs C) 7 hrs D) 15/2 hrs

Explanation:

$\inline {\color{Black} (A+B)'s\: 1\: hour\: work=\left ( \frac{1}{12}+\frac{1}{15} \right )=\frac{9}{60}=\frac{3}{20}}$$\inline {\color{Black} (A+C)'s\: 1\: hour\: work=\left ( \frac{1}{12}+\frac{1}{20} \right )=\frac{8}{60}=\frac{2}{15}}$  $\inline {\color{Black} Part \: \: filled \: in \: 2 \: hrs=\left ( \frac{3}{20} +\frac{2}{15}\right )=\frac{17}{60}}$

$\inline {\color{Black} Part \: \: filled \: in \: 6 \: hrs=(3\times \frac{17}{60})=\frac{17}{20}}$

$\inline {\color{Black} Remaining\; Part =\left ( 1-\frac{17}{20} \right )=\frac{3}{20}}$

Now, it is  the turn of A and B (3/20) part is filled by A and B in 1 hour.

therefore, Total  time taken to fill the tank =(6+1)hrs= 7 hrs

8 751
Q:

12 buckets of water fill a tank when the capacity of each tank is 13.5 liters. How many buckets will be needed to fill the same tank,if the capacity of each bucket is 9 liters?

 A) 8 B) 15 C) 16 D) 18

Explanation:

Capacity of the tank =(12 x 13.5) liters =162 liters.

Capacity of each bucket =9 liters

Number of buckets needed = 162/9 =18.

3 1190
Q:

One pipe can fill a tank  three times as fast as another pipe. If together the two pipes can fill the tank in 36 min, then the slower alone will be able to fill the tank in:

 A) 81 min B) 108 min C) 144 min D) 192 min

Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then,faster pipe will fill it in x/3 minutes.

$\inline \fn_jvn {\color{Black}\therefore\; \; \frac{1}{x}+\frac{3}{x}=\frac{1}{36}\: \: \: \: \Leftrightarrow \: \: \: \frac{4}{x}=\frac{1}{36}\: \: \: \Leftrightarrow \: \: \: x=144\: min}$

3 947
Q:

A water tank is two-fifth full.Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely?

 A) 6 min.to empty B) 6 min.to fill C) 9 min.to empty D) 9 min.to fill

Explanation:

Clearly,pipe B is faster than pipe A and so,the tank will be emptied.

part to be emptied = 2/5

part emptied by (A+B) in 1 minute=$\inline \fn_jvn {\color{Black}\left ( \frac{1}{6}-\frac{1}{10} \right )=\frac{1}{15}}$

$\inline \fn_jvn \Rightarrow \frac{1}{15}:\frac{2}{5}::1:x$

$\inline \fn_jvn \Rightarrow \frac{2}{5}\times 15=6mins$

so, the tank will be emptied in 6 min