10
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

Q:

A cistern has a leak which would empty the cistern in 20 minutes. A tap is turned on which admits 4 liters a minute into the cistern, and it is emptied in 24 minutes. How many liters does the cistern hold ?

 A) 360 lit B) 480 lit C) 320 lit D) 420 lit

Explanation:

1/k - 1/20 = -1/24

k = 120

120 x 4 = 480

Therefore, the capacity of the cistern is 480 liters.

1 6
Q:

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

 A) 85 min B) 92 min C) 187 min D) 144 min

Explanation:

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

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

1/x + 3/x = 1/36

4/x = 1/36 => x = 144 min.

1 14
Q:

Three taps I, J and K can fill a tank in 20,30and 40 minutes respectively. All the taps are opened simultaneously and after 5 minutes tap A was closed and then after 6 minutes tab B was closed .At the moment a leak developed which can empty the full tank in 70 minutes. What is the total time taken for the completely full ?

 A) 24.315 minutes B) 26.166 minutes C) 22.154 minutes D) 24 minutes

Explanation:

Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120
For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120
So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120
So remaining tank = 13/120
Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280
So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes

Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.

4 100
Q:

Pipe K fills a tank in 30 minutes. Pipe L can fill the same tank 5 times as fast as pipe K. If both the pipes were kept open when the tank is empty, how much time will it take for the tank to overflow ?

 A) 3 minutes B) 2 minutes C) 4 minutes D) 5 minutes

Explanation:

Let the total capacity of tank be 90 liters.
Capacity of tank filled in 1 minute by K = 3 liters.
Capacity of tank filled in 1 minute by L = 15 liters.
Therefore, capacity of the tank filled by both K and L in 1 minute = 18 liters.
Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.

4 132
Q:

Taps X and Y can fill a tank in 30 and 40 minutes respectively.Tap Z can empty the filled tank in 60 minutes.If all the three taps are kept open for one minute each,how much time will the taps take to fill the tank?

 A) 48min B) 72min C) 24min D) None of these

Explanation:

Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,

part filled by tap X in 1 minute = 1/30

part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,

part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute =

$\inline \fn_jvn \small \frac{1}{30}+\frac{1}{40}-\frac{1}{60}$

= 5/120 = 1/24

i.e., the tank can be filled in 24 minutes.