Cistern filled by A and B in 5 ninutes.
= 5 $\displaystyle \left(\frac{1}{20} + \frac{1}{25}\right) = \frac{9}{20}$
Part unfilled = $\displaystyle \left(1 - \frac{9}{20}\right) = \frac{11}{20}$
Now $\displaystyle \frac{1}{20}$ part is filled by A in 1 minute.
$\displaystyle \frac{11}{20}$ part wiil be filled by A in 20% of $\displaystyle \frac{11}{20}$ = 11 minutes.
Cistern filled in 5 ninutes.
= 5 $\displaystyle \left(\frac{1}{12} + \frac{1}{15}\right) = \frac{3}{4}$ of tank.
After 5 minutes, let the third pipe emptied be "t" minutes, then = t $\displaystyle \left(\frac{1}{12} + \frac{1}{15}\right) + \frac{3}{4} = \frac{t}{6}$
t = 45 minutes.
Water tank filled in 10 hours.
= 10 $\displaystyle \left(\frac{1}{15} + \frac{1}{20} - \frac{1}{25}\right) = \frac{23}{30}$
Remaining part = 1 - $\displaystyle \frac{23}{30} = \frac{7}{30}$
Now work done by (A + B) in 1 hour = $\displaystyle \left(\frac{1}{15} + \frac{1}{20}\right) = \frac{7}{60}$
It means $\displaystyle \frac{7}{60}$ part is filled in 1 hour.
So, $\displaystyle \frac{7}{30}$ part is filled in = $\displaystyle \frac{60}{7} \times \frac{7}{30}$ = 2 hours.
Cistern filled by (A + B) in 1 minute.
= $\displaystyle \frac{2}{15} + \frac{1}{5} = \frac{1}{3}$
Let the capacity of tank be "Z" litres.
In one hour, Inlet in litres + Z litres = outlet in litres.
= 60 $\displaystyle \times \frac{\text{Z}}{3}$ + Z = 14 x 60 = Z = 40.
Hence, the capacity of cistern = 40 litres.
Let the second pipe be turned of after "t" minutes.
Then, work done by (A + B) in x minutes is => t $\displaystyle \left(\frac{1}{24} + \frac{1}{32}\right) = \frac{7\text{t}}{92} => \left(1\right)$
Also, work done by A in (18-t) minutes = $\displaystyle \frac{18-\text{t}}{24} => \left(2\right)$
Now (1) + (2) => $\displaystyle \frac{7\text{t}}{96} + \frac{18-\text{x}}{24}$ = 1.
t = 8.
Work done in 3 minutes by 3 pipes.
Then, work done by (A + B) in x minutes is = $\displaystyle \frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{1}{60}$
It means $\displaystyle \frac{1}{60}$ of tank is filled in 3 minutes.
$\displaystyle \frac{55}{60}$ part of the cistern is filled in 3 x 55 or 165 minutes.
Out of remaining $\displaystyle \frac{5}{60}$ cistern, $\displaystyle \frac{1}{20}$ is filled by A in 1 minute, and remaining is
$\displaystyle \frac{5}{60} - \frac{1}{20}$ i.e. $\displaystyle \frac{1}{30}$ is filled by B in 1 minute.
Therefore answer is, 165 + 2 = 167 minutes.
Work done by both in 4 minutes.
= 4 $\displaystyle \left(\frac{1}{12} + \frac{1}{16}\right) = \frac{7}{12}$
Remaining part of cistern = $\displaystyle \left(1 - \frac{7}{12}\right) = \frac{5}{12}$
Now $\displaystyle \frac{5}{12}$ part is filled by P in $\displaystyle \frac{5}{12} \times \frac{12}{1}$ = 5 minutes.
$\displaystyle \frac{1}{7}$ of tank water leak in 8 hours.
So, full tank leak in 7 x 8 = 56 hours.
Descriptions | Status |
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Attempted Questions | |
Un-Attempted Questions | |
Total Correct Answers | |
Total Wrong Answers |
1. If pipe A can fill a tank in ‘n’ number of hours/min, then in 1 hour part of tank filled by A is $\displaystyle \frac{1}{\text{n}}$ part of the tank.
2. If, in one hour pipe A can fill $\displaystyle \frac{1}{\text{n}}$th part of the tank then, the tank completely filled in ‘n’ number of hours.
Note: Here ‘n’ denotes time; it can be hours/minutes.
3. If pipe A can fill a tank in ‘x’ hours while B can fill the same tank in ‘y’ hours. And Pipes A and B simultaneously can fill the tank in one hour = $\displaystyle \frac{\text{x} \ + \ \text{y}}{\text{x} \text{y}}$th part, then total work = $\displaystyle \frac{\text{x} \text{y}}{\text{x} \ + \ \text{y}}$ hours.
4. If pipes A and B working together can fill a work in ‘x’ hours, if A alone can fill the tank in ‘y’ hours, then B alone can fill the tank in = $\displaystyle \frac{1}{\text{x}} - \frac{1}{\text{y}} = \frac{\text{xy}}{\text{x} \ - \ \text{y}}$ hours.
5. If pipe A can fill a tank in x hours, pipe B can empty the full tank in y hours, If both the pipes are opened simultaneously, then the tank can be filled in = $\displaystyle \frac{1}{\text{x}} - \frac{1}{\text{y}} = \frac{\text{xy}}{\text{x} \ - \ \text{y}}$ hours.
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