### PIPES AND CISTERNS

1. Explanation :

Part filled by A in 1 hour = (1/36); Part filled by B in 1 hour = (1/45);
Part filled by (A + B) In 1 hour =(1/36)+(1/45)=(9/180)=(1/20)
Hence, both the pipes together will fill the tank in 20 hours.

1. Explanation :

Net part filled In 1 hour =(1/10)+(1/12)-(1/20)=(8/60)=(2/15).
The tank will be full in 15/2 hrs = 7 hrs 30 min.

1. Explanation :

let the reservoir be filled by first pipe in x hours.
Then ,second pipe fill it in (x+10)hrs.
Therefore (1/x)+(1/x+10)=(1/12)
$$\frac{(x+10+x)}{(x(x+10)}$$=(1/12).
$$x^2$$ –14x-120=0
(x-20)(x+6)=0
x=20 [neglecting the negative value of x]
so, the second pipe will take (20+10)hrs. (i.e) 30 hours to fill the reservoir

1. Explanation :

Workdone by the waste pipe in 1min
=(1/20)-(1/12)+(1/15) = -1/10 [negative sign means emptying] therefore the waste pipe will empty the full cistern in 10min

1. Explanation :

work done by the leak in 1 hour=(1/3)-(1/(7/2))=(1/3)-(2/7)=(1/21).
The leak will empty .the tank in 21 hours.

1. Explanation :

Work done by the two pipes in 1 hour =(1/14)+(1/16)=(15/112).
Time taken by these pipes to fill the tank = (112/15) hrs = 7 hrs 28 min. Due to leakage, time taken = 7 hrs 28 min + 32 min = 8 hrs
Work done by (two pipes + leak) in 1 hour = (1/8).
Work done by the leak m 1 hour =(15/112)-(1/8)=(1/112). Leak will empty the full cistern in 112 hours.

1. Explanation :

Part filled in 7 min. = 7*((1/36)+(1/45))=(7/20).
Remaining part=(1-(7/20))=(13/20).
Net part filled in 1min. when A,B and C are opened=(1/36)+(1/45)-(1/30)=(1/60).
Now,(1/60) part is filled in one minute.
(13/20) part is filled in (60*(13/20))=39 minutes.

1. Explanation :

let B be closed after x min. then , Part filled by (A+B) in x min. +part filled by A in (18-x)min.=1
Therefore x*((1/24)+(1/32))+(18-x)*(1/24)=1
(7x/96) + ((18-x)/24)=1.

7x +4*(18-x)=96.
Hence, be must be closed after 8 min.