**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.

**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.

**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

**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

**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.

**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.

**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.

**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.