A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at

 Short cut...... 

Given mee-:  A=3hr ;  B=4hr;  and c=1hr. 

Now. All a, b, c of LCM=12

become =3,4,1

Becose C=-1

Now. A+B=7

12-7=5

Now. if C=-12====: -12+5=7

And given back in. A=3p. m &

B=4p. m & C=5p. m

Now become3+4+5=12

And now come 7:12.     Ans. 

Listen here 2.2 hours = 2hrs + 0.2hours ... Then 0.2 hours = 2/10hrs = 1/5hrs = (1/5)x60min = 12 min...... So 2.2hours = 2hrs 12min... Now 5pm +2hr 12min = 7.12pm

Listen here 2.2 hours = 2hrs + 0.2hours ... Then 0.2 hours = 2/10hrs = 1/5hrs = (1/5)x60min = 12 min...... So 2.2hours = 2hrs 12min... Now 5pm +2hr 12min = 7.12pm

its not 2.2 hrs it is only 2.2

2.2*60+ = 132 minutes 

which gives 2hrs 12 min 

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Tell me if it 2.2 hour den how can be it 7.12

Hi Lal Raj,

Let total volume of cistern be x.
In one hour, the fraction of cistern that is filled by A = $$\frac{x}{3}$$
In one hour, the fraction of cistern that is filled by B = $$\frac{x}{4}$$
In one hour, the fraction of cistern that is filled by C = x

Total volume filled by 5 PM = $$\frac{2x}{3}$$ + $$\frac{x}{4}$$ = $$\frac{11x}{12}$$
Total decrease in volume per hour after 5 PM = x - $$\frac{x}{3}$$ - $$\frac{x}{4}$$ = $$\frac{5x}{12}$$
Time taken for the tank to become empty = $$\frac{\frac{11x}{12}}{\frac{5x}{12}}$$ = 2.2 hours

=> Time by which tank will be empty = 7:12 PM

Listen here 2.2 hours = 2hrs + 0.2hours ... Then 0.2 hours = 2/10hrs = 1/5hrs = (1/5)x60min = 12 min...... So 2.2hours = 2hrs 12min... Now 5pm +2hr 12min = 7.12pm

Listen here 2.2 hours = 2hrs + 0.2hours ... Then 0.2 hours = 2/10hrs = 1/5hrs = (1/5)x60min = 12 min...... So 2.2hours = 2hrs 12min... Now 5pm +2hr 12min = 7.12pm

2hours+.2=2×60÷10=12