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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
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
