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02 Dec 2020, 07:30
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C
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Difficulty:
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Question Stats:
60% (03:35) correct
40%
(03:23)
wrong
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Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hr, 10 hr and 15hr respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at (3/4)th of their normal rates for the 1st hour after which they supplied water at normal rate. Pipe C supplied water at (2/3)th of its normal rate for 1st 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled?
A. 1.05 hr
B. 2.05 hr
C. 3.05 hr
D. 3.15 hr
Е. 3.45 hr
Are You Up For the Challenge: 700 Level Questions
A. 1.05 hr
B. 2.05 hr
C. 3.05 hr
D. 3.15 hr
Е. 3.45 hr
Are You Up For the Challenge: 700 Level Questions
02 Dec 2020, 08:10
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Bunuel
LCM of the times - 5, 10 and 15 = 30 liters
We also observe that the question talks about 3/4 th and 2/3 rd filling rates
Thus, we assume the total volume of the tank to be a multiple of 30, 4 and 3 = 360 liters (to avoid any fractional values)
Normal filling rates of the 3 pipes would be: 360/5 = 72 liters/hour, 360/10 = 36 liters/hour, and 360/15 = 24 liters/hour
1st hour: Filling rate of A = 3/4 x 72 = 54 lit/h; B = 3/4 x 36 = 27 lit/h; C = 2/3 x 24 = 16 lit/h => Total quantity filled = 54 + 27 + 16 = 97 lit
2nd hour: Filling rate of A = 72 lit/h; B = 36 lit/h; C = 2/3 x 24 = 16 lit/h => Total quantity filled = 72 + 36 + 16 = 124 lit
Volume remaining = 360 - (97 + 124) = 139 lit
Final filling rate of A = 72 lit/h; B = 36 lit/h; C = 24 lit/h => Total quantity filled per hour = 72 + 36 + 24 = 132 lit/h
Thus, time required = 139/132 =~ 1 hour (slightly more than that)
[Actual value is 1.05 hours]
Total time required = 2 + 1.05 = 3.05 hours (slightly more than 3 hours)
Answer C
General Discussion
26 Jan 2026, 08:42
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To avoid working with fractions, we assume the total capacity of the tank is the Least Common Multiple (LCM) of the individual times (5, 10, and 15).
Total Capacity = 30 units.
Now, calculate the Normal Hourly Rates:
Efficiency of A (Ea) = 30/5 = 6 units/hr
Efficiency of B (Eb) = 30/10 = 3 units/hr
Efficiency of C (Ec) = 30/15 = 2 units/hr
Total Normal Efficiency (Et) = (6+3+2) = 11 units/hr
Now calculate Work Done in the 1st Hour
A and B work at 3/4 of their normal rates.
C works at 2/3 of its normal rate.
Work Done = [)6+3)*(3/4)] + [2*(2/3)]
Work Done = 97/12 units
Next calculate Work Done in the 2nd Hour
A and B return to normal rates ((6+3) = 9 units/hr).
C is still working at 2/3 rate( 4/3 units/hr).
Work Done = 9+(4/3) = 31/3 units
Total Work Done after 2 Hours and Remaining Work
Total Work (2 hrs) = (97/12) + (31/3) = 221/12 units
Remaining Work = 30 - (221/12) = 139/12 units
After 2 hours, all pipes (A, B, and C) work at their normal combined rate of 11 units/hr.
Time = (Remaining Work)/(Combined Normal Rate) = (139/12)/11 = 139/132 = 1.05 hours
Total Time = 1 + 1+ 1.05 = 3.05 hours
Total Capacity = 30 units.
Now, calculate the Normal Hourly Rates:
Efficiency of A (Ea) = 30/5 = 6 units/hr
Efficiency of B (Eb) = 30/10 = 3 units/hr
Efficiency of C (Ec) = 30/15 = 2 units/hr
Total Normal Efficiency (Et) = (6+3+2) = 11 units/hr
Now calculate Work Done in the 1st Hour
A and B work at 3/4 of their normal rates.
C works at 2/3 of its normal rate.
Work Done = [)6+3)*(3/4)] + [2*(2/3)]
Work Done = 97/12 units
Next calculate Work Done in the 2nd Hour
A and B return to normal rates ((6+3) = 9 units/hr).
C is still working at 2/3 rate( 4/3 units/hr).
Work Done = 9+(4/3) = 31/3 units
Total Work Done after 2 Hours and Remaining Work
Total Work (2 hrs) = (97/12) + (31/3) = 221/12 units
Remaining Work = 30 - (221/12) = 139/12 units
After 2 hours, all pipes (A, B, and C) work at their normal combined rate of 11 units/hr.
Time = (Remaining Work)/(Combined Normal Rate) = (139/12)/11 = 139/132 = 1.05 hours
Total Time = 1 + 1+ 1.05 = 3.05 hours
