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03 Jul 2024, 01:30
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Difficulty:
45%
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Question Stats:
75% (03:16) correct
25%
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Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours. Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours. Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank. What fraction of the tank was filled by Pump A?
A. \(\frac{1}{24}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
A. \(\frac{1}{24}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
03 Aug 2024, 04:45
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Official Solution:
Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours. Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours. Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank. What fraction of the tank was filled by Pump A?
A. \(\frac{1}{24}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
Let the time when all three pumps were working together be \(t\) hours. Then:
Pump A worked for \(t + 4\) hours and filled \(\frac{1}{12} * (t + 4)\) of the tank;
Pump B worked for \(t + 2\) hours and filled \(\frac{1}{5} * (t + 2)\) of the tank;
Pump C worked for \(t\) hours and filled \(\frac{1}{4} * t\) of the tank;
Hence, we'd have:
\(\frac{1}{12} * (t + 4) + \frac{1}{5} * (t + 2) + \frac{1}{4} * t = 1\)
Multiply the entire equation by 60 to eliminate fractions:
\(5(t + 4) +12(t + 2) + 15t= 60\)
\(32t + 44= 60\)
\(t=\frac{1}{2}\)
Therefore, Pump A filled \(\frac{1}{12} * (t + 4)=\frac{1}{12} * (\frac{1}{2} + 4)=\frac{3}{8}\) of the tank.
Answer: C
Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours. Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours. Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank. What fraction of the tank was filled by Pump A?
A. \(\frac{1}{24}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
Let the time when all three pumps were working together be \(t\) hours. Then:
Pump A worked for \(t + 4\) hours and filled \(\frac{1}{12} * (t + 4)\) of the tank;
Pump B worked for \(t + 2\) hours and filled \(\frac{1}{5} * (t + 2)\) of the tank;
Pump C worked for \(t\) hours and filled \(\frac{1}{4} * t\) of the tank;
Hence, we'd have:
\(\frac{1}{12} * (t + 4) + \frac{1}{5} * (t + 2) + \frac{1}{4} * t = 1\)
\(5(t + 4) +12(t + 2) + 15t= 60\)
\(32t + 44= 60\)
\(t=\frac{1}{2}\)
Answer: C
General Discussion
03 Jul 2024, 04:50
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Assuming Capacity of tank i.e. volume = 60 units (using LCM of given time)
Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours.
Rate of A = 60 / 12 = 5 units/hr
Rate of B = 60 / 5 = 12 units/hr
Rate of C = 60 / 4 = 15 units/hr
Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours.
volume of tank filled in first 4 hrs = 4*5 + 12*2 = 44 units
remaining volume = 60 - 44 = 16 units
Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank.
Time taken by all of them to fill remaining 16 units = 16 / (5+12+15) = 16/32 = 1/2 hr
work done by A in 1/2hr = 5/2 = 2.5 units
fraction of the tank was filled by Pump A = ((4*5) + 2.5) / 60 = 22.5 / 60 = 225/600 = 15/40 = 3/8
Answer C.
Pumps A, B, and C each fill a tank at their unique constant rates. Pump A alone can fill the tank in 12 hours. Pump B can complete the task in 5 hours, and Pump C in just 4 hours.
Rate of A = 60 / 12 = 5 units/hr
Rate of B = 60 / 5 = 12 units/hr
Rate of C = 60 / 4 = 15 units/hr
Initially, Pump A starts the process, operating alone for two hours. Then, Pump B joins, and they work together for the next two hours.
volume of tank filled in first 4 hrs = 4*5 + 12*2 = 44 units
remaining volume = 60 - 44 = 16 units
Finally, Pump C is also turned on, and all three pumps work together to finish filling the tank.
Time taken by all of them to fill remaining 16 units = 16 / (5+12+15) = 16/32 = 1/2 hr
work done by A in 1/2hr = 5/2 = 2.5 units
fraction of the tank was filled by Pump A = ((4*5) + 2.5) / 60 = 22.5 / 60 = 225/600 = 15/40 = 3/8
Answer C.
