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03 Jul 2024, 01:30
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C
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Difficulty:
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Question Stats:
76% (03:17) correct
24%
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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.
03 Jul 2024, 07:35
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Given that at different times a different amount of pumps pump, to solve this question it is best that one uses an LCM as the total units required to complete the job.
The time for each pump to fill the tank is: \(4 =(2^2)\),\(5\) & \(12 = (2^2*3)\). Therefore the LCM is \(2^2 *3 * 5 = 60\)
Pump A alone can fill the tank in 12 hours: which means that every hour it does \(\frac{1}{12}^{th}\) of the job or \(\frac{5}{60}\) which is \(5\) units an hour.
Pump B can complete the task in 5 hours: B does \(\frac{1}{5}\) of the job in an hour, or \(\frac{12}{60}\), which is \(12\) units an hour.
Pump C in just 4 hours: \(\frac{1}{4}\) an hour or \(\frac{15}{60}\) which is \(15\) units an hour.
Solving:
Pump A starts the process, operating alone for two hours: \(5*2 = 10\) with \(50\) units left to do.
Pump B joins, and they work together for the next two hours: \(2*(5+12) = 34\) of which \(10\) are by A, with \(16\) units left to do.
Pump C is also turned on, and all three pumps work together to finish filling the tank: \(16\) units completed by all three pumps will take \(\frac{16}{(5+12+15)} = 0.5\) hours. In other words, Pump A does \(\frac{5}{2}= 2.5\) units.
In total A does: \(\frac{(10+10+2.5)}{60} = \frac{22.5}{60}\)
multiply through by 2.
\(\frac{45}{120} \)
Divide through by 15
\(\frac{3}{8}\)
ANSWER C
The time for each pump to fill the tank is: \(4 =(2^2)\),\(5\) & \(12 = (2^2*3)\). Therefore the LCM is \(2^2 *3 * 5 = 60\)
Pump A alone can fill the tank in 12 hours: which means that every hour it does \(\frac{1}{12}^{th}\) of the job or \(\frac{5}{60}\) which is \(5\) units an hour.
Pump B can complete the task in 5 hours: B does \(\frac{1}{5}\) of the job in an hour, or \(\frac{12}{60}\), which is \(12\) units an hour.
Pump C in just 4 hours: \(\frac{1}{4}\) an hour or \(\frac{15}{60}\) which is \(15\) units an hour.
Solving:
Pump A starts the process, operating alone for two hours: \(5*2 = 10\) with \(50\) units left to do.
Pump B joins, and they work together for the next two hours: \(2*(5+12) = 34\) of which \(10\) are by A, with \(16\) units left to do.
Pump C is also turned on, and all three pumps work together to finish filling the tank: \(16\) units completed by all three pumps will take \(\frac{16}{(5+12+15)} = 0.5\) hours. In other words, Pump A does \(\frac{5}{2}= 2.5\) units.
In total A does: \(\frac{(10+10+2.5)}{60} = \frac{22.5}{60}\)
multiply through by 2.
\(\frac{45}{120} \)
Divide through by 15
\(\frac{3}{8}\)
ANSWER C
