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13 Dec 2020, 09:36
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C
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Pumps A and B, working together at their constant rates can fill \(\frac{3}{5}\) th of a tank in 2 hours. The rate of pump B is 25% greater than the rate of pump A. How many hours will it take pump A, working alone at its constant rate, to fill the entire tank?
(A) 6
(B) 7
(C) 7.5
(D) 8
(E) 9
(A) 6
(B) 7
(C) 7.5
(D) 8
(E) 9
26 Dec 2020, 08:09
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sjuniv32
Pumps A and B, working together at their constant rates can fill \(\frac{3}{5}\) th of a tank in 2 hours. The rate of pump B is 25% greater than the rate of pump A. How many hours will it take pump A, working alone at its constant rate, to fill the entire tank?
(A) 6
(B) 7
(C) 7.5
(D) 8
(E) 9
(A) 6
(B) 7
(C) 7.5
(D) 8
(E) 9
I find it easiest to assign values for this kind of work question.
The rate of pump B is 25% greater than the rate of pump A
Let pump A's rate = 4 liters per hour
So, pump B's rate = 5 liters per hour
So their COMBINED rate = 9 liters per hour
Pumps A and B, working together at their constant rates can fill \(\frac{3}{5}\) th of a tank in 2 hours.
Output = (rate)(time) = (9 liters per hour)(2 hours) = 18 liters
So, the volume in 3/5 of a tank = 18 liters
In other words, (3/5)(capacity of tank) = 18 liters
Multiply both sides of the equation by 5/3 to get: Capacity of tank = (18)(5/3) = 30 liters
How many hours will it take pump A, working alone at its constant rate, to fill the entire tank?
Pump A's rate = 4 liters per hour
Time = output/rate = 30/4 = 7.5
Answer: C
General Discussion
13 Dec 2020, 10:57
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Using RT=W equation
(Ra + Rb) *2 = 3/5
(Ra + Rb) = 3/10 ----(i)
Rb = 1.25*Ra---- (ii)
Substituting value Rb in eq (i)
2.25* Ra = 3/10
Ra = 2/15
Hence, time required by pump A alone is 15/2 = 7.5
C is correct
(Ra + Rb) *2 = 3/5
(Ra + Rb) = 3/10 ----(i)
Rb = 1.25*Ra---- (ii)
Substituting value Rb in eq (i)
2.25* Ra = 3/10
Ra = 2/15
Hence, time required by pump A alone is 15/2 = 7.5
C is correct
