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Pumps A and B, working together at their constant rates can fill 3/5 t

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Russ19
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Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   13 Dec 2020, 09:36
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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
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C
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BrentGMATPrepNow
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   26 Dec 2020, 08:09
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sjuniv32 wrote:
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


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
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   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
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   13 Dec 2020, 11:06
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Given: 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.
Asked: How many hours will it take pump A, working alone at its constant rate, to fill the entire tank?

Let rate of pump A be x tank/hour
Rate of pump B = 1.25x tank/hour

Their combined rate = 2.25x tank/hour = 4.5x tank/2 hours = 3/5 tank/2hours
x = (3/5)* (2/9) = 2/15

Time taken by pump A alone to fill the entire tank = 15/2 = 7.5 hours

IMO C
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   26 Dec 2020, 07:20
BrentGMATPrepNow, could you please resolve the problem above?
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   16 Jan 2021, 05:01
Where I am wrong?

A=x and B=1.25x

1/a + 1/b=3/10

1/a + 1/1.25a=3/10

2.25/1.25a=3/10

22.5=3.73a

a=6
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   16 Jan 2021, 05:01
BrentGMATPrepNow wrote:
sjuniv32 wrote:
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


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



Where I am wrong?

A=x and B=1.25x

1/a + 1/b=3/10

1/a + 1/1.25a=3/10

2.25/1.25a=3/10

22.5=3.73a

a=6
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink] Pumps A and B, working together at their constant rates can fill 3/5 t   09 May 2022, 06:55
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Re: Pumps A and B, working together at their constant rates can fill 3/5 t [#permalink]
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