Problem Solving

Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?

A. 9/29
B. 8/23
C. 3/8
D. 17/29
E. 3/4

The OA is the option A.

I am really confused here. Experts, may you clarify this question for me? I don't know how to solve it. Thanks in advanced.

Vincen wrote:Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?

A. 9/29
B. 8/23
C. 3/8
D. 17/29
E. 3/4

The OA is the option A.

I am really confused here. Experts, may you clarify this question for me? I don't know how to solve it. Thanks in advanced.

Time of P = (2/3) R & time of P = (3/4) Q
Let time taken by Pipe R = 9...then time taken by P = 6 .....then time taken by Q = 8

Total time taken to drain the tank = (6)(8)(9)/ [(6)(8)+(6)(9)+(8)(9)] = 72/29

proportion drain by Pipe Q = time taken by Q / total time = (8) / (72/29) = 29/9..........However is inverse the OA

If I use rates instead, I got the right answer as follows:

Rate of pipe Q = 1/8 ...Combines rate = 29/72

proportion drain by Pipe Q = rate of pipe Q / Combined rate = (1/8)/(29)(72) = 9/29

Where did I go wrong??? Is not time indicative the same as rate?

Vincen wrote:Pipe P can drain the liquid from a tank in 3/4 the time that it takes pipe Q to drain it and in 2/3 the time that it takes pipe R to do it. If all 3 pipes operating simultaneously but independently are used to drain liquid from the tank, then pipe Q drains what portion of the liquid from the tank?

A. 9/29
B. 8/23
C. 3/8
D. 17/29
E. 3/4

(P's time)/(Q's time) = 3/4 = 6/8.
(P's time)/(R's time) = 2/3 = 6/9.

Let:
P's time = 6 hours.
Q's time = 8 hours.
R's time = 9 hours.
Let the tank = the LCM of the 3 times = 72 liters.

Since P takes 6 hours to drain the 72-liter tank, P's rate = w/t = 72/6 = 12 liters per hour.
Since Q takes 8 hours to drain the 72-liter tank, Q's rate = w/t = 72/8 = 9 liters per hour.
Since R takes 9 hours to drain the 72-liter tank, R's rate = w/t = 72/9 = 8 liters per hour.
Combined rate for all 3 pumps = 12+9+8 = 29 liters per hour.
Of the 29 liters drained every hour when all 3 pumps work together, Q drains 9 liters.
Thus:
Fraction drained by Q = 9/29.

The correct answer is A.

proportion drain by Pipe Q = time taken by Q / total time = (8) / (72/29) = 29/9..........However is inverse the OA

Time and rate have a RECIPROCAL RELATIONSHIP.
If Mary's rate is 2 times John's rate, then Mary time is 1/2 John's time.
If Mary's rate is 3 times John's rate, then Mary time is 1/3 John's time.
If Mary's rate is 4 times John's rate, then Mary time is 1/4 John's time.

Since the TIME RATIO for Q and all 3 pumps = 29/9, the RATE RATIO for Q and all 3 pumps is the RECIPROCAL:
9/29.