A small water pump would take 2 hours to fill an empty tank. A larger

Hi, the logic I used was pretty much the same as yours, still Idk where I went wrong...

small pump fills 1 tank in 2 hours -> 1/2 tank in 1 hour -> 1/4 tank in 1/2 hour
big pump fills 1 tank in 1/2 hour

1/4 + 1 = 5/4 work which will take 4/5 hours

Hi sarathvr,

With the work that you've done, you have calculated how much of the tank would be filled in 1/2 an hour by the two pumps (working together).

In 1/2 hour:
The small pump would fill 1/4 of the tank
The large pump would fill 1/1 of the tank

Combined, that is 1/4 + 4/4 = 5/4 of the tank in 1/2 an hour. This means that the two pumps would OVER-FILL the tank in 1/2 an hour. By extension, it would take LESS than 1/2 an hour to fill the tank, so the answer CAN'T be 4/5 of an hour.

If you're going to use this approach, then you would have to MULTIPLY both sides by 4/5....

(5/4 of a tank)(4/5) = (1/2 hour)(4/5)
20/20 of a tank = 4/10 of an hour
1 tank = 2/5 of an hour

GMAT assassins aren't born, they're made,
Rich

OK yes! I didn't realise the over-full thing, so yeah that made sense.

But I still didn't understand the logic behind multiplying by 4/5 :/

Hi sarathvr,

The math that you did calculated the time it takes to fill 5/4 of a tank, but the question asks you to fill 4/4 of a tank. Mathematically, since we're dealing with rates, to decrease 5/4 to 4/4 we have to multiply by 4/5. Since we're multiplying one side of the 'equation' by 4/5, we have to multiply the other side by 4/5 also...

(5/4 of a tank)(4/5) = (1/2 hour)(4/5)
20/20 of a tank = 4/10 of an hour
1 tank = 2/5 of an hour

GMAT assassins aren't born, they're made,
Rich

Efficiency of small water pump \(= 50\)%, i.e., it can fill \(50\)% of the tank in an hour.

Similarly, efficiency of the larger water pump \(= 200\)%, i.e, it can fill the tank twice in an hour.

So, the time taken to fill the tank, which is the inverse of efficiency, when both work simultaneously \(= \frac{100}{250} = \frac{20}{50} = \frac{2}{5}\).

Time taken per hour by small pump \(= \frac{1}{2}\)

Time taken per hour by large pump \(= 2\)

Total time taken \(= \frac{1}{2} + 2 = \frac{4+1}{2} = \frac{5}{2}\)

Total work done \(= \frac{2}{5}\)

Answer (C)...

Before you launch into the math, take a look at the answer choices. Only two answer choices are possible - B and C. Depending on time, you might want to guess quickly and move on. Here is how you know.

If the two pumps are working together, they will complete the job faster than either pump working alone. So, the two pumps will fill the tank faster than 0.5 hours. Eliminate D and E

Consider this - two big pumps fill the tank in half of the time that one big pump takes to fill the tank. So, two big pumps fill the tank in 0.25 hours. The small pump is slower, so a small pump plus a big pump must take longer than 0.25 hours. Eliminate A.

Depending one time, guess or do enough math to eliminate B or C.

Jayson Beatty
Indigo Prep

Given:
1. A small water pump would take 2 hours to fill an empty tank.
2. A larger pump would take 1/2 hour to fill the same tank.

Asked: How many hours would it take both pumps, working at their respective constant rates, to fill the empty tank if they began pumping at the same time?

1. A small water pump would take 2 hours to fill an empty tank.
In 1 hour, small water pump will fill = 1/2 of tank

2. A larger pump would take 1/2 hour to fill the same tank.
In 1 hour, small water pump will fill = 2 of tank

Working together, in 1 hour small & large pump will fill = 2.5 of tank
Number of hours required to fill the empty tank together = 1/2.5 = 2/5 hours

IMO C

Solution:

Recall that rate = work/time. Therefore, the rate of the first pump is 1/2. The rate of the larger pump is 1 / (1/2) = 2.

The combined rate of the two pumps is 1/2 + 1/(1/2) = 1/2 + 2 = 5/2. Since time is the inverse of rate, it will take them 1/(5/2) = 2/5 of an hour to fill the empty tank working simultaneously.

Answer: C

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