At a constant Rate of flow, it takes 20 minutes to fill a swimming poo

can be done in a few ways.

The std way

time taken to fill the pool by Large Hose =20 minutes =L or 1/L=1/20 similarly 1/S=1/30
simultaneously it will take 1/L+1/S=1/20+1/30=5/60=12 minutes

efficiency way (not necessarily the efficient way)

let l and s denote efficiency of large and small hose
efficiency = 100/time take to fill the pool=100/30=3.33% for smaller hose
similarly 100/20=5% for the larger hose
working together 8.33% per minute
time taken = 100/8.33 ===12.0048 ~ 12 minutes

But how do we know when we have to convert minutes to hours? What is an example of a more twisted problem?

HI sagnik242,

The simple answer to your question is that we have to follow whatever 'instructions' we're given in the prompt. Here, we're given the two rates (time to fill the pool) in terms of 'minutes' and we're asked for an answer in 'minutes.' As such, there are no 'twists' or extra steps (and converting minutes to hours would like make all the work take longer).

If you're looking for more-complex Work questions, then you can search for them in the Quant Forums here.

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

I had approached the question by picking numbers, but not sure if this is correct - can someone please verify if I what did below is correct or not?

Since in the question, they don't tell us the quantity to fill the pool with each hose, I just chose 60 liters as an estimate, therefore:

For large hose: 60L /20 min = 3L in 1 min
For small hose: 60L/30 min = 2L in 1 min

Combined rates: 3L + 2L = 5L in 1 min together, so if we wanted to find how much it would take to fill up to the original 60L, that would mean you multiplied 5L by 12, therefore, 60 L would fill up in 12 min with the two hoses. Does this make sense or is the approach of picking numbers incorrect?

Thanks!

Hi infinitemac,

YES - your approach (TESTing VALUES) works just fine on this question. Since the prompt tells us nothing about the actual rates of the two hoses nor about the volume of the swimming pool, you can choose whatever values you like - as long as they properly account for the times that it takes the two hoses to fill the pool individually.

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

20*30/(20+30)= 12 mins

Let's assign a nice value to the volume of the pool. We want a volume that works well with the given information (20 minutes and 30 minutes).
So, let's say the pool has a total volume of 60 gallons

It takes 20 minutes to fill a swimming pool with a LARGE hose
In other words, the LARGE hose can pump 60 gallons of water in 20 minutes
So, the RATE of the large hose = 3 gallons per minute

It takes 30 minutes to fill a swimming pool with a SMALL hose
In other words, the SMALL hose can pump 60 gallons of water in 30 minutes
So, the RATE of the small hose = 2 gallons per minute

So, the COMBINED rate of BOTH pumps = 3 gallons per minute + 2 gallons per minute = 5 gallons per minute

How many minutes will it take to fill the pool when both hoses are used simultaneously?
We need to pump 60 gallons of water, and the combined rate is 5 gallons per minute
Time = output/rate
= 60/5
= 12 minutes

Answer: B

Cheers.
Brent

We are given that the large hose takes 20 minutes to fill a swimming pool and that the small hose takes 30 minutes to fill the same swimming pool.

Since rate = work/time, the rate of the large hose is 1/20 and the rate of the small hose is 1/30.

We need to determine the number of minutes, when working simultaneously, it would take both hoses to fill the swimming pool. To solve, we can use the combined worker formula.

work of the large hose + work of the small hose = total work

Since the pool is being filled, the total work completed is 1 job. We can also let the time worked, in minutes, for both hoses, equal the variable t. Using the formula work = rate x time, we express the work of each hose as follows:

work of the large hose = (1/20)t

work of the small hose = (1/30)t

Lastly, we can determine t:

work of the small hose + work of the large hose = 1

(1/20)t + (1/30)t = 1

(3/60)t + (2/60)t = 1

(5/60)t = 1

t = 1/(5/60)

t = 60/5

t = 12

Alternate Solution:

In one minute, the larger hose can fill 1/20 of the pool, and the smaller hose can fill 1/30 of the pool. If they are used simultaneously, they fill 1/20 + 1/30 = 5/60 = 1/12 of the pool in one minute. If they fill 1/12 of the pool in one minute, they will fill 12/12 of the pool in 12 minutes.

Answer: B

CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)
i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

20L = 30S
i.e. L = 1.5S
Both Hoses = L+S = 1.5S+S = 2.5S

i.e. \(\frac{(S * 30)}{1} = \frac{(2.5S * T_2)}{1}\)

i.e. \(T_2 = 30/2.5 = 12\)

Answer: option B

Let the capacity of the swimming pool be 60 units

So, Efficiency of the Large hose is 3 units/min & The efficiency of the Small hose is 2 units/min

So, Time required to completely fill the Pool when both the hose's are in operation is 60/(3 + 2) = 12 Min, Answer must be (B)

Volume filled by hose A in 1 min = 1/20
Volume filled by hose B in 1 min = 1/30
Both (A+B) = 1/20+1/30 = 5/60 = 1/12
Thus, 12 minutes.

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