adkikani wrote:

pushpitkc **Quote:**

I prefer numbers to solve such problems. Assume that the tank has 180 units of water.

Is this 'random' number LCM of answer choices?

**Quote:**

Working alone, the smaller pump takes twice as long as the large pump to fill the empty tank.

So if the larger pump fill(s) 2x units, the smaller pump must be filling x units.

Now, 2x+x = 3x = 30 -> x = 10(rate at which smaller pump fills the tank)

Why do we bring an additional variable x? If we know that Small pump takes twice the time to

empty the tank than large pump, then can I infer RATE of large pump must be double than that of small

pump, summing total rate as 1/6 (total time is 6 hours)

Let 2x: rate of larger pump; x: rate of smaller pump then

3x= 1/6 or x = 18 I assume that x in your solution can not be rate since I do not have a time unit (min or hr) in denominator.

Let me know if my approach is correct.

Hi

adkikaniYour first observation is spot on. It is generally the LCM of the numbers when

the individual times are given. In this case, we are given the detail of the time

taken by both the pumps(6 hours) to fill the pump. So I went with the number

180.

As for the latter part of my solution, I am using the rate at which the pumps fill

the tank. If x units is the rate at which the small pump fills the tank, the larger

pump will fill 2x units respectively.

This is why we equate the sum to 30(which is calculated from the rate at which

both the pumps fill in an hour). Once we calculate the individual rate of the

smaller tank to be 10 units, the time taken is \(\frac{180}{10}\) or 10.

Hope this helps you!

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