Working alone, a small pump takes twice as long as large pump takes to

adkikani

See calculation mistake 1/6 * 2/3 = 1/9 not 1/6.

Hope it helps.

Hi adkikani

What do you mean by emptying the tank. I did not get it.?

Also re-visit your calculation for the highlighted part. You made a mistake there, hence you are getting a different answer.

niks18



Just as we add two rates for two pumps for getting combined rates, can we add individual time to get TOTAL time
taken by both pumps together to empty the tank?

Hi adkikani

Ok I got your question. in this scenarios, you cannot simply add the individual times to get total time. Because the rate at which both work is different. Hence Larger pump will empty more water at a faster rate than the smaller pump. Hence smaller pump will have to make "LESS EFFORT" i.e. it will have to work for "LESS TIME". So adding individual times will give you "HIGHER" working hours.

That's why we first calculate the rates and then add. same has been done here. As you can see the individual times are 9 & 18 but when they work together they complete the job in 6 hours only because larger pump works at a faster rate.

I prefer numbers to solve such problems. Assume that the tank has 180 units of water.
Since working together they can fill the tank in 6 hours, their combined rate - 30 units.

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)

Therefore, the smaller tank must take \(\frac{180}{10}\) or 18 hours(Option E) to fill the tank.

pushpitkc



Is this 'random' number LCM of answer choices?



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.

Time is Inversely related to efficiency (The more the efficiency the lesser the time required)

So, 3e = 18

Or, e = 6 and total work is 18 units

So, Time required for the small pump will be (E) 18 hours.

Hi adkikani

Your 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!

1/x + 1/2x = 1/6
3/2x = 1/6
x = 18/2 = 9

2x = 18

IMO E

Bumping for forum quiz and test.

Let's do this:

If the big one works at a 2x speed than the small one, the 6 hours must be divided as follows:
- 4 hours for the bigger
- 2 hours for the smaller

And so:

Rate bigger =\(\frac{ 4 }{ 6 }=\frac{ 2 }{ 3 }\)
Rate smaller =\(\frac{ 2 }{ 6 }=\frac{ 1 }{ 3 }\\
\)
Needed hours for filling the tank using the smaller:

\(3 * 6 = 18\)

Um, that's there 1/x=1/6*2/3=1/6 where you made an error.
that should have been 1/9 as per your calculations

Given>
P1=1/2R
P2=R
So
1/2R+R=1/6
R=1/9

T2=W/R=1/R=9
T1=W/(1/2R)=18