Working alone at its constant rate, pump X pumped out of the water

Question

Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

A) 6

B) 12

C) 15

D) 18

E) 24

(A) 6
(B) 12
(C) 15
(D) 18
(E) 24

I had trouble solving these kind of questions in less than 3 minutes. If you guys have any inputs on how to solve it quickly, please share them.

chetan2u · Accepted Answer

Hi,
 when you look at the Q, you realize speed of all three combined and speed of two pump individually is given..
convert all into time for entire work-- 
 1) pump X 
it fills up 1/4 in 2 hr, so it will fill up entire into 2*4= 8 hr
 1 hr work = 1/8

2) combined 
combined three do 1-1/4 = 3/4 in 3 hr, so all three will fill entire tank in 3*4/3=4hr
 1 hr work = 1/4

3) pump y 
pump y can do 3/4 in 18 hr
so it will do complete in 18*4/3=24h...
 1 hr work = 1/24

combined 1 hr work should be sum of 1 hr work of x,y and z
so \frac{1}{4}= \frac{1}{8} +\frac{1}{24} +\frac{1}{z} ..
 \frac{1}{z}=\frac{1}{4}- (\frac{1}{8} + \frac{1}{24}) =\frac{(6-3-1)}{24}= \frac{2}{24} =\frac{1}{12} ...
 so z can do it in 12 hr 
B

Kurtosis · Answer

Rate of pump X = 1/8

3 hours are required to pump out the remaining (3/4)ths of tank --> 1 hr to pump out 1/4

Rate of X + Rate of Y + Rate of Z = 1/4

Rate of Y + Rate of Z = 1/4 - 1/8 = 1/8

Y takes 18 hours to pump out the remaining (3/4)ths of tank --> 6 hrs per (1/4)ths --> 24 hrs to pump out fully.

Rate of Y = 1/24

1/24 + Rate of Z = 1/8

Rate of Z = 1/8 - 1/24 = 1/12

Time required to pump out all the water by Z = 12 hrs

Answer: B

KarishmaB · Answer

Take one line of the question at a time:

"pump X pumped out ¼ of the water in a tank in 2 hours." 
1/4th was pumped out in 2 hrs. So entire 4/4 tank would be pumped out in 4*2= 8 hrs. Hence rate of work of X is 1/8th tank every hr.

"Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours."
The 3 pumped 3/4th tank in 3 hrs. So they pumped 1/4th tank every hour.

Note here that the rate of tank X is 1/8 and that of all 3 tanks is 1/4. So basically rate of tank Y + rate of tank Z is 1/8 tank per hour.

"If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water,"
Rate of pump Y = (3/4)/18 = 1/24

So out of a total rate of 1/8 for pumps Y and Z, 1/24 (i.e. a third) belongs to pump Y. So two thirds i.e. 2/24 = 1/12 would belong to pump Z.
So pump Z will take 12 hrs to pump out the water.

Answer (B)

ScottTargetTestPrep · Answer

We are given that Working alone at its constant rate, pump X pumped out 1/4 of the water in a tank in 2 hours. Since rate = work/time, the rate of pump X is (1/4)/2 = 1/8.

Since 1/4 of the water is pumped out of the tank, 3/4 is left to be pumped out.

We are also given that all 3 pumps pumped the remaining 3/4 of the water out in 3 hours; thus the combined rate of all three pumps is (3/4)/3 = 3/12 = 1/4.

We are finally given that pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water. Thus, the rate of pump Y = (3/4)/18 = 3/72 = 1/24.

If we let z = the time it takes pump Z to pump out all the water, then the rate of pump Z = 1/z and create the following equation:

1/8 + 1/24 + 1/z = 1/4

Multiplying the entire equation by 24z gives us:

3z + z + 24 = 6z

24 = 2z

12 = z

Answer: B

aceGMAT21 · Answer

Solving this question quickly requires to go through each statement thoroughly and interpret an equation from all of those. I am following the same strategy to solve this question,

This means that pump X can pump out the entire water in a tank in 8 hours.

The rest of the water here means 3/4th of the water in the tank. Thus, the complete water will be pumped out by (x+y+z) in 4 hours.

This means that 3/4th of the water in the tank is pumped out in 18 hours. Thus, the entire water will be pumped out by Y in 24 hours.

Q: how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

1/X + 1/Y + 1/Z = 1/4
1/8 + 1/24 + 1/Z = 1/4

1/Z = 1/4 - 1/8 - 1/24 = (6 - 3 -1)/24 = 2/24 = 1/12

Thus, 12 hours is the answer! Bingo.

OCDianaOC · Answer

Everything makes sense except for the 1/4 at the end. Where did that come from? The question is asking how many hours it would've taken pump z to pump ALL the water, so wouldn't that be a whole number? 1?

1/8 + 1/24 + 1/z = 1?

Why would it be 1/4 instead? That is only a quarter of the water (which is how much X pumped out)...

Please help!

amanvermagmat · Answer

Hi

Here's how the 1/4 on the Right hand side comes from:

In 2 hours, X pumped out 1/4 of water. So how much water is left (in fraction) = 3/4 (three-fourth).
Now all three pumps (X, Y, Z) pumped out remaining water (3/4 th) in 3 hours. See, since they pumped 3/4 th water in 3 hours, it means they will pump out 1/4 th water in 1 hour, OR you could say together the three pumps will pump out complete water in 1*4 = 4 hours

(1/4 water in = 1 hour
full water in = 1*4 = 4 hours)

This means, when X, Y, Z are working together - they take 4 hours to pump out full tank, Or you could say they pump out 1/4 of water in 1 hour. 
Thus, their respective waters pumped out in 1 hour would equal to 1/4 of tank drained out.

So, per hour work of X + per hour work of Y + per hour work of Z = 1/4

CrackverbalGMAT · Answer

Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

A) 6

B) 12

C) 15

D) 18

E) 24

Work = Rate * Time

Rate of ( X + Y + Z ) = Rate of X + Rate of Y + Rate of Z

Let's use these 2 basic concepts to solve this question.

Most of the above explanations have used the fractional method. So, let's try the  LCM approach  here.

X pumped out ¼ of the water in a tank in 2 hours

i.e X pumped out full water in 2* 4 hours = 8 hours

Let's assume that total volume = 8 L

Why should I select 8 as the total volume?
 The reason is very simple. A number that is divisible by 8. You can also try 16 or 24 .. . The final answer will not change. Remember, in fraction method we are assuming total volume as 1 . Taking LCM as the total volume can help you to reduce the calculation time.

Rate of pump X = 8 L/ 8 hrs = 1L/hr

X pumped out 2 L in 2hrs, so the remaining water in the tank = 8-2 = 6 L

X, Y, and Z working together pumped out the rest of the water (6L) in 3 hours.

Rate of ( X + Y + Z ) = 6/3 = 2 L/hr

If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water.

i.e Pump Y alone can pump out the rest of the water (6L) in 18 hours

Rate of Y = 6/18 = 1/3 L/hr

Rate of ( X + Y + Z ) = Rate of X + Rate of Y + Rate of Z

2 = 1 + 1/3 + Rate of Z

Rate of Z = 2/3 L/hr

how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

Time taken by pump Z = Work/Rate of Z =  8/\frac{2}{3}=  4 *3 =12 hours

Option B is the answer.

Thanks,
Clifin J Francis,
GMAT QUANT SME

sauravhippargi · Answer

Let the total volume of the tank be 120 lt
X pumps out  \frac{1}{4}  of the volume in 2 hours i.e., 30 lt in 2 hours.
So rate of X = 15 lt/hr

Remaining volume = 90 lt
Y can pump out  \frac{3}{4}  of the volume or 90 lt in 18 hours.
So rate of Y = 5 lt/hr

Together X, Y, and Z can pump out  \frac{3}{4}  of the volume or 90 lt in 3 hours.
So Rate (X+Y+Z) = 30 lt/hr

15+5+Z = 30 -> Z = 10

Z can pump out the entire volume in  \frac{120}{10}  =  12 hours 
Answer:  (B)

ThatDudeKnows · Answer

I hate fractional rates. Boy, wouldn't it be great if we knew how big the tank is and the volume just happened to be a number that's easy to work with? Let's make it up!! If there's a piece of information missing from a question that you can make up, I call that a "Hidden Plug In" question, and they're really common on rates questions as a way to avoid having to work with fractional rates. If you search for ThatDudeKnowsHiddenPlugIn, you'll find other examples.

Let's call the tank 8 gallons.

Pump X pumped 1/4 in 2 hours. So, 2 gallons in 2 hours. X pumps 1 gallon per hour.

We are left with 6 gallons. X, Y, and Z working together knock that out in 3 hours, so all three together pump 2 gallons per hour. X accounts for 1 gallon per hour, so Y+Z are the other 1 gallon per hour.

Y would have take 18 hours to do the 6 gallons, so Y does 1 gallon every 3 hours, or 1/3 of a gallon per hour.

Y+Z is 1 gallon per hour and Y is 1/3 of a gallon per hour, so Z is 2/3 of a gallon per hour.

How long would it take Z to pump 8 gallons? 8 divided by 2/3 = 8*(3/2) = 24/2 = 12.

Answer choice B.

ThatDudeKnowHiddenPlugIn

mcelroytutoring · Answer

Attached is a visual that should help.

Please note that the "Z" I am solving for in the combined time equation (9 hours) is the time it would take for pump Z to complete the REMAINING 3/4 of the job, not the time it would take for pump Z to remove ALL the water by itself, which is what the question is asking for. Read carefully!

ccoinflip · Answer

I think the best way here is to aggregate everything for how much time it would take the pumps to do full job, not just 3/4, and in that way you can use algebra to figure out all the other values quite easily

Lodz697 · Answer

Hello!

I know that the question asks to calculate the time needed by Z to empty the tank, but if I want to calculate the time to empty the tank just for the remaining water (3/4) I should have done the following:

(3/4)/(1/12) = 9

Am I right?

Bunuel · Answer

Yes. If Z takes 12 hours to empty the whoe tank, then to empty 3/4 of the tank, it would need 3/4*12 = 9 hours.

Lodz697 · Answer

Thanks Bunuel, you're so punctual as always.

Manbhav7 · Answer

Let work done = 4 Units (since X did 1/4 Work which will make it 1 unit after cancelling out)

Rate of X = 1 Unit/ 2h = 1/2 ----(1)

Combined Rate (X,Y,Z) = Remaining work/ Time taken = 3 Units/ 3h = 1 Unit/h ----(2)

Given, Y can do remaining work (i.e. 3 units) in 18h, >> Rate of Y = 3/18 = 1/6 Unit/h----(3)

Combined Rate = Rx + Ry + Rz 
1 Unit/ h = 1/2 + 1/6 + Rz (From 1,2,3 Statements)
=> Rz = 1/3
Now Time Taken by Z = Total Work/ Rz = 4 units/ 1/3 = 12 hours

weirdapoet · Answer

I assumed full of water in the tank is 36 
=> 1/4 of tank = 9 and the remain (3/4) equals 27 
X finished 1/4 of tank in 2hrs => rate of X (in 1hr) = 9/2 = 4.5 
Y finished 3/4 of tank in 18hrs => rate of Y (in 1hr) = 27/18 = 1.5 
X,Y,Z finished 3/4 of tank in 3hrs => rate of 3 pumps = 27/3 = 9 
=> rate of Z = 9 - 4.5 - 1.5 = 3 
=> time for Z to pump out all of water in the tank = 36/3 = 12hrs 
=> Answer is B 
Kinda of less using fraction :")

Usernamevisible · Answer

Fast GMAT Shortcut
X pumps 1/4 in 2 hours.
Therefore in 3 hours X pumps 3/8.
Three pumps together pump 3/4 in 3 hours.
So Y + Z pump:
3/4 - 3/8 = 3/8 in 3 hours.
Y + Z rate = 1/8.
Y pumps 3/4 in 18 hours.
Y rate = 1/24.
Z rate = 1/8 - 1/24 = 1/12.
Z alone needs 12 hours for the whole tank.
Answer = 12.

Let total water = 1.
Pump X pumps 1/4 of the tank in 2 hours.
Therefore X pumps 1/8 of the tank per hour.
In 3 hours, X would pump:
3 × (1/8) = 3/8
---------------

After the first 1/4 is removed, the remaining water is:
1 - 1/4 = 3/4
The three pumps remove this 3/4 in 3 hours.
Therefore, during those 3 hours:
Y + Z pump:
3/4 - 3/8 = 3/8
So Y + Z rate:
(3/8) ÷ 3 = 1/8 tank per hour
-----------------------------

Pump Y alone would take 18 hours to pump out the remaining 3/4 of the tank.
Therefore Y's rate:
(3/4) ÷ 18 = 1/24 tank per hour
-------------------------------

Z's rate:
1/8 - 1/24
= 3/24 - 1/24
= 2/24
= 1/12 tank per hour
--------------------

Time for Z alone to pump the entire tank:
1 ÷ (1/12) = 12 hours
---------------------

## Answer: B. 12

LCM Method
Let total water = 144 units.
Pump X pumps 1/4 of the tank in 2 hours.
1/4 of 144 = 36 units.
Therefore X's rate = 36 ÷ 2 = 18 units/hr.
After X pumps out the first 1/4, the remaining water is:
144 - 36 = 108 units.
The three pumps together remove these 108 units in 3 hours.
Therefore:
X + Y + Z rate = 108 ÷ 3 = 36 units/hr.
Pump Y alone would take 18 hours to pump out the remaining 108 units.
Therefore:
Y rate = 108 ÷ 18 = 6 units/hr.
Find Z's rate:
Z = (X + Y + Z) - X - Y
Z = 36 - 18 - 6
Z = 12 units/hr.
Time for Z alone to pump the entire tank:
144 ÷ 12 = 12 hours.
Answer: B. 12
Ultra-Fast LCM Version
Total work = 144 units.
X: 36 units in 2 hr ⇒ rate = 18.
Remaining = 108 units.
X + Y + Z: 108 units in 3 hr ⇒ rate = 36.
Y: 108 units in 18 hr ⇒ rate = 6.
Z = 36 - 18 - 6 = 12.
Time for Z alone = 144 ÷ 12 = 12 hr.
Answer = 12.

Working alone at its constant rate, pump X pumped out of the water

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Working alone at its constant rate, pump X pumped out of the water

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