Two pipes can fill a tank in 24 min and 32 min respectively.

Question

Two pipes can fill a tank in 24 min and 32 min respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min ?

A. 4 min
B. 6 min
C. 9 min
D. 8 min
E. 12 min

mrinal2100 · Accepted Answer

D)

A worked for 18 mins so amount of work done in 18 mi ns.
1/24(18)=3/4
work left=1-3/4=1/4

complete work is done by B in 32 m ins
so 1/4 work will be done in =1/4*32=8 mins

shrouded1 · Answer

Lets say that you shut down pipe B after x mins. And that pipe A runs for exactly 18mins.

Fraction of tank filled in x mins = (x/24) + (x/32)
  The logic being that if a pipe fills a tank in m minutes, running for n minutes (n<m) , it will fill n/m of the tank  
The rest of the tank is filled by A alone in 18-x mins
Therefore, (18-x)/24 = 1 - (x/24 + x/32)
3/4 - x/24 = 1 - x/24 - x/32
x/32 = 1/4
x = 8min

Answer is (d)

ankitranjan · Answer

I think question is incomplete as there is no reference for B.

jimmy86 · Answer

The answer that i have is this :

Let B be closed after x minutes Then,
part filled by(A+B) in x min + part filled by A in (18-x) min =1

thus x*(1/24+1/32) + (18-x)*1/24 = 1

7x/96+(18-x)/24 =1

7x+4(18-x)=96

x=8

Hence B is 8 min

But is there a simpler way to do this ????? :?:

PTK · Answer

Please state the source of the question.

jimmy86 · Answer

Its from a school math book that i am studying for basics. Why?

jimmy86 · Answer

Agree that A worked for last 18 min But the 1/4th work left was done by A and B both working together. Thus they work at a combined rate of 7/96. Thus time would be 1/4*96/7 = 3.4 approx . Am still confused :shock:

jimmy86 · Answer

Yes i get it .Thanks a LOT!!! :-D

Any idea where i can get more practice and theory on RATE Problems?

Bunuel · Answer

Some work problems with solutions: time-n-work-problem-82718.html?hilit=reciprocal%20rate facing-problem-with-this-question-91187.html?highlight=rate+reciprocal what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal gmat-prep-ps-93365.html?hilit=reciprocal%20rate questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate a-good-one-98479.html?hilit=rate solution-required-100221.html?hilit=work%20rate%20done work-problem-98599.html?hilit=work%20rate%20done Also check: PS work problem: search.php?search_id=tag&tag_id=66 DS work problems: search.php?search_id=tag&tag_id=46 And: work-and-rate-gmat-questions-word-problems-master-list-79787.html Hope it helps.

GMATGuruNY · Answer

When the job is undefined, we can plug in our own number for the job. Let tank = 96.

Rate for A = w/t = 96/24 = 4/min
Rate for B = w/t = 96/32 = 3/min
Combined rate is 4+3 = 7/min

Now let's plug in the answer choices, which represent how long the tanks work together.

Answer choice C: 
w = r*t = 7*9 = 63.
Remaining work = 96-63 = 33, which is not divisible by A's rate of 4.
Eliminate C.

Answer choice D:
w = r*t = 7*8 = 56.
Remaining work = 96-56 = 40.
For A alone, t = w/r = 40/4 = 10.
Total time = 8+10 = 18. Success!

The correct answer is  D .

Salvetor · Answer

Pipe A will run for 18 minutes

Sp pipe A will do = 18/24 = 3/4 portion of the work

left work is 1/ 4

Just assume that what will be 1/4 if multiplied with 1/32 (B's rate )

Only 8 will make it 1/4

So after 8 minutes B should be stopped

Answer : D

gracie · Answer

18/24+(18-x)/32=1
x=10 minutes
18-10=8 minutes

mvictor · Answer

good question..this is how I solved..
1 pipe in 8 minutes fills 1/3 of the tank
2 pipe in 8 minutes fills 1/4 of the tank
after 8 minutes, we'll have 7/12 of the tank.
suppose we stop the second pipe.
after another 8 minutes, the tank fills 1/3, or 4/12. total, we have 11/12 of the tank after 16 minutes.

we know that tank 1 fills 4/12 in 8 minutes, then 2/12 in 4 minutes, and 1/12 in 2 minutes.
we can thus see, that we need to stop the second pipe after 8 minutes. then 1 alone in 10 minutes will fill the tank.

rsh12 · Answer

Time taken by A - 24mins
Rate A - 1/24

Time taken by B - 32mins
Rate B - 1/32

Work time equation

18/24 +(18-X)/32 = 1

Solving X = 10

Thus, B can be closed after (18-x)=8mins

Anigr16 · Answer

Assume total unit = 96
Pipe A does 96/24=4 units
Pipe B does 96/32=3 units
Total effort =7 units
Let required time be x
Work done by (A+B)= 7x
Work done by only A after pipe B is closed = 4(18-x)
Hence,  7x + 4(18-x)=96 
x=8
Ans D

Abhishek009 · Answer

Let the capacity of the tank be 96 units

Efficiency of A is 4 units/min
Efficiency of B is 3 units/min

Now, A works for full 18 minutes , so A fills up 72 units and the rest ( 24 Units ) is filled up by B

SO, Time taken by B to fill up 24 units is 8 minutes, thus Answer must be (D)

EgmatQuantExpert · Answer

Solution

Given  
 • Two pipes can fill a tank in 24 min and 32 min respectively.

To find  
 • The time after which B should be closed so that the tank is full in 18 min.

Approach and Working out

From the question, it seems one pipe is A and another is B.
So, pipe A will be open for complete 18 mins and Pipe B will be open for some time.

Let, the capacity of the tank = LCM (24, 32) = 96 Units
 • So, capacity of pipe A = 96/24 = 4 units per min
• Capacity of pipe B = 96 /32 = 3 units per mint

Pipe A is open for 18 min. So, 18 * 4 = 72 units of tank will get filled.
 • Remaining (96-72) =24 units will be filled by pipe B.

Time taken by pipe B to fill 24 units = 24/3 = 8 min. 
So, pipe B will be open for 8 mins.

Thus, option D is the correct answer.

Correct Answer: Option D

Dipu10 · Answer

We first calculated the total work done by A- which is 18/24 = 3/4

so remaining 1/4 part is done by B alone, with the rate of 1/32

Thus , 1/4 = x* 1/32
=> x = 8 , i.e., B worked for 8 mins. so ans is 8.

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Two pipes can fill a tank in 24 min and 32 min respectively.

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