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Re: PS(RATE Problem):Pipes [#permalink]
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jimmy86 wrote:
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 fill in 18 min ?

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

I have the answer with me but need an easier solution. Please help with an explanation.


I think question is incomplete as there is no reference for B.
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Re: PS(RATE Problem):Pipes [#permalink]
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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 ????? :?:
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Re: PS(RATE Problem):Pipes [#permalink]
Please state the source of the question.
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Re: PS(RATE Problem):Pipes [#permalink]
Pkit wrote:
Please state the source of the question.



Its from a school math book that i am studying for basics. Why?
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Re: PS(RATE Problem):Pipes [#permalink]
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mrinal2100 wrote:
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


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:
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Re: PS(RATE Problem):Pipes [#permalink]
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shrouded1 wrote:
jimmy86 wrote:
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

I have the answer with me but need an easier solution. Please help with an explanation.



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)



Yes i get it .Thanks a LOT!!! :-D
Any idea where i can get more practice and theory on RATE Problems?
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Re: PS(RATE Problem):Pipes [#permalink]
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jimmy86 wrote:
shrouded1 wrote:
jimmy86 wrote:
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

I have the answer with me but need an easier solution. Please help with an explanation.



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)



Yes i get it .Thanks a LOT!!! :-D
Any idea where i can get more practice and theory on RATE Problems?


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.
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Re: PS(RATE Problem):Pipes [#permalink]
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jimmy86 wrote:
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

I have the answer with me but need an easier solution. Please help with an explanation.


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 .
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
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jimmy86 wrote:
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


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
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
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18/24+(18-x)/32=1
x=10 minutes
18-10=8 minutes
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
jimmy86 wrote:
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


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.
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
jimmy86 wrote:
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


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
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
jimmy86 wrote:
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


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
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
jimmy86 wrote:
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

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)
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
Expert Reply

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
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Re: Two pipes can fill a tank in 24 min and 32 min respectively. [#permalink]
jimmy86 wrote:
mrinal2100 wrote:
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


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:



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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