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Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor

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Bunuel
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Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   22 Apr 2020, 05:55
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Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, working together, can fill the tank in 8 hours. In how many hours, working alone, can pipe B fill the tank ?

A. 8 hrs
B. 12 hrs
C. 16 hrs
D. 20 hrs
E. 24 hrs
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E
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   22 Apr 2020, 06:03
Answer is 24 hours.
Let the total capacity is 12litre.
A fill 1litre/hour
12litre /12 hour.
A + B filled it in 8 hours
That means A could only fill 8 litres. Rest 4(12-8) litres is filled by B in 8 hours
That means B fill 0.5 litre per hour.
It wil take 12/0.5 hours = 24 hours for B to fill the tank own it's own.
OA is E

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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   23 Apr 2020, 04:35
Bunuel wrote:
Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, working together, can fill the tank in 8 hours. In how many hours, working alone, can pipe B fill the tank ?

A. 8 hrs
B. 12 hrs
C. 16 hrs
D. 20 hrs
E. 24 hrs


rate of A = 1/12
rate of B = x
together rate of A & B =
x+12/12x = 1/8
solve for x = 24 hrs
OPTION E
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   23 Apr 2020, 04:41
Rate of B alone = 1/((1/8)-(1/12))
=24 hrs

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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   23 Apr 2020, 07:57
SOLUTION

We can set-up the equation:

\(\frac{1}{A} +\frac{1}{x} = 8\)

\(\frac{1}{A} = \frac{1}{12}\)

\(\frac{1}{12}+ \frac{1}{x} = 8\)

\(2x+24=3x\)
\(x=24\)

ANSWER E
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   27 Apr 2020, 04:01
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Bunuel wrote:
Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, working together, can fill the tank in 8 hours. In how many hours, working alone, can pipe B fill the tank ?

A. 8 hrs
B. 12 hrs
C. 16 hrs
D. 20 hrs
E. 24 hrs


The rate of pipe A is 1/12, and the rate of both pipes is 1/8. Letting B = the number of hours it takes pipe B to fill the tank working alone, we see that pipe B’s rate is 1/B. Thus, we can create the equation:

1/12 + 1/B = 1/8

Multiplying by 24B, we have:

2B + 24 = 3B

24 = B

Answer: E
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   12 May 2020, 20:45
Bunuel wrote:
Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, working together, can fill the tank in 8 hours. In how many hours, working alone, can pipe B fill the tank ?

A. 8 hrs
B. 12 hrs
C. 16 hrs
D. 20 hrs
E. 24 hrs


Let work be 96 units.

A can do 96 units in 12 hours. That's 8 units per hour.

A+B can do 96 units in 8 hours. That's 12 units per hour.

That means B alone is doing 4 units per hour but we need to find out the time.

Time taken by B alone = 96/4=24 hours

Answer E
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   28 Jun 2020, 06:27
IMO-E

Pipe Efficiency Time. Total work
A. 4 12. 48
A+B. 6 8. 48

Total work is equal to LCM OF 12 & 8
Efficiency of B=. 6-4
=2
Time taken by B. =Total work/efficiency
=48/2
=24 hour

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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   28 Jun 2020, 06:38
Bunuel wrote:
Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, working together, can fill the tank in 8 hours. In how many hours, working alone, can pipe B fill the tank ?

A. 8 hrs
B. 12 hrs
C. 16 hrs
D. 20 hrs
E. 24 hrs

Ra=1/12
Ra+Rb=1/8
Rb=1/8-1/12
Rb=1/24
E:
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink] New post   27 Apr 2023, 05:59
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Re: Pipe A, working alone, can fill a tank in 12 hours. Pipes A and B, wor [#permalink]
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