Pipe A, B and C are kept open and together fill a tank in t minutes. P

Question

Pipes A, B and C together filled a tank in t minutes working in the following manner:

Pipe A was open for t minutes;
Pipe B was open for the first 10 minutes and then closed. 
Two minutes after pipe B was closed, pipe C was opened and is kept open till the tank was full.

Each pipe filled an equal share of the tank. If pipes A and B, both opened, can fill the tank in t minutes. How long will it take pipe C alone to fill the tank ?

A. 36
B. 27
C. 24
D. 20
E. 18

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gurmukh · Accepted Answer

Pipe A can fill 1/3rd tank in t minutes
So, pipe B can fill 2/3rd of tank in t minutes.
Since 1/3rd tank pipe B is filling in 10 minutes so 2/3rd it will fill in 20 minutes.
Hence t =20 minutes.
Pipe C is open for 20 -12 =8 minutes
In 8 minutes pipe C fill 1/3rd of the tank, so full tank it will fill in 24 minutes.

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nick1816 · Answer

Efficiency of Pipe A = A; Efficiency of Pipe B = B; and Efficiency of Pipe C = C

A*t + B*10 + C*(t-12) = 100

A*t= B*10 = C(t-12) = 100/3

Also, A*t + B*t = 100

As A*t = 100/3, B*t=200/3

As B*10= 100/3; t= 20

C(t-12) = 100/3 
C*8 = 100/3

100/C = 24 mins

Solenja · Answer

could you please explain howd you got 100 in the answer.

youcouldsailaway · Answer

We can solve this using the efficiency method.

Remember, the efficiency method equates the work, i.e, given A's efficiency to be so and so and given B's efficiency to be so and so, the work done is the same.

Now, let's assume efficiency of pipe A to be a, of pipe B to be b, and of pipe C to be c.

at = 10b = c(t-12) = \frac{1}{3}

Now,

at + bt = 1

But  at = \frac{1}{3}

Therefore,  \frac{1}{3} + bt = 1

Thus,  bt  =  \frac{2}{3}

But  10b = \frac{1}{3}; b = \frac{1}{30}

\frac{1}{30} * t = \frac{2}{3}
t = 20
 
Time taken by C to fill up  \frac{1}{3}rd  of the tank is  t-12 = 20-12 = 8 minutes . 
Therefore, to fill the whole tank, C will take  8*3 = 24 minutes .

adewale223 · Answer

Rate of A = 1/3t since each pipe pumped same amount into tank

B = 1/3 x 1/10

B = 1/30

C = 1/3(t - 12)

A and B can fill tank in t minutes

A + B = 1/t

1/3t + 1/30 = 1/t

10 + t = 30

t = 20

Time it will take C alone to fill tank

Sub for t in equation below

C = 1/3(t - 12)

= 1/24

Answer C

Adewale Fasipe, GMAT quant instructor (YT channel Quant Clinic), Lagos Nigeria

nandini14 · Answer

I think this is the most fastest and easiest way!

Priyank1905 · Answer

Here is How I tried and solved this sum :

A + B + C = t min

Pipe A = t minutes
Pipe B = 10 minutes 
Pipe C = (t-12) minutes

Let the total capacity of the tank be 1 unit, Since we know that all pipes filled equal share of tank we can assume that A B and C all three filled 1/3 unit each.

Since it is mentioned that A + B = t minutes and also that each pipe filled equal share of tank so in this case we can say that A and B both filled the tank in 1/2 time 
we can say that, 
B=1/2t ---- 1 equation
B= 10 min ---- 2 equation

By using equation 1 and 2
T= 20 min

Now, C fills 1/3 unit of tank in (t-12) min so for full tank it will take 3 * (t-12) min 
(t-12) = 20-12 = 8 
3*8= 24

C is the correct answer.

I hope I solved the question in the right way.

malfoy · Answer

A,B,C Working together can fill the tank in t minutes, also A was on for entire duration of t minutes - a key detail which I overlooked and assumed the time each pipe was running was mutually exclusive, it is not really the case.

Also, A & B can fill the tank in t minutes, B ran only for first 10 mins and each PIPE FILLED EQUAL PORTION OF THE TANK
With this given information we can safely determine:
1. Each Pipe filled  \frac{1}{3}  parts of the tank => B can fill the tank in 30 mins and A can fill the tank in 3t mins working alone at its constant rate.
2. Using our findings from (1) we can calculate value of t, as t = combined time taken by pipes A & B to fill the tank together

=>  \frac{30*3t}{3t+30 } = t

=> \frac{90t}{3t+30 } = t

=> \frac{30t}{t+10 } = t

=>  t^2+10t = 30t

=>  t^2-20t = 0

=>  t(t-20) = 0

=>  t = 0   or  t = 20

Obviously t can't be 0 so t = 20

Now a key detail in the question was C started 2 mins after B was closed which means C started 12 mins after B was first opened (∵ B ran for 10 mins)
We know the tank was filled in 20 mins so C only ran of 20-12 = 8 mins

Which using information as stated in (1) means C filled  \frac{1}{3}   part of tank in 8 mins that means it could have filled the entire tank in 3*8 = 24 mins by itself. Option (C)

Hope this clears things! Cheers

KarishmaB · Answer

Responding to a pm: Solving it with logic All 3 pipes did equal work so pipe B did 1/3rd work in 10 mins. If only pipes A and B worked for t mins, they would complete the work so pipe A would still do 1/3 work (in t mins) so B would do the leftover 2/3 work in t mins. Then t must be 20 mins. So Pipe B took 10 mins for 1/3rd work and pipe C took 8 mins (t - 10 - 2 mins) for 1/3rd work. This means C will take 8*3 = 24 mins for full 1 work alone. Answer (C) Work Rate Concept video: https://youtu.be/88NFTttkJmA

Pipe A, B and C are kept open and together fill a tank in t minutes. P

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GMAT Problem Solving (PS) Questions

Pipe A, B and C are kept open and together fill a tank in t minutes. P

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