Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectiv

Question

Competition Mode Question

Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectively. Pipes B and C are kept open for 10 minutes, and then Pipe B is shut while Pipe A is opened. Pipe C is closed 10 minutes before the tank overflows. How long does it take to fill the tank?

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

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unraveled · Accepted Answer

Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectively. Pipes B and C are kept open for 10 minutes, and then Pipe B is shut while Pipe A is opened. Pipe C is closed 10 minutes before the tank overflows. How long does it take to fill the tank?

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

Let the total units of tank to be filled = 120 units
Speed of pipe A =  \frac{120}{30}  = 4 units/minute
Speed of pipe B =  \frac{120}{60}  = 2 units/minute
Speed of pipe C =  \frac{120}{120}  = 1 unit/minute
So, as per question, this is how tank's is filled:
Pipe B and Pipe C + Pipe C and Pipe A + Pipe A
 ...(10 minutes)... + ... (x minutes) ... + .. 10 minutes ..

Units filled in first 10 minutes = (2 + 1) * 10 = 30 
Units filled in last 10 minutes = 4 * 10 = 40
Units left to be filled by Pipe C and Pipe A = 120 - 40 - 30 = 50
Time taken(x) by Pipe C and Pipe A to fill 50 units =  \frac{50}{(1+4)}  = 10 minutes
Total time = 10 + x + 10 = 10 + 10 + 10 = 30 minutes

Answer B.

ArunSharma12 · Answer

Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectively. Pipes B and C are kept open for 10 minutes, and then Pipe B is shut while Pipe A is opened. Pipe C is closed 10 minutes before the tank overflows. How long does it take to fill the tank?

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

amount of work done by B & C in  10  minutes:
 \frac{1}{B}+\frac{1}{C}=\frac{1}{60}+\frac{1}{120}=\frac{1}{40} 
 10*(\frac{1}{B}+\frac{1}{C})=\frac{10}{40}= \frac{1}{4}

remaining work =  \frac{3}{4}

amount of work done by A & C:
 \frac{1}{A}+\frac{1}{C}=\frac{1}{30}+\frac{1}{120}=\frac{1}{24}

A & C will complete the remaining work in  \frac{3*24}{4} =18 minutes.

C is closed 10 minutes before Tank fills, so A & C run for 8 minutes.
amount of work done by A & C in  8  minutes:
 8*(\frac{1}{A}+\frac{1}{C})=\frac{1*8}{24} = \frac{1}{3}

remaining work =  \frac{3}{4}-\frac{1}{3}=\frac{5}{12} 
A will take  \frac{5*30}{12}  =  12.5  minutes

total time taken = 30.5 minutes
Ans: B

nick1816 · Answer

Suppose Capacity of tank is LCM(30, 60 120) = 120 units

Pipe A can fill = 4 units/min
Pipe B can fill = 2 units/min
Pipe C can fill = 1 units/min

Time taken by the pipes to fill the tank under given constraints = 't'

4(t-10)+2*10+1*(t-10) = 120

5(t-10) =100

t=30 min

yashikaaggarwal · Answer

Tank will be filled in 30 min. 
Take LCM of A,B,C 30,60,120
Which is 120.
So the pipes fill 120 litre in their respective time
Pipe A fill 4 litre/min
Pipe B fill 2 litre/min
Pipe C fill 1 litre/min

Now. Pipe B and C are kept open for 10 min.
B will fill 20 litre water and C will fill 10 litre water.
Total 30 -----(1)
Now B is shut close and A is opened simultaneously while C is already running.
C is shut close after 10 min so A filled 4*10=40 litres
C filled 1*10=10 litres
Total 50 -----(2)

From equation 1&2 we know 80/120 of tank is already filled.
Remaining 40 is to be filled by A only.
So 40/4=10 min.
So total time taken by the pipes to fill tank is 30 min.

OA is B

Posted from my mobile device

eakabuah · Answer

Alternatively

Let the total time taken to fill the tank be  x 
Pipe A worked for  x-10  minutes because the first 10 minutes that B and C worked, A was off.
Pipe C also worked for  x-10  minutes because it was closed 10 minutes before the tank got full.
Pipe B worked for 10 minutes. It only worked for the first 10 minutes.

\frac{x-10}{30}+\frac{10}{60}+\frac{x-10}{120}=1 
 4x-40+20+x-10=120 
 5x=150 
 x=30

The answer is B.

GMATGuruNY · Answer

Let the tank = 120 gallons, implying the following rates:
A -->  \frac{work}{time} = \frac{120}{30} = 4  gallons per minute
B -->  \frac{work}{time} = \frac{120}{60} = 2  gallons per minute
C -->  \frac{work}{time} = \frac{120}{120} = 1  gallon per minute

Pipes B and C are kept open for 10 minutes. 
Work produced by B and C in  10 minutes  = (combined rate for B+C)(time) = (2+1)(10) = 30 gallons

Pipe B is shut while Pipe A is opened. Pipe C is closed 10 minutes before the tank overflows. 
In the last  10 minutes , A works alone.
Work produced by A alone in the last 10 minutes = (A's rate)(time) = 4*10 = 40 gallons

Remaining work = tank - (gallons filled by B and C) - (gallons filled by A alone) = 120-30-40 = 50 gallons

Pipes B and C are kept open...Pipe B is shut while Pipe A is opened. 
Time for A and C to fill the remaining 50 gallons  = \frac{remaining-work}{combined-rate-for-A-and-C} = \frac{50}{4+1} = . 10 minutes

Total time = sum of the blue times above = 10+10+10 = 30 minutes

B

madzaka · Answer

A - B - C
 0 - 1/6 - 1/12   - 10min
 x/2 - 0 - x/2  - Ymin
 1/3 - 0 - 0  - 10min

(I wanted to make it look something like a matrix but it doesnt look good, sorry)

We know 
 1/6+1/12+1/3  is the work done in 20 min for the first 10 and last 10 min. which makes 7/12 of the work

the work in the middle section is 5/12
 4*A+C=120 min (5 times the work)
120/5=24 (A+C to do 1 Work)
1-24
Y-5/12

Y=10 thus answer B

Answer B

exc4libur · Answer

rate*time=work
a=1/30, b=1/60, c=1/20
10(b+c)+x(c+a)+10a=1
10(1/60+1/120)=10(3/120)
x(1/120+1/30)=x(5/120)
10a=10(1/30)
30/120+5x/120+10/30=1
5x/120=1-30/120-40/120
x=(50/120)*120/5=10
time=10+x+10=30

Ans (B)

freedom128 · Answer

Speed of pipes A, B, and C is 1/30, 1/60 and 1/120 per mins respectively.

It take   T mins   to fill the whole tank

Pipes B and C are kept open for 10 minutes ---> This will fill (1/60+1/120)*10 mins = 1/4  ...(1) 
Pipes A are kept open for (T-10) minutes ---> This will fill (1/30)*(T-10) mins = (1/30)*T -1/3  ...(2) 
Pipes C are kept open for (T-10-10)=(T-20) minutes ---> This will fill (1/120)*(T-20) mins = (1/120)*T -1/6  ...(3)

(1)+(2)+(3) = 1 
1/4 + T/30 -1/3 + T/120 -1/6 = 1
T*(5/120) = 1 +1/4 = 5/4
 T = 30 mins

FINAL ANSWER IS (B)

monikakumar · Answer

Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectively. Pipes B and C are kept open for 10 minutes, and then Pipe B is shut while Pipe A is opened. Pipe C is closed 10 minutes before the tank overflows. How long does it take to fill the tank?

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes

A. 28 minutes
B. 30 minutes
C. 36 minutes
D. 40 minutes
E. 60 minutes
As rate = 1/30, Bs rate = 1/60, Cs rate = 1/120
so,
10/60+(t-10)/120+(t-10)/30=1
on solving, we get,
t=30
Ans B

RajatGMAT777 · Answer

A very easy way to solve the problem
Note - The time taken has to be greater than 20 minutes as we are itself mentioned 2 time frames when the pumps were used (For beginners) 
Total work = Work done by B and C(10 minutes) + Work Done by A and C(time is unknown) + Work done by A Alone(for 10 minutes)
Amount Work done when B and C worked is 10*(\frac{1}{60} + \frac{1}{120}) = 1/4
Amount of work done by A alone = 10*\frac{1}{30} = 1/3
Work done by A and C = 1-(1/4+1/3) = 5/12
So time required to do 5/12 work is given as Work done = Rate * Time 
 5/12 = (\frac{1}{30} + \frac{1}{120}) *t
 5/12 = 5/120 *t
 t = 10 minutes 
So total = 30 minutes
 Bunuel   KarishmaB 
This is one of the approaches but did I took more time to reach the answer, took around ~ 1 minute 10 sec

KarishmaB · Answer

This is how I would solve it too. 
Last 10 mins, only A works. Since A takes 30 mins alone, it does 1/3rd of the work in last 10 mins. 
We ar leeft with 2/3rd of the work. First 10 mins, B and C work together. Their combined rate is 1/60 + 1/120 = 1/40 so they complete 1/4th of the work. 
Work left is 1 - 1/3 - 1/4 = 5/12. This is done by A and C. Their combined rate is 1/30 + 1/120 = 1/24.
Time taken = (5/12) / (1/24) = 10 mins

Total time taken is 10 + 10 + 10 = 30 mins.

Many people prefer the units of work approach since it gets rid of fractions. If you are comfortable with fractions, you can do this calculation in your head as you read the question. Of course either works and time taken to solve with either strategy will vary person to person. If you are taking less than 2 mins on average for a question, you needn't worry. Some questions may need more than 2 mins but someone who gets those toughies is at a level where he/she solves simpler questions within seconds. So it all balances out.

dolor · Answer

is the pipe A not kept open the whole time after B was shut? please explain

Krunaal · Answer

Yes, Pipe A is kept open after B is shut.

Let's suppose capacity of tank is 240 lts

R 
 T 
 W 
 
  A 
 8 
 30 
 240 
 
  B 
 4 
 60 
 240 
 
  C 
 2 
 120 
 240

B + C work for 10 mins => 6*10 = 60 lts; they fill 60lts ..........(1)

Now, B is closed and A is open => A + C = 10

We are told that 10 mins before tank overflows, C is closed, which means for the last 10 minutes only A was open, and it filled 8 * 10 = 80 lts ....... (2)

From (1) and (2), we can see that 60 + 80 = 140 lts were filled by these two operations, therefore the remaining 100 lts was filled when A+C were working, time A+C worked together = 100/10 = 10 mins

R 
 T 
 W 
 
  B+C 
 6 
 10 
 60 
 
  A+C 
 10 
 10 
 100 
 
  A 
 8 
 10 
 80

So, from above we can see that it took 30 mins to fill 240 lts.

Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectiv

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GMAT Problem Solving (PS) Questions

Pipes A, B, and C can fill a tank in 30, 60, and 120 minutes respectiv

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