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Three pipes – A, B, and C – are attached to a tank. Pipe A alone can

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Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post Updated on: 13 Aug 2018, 05:59
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Difficulty:

  95% (hard)

Question Stats:

45% (03:04) correct 55% (02:53) wrong based on 197 sessions

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Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #3

Three pipes – A, B, and C – are attached to a tank. Pipe A alone can fill the empty tank in 6 hours, pipe B alone can empty the full tank in 8 hours and pipe C alone can fill the empty tank in 12 hours. At 9 am pipe A is opened. One hour after that pipe B is opened and one hour from that pipe C is opened. At what time the tank will be full?

    A. 11:40 am
    B. 12:27 pm
    C. 1:40 pm
    D. 5 pm
    E. 5:20 pm


To solve question 4: Question 4

To read the article: Solve Time and Work Problems Efficiently using Efficiency Method!

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Originally posted by EgmatQuantExpert on 23 May 2018, 04:15.
Last edited by EgmatQuantExpert on 13 Aug 2018, 05:59, edited 2 times in total.
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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post 23 May 2018, 06:00
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EgmatQuantExpert wrote:
Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #3

Three pipes – A, B, and C – are attached to a tank. Pipe A alone can fill the empty tank in 6 hours, pipe B alone can empty the full tank in 8 hours and pipe C alone can fill the empty tank in 12 hours. At 9 am pipe A is opened. One hour after that pipe B is opened and one hour from that pipe C is opened. At what time the tank will be full?

    A. 11:40 am
    B. 12:27 pm
    C. 1:40 pm
    D. 5 pm
    E. 5:20 pm


We need to understand a few things when we read the question stem
1. Pipes A and C fill the tank in 6 and 12 hours. These are outlet pipes
2. Pipe B can empty the full tank in 8 hours. This is an inlet pipe.

Let's assume the total capacity of the tank to be LCM(6,8,12) = 24 units.

Individual rates:
Pipe A: +4 units/hour | Pipe B: -3 units/hour | Pipe C: +2 units/hour

The pipe A works alone for an hour filing the tank by 4 units. During the second hour,
when pipe B is also open, there is a net increase of +4-3 or 1 unit. In the first 2 hours,
there is a net increase of 5 units of water in the tank. From the third hour, the 3 pipes
are open. If all 3 pipes are open, we have an increase of +4-3+2 units(+3 units) per hr.

To fill the remaining 19 units, we would need an additional \(\frac{19}{3} = 6\frac{1}{3}\) hour.
The total time taken to fill the empty tank will be \(2 + 6\frac{1}{3}\) or \(8\frac{1}{3}\) hours or 8 hours and 20 minutes

Therefore, the time at which the tank is full is 5:20 PM(Option E), 8 hours 20 minutes after the first pipe is opened.
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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post 26 May 2018, 13:19
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Solution



Given:
In this question, it is given that there are three pipes – A, B, and C – attached to a tank.
    • A alone can fill the empty tank in 6 hours
    • B alone can empty the full tank in 8 hours
    • C alone can fill the empty tank 12 hours
On a certain day, pipe A is opened first at 9 am
    • One hour from that, i.e. at 10 am, pipe B is opened
    • One hour from that, i.e. at 11 am, pipe C is opened

To find:
We need to find out the exact time at which the tank will become full

Approach and Working:
Let us assume the capacity of the tank = LCM (6, 8, 12) = 24 units
    • In 6 hours, A can fill 24 units
      • Therefore, in 1 hour, A can fill \(\frac{24}{6}\) = 4 units
    • In 8 hours, B can empty 24 units
      • Therefore, in 1 hour, B can empty = \(\frac{24}{8}\) = 3 units
    • In 12 hours, C can fill 24 units
      • Therefore, in 1 hour, C can fill = \(\frac{24}{12}\) = 2 units
It is given that at 9 am only pipe A is opened
    • In the first 1 hour, pipe A can fill 4 units
      • Hence, remaining volume to fill = (24 – 4) units = 20 units
At 10 am pipe B is opened
    • In the second hour, total water filled by A and B together = (4 – 3) unit = 1 unit
      • Hence, remaining volume to fill = (20 – 1) = 19 units
At 11 am pipe C is opened
    • Third hour onwards total water filled by all A, B, and C together in 1 hour = [4 + (-3) + 2] units = 3 units
      • Therefore, to fill the remaining tank, they will take = \(\frac{19}{3}\) hours = \(\frac{19}{3}\) * 60 minutes = 380 minutes = 6 hours 20 minutes
Hence, the tank will get filled in another 6 hours 20 minutes time, started from 11 am, or at 5:20 pm.

Hence, the correct answer is option E.

Answer: E

Important Observation


    • It is observable that only 3rd hour onwards all 3 pipes are working together. In the 1st hour only pipe A is working, whereas in the 2nd hour both pipe A and B are working but not C. Therefore, one needs to separately count these two hours and the work done in this time span.
    • Pipe B is doing negative work, opposite to what pipe A and pipe C are doing.

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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post 26 May 2018, 13:25
LCM Method:-

Pipe A can alone fill the empty tank in 6 hours.

Pipe B can empty the tank in 8 hours.

Pipe C can alone fill the empty tank in 12 hours.

LCM of 6, 8 and 12 is 24 [assume the capacity of tank is 24 units]

Therefore:-

Pipe A fills (24/6=4 units) of the empty tank in one hours.

Pipe B empties (24/8=3 units) of the tank in one hours.

Pipe C fills (24/12=2 units) of the empty tank in one hours.

Required:-
At 9 am pipe A is opened. One hour after that pipe B is opened and one hour from that pipe C is opened. At what time the tank will be full?

So from 9 am to 10 am only Pipe A is filling up the tank, i.e. 4 units.
From 10 am to 11 am Pipe A will fill the tank by 4 units and Pipe B empties it by 3 units, i.e. 4 units - 3 units = 1 units per hour
From 11 AM Pipes A & C are filling up the tank of (4+2) units per hour and Pipe B is emptying the tank @ 3 units per hour. Effective rate of filling up the tank is 3 units per hour

Kindly note that uptill 11 am; 4 units + 1 units = 5 units of tank has already been filled in.
For the remaining 19 units (i.e. 24 units - 5 units), i.e. When Pipe A, B & C work together will fill the remaining tank at the rate of 3 units per hour. Therefore it will be 19/3 hours. Therefore the time taken after 11 am to fill the remaining 19 units will be 6 hours and 20 minutes (i.e. 6 hours & 1/3*60 minutes) hence the time will be 5:20 PM

Ans: Option E
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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post 27 May 2018, 15:12
Wow this is a very tricky one. The empty vs full thing really threw me off. I just spent the last 15 mins trying to figure out why my answer wasn't matching up with yours. I kept over-looking the negative work that pump B was doing. I'm hoping on the GMAT they would bold the empty part, otherwise that is something I'd definitely overlook, being so accustomed to more typical rate problems!
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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post 27 May 2018, 20:59
bp2013 wrote:
Wow this is a very tricky one. The empty vs full thing really threw me off. I just spent the last 15 mins trying to figure out why my answer wasn't matching up with yours. I kept over-looking the negative work that pump B was doing. I'm hoping on the GMAT they would bold the empty part, otherwise that is something I'd definitely overlook, being so accustomed to more typical rate problems!


I don't think that they would bold the "empty" part as I did while explaining my solution. We will have to be more careful while reading questions. If we don't read the questions very carefully then we will miss minute details and this how & when GMAT would trick/trap us. The trap answers are sitting right there!! Be careful while while reading questions!! You ain't gonna get any bold/highlighted stuff. In fact the questions will written in a way to do just the opposite where you miss out on the most "important" stuff. Wishing you a happy and fruitful preparation. Develop good habit patterns/strategies during your practice!!
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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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New post 01 Jun 2018, 05:08
EgmatQuantExpert wrote:
Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #3

Three pipes – A, B, and C – are attached to a tank. Pipe A alone can fill the empty tank in 6 hours, pipe B alone can empty the full tank in 8 hours and pipe C alone can fill the empty tank in 12 hours. At 9 am pipe A is opened. One hour after that pipe B is opened and one hour from that pipe C is opened. At what time the tank will be full?

    A. 11:40 am
    B. 12:27 pm
    C. 1:40 pm
    D. 5 pm
    E. 5:20 pm


To solve question 4: Question 4

To read the article: Solve Time and Work Problems Efficiently using Efficiency Method!




In these question types, its very crucial to identify Positive Work & Negative Work, before establishing relations given in the question.

Given:

Pipe A alone can fill the tank in 6 hours
Pipe B alone can empty the tank in 8 hours
Pipe C alone can fill the tank in 12 hours

Next crucial step, is to check whether the Total Work is defined with a specific Quantity. If not then to establish the given relations in the questions we can assume the work to be a VERY EASY TO WORK NUMBER.

EASY to Work number in most cases, for such question types, is usually the LCM of given rates.

In this case the LCM of rates (6,8,12) is 24. However, personally i would not use 24 as the Total capacity of the tank, i would rather choose 120(which is a multiple of 24) for ease of calculations.

Multiples of 10 are easier to work & they put me at ease. Imagine trying to solve this question under exam pressure, won't you welcome any possible comfort in the calculations? I definitely would.

Back to the question,

Assuming Total capacity of the Tank as 120 litres.

Pipe A alone can fill the Tank in 6 hours, hence rate of filling of A = 120/6 = 20 litres/hour

Pipe B alone can empty the Tank in 8 hours, hence rate of emptying of B = 120/8 = 15 litres/hour

Pipe C alone can fill the Tank in 12 hours, hence of rate of filling of C = 120/12 = 10 litres/hour

Now,

Activity 1
Pipe A is opened at 9 am, for one hour, till 10 am.

Pipe A alone for one hour will fill the Tank with 20 litres

Hence the Tank still has 120-20 = 100 litres left to be filled.

Activity 2
Pipe B is opened at 10 am, for one hour, till 11 am.

Pipe A filling the Tank & Pipe B Emptying the Tank for one hour.

Hence, 20 litres filled in & 15 litres emptied out in one hour, has a resultant effect of (20 - 15) = 5 litres filled in.

That means the Tank still has 100 - 5 = 95 litres to be filled in

Activity 3
Pipe C is opened at 11 am, now all pipes are open till the Tank is full. The three pipes together have to give a net effect of filling 95 litres.

Pipe A, B & C working together for one hour will have a resultant effect of (20-15+10) = 15 litres filled in one hour.

Hence in 6 hours, together they will fill 90 litres.

The remaining 5 litres will take (5*60)/15 = 20 mins, since all pipes together fill 15 litres per 60 mins

Therefore with all the pipes opened, the tank will fill the remaining 95 litres in 6 hours & 20 mins.

Now Total Time Taken to fill the Tank = 1 Hour (Pipe A) + 1 Hour (Pipe A+Pipe B) + 6 Hours 20 mins (Pipe A + Pipe B + Pipe C) = 8 hours 20 mins

Therefore the Time at which the Tank will be full is 9 am + 8 hours 20 mins = 5:20 pm.

Answer E.
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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can  [#permalink]

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Re: Three pipes – A, B, and C – are attached to a tank. Pipe A alone can   [#permalink] 30 Oct 2019, 09:58
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