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Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec

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Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post Updated on: 03 Sep 2019, 03:40
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Question Stats:

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e-GMAT Question of the Week #24

Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respectively. All the three taps are used simultaneously to fill the tank. But, due to a leak at the bottom of the tank, only \(\frac{1}{3}\)rd of the tank was filled by the time it was supposed to be full. In how much time could the leak empty a full tank?

    A. \(\frac{75}{31}\)

    B. \(\frac{150}{31}\)

    C. \(\frac{225}{31}\)

    D. \(\frac{15}{2}\)

    E. \(\frac{450}{31}\)


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Originally posted by EgmatQuantExpert on 23 Nov 2018, 06:57.
Last edited by Bunuel on 03 Sep 2019, 03:40, edited 2 times in total.
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 28 Nov 2018, 05:55
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Solution


Given:
    • Three pipes, A, B and C can fill a tank in 10, 15 and 25 days respectively
    • All the three taps are used simultaneously to fill the tank
    • Only \(\frac{1}{3}^{rd}\) of the tank was filled by the time it was supposed to be full, due to a leak

To find:
    • The time in which the leak can empty a full tank

Approach and Working:
    • Three taps together can fill the tank in \(\frac{1}{((1/10) + (1/15) + (1/25))} = \frac{150}{31}\) days
      o But only \(\frac{1}{3}^{rd}\) of the tank was filled in \(\frac{150}{31}\) days, which implies that the leak can empty \(\frac{2}{3}^{rd}\) of the tank in \(\frac{150}{31}\) days

    • Thus, it can empty a full tank in \((\frac{150}{31}) * (\frac{3}{2})\) days = \(\frac{225}{31}\) days

Hence the correct answer is Option C.

Answer: C

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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 23 Nov 2018, 10:03
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This took me a little over 3 minutes but...

First let's figure out how much time it SHOULD take the pipes to fill the tank:

Individual rates of the three pipes are 1/10, 1/15, and 1/25, respectively.

Using W = R*T, and setting W = 1, we know that together, the three pipes can fill the tank in 1/((15+10+6)/150), or 150/31 units of time.

Next we need to find the rate that the tank is leaking. We know that 1/3 of the tank is left after this period (150/31) of time, so we set a new equation with the W = R*T formula:

(1/3) = ((31/150) - (leaking rate)) * (150/31)

After some "quick" algebra, we identify that:

31/450 = 31/150 - leaking rate

Leaking rate = (93/450) - (31/450) = 62/450 or 31/225

FINALLY, we need to know how long it would take the lead to drain an entire tank:

Back to W=R*T, we know that if unit of work is 1, then R and T are reciprocals of each other, so t=225/31 or ANSWER C

I really hope that's right...

Posted from my mobile device
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 14 Dec 2018, 10:00
Hi
Thank you for the solution.
But can you please explain the highlighted part?
Thanks again!

EgmatQuantExpert wrote:

Solution


Given:
    • Three pipes, A, B and C can fill a tank in 10, 15 and 25 days respectively
    • All the three taps are used simultaneously to fill the tank
    • Only \(\frac{1}{3}^{rd}\) of the tank was filled by the time it was supposed to be full, due to a leak

To find:
    • The time in which the leak can empty a full tank

Approach and Working:
    • Three taps together can fill the tank in \(\frac{1}{((1/10) + (1/15) + (1/25))} = \frac{150}{31}\) days
      o But only \(\frac{1}{3}^{rd}\) of the tank was filled in \(\frac{150}{31}\) days, which implies that the leak can empty \(\frac{2}{3}^{rd}\) of the tank in \(\frac{150}{31}\) days

    Thus, it can empty a full tank in \((\frac{150}{31}) * (\frac{3}{2})\) days = \(\frac{225}{31}\) days
Hence the correct answer is Option C.

Answer: C

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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 15 Dec 2018, 06:44
EgmatQuantExpert wrote:
Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respectively. All the three taps are used simultaneously to fill the tank. But, due to a leak at the bottom of the tank, only \(\frac{1}{3}\)rd of the tank was filled by the time it was supposed to be full. In how much time could the leak empty a full tank?

    A. \(\frac{75}{31}\)

    B. \(\frac{150}{31}\)

    C. \(\frac{225}{31}\)

    D. \(\frac{15}{2}\)

    E. \(\frac{450}{31}\)


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TIme taken : 1 min 16 seconds
used the choices :
approach :
first calculate the together rate = \(\frac{150}{3}\)
check the choices ...option B has \(\frac{150}{3}\) then the tank would have filled halfway ...because the leak and the filling pipes are working at the same rate and hence the work done by them will be equal ...but we are given work done is less than half...eliminate B
( proabibility of answering correct = 1/4)

from the above cognition we can undertsnad that the empty rate is greater than\(\frac{150}{3}\)
eliminate A
(P(correct answer) =1/3)

answer choice C is just plain wrong as we have a prime number in the denominator so no matter what we do we still are going to end up up 31
eliminate D
P(correct answer(=1/2

Option E = \(\frac{450}{3}\) ...this is 3 times the rate of together filling... if the empty rate were 3 times the filling the rate then the tank may not even be filled 10 percent let alone 33%

only plausible answer choice is C
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 11 Jan 2019, 20:59
total time: 150/31 is required to empty the tank to 2/3 or to fill the tank to 1/3

time: 150/3

work is done: 2/3

2/3 work is done in 150/31(time)

complete work is done: 150/31*3/2 = 225/31


let me know if it is helpful to anyone
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 31 Aug 2019, 23:17
Does anybody have the LCM approach solution for this one?
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 03 Sep 2019, 03:38
EgmatQuantExpert wrote:
e-GMAT Question of the Week #24

Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respectively. All the three taps are used simultaneously to fill the tank. But, due to a leak at the bottom of the tank, only \(\frac{1}{3}\)rd of the tank was filled by the time it was supposed to be full. In how much time could the leak empty a full tank?

    A. \(\frac{75}{31}\)

    B. \(\frac{150}{31}\)

    C. \(\frac{225}{31}\)

    D. \(\frac{15}{2}\)

    E. \(\frac{450}{31}\)


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Combined rate of all 3 pipes = 1/10 + 1/15 + 1/25 = 31/150 tank/min (taking LCM of the time)

Time in which the 3 pipes will fill the tank = 150/31 mins

Negative work done by the leak in 150/31 mins = (2/3)rd of the tank
Rate of work of tank = (2/3)/(150/31) = (31/225)th tank per min
Time taken to empty 1 full tank = 225/31 mins

Answer (C)
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 03 Sep 2019, 08:01
EgmatQuantExpert wrote:
e-GMAT Question of the Week #24

Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respectively. All the three taps are used simultaneously to fill the tank. But, due to a leak at the bottom of the tank, only \(\frac{1}{3}\)rd of the tank was filled by the time it was supposed to be full. In how much time could the leak empty a full tank?

    A. \(\frac{75}{31}\)

    B. \(\frac{150}{31}\)

    C. \(\frac{225}{31}\)

    D. \(\frac{15}{2}\)

    E. \(\frac{450}{31}\)


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Let the capacity of the tank be 150 units

Efficiency of the 3 taps are 15 , 10 & 6

Combined efficiency of the 3 taps are 31 units/min

Time required to fill the entire tank is 150/31 but by that time only 50 units were filled

So, (31 - x)*150/31 = 50

Or, 3(31 - x) = 31

Or, 93 - 3x = 31

Or, x = 62/3

So, Time required to empty the tank is 150*3/62 = 450/62 = 225/31 , Answer must be (C)
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 03 Sep 2019, 10:12
EgmatQuantExpert wrote:
e-GMAT Question of the Week #24

Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respectively. All the three taps are used simultaneously to fill the tank. But, due to a leak at the bottom of the tank, only \(\frac{1}{3}\)rd of the tank was filled by the time it was supposed to be full. In how much time could the leak empty a full tank?

    A. \(\frac{75}{31}\)

    B. \(\frac{150}{31}\)

    C. \(\frac{225}{31}\)

    D. \(\frac{15}{2}\)

    E. \(\frac{450}{31}\)


Image



Let \(A\), \(B\) and \(C\) be time that pipes \(A\), \(B\) and \(C\) takes to fill a tank, respectively.

We can figure out the time \(X\) that \(A\) and \(B\) take to fill a tank together using a formula \(\frac{1}{X} = \frac{1}{A} + \frac{1}{B}\).
We can figure out the time \(T\) that \(A\), \(B\) and \(C\) take to fill a tank together using a formula \(\frac{1}{T} = \frac{1}{A} + \frac{1}{B} + \frac{1}{C}\).

Then we have \(\frac{1}{T} = \frac{1}{10} + \frac{1}{15} + \frac{1}{25} = \frac{31}{150}\) or \(T = \frac{150}{31}\).

Since the tank is leaking and \(\frac{1}{3}\) of the tank was filled, the time to fill the tank is \(3T = \frac{450}{31}\).
Let \(Y\) be the sinking speed of the tank.
\(\frac{31}{150} - \frac{1}{Y} = \frac{1}{3T} = \frac{31}{450}\).
Then \(\frac{1}{Y} = \frac{31}{150} - \frac{31}{450} = \frac{62}{450} = \frac{31}{225}\)
Thus we have \(Y = \frac{225}{31}\).

Therefore, the answer is C.
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec  [#permalink]

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New post 26 Sep 2019, 07:54
Anyone has taken the approach:
1/10+1/15+1/25 - 1/x = 1/3?

Where is this approach going wrong?
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Re: Three pipes A, B and C can fill a tank in 10, 15 and 25 minutes respec   [#permalink] 26 Sep 2019, 07:54
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