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# Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2577
Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)  [#permalink]

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23 Nov 2018, 05:57
17
00:00

Difficulty:

85% (hard)

Question Stats:

53% (03:04) correct 47% (02:44) wrong based on 189 sessions

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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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Joined: 04 Jan 2015
Posts: 2577
Re: Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)  [#permalink]

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28 Nov 2018, 04:55
1
4

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.

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##### General Discussion
Intern
Joined: 20 Aug 2018
Posts: 4
Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)  [#permalink]

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23 Nov 2018, 09:03
1
3
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
Intern
Joined: 21 Mar 2016
Posts: 15
Re: Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)  [#permalink]

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14 Dec 2018, 09:00
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.

Director
Joined: 20 Sep 2016
Posts: 592
Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)  [#permalink]

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15 Dec 2018, 05:44
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. $$\frac{450}{31}$$

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

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

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
Intern
Status: when you say,"I can or I can't", Both times you are right!
Joined: 26 Nov 2018
Posts: 9
Re: Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)  [#permalink]

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11 Jan 2019, 19:59
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
Re: Question of the Week- 24 (Three pipes, A, B , C can fill a tank in...)   [#permalink] 11 Jan 2019, 19:59
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