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

Question

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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e-GMAT Question of the Week #24

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Official Answer

C

A.  \frac{75}{31}

B.  \frac{150}{31}

C.  \frac{225}{31}

D.  \frac{15}{2}

E.  \frac{450}{31}

e-GMAT          Question of the Week #24

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EgmatQuantExpert · Accepted Answer

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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kilgo16 · Answer

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

iiindiangirl · Answer

Hi 
Thank you for the solution.
But can you please explain the highlighted part?
Thanks again!

AdityaHongunti · Answer

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

Rupesh1Nonly · Answer

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

menonpriyanka92 · Answer

Does anybody have the LCM approach solution for this one?

KarishmaB · Answer

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)

Abhishek009 · Answer

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)

MathRevolution · Answer

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.

PinakiRDas · Answer

Anyone has taken the approach:
1/10+1/15+1/25 - 1/x = 1/3?

Where is this approach going wrong?

Can,Will · Answer

" due to a leak at the bottom of the tank, only 1/3 rd of the tank was filled by the time it was supposed to be full."
if only the 1/3 rd of the tank was filled then why dont we take work done by taps as 1/3

Bunuel · Answer

Could you please elaborate on what you mean by 'why don't we take the work done by taps as 1/3'? Which solution are you referring to? Where or how should we consider 'the work done by taps as 1/3'?

itachi_gmat · Answer

The combined rate of filling 1 tank = 3/150.

Now, use R*T=W: 3/150*t = 1, and the time calculated to fill the full tank is 150/31 minutes.

Now at the same time the tank is emptied 2/3rd we need to find the rate of leakage using R*T =W

R*150/31=2/3, R = 31/225.

this is the rate at which the 2/3rd of the tank is emptied now to find the time taken by leak to empty the full tank.

R*T = W: 31/225*t = 1:- t= 225/31.

Kinshook · Answer

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?

Work completed with a minute = 1/10 + 1/15 + 1/25 = (15+10+6)/150 = 31/150

Let the time required by the leak to empty a full tank be x minutes

1/10 + 1/15 + 1/25 - 1/x = 31/150 - 1/x = 31/450

1/x = 31/150 - 31/450 = 62/450 = 31/225

The time required by the leak to empty a full tank = 225/31 minutes

IMO C

KevoK · Answer

Okay here's my two cents on this:

10, 15 and 25 min so we're back to using LCM(10,15,25), which is 150 Liters

In order to fill up this tank, the work rate of each pipes would be:

A = \frac{150 }{ 10} = 15

B = \frac{150 }{ 15} = 10

C = \frac{150 }{ 25} = 6

The time it takes to fill up the entire tank is  \frac{150}{(10+15+6)} = \frac{150}{31}

Since the number is a bit wonky, I decided to approach it from the drain hole perspective:

Because there's only  \frac{1}{3}  of the tank, the hole drained  \frac{2}{3} of the tank (which is  150 * \frac{2}{3} = 100 L ) in  \frac{150}{31}  minutes

So the Rate of draining is   \frac{ 100 * 31 }{ 150}  (R = W / T formula), which can be simplified relatively quickly into  \frac{62}{3}

Now we just have to find the time to drain the entire tank,
which is :  \frac{150 }{ (62/3)}

\frac{225 }{ 31}

C

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

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