Problem Solving

There is a leak in the bottom of a cistern. When the cistern is thoroughly repaired, it would be filled 1(3/2) hrs. Before getting repaired, it takes half an hour longer. If the cistern is full, how long would the leak take to empty the cistern?
(1) 24 hrs
(2) 28 hrs
(3) 32 hrs
(4) 36 hrs
(5) 40 hrs

OE: B

Though I was able to solve the problem, I would like the experts to teach me some shortcuts in such problems.


@TheExperts : Could U please explain me the solution posted in the image.

Thank U.

I believe that the problem should read as follows:

There is a leak in the bottom of a cistern. When the cistern is thoroughly repaired, it would be filled in 3.5 hrs. Before getting repaired, it takes half an hour longer. If the cistern is full, how long would the leak take to empty the cistern?
(1) 24 hrs
(2) 28 hrs
(3) 32 hrs
(4) 36 hrs
(5) 40 hrs

Let P = the pipe that FILLS the cistern and L = the leak that DRAINS the cistern.
When elements COMPETE, SUBTRACT their rates.
Since P FILLS the cistern, while L DRAINS the cistern, P and L are COMPETING.
Thus, when P and L operate together, the resulting rate = P-L.

When the cistern is thoroughly repaired, it would be filled in 3.5 hrs.
Thus, the time for P alone = 3.5 hours.

Before getting repaired, it takes half an hour longer.
Thus, the time for P-L = 4 hours.

The time ratio for P/(P-L) = 3.5/4 = 35/40.
Since time and rate have a RECIPROCAL RELATIONSHIP, the rate ratio for P/(P-L) = 40/35.
Implication:
For every 40 liters input by P, L drains the cistern 5 liters, with the result that P-L = 40-5 = 35.

Thus, the rate ratio for P/L = 40/5 = 8/1.
Since L operates 1/8 AS FAST as P, L's number of hours is 8 TIMES P's number of hours.
Thus, L's time = (8)(3.5) = 28 hours.

The correct answer is B.

There is a leak in the bottom of a cistern. When the cistern is thoroughly repaired, it would be filled in 3.5 hrs. Before getting repaired, it takes half an hour longer. If the cistern is full, how long would the leak take to empty the cistern?

(1) 24 hrs
(2) 28 hrs
(3) 32 hrs
(4) 36 hrs
(5) 40 hrs

An alternate approach is to PLUG IN THE ANSWERS, which represent the time for L to empty the cistern.
When the cistern is thoroughly repaired, it would be filled in 3.5 hrs. Before getting repaired, it takes half an hour longer.
Implication:
When the correct answer is plugged in, P-L will take 4 hours to fill the cistern.

B: 28 hours
Let the cistern = 28 liters.
Since L takes 28 hours to empty the cistern, L's rate = w/t = 28/28 = 1 liter per hour.
Since P takes 7/2 hours to fill the cistern, P's rate = w/t = 28/(7/2) = (28)(2/7) = 8 liters per hour.
Since the rate when P and L operate together = P-L = 8-1 = 7, the time for P-L to fill the cistern = w/r = 28/7 = 4 hours.
Success!

The correct answer is B.

One more approach:

Let the volume of the cistern = the product of the two times = (3.5)(4) = 14 liters.
Since the repaired cistern is filled in 3.5 hours, the rate for the pipe = w/t = 14/(3.5) = 4 liters per hour.
Since the leak increases the time by 1/2 hour to 4 hours, the rate for the pipe and leak together = w/t = 14/4 = 3.5 hours.
Since the leak reduces the rate by 1/2 liter per hour, the rate for the leak = 1/2 liter per hour.
At rate of 1/2 liter per hour, the time for the leak to empty the cistern = w/r = 14/(1/2) = 28 hours.

The correct answer is B.

The time ratio for P/(P-L) = 3.5/4 = 35/40.
Since time and rate have a RECIPROCAL RELATIONSHIP, the rate ratio for P/(P-L) = 40/35.
Implication:
For every 40 liters input by P, L drains the cistern 5 liters, with the result that P-L = 40-5 = 35.

Thus, the rate ratio for P/L = 40/5 = 8/1.
Since L operates 1/8 AS FAST as P, L's number of hours is 8 TIMES P's number of hours.
Thus, L's time = (8)(3.5) = 28 hours.

Hi GMATGuruNY,

Can you please explain the above part.

Thanks,

Rashi

Hi GMATGuruNY,

Can you please explain the above part.

Thanks,

Rashi

Rate and time are RECIPROCALS.
If Adam's rate is 1/2 Bob's rate, Adam's time is TWICE Bob's time.
If Adam takes 3 TIMES as long as Bob, then Adam's rate is 1/3 Bob's rate.

When the cistern is repaired, P takes 3.5 hours to fill the cistern.
When the cistern has a leak, P-L requires 4 hours to fill the cistern.
Thus, the time ratio for P and P-L = (3.5 hours)/(4 hours) = (7 hours)/(8 hours).
Since time and rate are reciprocals, the rate ratio for P and P-L is the reciprocal of their time ratio:
P/(P-L) = (8 liters per hour)/(7 liters per hour).

Since L reduces the input rate from 8 liters per hour to 7 liters per hour, L must drain 1 liter per hour for every 8 liters input by P.
Thus, the rate ratio for P and L = (8 liters per hour)/(1 liter per hour).

Since the rate ratio for P and L = (8 liters per hour)/(1 liter per hour) -- and rate and time are reciprocals -- the time ratio for P and L = (1 hour)/(8 hours).
The resulting time ratio indicates that L's time must be 8 times P's time.
Since P's time = 3.5 hours, we get:
L's time = (8)(3.5) = 28 hours.

GMATGuruNY wrote:I believe that the problem should read as follows:

There is a leak in the bottom of a cistern. When the cistern is thoroughly repaired, it would be filled in 3.5 hrs. Before getting repaired, it takes half an hour longer. If the cistern is full, how long would the leak take to empty the cistern?
(1) 24 hrs
(2) 28 hrs
(3) 32 hrs
(4) 36 hrs
(5) 40 hrs

Let P = the pipe that FILLS the cistern and L = the leak that DRAINS the cistern.
When elements COMPETE, SUBTRACT their rates.
Since P FILLS the cistern, while L DRAINS the cistern, P and L are COMPETING.
Thus, when P and L operate together, the resulting rate = P-L.

When the cistern is thoroughly repaired, it would be filled in 3.5 hrs.
Thus, the time for P alone = 3.5 hours.

Before getting repaired, it takes half an hour longer.
Thus, the time for P-L = 4 hours.

The time ratio for P/(P-L) = 3.5/4 = 35/40.
Since time and rate have a RECIPROCAL RELATIONSHIP, the rate ratio for P/(P-L) = 40/35.
Implication:
For every 40 liters input by P, L drains the cistern 5 liters, with the result that P-L = 40-5 = 35.

Thus, the rate ratio for P/L = 40/5 = 8/1.
Since L operates 1/8 AS FAST as P, L's number of hours is 8 TIMES P's number of hours.
Thus, L's time = (8)(3.5) = 28 hours.

The correct answer is B.

Dear GMAT Guru,

I'm confused about this type of work problems.

1- I want to understand the logic in this problem. The filling is 8 littler per hour and the leakage is 1 liter per hour. It means that the cistern won't be empty. How come the tank is full and we still filling it?

2- Can we solve the problem as follows

2/7 - 1/4 = 1/28........So it will take 28 hrs. I applied the rule blindly although I do not understand logic as stated above.

3- As state above :
"When elements COMPETE, SUBTRACT their rates.
Since P FILLS the cistern, while L DRAINS the cistern, P and L are COMPETING.
Thus, when P and L operate together, the resulting rate = P-L."

As I know we can subtract or sum RATES Only but then you mentioned 'Thus, the time for P-L = 4 hours.' So you subtract times NOT rates. Is it acceptable?


Thanks

Mo2men wrote:1- I want to understand the logic in this problem. The filling is 8 littler per hour and the leakage is 1 liter per hour. It means that the cistern won't be empty. How come the tank is full and we still filling it?

The prompt present hypotheticals.
WITHOUT THE LEAK, the input pipe can fill the cistern at a rate of 8 liters per hour.
WITH THE LEAK, the input pipe can fill the cistern at a rate of 7 liters per hour, since the leak outputs 1 liter per hour.

2- Can we solve the problem as follows

2/7 - 1/4 = 1/28........So it will take 28 hrs. I applied the rule blindly although I do not understand logic as stated above.

This approach is perfect.
Let the cistern = 1.
Since the pipe takes 3.5 hours to fill the cistern, the rate for the pipe alone = 1/(3.5) = 2/7.
Since the pipe and the leak together require 4 hours to fill the cistern, the rate for the pipe with the leak = 1/4.
Since the leak reduces the rate from 2/7 to 1/4, the rate for the leak alone = 2/7 - 1/4 = 1/28.
Thus, the time for the leak to drain the cistern = the reciprocal of the leak's rate = 28 hours.

3- As state above :
"When elements COMPETE, SUBTRACT their rates.
Since P FILLS the cistern, while L DRAINS the cistern, P and L are COMPETING.
Thus, when P and L operate together, the resulting rate = P-L."

As I know we can subtract or sum RATES Only but then you mentioned 'Thus, the time for P-L = 4 hours.' So you subtract times NOT rates. Is it acceptable?

Thanks

In my initial solution, the time for P-L ≠ (P's time alone) - (L's time alone).
Rather, the time for P-L = the number of hours for the pipe and the leak TOGETHER.
The purpose of the subtraction sign is to indicate that the pipe and the leak are COMPETING when they work together.

GMATGuruNY wrote:

Mo2men wrote:1- I want to understand the logic in this problem. The filling is 8 littler per hour and the leakage is 1 liter per hour. It means that the cistern won't be empty. How come the tank is full and we still filling it?

The prompt present hypotheticals.
WITHOUT THE LEAK, the input pipe can fill the cistern at a rate of 8 liters per hour.
WITH THE LEAK, the input pipe can fill the cistern at a rate of 7 liters per hour, since the leak outputs 1 liter per hour.

2- Can we solve the problem as follows

2/7 - 1/4 = 1/28........So it will take 28 hrs. I applied the rule blindly although I do not understand logic as stated above.

This approach is perfect.
Let the cistern = 1.
Since the pipe takes 3.5 hours to fill the cistern, the rate for the pipe alone = 1/(3.5) = 2/7.
When the pipe and the leak together require 4 hours to fill the cistern, the rate for the pipe with the leak = 1/4.
Since the leak reduces the rate from 2/7 to 1/4, the rate for the leak alone = 2/7 - 1/4 = 1/28.
Thus, the time for the leak to drain the cistern = the reciprocal of the leak's rate = 28 hours.

3- As state above :
"When elements COMPETE, SUBTRACT their rates.
Since P FILLS the cistern, while L DRAINS the cistern, P and L are COMPETING.
Thus, when P and L operate together, the resulting rate = P-L."

As I know we can subtract or sum RATES Only but then you mentioned 'Thus, the time for P-L = 4 hours.' So you subtract times NOT rates. Is it acceptable?

Thanks

In my initial solution, the time for P-L ≠ (P's time alone) - (L's time alone).
Rather, the time for P-L = the number of hours for the pipe and the leak TOGETHER.
The purpose of the subtraction sign is to indicate that the pipe and the leak are COMPETING when they work together.

Dear GMATGuru,

Thanks for detailed reply. one more question. In the prompt, 'if the cistern is full,....' does it make any difference if it says half of cistern is full? If it has difference, hoe to approach in that case??

Thanks

Mo2men wrote:Dear GMATGuru,

Thanks for detailed reply. one more question. In the prompt, 'if the cistern is full,....' does it make any difference if it says half of cistern is full? If it has difference, hoe to approach in that case??

Thanks

Since the leak requires 28 hours to empty a full cistern, it would require half this time -- 14 hours -- to empty a half-full cistern.