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Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 08:00
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39% (01:57) correct 61% (02:39) wrong based on 194 sessions
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Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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Updated on: 26 Jul 2019, 23:01
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Tap A can fill a tank in 10 hours -> In 1 hour it fills +10% of the tank. Tap B can fill it in 20 hours -> In 1 hour it fills +5% of the tank. Tap C can empty the tank in 30 hours -> In 1 hour it fills -\((\frac{10}{3}\)) % of the tank.
Each tap is opened, one by one, for exactly one hour and then closed. Therefore in every 3 hours: 10% + 5% - \(\frac{10}{3}\)% = \(\frac{35}{3}\)% of the tank gets filled.
The tank is already \(\frac{1}{4}\) full. Therefore, we only have \(\frac{3}{4}\) i.e. 75% of the tank to fill.
In 18 hours, it fills: \(\frac{35}{3}\)% * 6 = 70% In the 19th hour, Tap A will start filling which will fill +10% in one hour. However, we only need +5% to make it 75%. So Tap A should fill +5 % in ½ hour i.e. 30 minutes.
Therefore, it will take 18 hours 30 minutes to completely fill the tank, after which it will begin to overflow.
Answer D
Originally posted by Sayon on 25 Jul 2019, 08:19.
Last edited by Sayon on 26 Jul 2019, 23:01, edited 4 times in total.
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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Updated on: 25 Jul 2019, 23:30
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IMO D
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
1/a+1/b-1/c will give the solution.
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Sol:
(1/10)+(1/20)-(1/30)=(7/60)
now this represents a unit of three hours,every three hours this much will be the increase in the water,
from smart picking of numbers lets wee if this unit is repeated 6 times we get (7/60)*6=42/60 then 42/60 +1/4 =57/60
so the let is 1-57/60=3/60 or 1/20
then we know that A will start after this so we know
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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Updated on: 26 Jul 2019, 08:38
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Ra = 1/10 Rb= 1/20 Rc= -1/30 per hour net rate ; 0.1+0.05-0.03 ; .12 let total tank capacity be 60 so if its 1/4th full ; left with 45 so time it would take to fill ~45/.12 ; 375 mins ~ 6 hours 25 mins but since each tap is opened once for an hour so total time 18 hours 25 mins overflow would be at 18 hours 30 mins IMO D
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Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 08:30
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A -10 B -20 C -30 taking LCM lets say total work is 60. So in 3 hours the work done by the taps will be 6+3-2 (- because of the emptying by C )=7
Now 1/4th is already filled so remain work=60*3/4=45 42 part of work can be done in 6*7 i.e 18 hours(remember 7 is work done in 3 hours) now to do remaining part of work i.e 3 A will need 30 mins as he does 6 parts in 1 hour.
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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Updated on: 25 Jul 2019, 20:06
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Each 3 hours the tank is going to be fill in the rate of: 1/10 + 1/20 - 1/30 = 6/60 + 3/60 - 2/60 = 7/60
Then, we know that the tank is already 1/4 full, so to fill up the tank we will need: (3/4)/(7/60) = 3/4 x 60/7 = 45/7= 19 2/7 but actually the question is when does the tank start overflowing?
We can try and see what happens after 18 hours, so we have that:
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Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 08:31
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Rate of filling water = (1/10 + 1/20 - 1/30) = 7/60 Now the taps are opened only when 1/4 tank is full so the remaining tank to be filled is 3/4 of capacity. Hence time taken = 7/60 * 4/3 = 6 hours 25 mins.
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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Updated on: 25 Jul 2019, 08:37
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Quote:
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Say volume of tank =60 Rate of A=60/10=6 Rate of B=60/20=3 Rate of C=-60/30=-2 Now,its 1/4th filled =60/4=15l
In 3 hours, Rate=6+3-2=7l is filled in tank In every 3 hours 7l is added to tank
In 18 hours= 7*6=42l is added to tank Tank is filled with 15+42=57 l
Next, when Tap A is opened it'll take 30 min to fill 3L as its speed is 6l/hr
So, complete will be filled in 18 hr 30min Hence D
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Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 08:33
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
Okay to solve this question take the question in the multiple of 3. A = 1/10, B =1/20 & C=1/30 . So together they can fill 7/60th of the tank in 3 hours. 3/4th of the tank will be filled in ... So the total tank will be in 19.288 hours.
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 08:59
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Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
Say, the full tank contains 60 liters Then the performance of A=6, the performance of B=3, the performance of C=2, Wt have 60*1/4 =15 liters in the tank 15+6+3-2.....15+7*6=57 and the last can A will work for half an hour to hill 3 liters. Thus we have 6*3+ 30 min=18h 30 min
Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 09:04
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Quote:
Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
Let us write down what we are given: Tap A fills the pool for 10 hours Tap B fills the pool for 20 hours Tap C empties the pool for 30 hours Pool is 1/4 full Taps work each 1 hour separately. We need to find time \(t\) when the pool will start to overflow.
Let us write down the work which will be done by all three taps in 1 run (3 consecutive hours): \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{10}+\frac{1}{20}+(-\frac{1}{30})=\frac{6+3-2}{60}=\frac{7}{60}\) We write down C with minus, since it actually empties the tank, and creates negative effect to the work done by A and B. The tank is 1/4 full, so we need to fill \(1-\frac{1}{4}=\frac{3}{4}=\frac{45}{60}\) of a pool. Now, let us find the number of hours to fill the pool: \(\frac{45}{7}=6 + 3\), where 3 is a remainder of work to be done. Do not forget that it is actually not 6 hours, but 6 cycles each of 3 hours, thus, \(6*3=18\) hours. Now we need to find out, what time it takes to fill the remaining \(\frac{3}{60}\) of the pool. We know that the taps work in order starting from A and A fills the pool in 10 hours so it can perform \(\frac{6}{10}\) of a job in one hour. \(\frac{6}{10} : \frac{3}{10} = \frac{1}{2}\) hour \(18 + 1/2 = 18 1/2\) hours = \(18\) hours \(30\) minutes.
Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 09:04
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Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
A. 6 hours 25 minutes B. 6 hours 30 minutes C. 18 hours 25 minutes D. 18 hours 30 minutes E. 19 hours 18 minutes
Given: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C). Asked: After how long will the tank begin to overflow?
Tap A can fill a tank in 10 hours => Tap A will fill 1/10 tank in an hour Tap B can fill it in 20 hours => Tap B will fill 1/20 tank in an hour An outlet, Tap C, can empty the tank in 30 hours => Tap C will empty 1/30 tank in an hour
Taps start working in alphabetic order (A, B, and then C) => Within 3 hours they together fill \(\frac{1}{10}+\frac{1}{20}-\frac{1}{30} = \frac{7}{60}\) tank
If the tank is \(\frac{1}{4}\)th full => After 3x hours it will be \(\frac{1}{4} + \frac{7x}{60} = \frac{15+7x}{60}\) full \(\frac{15+7x}{60}<1\) => 15+7x<60 =>7x<45 =>x=6 After 3x = 18 hours tank will be \(\frac{57}{60} full or \frac{3}{60} = \frac{1}{20}\) empty
Now tank A will be in operation and will fill \(\frac{1}{20}\)tank in \(\frac{1}{2}\) hours (30 minutes)
=> Tank will overflow in 18 hours 30 minutes.
IMO D
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Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 09:05
E I think.
Assume V is the volume of the tank, it is initially filled for 1/4th so remaining to be filled is 3V/4.
Assume all three taps work together to fill the tank, time to fill the tank now is 3v/4t = V/10+V/20 - V/30 or t = 45/7 ~ 6 hrs 25 mins
However, question says each tap works one-by-one, so total time to fill in will be 3 times of what is calculated above, so total time is 3*45/7 hrs ~ 19 hrs 18 mins
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Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 09:17
We need to fill 75% of the tank In 1 hour A fills 10% of the tank, B fills 5% and C removes 3.3% from the tank
Working one after the other, in 3 hours total 10%+5%-3.3% = 11.7% approx will be filled Working at this rate (one after the other) in 18 hours, tank filled is = 6 x 11.7% = 70.2% Remaining work to be done is 75% - 70.2% = 4.8% which will be done by A alone
A can complete 10% of work in 1 hour, so 5% work is completed in 30 mintes We require less 5% of work to be done, which means less than 30 minutes are required.
Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 09:18
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Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?
A fills 1/10 of tank in 1 hour B fills 1/20 of tank in 1 hour C empties 1/30 of tank in 1 hour so in 3 hours , with each 1 hour of work completes = 1/10 + 1/20 -1/30 = 7/60 work
for 6 hours --> 7/60 * 2 = 7/30 for 18 hours ---> 7/60 * 6 = 7/10 no A has to work little more to fill 3/4 of the tank ---> 3/4 - 7/10 = 30-28 / 40 = 2 /40 = 1/20 A has to finish 1/20 work , --> it takes = 30 minutes
so in total it takes 18 hours , 30 min to fill 3/4 of the tank--Option D is the answer
Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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25 Jul 2019, 09:32
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IMO-D
A fill - 10 hrs B fill - 20 hrs C empty - 30 hrs
Let the tank has a capacity of 60 Units [ LCM (10,20,30) ]
A/Q , Tank already full = 1/4 th = 1/4 x 60 = 15 Units
Tank to be filled before overflow = ( 60-15 ) = 45 Units
Note: Tap are opened at an interval of one hour & operated for only one hour.
In one hour, A fill = 60/10 = 6 units || B fill = 60/20 = 3 units || C empty = 60/30 = 2 units
In 3 hours combined , tank fill= (6+3-2) = 7 units In 18 hrs tank fill = 6 x 7 = 42 units Remaining tank = (45 - 42) units = 3 units Now Tap A turn it is, Time req. for remaining = 3/6 = 0.5 hrs = 30 min
Total time = 18 hrs + 30 mins
gmatclubot
Re: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,
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39% (01:57) correct 
