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Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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22 Jan 2017, 00:00
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Q.) Operating at a constant rate, 2 pumps of Type 1 can completely fill a reservoir with water in 24 hours. A pump of Type 2 can also fill the same reservoir with water but it operates at a different constant rate. At 1 PM on 21 March, 3 pumps of Type 2 start filling the empty reservoir. At 3 PM, 6 pumps of Type 1 also start filling the reservoir with water. The reservoir is completely filled with water at 7 PM. If the 6 Type 1 pumps were not started, at what time would the reservoir have been completely filled with water? A. 1 AM on 22 March B. 7 AM on 22 March C. 1 AM on 23 March D. 7 AM on 23 March E. 7 AM on 25 March Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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Updated on: 27 Jan 2017, 21:51



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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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22 Jan 2017, 11:42
EgmatQuantExpert wrote: Q.) Operating at a constant rate, 2 pumps of Type 1 can completely fill a reservoir with water in 24 hours. A pump of Type 2 can also fill the same reservoir with water but it operates at a different constant rate. At 1 PM on 21 March, 3 pumps of Type 2 start filling the empty reservoir. At 3 PM, 6 pumps of Type 1 also start filling the reservoir with water. The reservoir is completely filled with water at 7 PM. If the 6 Type 1 pumps were not started, at what time would the reservoir have been completely filled with water? A. 1 AM on 22 March B. 7 AM on 22 March C. 1 AM on 23 March D. 7 AM on 23 March E. 7 AM on 25 March Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Let rate of Type 2 =B & Type 1 =A thus 2 type 1 can fill with rate= 1/24 thus A=1/48 similarly work done by 3 Pump Type 2 is 3/B thus after starting type2 1 PM it worked alone upto 3PM(2 hrs) work done by type2 for 2 hrs = (2*3)/B for the next 4 hrs (7PM) both types of pump run. contribution of 6 pump type 1 for 4 hrs = (6*4)/48 =1/2 contribution of 3 pump type 2 for 4 hrs = (3*12)/B all three combined = work done by type2 for initial 2 hrs + contribution of 6 pump type 1 for 4 hrs + contribution of 6 pump type 2 for 4 hrs 6/B + 12/B + 1/2 = 1 18/B = 1/2 or ( 3*6)/B = 1/2 3/B = 1/12 thus 3 type 2 pump can fill in 12 hrs 1 PM on 21st March +12 hrs =1 AM on 22nd march Ans A



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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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22 Jan 2017, 22:11
2 pumps fill empty reservoir in 24 hours 6 pumps fill in 6 hours
Type 1 worked from 3 to 7 PM 4 hours hence type 1 completed the 4/6 work of the total work.
Work remaining done by type 2 pumps = 12/3 =1/3
now type 2 pumps worked for 6 hours 3 pumps completed 1/3 work in 6 hours therefore they would have taken 6*3 hours if type 1 did not start. which means 7 AM on march 22..Answer B



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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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23 Jan 2017, 21:58
ravisinghal wrote: 2 pumps fill empty reservoir in 24 hours 6 pumps fill in 6 hours
Type 1 worked from 3 to 7 PM 4 hours hence type 1 completed the 4/6 work of the total work.
Work remaining done by type 2 pumps = 12/3 =1/3
now type 2 pumps worked for 6 hours 3 pumps completed 1/3 work in 6 hours therefore they would have taken 6*3 hours if type 1 did not start. which means 7 AM on march 22..Answer B Hey ravisinghal, There are few mistakes in your solution. If 2 pumps fill in 24 hours How can 6 pumps fill in 6 hours? I think you have made a calculation mistake here. The correct way is  If 2 pumps fill in 24 hours 1 pump will fill in 48 hours 6 pumps will fill in 8 hours This is one of the mistakes. Kindly go through your solution again and do try to solve once more. Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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24 Jan 2017, 08:42
Operating at a constant rate, 2 pumps of Type 1 can completely fill a reservoir with water in 24 hours. A pump of Type 2 can also fill the same reservoir with water but it operates at a different constant rate. At 1 PM on 21 March, 3 pumps of Type 2 start filling the empty reservoir. At 3 PM, 6 pumps of Type 1 also start filling the reservoir with water. The reservoir is completely filled with water at 7 PM. If the 6 Type 1 pumps were not started, at what time would the reservoir have been completely filled with water?
A. 1 AM on 22 March B. 7 AM on 22 March C. 1 AM on 23 March D. 7 AM on 23 March E. 7 AM on 25 March
rate of 1 T1 pump=1/(2*24)=1/48 in 4 hours 6 T1 pumps fill 24/48=1/2 of reservoir in 6 hours 3 T2 pumps fill the other 1/2 of reservoir thus, it would take 3 T2 pumps 12 hours to fill reservoir alone 1 AM on 22 March A



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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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27 Jan 2017, 21:50
Hey, PFB the official solution. Given:• Let the total volume of the reservoir be V cubic units • Time taken by 2 "Type 1" pumps to fill V volume = 24 hours
o So, Time taken by 1 "Type 1" pump to fill V volume = 24 x 2 = 48 hours o Filling Rate of Type 1 pump = \(\frac{V}{48}\) volume per hour • Let one "Type 2" pump take t hours to fill V volume
o So, Filling Rate of "Type 2" pump = \(\frac{V}{t}\)volume per hour • 1 PM – 3 PM
o 3 "Type 2" pumps start filling the empty reservoir • 3 PM – 7 PM
o 3 "Type 2" pumps + 6 "Type 1" pumps fill the reservoir to fullness To find: • At what time would 3 "Type 2" pumps alone have filled the reservoir? Approach:1. (Time at which 3 "Type 2" pumps alone would have filled the reservoir) = 1 PM + (Volume to be filled/Filling rate of 3 "Type 2" pumps) hours
o = 1 PM + \(\frac{V}{(3*\frac{V}{t})}\) hours o = 1 PM +\(\frac{t}{3}\) hours o So, to answer the question, we need to find the value of t 2. We’re told that 3 "Type 2" pumps operating for 6 hours (from 1 PM to 7 PM) and 6 Type 1 pumps operating for 4 hours (from 3 PM to 7 PM) fill V cubic units of volume
o We’ll use this information to find the value of t. Working Out:• Finding the value of t
o (Volume filled by 3 "Type 2" pumps in 6 hours) + (Volume filled by 6 "Type 1" pumps in 4 hours) = (V, the total Volume of the Reservoir) o 3(Volume filled by 1 "Type 2" pump in 6 hours) + 6(Volume filled by 1 "Type 1" pump in 4 hours) = V o \(3*(\frac{V}{t}*6) + 6*(\frac{V}{48}*4) = V\) o \(\frac{18}{t} + \frac{1}{2} = 1\) o \(\frac{18}{t} = \frac{1}{2}\) o So, t = 36 hours • Finding the required time
o (Time at which 3 "Type 2" pumps alone would have filled the reservoir) = 1 PM + \(\frac{t}{3}\)hours o = 1 PM + \(\frac{36}{3}\) hours o = 1 PM + 12 hours o = 1 AM the next day Looking at the answer choices, we see that the correct answer is Option AThanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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27 Jan 2017, 22:06
Alternative method – (without using variables)We are given:• 2 “Type 1” pumps can fill the reservoir in 24 hours. • At 1 pm –
o 3 pumps of “Type 2” start filling the reservoir and continue filling it till 7 pm. o That means it fills the reservoir for (71) = 6 hours • At 3 pm
o 6 pumps of “Type 1” also start filling the reservoir and it is also open till 7 pm. o That means it fills the reservoir for = (73) = 4 hours Working out:• Now 2 pumps of “Type 1” fill the reservoir in 24 hours. • 1 pump of “Type 1” can fill the reservoir in 48 hours. • 6 pumps of “Type 1” can fill the reservoir in 8 hours.
o But 6 reservoirs were open for 4 hours only. o So what can we conclude from this? o The reservoir was half filled by these 6 reservoirs. • The other half was filled by 3 pumps of “Type 2” which was open for 6 hours. • Therefore, to completely fill the reservoir, it will take double the time, which is 6 x 2 hours . So total time taken will be (1 pm + 12 hours) = 1 AM of 22nd March. Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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30 Oct 2017, 01:29
Tricky! I got A.
2 T1 can fill in 24 hours. so 6 T1 can fill in 8 hours.
T2 rate is unknown.
3 T2 pumps start at 1pm and work till 7pm. at 3pm joined by 6 T1 pumps, which work for 4 hours.
if 6 T1 can fill the full tank in 8 hours. In 4 hours, half of the tank was filled by these pumps. which means, the other half of the tank was filled by 3 T2 pumps which worked for 6 hours in total.
so 3 T2 can fill half in 6 hours, therefore full tank in 12 hours.
1pm+ 12 hours= 1 am on 22nd



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Re: Word Problem  Operating at a constant rate, 2 pumps of Type 1 can com
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10 Nov 2017, 10:48
Once you find that the work done by both these types of pumps is the same, i.e 1/2, we can equate them. Rate of type 1 * time the type 1 pumps worked = Rate of type 2 * time the type 2 pumps worked. \(\frac{1}{8}*4=\frac{3}{x}*2\) Thus, x =12. Rate of type 2 pump is \(\frac{1}{12}\)
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