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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?

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Re: Word Problem - Operating at a constant rate, 2 pumps of Type 1 can com [#permalink]

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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?

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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 type-2 1 PM it worked alone upto 3PM(2 hrs) work done by type-2 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 type-2 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

Re: Word Problem - Operating at a constant rate, 2 pumps of Type 1 can com [#permalink]

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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 = 1-2/3 =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

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 = 1-2/3 =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.

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Word Problem - Operating at a constant rate, 2 pumps of Type 1 can com [#permalink]

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24 Jan 2017, 08:42

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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

• 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 A

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• 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 (7-1) = 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 = (7-3) = 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.

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Re: Word Problem - Operating at a constant rate, 2 pumps of Type 1 can com [#permalink]

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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.