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

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

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22 Jan 2017, 00:00
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3
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Difficulty:

85% (hard)

Question Stats:

55% (03:07) correct 45% (03:20) wrong based on 132 sessions

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

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Updated on: 27 Jan 2017, 21:51
The official solution has been posted. Looking forward to a healthy discussion..
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Originally posted by EgmatQuantExpert on 22 Jan 2017, 00:02.
Last edited by EgmatQuantExpert on 27 Jan 2017, 21:51, edited 1 time in total.
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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?

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,
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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

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

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23 Jan 2017, 21:58
1
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 = 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.

Thanks,
Saquib
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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
1
1
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  [#permalink]

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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 A

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

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27 Jan 2017, 22:06
4
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 (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.

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  [#permalink]

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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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Re: Word Problem - Operating at a constant rate, 2 pumps of Type 1 can com &nbs [#permalink] 10 Nov 2017, 10:48
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