Operating at a constant rate, 2 pumps of Type 1 can completely fill

Question

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

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Saquib
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EgmatQuantExpert · Accepted Answer

Alternative method – (without using variables)

We are given:

Working out:

So what can we conclude from this?

The reservoir was half filled by these 6 reservoirs

.

So total time taken will be (1 pm + 12 hours) = 1 AM of 22nd March.

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So total time taken will be (1 pm + 12 hours) = 1 AM of 22nd March. Thanks, Saquib Quant Expert e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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EgmatQuantExpert · Answer

The official solution has been posted. Looking forward to a healthy discussion..

rohit8865 · Answer

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

ravisinghal · Answer

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

EgmatQuantExpert · Answer

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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gracie · Answer

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

EgmatQuantExpert · Answer

Hey,

PFB the official solution.

Given:

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:

o We’ll use this information to find the value of t.

Working Out:

Finding the value of t

Finding the required time

o (Time at which 3 "Type 2" pumps alone would have filled the reservoir) = 1 PM + \(\frac{t}{3}\)hours

Looking at the answer choices, we see that the correct answer is Option A

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Saquib
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hanyhamdani · Answer

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

sasyaharry · Answer

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}

PSKhore · Answer

3 pumps of type 2 work for 6 hours and 6 pumps of type 1 work for 4 hours

3*6/x + 6*4/48 = 1
1/x = 1/36

Total time taken by 3 pumps of type 2 to complete the entire work

3*t/36 = 1
t=12 hours

Hence, 1 AM on 22nd March

Operating at a constant rate, 2 pumps of Type 1 can completely fill

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Operating at a constant rate, 2 pumps of Type 1 can completely fill

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