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555-605 Level|   Math Related|         
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 19:52
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.

estimate range of storage : 1250 - 1350 (1300 +- off up to 50 liters)

Estimate of time 3.5 - 4.5 hours (4 hours +- 0.5 hours)

Maximum: highest amount / least of hours : 1350/3.5

Minimum : lowest amount of storage / highest amount of hours = 1250/4.5
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estimated volume of water pumped = 1300 ltr with margin of error of 50 ltr
estimated time taken = 4 hours with margin of error of 0.5 hours

rate of water pumping = volume of water pumped in ltrs/ time taken in hours

we will get the lowest possible actual pumping rate by making numerator smaller and denominator larger

volume of water pumped in ltrs = 1300-50 = 1250 ltr
time taken for the same = 4 + 0.5 = 4.5 hrs

minimum rate = 1250/4.5

we will get the highest possible actual pumping rate by making numerator bigger and denominator smaller

volume of water pumped in ltrs = 1300+50 = 1350 ltr
time taken for the same = 4 - 0.5 = 3.5 hrs

maximum rate = 1350/3.5


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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 21:50
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
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1300 L off by up to 50 L => Range: 1250 L to 1350 L
4 h off by up to 0.5h => Range: 3.5 h to 4.5 h

Pumping rate: Litres of water pumped/ Time taken

Now, we know that for the fraction to have the minimum value, the numerator should be as small as possible and the denominator should be as large as possible.
=> 1250/4.5

Now, we know that for the fraction to have the maximum value, the numerator should be as large as possible and the denominator should be as small as possible.
=> 1350/3.5

Minimum: 1250/4.5
Maximum: 1350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 22:32
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the range of volume could be from 1250-1350 liters, and the range of time could be from 3.5-4.5 hours.
Minimum = 1250/4.5 (smaller numerator divide by larger denominator) and maximum =1350/3.5 (larger numerator divide by smaller denominator)
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 23:10
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Correct Answer: Minimum rate= 1,250/4.5 & Maximum rate= 1,350/3.5

Minimum Volume= 1250
Maximum Volume= 1350
Minimum Hour= 3.5
Maximum Hour= 4.5

Now let's check minimum and maximum rate:

1) Minimum rate:

Here we will take

Minimum Volume= 1250
Maximum Hour= 4.5
Rate= 1250 liters / 4.5 hours
= 277.78 liter per hour

2) Maximum rate:

Here we will take

Maximum Volume= 1350
Minimum Hour= 3.5
Rate= 1350 liters / 3.5 hours
= 385.71 liter per hour
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 12 Jul 2025, 00:04
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MinMax
Volume Estimate12501350
Time Estimate3.54.5

Minimum pumping rate:

\(\frac{\text{Min Volume}}{\text{Max time}}=1250/4.5\)

Maximum pumping rate:

\(\frac{\text{Max Volume}}{\text{Min time}}=1350/3.5\)


Minimum: \(1250/4.5\)

Maximum: \(1350/3.5\)
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Given:
1)Estimated volume is 1300 litres with an error upto 50 litres
2)Estimated time is 4 hours with an error upto 0.5hrs

Pumping rate = \(\frac{Total Volume}{Total time}\)

Minimum pumping rate:
In a fraction to get minimum value numerator should be minimum possible value and denominator should be max possible value
Therefore,
Minimum possible volume = 1300-50= 1250
Maximum possible time = 4+ 0.5= 4.5
Minimum rate = \(\frac{Min Volume}{Max time}\)= \(\frac{1250}{4.5}\)

Maximum pumping rate:
In a fraction to get maximum value numerator should be maximum possible value and denominator should be minimum possible value
Maximum possible volume = 1300+50= 1350
Minimum possible time = 4-0.5= 3.5
Maximum rate = \(\frac{Max Volume}{Min time}\)= \(\frac{1350}{3.5}\)
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Possible Volume: 1300+50 = 1350 or 1300-50=1250;

Possible Time: 4 + 0.5 = 4.5 Hours or 4 - 0.5 = 3.5 Hours

lowest possible actual pumping rate (in liters per hour) = 1250/4.5;

highest possible actual pumping rate (in liters per hour) = 1350/3.5;
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 12 Jul 2025, 01:24
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1300 liters in 4 hrs

volume may be off up to 50, then the range of volume, 1250 to 1350
time may be off by up to 0.5 hours, with a range of time, 3.5 to 4.5 hr

to minimize the lowest numerator with the highest denominator. then. 1250/4.5 minimum
to maximize the highest numerator with the lowest denominator, then 1350/3.5 Maximum
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To get the minimum pumping rate:
Use the lowest volume and longest time
Minimum rate -> 1250 liters / 4.5 hours = 1250/4.5

To get the maximum pumping rate:
Use the highest volume and shortest time
Maximum rate -> 1350 liters / 3.5 hours = 1350/3.5
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Bunuel
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
Show more

Volume = 1300 liters
Volume variable = +-50 liters

Time = 4 hours
Time variable = +-0.5 hours

Highest pumping rate = Highest Volume / Lowest time
Highest pumping rate = {1300 + 50} / {4-0.5}
Highest pumping rate = Maximum = 1350/3.5

Lowest pumping rate = Lowest Volume / Highest time
Lowest pumping rate = {1300 - 50} / {4+0.5}
Lowest pumping rate = Minimum = 1250/4.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 12 Jul 2025, 02:13
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Given,
1,300 liters of water into a storage tank in 4 hours.
Volume estimate may be off by up to and including 50 liters,
Time estimate may be off by up to and including 0.5 hours.

Case 1 - Minimum the lowest possible actual pumping rate (in liters per hour)
For this Minimize volume of water delivered and maximize time taken,
Volume = 1300-50 = 1250
Time = 4+0.5 = 4.5
hence Rate = 1250/4.5

Case 2 - Maximum the highest possible actual pumping rate (in liters per hour)
For this Maximize volume of water delivered and minimize time taken,
Volume = 1300+50 = 1350
Time = 4-0.5 = 3.5
hence Rate = 1350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 12 Jul 2025, 02:35
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Bunuel
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
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Range of liters: 1300 +/- 50
Range of time: 4 +/- 0.5

Minimum= 1250/4.5
Max = 1350/3.5
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Bunuel
 


This question was provided by GMAT Club
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
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We are looking for the pumping rate = Liters pumped/Est time.

So to maximize the pumping rate, We need to get max(Numerator) and min(Denominator)

Maximum : Max( Liters pumped) / Min( Est time) => 1350/3.5.

Similarly, Minimum : Min( Liters pumped) / Max(Est time) = 1250/4.5

IMO Minimum : 1250/4.5 , Maximum : 1350/3.5.
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Maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours
Volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours

Estimated Volume = 1300 litres
Possible error in volume calculation = +- 50 litres
=> Actual volume is between 1250 litres and 1350 litres

Possible error in time calculation = +- 30 mins
=> Actual time is between 3.5hrs and 4.5 hrs

Pumping rate = Volume / Time
Minimum Possible Rate = Least Volume / Most Time = 1250 / 4.5
Maximum Possible Rate = Most Volume / Least Time = 1350 / 3.5

Minimum = 1250 / 4.5
Maximum = 1350 / 3.5
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The answer is
For Minimum: 1250/3.5
As the stem says that volume estimate may be off by up to and including 50 liters so 1300-50=1250 and time estimate may be off by up to and including 0.5 hours. So 4-0.5=3.5
For Maximum:1350/4.5
As the stem says that volume estimate may be off by up to and including 50 liters so 1300+50=1350 and time estimate may be off by up to and including 0.5 hours. So 4+0.5=4.5
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Bunuel
 


This question was provided by GMAT Club
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
Show more

Max Volume = 1350
Min Volume = 1250

Min Time = 3.5
Max Time = 4.5

Minimum = Min Volume / Max Time

= 1250/4.5

Maximum = Max Volume / Min Time

= 1350/3.5

Answer:
Minimum = 1250/4.5
Maximum = 1350/3.5
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Bunuel
 


This question was provided by GMAT Club
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
Show more
IMO , FROM QUESTION WE CAN KNOW THAT VOLUME ESTIMATION CAN GO WRONG BY +_50 AND TIME ESTIMATION CAN GO WRONG BY +_0.5

So for the range of values which can come is
[*]Volume range: 1,250 to 1,350 liters
[*]Time range: 3.5 to 4.5 hours

from here to create a maximum value we need to keep volume of water value max and time least. SO, it will be 1350/3.5.
and for minima we need to keep volume low aswell as time max to get a least value so, 1250/4.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 12 Jul 2025, 04:40
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Given,
Estimated total volume = 1300 l
Time taken = 4 Hrs.
Error range:
Volume = +/- 50 l
Time = +/- 0.5 Hrs.
Volume range = [1250, 1350]
Time range = [3.5, 4.5]

To find:
Minimum: the lowest possible actual pumping rate (in liters per hour)
= Minimum volume/ maximum time
Maximum: the highest possible actual pumping rate (in liters per hour)
= Maximum volume/ Minimum time

Solution:
Minimum Rate = 1250/4.5
Maximum Rate = 1350/3.5

Ans:
Minimum = 1250/4.5
Maximum = 1350/3.5


Bunuel
 


This question was provided by GMAT Club
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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.
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