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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 08:00
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Minimum: 1,250/4.5 Maximum: 1,350/3.5
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75% (01:35) correct 25% (01:52) wrong based on 561 sessions

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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.
Minimum Maximum
1,300/3
1,350/3.5
1,250/3.5
1,350/4.5
1,250/4.5
1,250/5
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Minimum: 1,250/4.5 Maximum: 1,350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 08:30
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Bunuel
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 Official Explanation:



  • The actual volume is between 1,250 and 1,350 liters, inclusive.
  • The actual time is between 3.5 and 4.5 hours, inclusive.

For the minimum possible actual pumping rate (in liters/hour), use the lowest volume (numerator) and the highest time (denominator):

1,250/4.5 liters/hour

For the maximum possible actual pumping rate (in liters/hour), use the highest volume (numerator) and the lowest time (denominator):

1,350/3.5 liters/hour
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 08:27
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volume = 1300 L
the delta volume could be +/- 50 L

The time is 4hrs
delta time could .5 hours

for minimum rate the volume will be least, and time will be max

volume= (1300-50)/(4+0.5)
1250/4.5

similarly for maximum rate:
volume max and time minimum

vol=1350/(4-0.5) = 1350/3.5

(It seems simple hope i am not missing anything on this)
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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, 08:40
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The pipe’s volume estimate can vary between 1,250 liters (1,300 minus 50) and 1,350 liters (1,300 plus 50), while the time estimate can range from 3.5 hours (4 minus 0.5) to 4.5 hours (4 plus 0.5). To find the lowest possible pumping rate, we take the smallest volume and the longest time, which gives 1,250 liters divided by 4.5 hours. For the highest possible pumping rate, we use the largest volume and the shortest time, which is 1,350 liters divided by 3.5 hours. Therefore, the minimum rate corresponds to 1,250 over 4.5, and the maximum rate corresponds to 1,350 over 3.5.

For maximizing any fraction: Maximize numerator , Minimize denominator

For minimizing any fraction: Minimize numerator , Maximize denominator
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 08:43
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As given in the question, actual pumping rate= (total liters)/(total hours)
To get the highest possible pumping rate, we want as much water as possible in as little time as possible. so you add 50 liters to the estimated volume and subtract 0.5 hour from the estimated time.
To get the lowest possible rate, you do the reverse: subtract 50 liters from the estimated volume and 0.5 hour to the estimated time.
Hence Minimum= (1300-50)/(4+0.5)=1250/4.5
Maximum= (1300+50)/(4-0.5)=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, 08:50
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Given information:
* Estimated volume: 1,300 liters (may be off by up to 50 liters)
* Estimated time: 4 hours (may be off by up to 0.5 hours)
This means the actual volume could be between 1,250 and 1,350 liters, and the actual time could be between 3.5 and 4.5 hours.
To find the pumping rate, we use the formula:
Rate = Volume ÷ Time
For the minimum rate, we need the smallest possible volume divided by the largest possible time:
Minimum Rate = 1,250 ÷ 4.5 = 1,250/4.5 liters per hour
For the maximum rate, we need the largest possible volume divided by the smallest possible time:
Maximum Rate = 1,350 = 3.5 = 1,350/3.5 liters per hour
Therefore, the minimum possible pumping rate is 1,250/4.5 liters per hour, and the maximum possible pumping rate is 1,350/3.5 liters per hour.
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 08:58
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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
clearly we can see that the highest and lowest volume is equal to 1300+50 and 1300-50 and longest and shortest time is 4+0.5 and 4-0.5 therefore minimum rate is minimum litre/longest time ie 1250/4.5 amd maximum rate is highest litres/shortest time ie 1350/4.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:03
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V = 1300L
t = 4hrs
V_corr = 50L
t_corr = 0.5hrs

min Rate = min V / max t = 1250/4.5
max Rate = max V / min t = 1350/3.5

Thus Answer is : 1250/4.5 , 1350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:09
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total qty 1300 liters
time 4 hours
minimum pumping rate

1300-50 ; 1250
time 4 hours and 0.5 hours error ; 4.5
1250/4.5

maximum pumping rate
1300+50 ; 1350
time 4-0.5 ; 3.5 hours
1350/3.5

Minimum 1250/4.5 in liters per hour ; Maximum 1350/3.5 in liters per hour


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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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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:12
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Rate=work/time

To maximize the pumping rate we need to increase numerator and decrease denominator.So its 1350/3.5
To minimize the pumping rate we need to decrease numneraot and increase denominaor.So its 1250/4.5

So answers are
Maximum: 1350/3.5
Minimum: 1250/4.5
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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, 09:16
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Ans: Min = 1250/4.5 and Max = 1350/3.5

Max rate means less water more time = 1300-50 = 1250 liter water and 4 + 0.5 hrs = 4.5 hrs
Max rate means more water in less time = 1300+50 = 1350 liter water and 4 - 0.5 hrs = 3.5 hrs
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:16
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Estimated volume pumped = 1300 liters. .....(+-50)
Estimated time = 4 hours.......(+-0.5 hrs)

To get Minimum Rate, select Smallest possible volume and Largest possible time
Smallest Volume = 1300 - 50 = 1250
Largest Time = 4 + 0.5 = 4.5
Minimum Rate = 1250/4.5

To get Maximum Rate, select Largest possible volume and Smallest possible time
Largest Volume = 1300 +50= 1350
Smallest Time = 4-0.5=3.5
Maximum Rate = 1350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:18
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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.

Minimum pumping rate (in liters per hour) = (1300-50)/(4.5) = 1250/4.5 liters/hour
Maximum pumping rate (in liters per hour) = (1300+50)(3.5) = 1350/3.5 liters/hour

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.

MinimumMaximum
1250/4.51350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:23
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To find the minimum pumping rate, we need the minimum possible volume and the maximum possible time.
Minimum volume = 1300−50=1250 liters
Maximum time = 4+0.5=4.5 hours
Minimum pumping rate = 1250/4.5 liters/hour

To find the maximum pumping rate, we need the maximum possible volume and the minimum possible time.
Maximum volume = 1300+50=1350 liters
Minimum time = 4−0.5=3.5 hours
Maximum pumping rate = 1350/3.5 liters/hour

Therefore:
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, 09:33
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Given

Total water pumped = 1300 liters; error margin +/- 50 liters
Total time taken = 4 hours; error margin +/- 0.5 hours

Asked:

Minimum Possible Actual Pumping Rate
=> Minimum possible water pumped / Maximum possible time = 1250 / 4.5 liters/hours

Maximum Possible Actual Pumping Rate
=> Maximum possible water pumped / minimum possible time = 1350 / 3.5 liters/hours

Hence, for each respective column, these figures can be checked.

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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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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:33
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Okay
So 1250 <=volume <= 1350
and 3.5<=time<=4.5

and we can get min be = min/max = 1250/4.5
Now we can get max by = (max)/min = 1350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:35
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We can derive from the sum the minimum and the maximum value
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minimum:
min vol., max time = 1300-50 / 4+0.5 = 1250/4.5
maximum:
max vol., min time = 1300+50/4-0.5 = 1350/3.5
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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that 11 Jul 2025, 09:48
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Volume pumped = 1250<=V<=1350.
Time = 3.5<=T<=4.5

Min actual pumping rate = Min V/Max T = 1250/4.5

Max actual pumping rate = MaxV/Min T = 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, 09:50
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Volume can range from 1250 L to 1350 L
Time can range from 3.5 hrs to 4.5 hrs

So minimum rate will be one with lowest volume and highest time ie 1250/4.5
and Maximum rate will be one with highest volume and lowest time ie 1350/3.5

Hence answer is:
Minimum 1250/4.5
Maximum 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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